\(\int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [580]

Optimal result

Integrand size = 21, antiderivative size = 257 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {5 \left (4 a^2+b^2\right ) \sec (c+d x)}{2 b^5 d}+\frac {5 \sec ^3(c+d x)}{6 b^3 d}-\frac {5 a \left (4 a^2+3 b^2\right ) \text {arcsinh}(\tan (c+d x)) \sec (c+d x)}{2 b^6 d \sqrt {\sec ^2(c+d x)}}-\frac {5 \sqrt {a^2+b^2} \left (4 a^2+b^2\right ) \text {arctanh}\left (\frac {b-a \tan (c+d x)}{\sqrt {a^2+b^2} \sqrt {\sec ^2(c+d x)}}\right ) \sec (c+d x)}{2 b^6 d \sqrt {\sec ^2(c+d x)}}-\frac {5 a \sec (c+d x) \tan (c+d x)}{b^4 d}-\frac {\sec ^5(c+d x)}{2 b d (a+b \tan (c+d x))^2}+\frac {5 a \sec ^3(c+d x)}{2 b^3 d (a+b \tan (c+d x))} \] Output:

5/2*(4*a^2+b^2)*sec(d*x+c)/b^5/d+5/6*sec(d*x+c)^3/b^3/d-5/2*a*(4*a^2+3*b^2 
)*arcsinh(tan(d*x+c))*sec(d*x+c)/b^6/d/(sec(d*x+c)^2)^(1/2)-5/2*(a^2+b^2)^ 
(1/2)*(4*a^2+b^2)*arctanh((b-a*tan(d*x+c))/(a^2+b^2)^(1/2)/(sec(d*x+c)^2)^ 
(1/2))*sec(d*x+c)/b^6/d/(sec(d*x+c)^2)^(1/2)-5*a*sec(d*x+c)*tan(d*x+c)/b^4 
/d-1/2*sec(d*x+c)^5/b/d/(a+b*tan(d*x+c))^2+5/2*a*sec(d*x+c)^3/b^3/d/(a+b*t 
an(d*x+c))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.08 (sec) , antiderivative size = 688, normalized size of antiderivative = 2.68 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (\frac {6 b^2 \left (a^2+b^2\right )^2 \sin (c+d x)}{a}+\frac {6 (a-i b) (a+i b) b \left (8 a^2-b^2\right ) (a \cos (c+d x)+b \sin (c+d x))}{a}+2 b \left (36 a^2+13 b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2+60 \sqrt {a^2+b^2} \left (4 a^2+b^2\right ) \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right ) (a \cos (c+d x)+b \sin (c+d x))^2+30 a \left (4 a^2+3 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2-30 a \left (4 a^2+3 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2+\frac {b^2 (-9 a+b) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 b^3 \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 b \left (36 a^2+13 b^2\right ) \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {2 b^3 \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {b^2 (9 a+b) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {2 b \left (36 a^2+13 b^2\right ) \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}\right )}{12 b^6 d (a+b \tan (c+d x))^3} \] Input:

Integrate[Sec[c + d*x]^7/(a + b*Tan[c + d*x])^3,x]
 

Output:

(Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])*((6*b^2*(a^2 + b^2)^2*Si 
n[c + d*x])/a + (6*(a - I*b)*(a + I*b)*b*(8*a^2 - b^2)*(a*Cos[c + d*x] + b 
*Sin[c + d*x]))/a + 2*b*(36*a^2 + 13*b^2)*(a*Cos[c + d*x] + b*Sin[c + d*x] 
)^2 + 60*Sqrt[a^2 + b^2]*(4*a^2 + b^2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/S 
qrt[a^2 + b^2]]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2 + 30*a*(4*a^2 + 3*b^2) 
*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a*Cos[c + d*x] + b*Sin[c + d*x] 
)^2 - 30*a*(4*a^2 + 3*b^2)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a*Cos 
[c + d*x] + b*Sin[c + d*x])^2 + (b^2*(-9*a + b)*(a*Cos[c + d*x] + b*Sin[c 
+ d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*b^3*Sin[(c + d*x)/ 
2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/ 
2])^3 + (2*b*(36*a^2 + 13*b^2)*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c 
+ d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (2*b^3*Sin[(c + d*x)/2] 
*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2] 
)^3 + (b^2*(9*a + b)*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2 
] + Sin[(c + d*x)/2])^2 - (2*b*(36*a^2 + 13*b^2)*Sin[(c + d*x)/2]*(a*Cos[c 
 + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/(12*b 
^6*d*(a + b*Tan[c + d*x])^3)
 

Rubi [A] (warning: unable to verify)

Time = 0.48 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.87, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3992, 492, 590, 25, 27, 682, 27, 719, 222, 488, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^7}{(a+b \tan (c+d x))^3}dx\)

\(\Big \downarrow \) 3992

\(\displaystyle \frac {\sec (c+d x) \int \frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{(a+b \tan (c+d x))^3}d(b \tan (c+d x))}{b d \sqrt {\sec ^2(c+d x)}}\)

\(\Big \downarrow \) 492

\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \int \frac {b \tan (c+d x) \left (\tan ^2(c+d x)+1\right )^{3/2}}{(a+b \tan (c+d x))^2}d(b \tan (c+d x))}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\)

\(\Big \downarrow \) 590

\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {\left (\tan ^2(c+d x)+1\right )^{3/2} (4 a+b \tan (c+d x))}{3 (a+b \tan (c+d x))}-\int -\frac {\left (b^2-4 a b \tan (c+d x)\right ) \sqrt {\tan ^2(c+d x)+1}}{b^2 (a+b \tan (c+d x))}d(b \tan (c+d x))\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\int \frac {\left (b^2-4 a b \tan (c+d x)\right ) \sqrt {\tan ^2(c+d x)+1}}{b^2 (a+b \tan (c+d x))}d(b \tan (c+d x))+\frac {\left (\tan ^2(c+d x)+1\right )^{3/2} (4 a+b \tan (c+d x))}{3 (a+b \tan (c+d x))}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {\int \frac {\left (b^2-4 a b \tan (c+d x)\right ) \sqrt {\tan ^2(c+d x)+1}}{a+b \tan (c+d x)}d(b \tan (c+d x))}{b^2}+\frac {\left (\tan ^2(c+d x)+1\right )^{3/2} (4 a+b \tan (c+d x))}{3 (a+b \tan (c+d x))}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\)

\(\Big \downarrow \) 682

\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {\frac {1}{2} b^2 \int \frac {2 \left (\left (\frac {2 a^2}{b^2}+1\right ) b^4-a b \left (4 a^2+3 b^2\right ) \tan (c+d x)\right )}{b^4 (a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))+\sqrt {\tan ^2(c+d x)+1} \left (b^2 \left (\frac {4 a^2}{b^2}+1\right )-2 a b \tan (c+d x)\right )}{b^2}+\frac {\left (\tan ^2(c+d x)+1\right )^{3/2} (4 a+b \tan (c+d x))}{3 (a+b \tan (c+d x))}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {\frac {\int \frac {b^2 \left (2 a^2+b^2\right )-a b \left (4 a^2+3 b^2\right ) \tan (c+d x)}{(a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))}{b^2}+\sqrt {\tan ^2(c+d x)+1} \left (b^2 \left (\frac {4 a^2}{b^2}+1\right )-2 a b \tan (c+d x)\right )}{b^2}+\frac {\left (\tan ^2(c+d x)+1\right )^{3/2} (4 a+b \tan (c+d x))}{3 (a+b \tan (c+d x))}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {\frac {\left (a^2+b^2\right ) \left (4 a^2+b^2\right ) \int \frac {1}{(a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))-a \left (4 a^2+3 b^2\right ) \int \frac {1}{\sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))}{b^2}+\sqrt {\tan ^2(c+d x)+1} \left (b^2 \left (\frac {4 a^2}{b^2}+1\right )-2 a b \tan (c+d x)\right )}{b^2}+\frac {\left (\tan ^2(c+d x)+1\right )^{3/2} (4 a+b \tan (c+d x))}{3 (a+b \tan (c+d x))}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {\frac {\left (a^2+b^2\right ) \left (4 a^2+b^2\right ) \int \frac {1}{(a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))-a b \left (4 a^2+3 b^2\right ) \text {arcsinh}(\tan (c+d x))}{b^2}+\sqrt {\tan ^2(c+d x)+1} \left (b^2 \left (\frac {4 a^2}{b^2}+1\right )-2 a b \tan (c+d x)\right )}{b^2}+\frac {\left (\tan ^2(c+d x)+1\right )^{3/2} (4 a+b \tan (c+d x))}{3 (a+b \tan (c+d x))}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\)

\(\Big \downarrow \) 488

\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {\frac {-\left (a^2+b^2\right ) \left (4 a^2+b^2\right ) \int \frac {1}{\frac {a^2}{b^2}-b^2 \tan ^2(c+d x)+1}d\frac {1-\frac {a \tan (c+d x)}{b}}{\sqrt {\tan ^2(c+d x)+1}}-a b \left (4 a^2+3 b^2\right ) \text {arcsinh}(\tan (c+d x))}{b^2}+\sqrt {\tan ^2(c+d x)+1} \left (b^2 \left (\frac {4 a^2}{b^2}+1\right )-2 a b \tan (c+d x)\right )}{b^2}+\frac {\left (\tan ^2(c+d x)+1\right )^{3/2} (4 a+b \tan (c+d x))}{3 (a+b \tan (c+d x))}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sec (c+d x) \left (\frac {5 \left (\frac {\frac {-a b \left (4 a^2+3 b^2\right ) \text {arcsinh}(\tan (c+d x))-b \sqrt {a^2+b^2} \left (4 a^2+b^2\right ) \text {arctanh}\left (\frac {b^2 \tan (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^2}+\sqrt {\tan ^2(c+d x)+1} \left (b^2 \left (\frac {4 a^2}{b^2}+1\right )-2 a b \tan (c+d x)\right )}{b^2}+\frac {\left (\tan ^2(c+d x)+1\right )^{3/2} (4 a+b \tan (c+d x))}{3 (a+b \tan (c+d x))}\right )}{2 b^2}-\frac {\left (\tan ^2(c+d x)+1\right )^{5/2}}{2 (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\)

Input:

Int[Sec[c + d*x]^7/(a + b*Tan[c + d*x])^3,x]
 

Output:

(Sec[c + d*x]*(-1/2*(1 + Tan[c + d*x]^2)^(5/2)/(a + b*Tan[c + d*x])^2 + (5 
*(((4*a + b*Tan[c + d*x])*(1 + Tan[c + d*x]^2)^(3/2))/(3*(a + b*Tan[c + d* 
x])) + ((-(a*b*(4*a^2 + 3*b^2)*ArcSinh[Tan[c + d*x]]) - b*Sqrt[a^2 + b^2]* 
(4*a^2 + b^2)*ArcTanh[(b^2*Tan[c + d*x])/Sqrt[a^2 + b^2]])/b^2 + ((1 + (4* 
a^2)/b^2)*b^2 - 2*a*b*Tan[c + d*x])*Sqrt[1 + Tan[c + d*x]^2])/b^2))/(2*b^2 
)))/(b*d*Sqrt[Sec[c + d*x]^2])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 488

Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ 
Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ 
[{a, b, c, d}, x]
 

rule 492

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 1))), x] - Simp[2*b*(p/(d*(n + 1)) 
)   Int[x*(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, 
d, n}, x] && GtQ[p, 0] && (IntegerQ[p] || LtQ[n, -1]) && NeQ[n, -1] &&  !IL 
tQ[n + 2*p + 1, 0] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
 

rule 590

Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-(c + d*x)^(n + 1))*(a + b*x^2)^p*((c*(2*p + 1) - d*(n + 1)*x)/(d^2*( 
n + 1)*(n + 2*p + 2))), x] + Simp[2*(p/(d^2*(n + 1)*(n + 2*p + 2)))   Int[( 
c + d*x)^(n + 1)*(a + b*x^2)^(p - 1)*(a*d*(n + 1) + b*c*(2*p + 1)*x), x], x 
] /; FreeQ[{a, b, c, d}, x] && GtQ[p, 0] && LtQ[n, -1] &&  !ILtQ[n + 2*p + 
1, 0]
 

rule 682

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p 
+ 1) + g*c*e*(m + 2*p + 1)*x)*((a + c*x^2)^p/(c*e^2*(m + 2*p + 1)*(m + 2*p 
+ 2))), x] + Simp[2*(p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)))   Int[(d + e*x) 
^m*(a + c*x^2)^(p - 1)*Simp[f*a*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f* 
d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))*x, x], x 
], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] ||  ! 
RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (Intege 
rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 719

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3992

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[Sec[e + f*x]/(b*f*Sqrt[Sec[e + f*x]^2])   Subst[Int[( 
a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b 
, e, f, n}, x] && NeQ[a^2 + b^2, 0] && IntegerQ[(m - 1)/2]
 

Maple [A] (verified)

Time = 291.07 (sec) , antiderivative size = 444, normalized size of antiderivative = 1.73

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (\frac {\frac {b^{2} \left (7 a^{4}+5 b^{2} a^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a}+\frac {b \left (8 a^{6}-9 a^{4} b^{2}-15 a^{2} b^{4}+2 b^{6}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a^{2}}-\frac {b^{2} \left (25 a^{4}+23 b^{2} a^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-4 a^{4} b -\frac {7 a^{2} b^{3}}{2}+\frac {b^{5}}{2}}{\left (a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{2}}-\frac {5 \left (4 a^{4}+5 b^{2} a^{2}+b^{4}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{6}}+\frac {1}{3 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-3 a +b}{2 b^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-12 a^{2}+3 a b -5 b^{2}}{2 b^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {5 a \left (4 a^{2}+3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{6}}-\frac {1}{3 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {3 a +b}{2 b^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {12 a^{2}+3 a b +5 b^{2}}{2 b^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {5 a \left (4 a^{2}+3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{6}}}{d}\)	\(444\)
default	\(\frac {-\frac {2 \left (\frac {\frac {b^{2} \left (7 a^{4}+5 b^{2} a^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a}+\frac {b \left (8 a^{6}-9 a^{4} b^{2}-15 a^{2} b^{4}+2 b^{6}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a^{2}}-\frac {b^{2} \left (25 a^{4}+23 b^{2} a^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-4 a^{4} b -\frac {7 a^{2} b^{3}}{2}+\frac {b^{5}}{2}}{\left (a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{2}}-\frac {5 \left (4 a^{4}+5 b^{2} a^{2}+b^{4}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{6}}+\frac {1}{3 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-3 a +b}{2 b^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-12 a^{2}+3 a b -5 b^{2}}{2 b^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {5 a \left (4 a^{2}+3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{6}}-\frac {1}{3 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {3 a +b}{2 b^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {12 a^{2}+3 a b +5 b^{2}}{2 b^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {5 a \left (4 a^{2}+3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{6}}}{d}\)	\(444\)
risch	\(\frac {60 a^{4} {\mathrm e}^{9 i \left (d x +c \right )}-15 b^{4} {\mathrm e}^{9 i \left (d x +c \right )}+240 a^{4} {\mathrm e}^{7 i \left (d x +c \right )}-20 b^{4} {\mathrm e}^{7 i \left (d x +c \right )}+360 a^{4} {\mathrm e}^{5 i \left (d x +c \right )}+22 b^{4} {\mathrm e}^{5 i \left (d x +c \right )}+15 a^{2} b^{2} {\mathrm e}^{9 i \left (d x +c \right )}+140 a^{2} b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+140 a^{2} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+15 a^{2} b^{2} {\mathrm e}^{i \left (d x +c \right )}-15 b^{4} {\mathrm e}^{i \left (d x +c \right )}+240 a^{4} {\mathrm e}^{3 i \left (d x +c \right )}+60 a^{4} {\mathrm e}^{i \left (d x +c \right )}-20 b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+250 a^{2} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+180 i a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}-90 i a^{3} b \,{\mathrm e}^{9 i \left (d x +c \right )}-60 i a \,b^{3} {\mathrm e}^{9 i \left (d x +c \right )}-180 i a^{3} b \,{\mathrm e}^{7 i \left (d x +c \right )}+60 i a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}+90 i a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-100 i a \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+100 i a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +a \right )^{2} d \,b^{5}}-\frac {10 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,b^{6}}-\frac {15 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d \,b^{4}}+\frac {10 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \,b^{6}}+\frac {15 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d \,b^{4}}+\frac {10 \sqrt {a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -b}{\sqrt {a^{2}+b^{2}}}\right ) a^{2}}{d \,b^{6}}+\frac {5 \sqrt {a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -b}{\sqrt {a^{2}+b^{2}}}\right )}{2 d \,b^{4}}-\frac {10 \sqrt {a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a -b}{\sqrt {a^{2}+b^{2}}}\right ) a^{2}}{d \,b^{6}}-\frac {5 \sqrt {a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a -b}{\sqrt {a^{2}+b^{2}}}\right )}{2 d \,b^{4}}\)	\(693\)

Input:

int(sec(d*x+c)^7/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-2/b^6*((1/2*b^2*(7*a^4+5*a^2*b^2-2*b^4)/a*tan(1/2*d*x+1/2*c)^3+1/2*b 
*(8*a^6-9*a^4*b^2-15*a^2*b^4+2*b^6)/a^2*tan(1/2*d*x+1/2*c)^2-1/2*b^2*(25*a 
^4+23*a^2*b^2-2*b^4)/a*tan(1/2*d*x+1/2*c)-4*a^4*b-7/2*a^2*b^3+1/2*b^5)/(a* 
tan(1/2*d*x+1/2*c)^2-2*b*tan(1/2*d*x+1/2*c)-a)^2-5/2*(4*a^4+5*a^2*b^2+b^4) 
/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2)) 
)+1/3/b^3/(tan(1/2*d*x+1/2*c)+1)^3-1/2*(-3*a+b)/b^4/(tan(1/2*d*x+1/2*c)+1) 
^2-1/2*(-12*a^2+3*a*b-5*b^2)/b^5/(tan(1/2*d*x+1/2*c)+1)-5/2*a*(4*a^2+3*b^2 
)/b^6*ln(tan(1/2*d*x+1/2*c)+1)-1/3/b^3/(tan(1/2*d*x+1/2*c)-1)^3-1/2*(3*a+b 
)/b^4/(tan(1/2*d*x+1/2*c)-1)^2-1/2*(12*a^2+3*a*b+5*b^2)/b^5/(tan(1/2*d*x+1 
/2*c)-1)+5/2*a*(4*a^2+3*b^2)/b^6*ln(tan(1/2*d*x+1/2*c)-1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 564 vs. \(2 (237) = 474\).

Time = 0.18 (sec) , antiderivative size = 564, normalized size of antiderivative = 2.19 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {4 \, b^{5} + 30 \, {\left (4 \, a^{4} b + a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} + 20 \, {\left (2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left ({\left (4 \, a^{4} - 3 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (4 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (4 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - 15 \, {\left ({\left (4 \, a^{5} - a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (4 \, a^{4} b + 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left ({\left (4 \, a^{5} - a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (4 \, a^{4} b + 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 10 \, {\left (a b^{4} \cos \left (d x + c\right ) - 6 \, {\left (3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (2 \, a b^{7} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + b^{8} d \cos \left (d x + c\right )^{3} + {\left (a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{5}\right )}} \] Input:

integrate(sec(d*x+c)^7/(a+b*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

1/12*(4*b^5 + 30*(4*a^4*b + a^2*b^3 - b^5)*cos(d*x + c)^4 + 20*(2*a^2*b^3 
+ b^5)*cos(d*x + c)^2 + 15*((4*a^4 - 3*a^2*b^2 - b^4)*cos(d*x + c)^5 + 2*( 
4*a^3*b + a*b^3)*cos(d*x + c)^4*sin(d*x + c) + (4*a^2*b^2 + b^4)*cos(d*x + 
 c)^3)*sqrt(a^2 + b^2)*log(-(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2) 
*cos(d*x + c)^2 - 2*a^2 - b^2 + 2*sqrt(a^2 + b^2)*(b*cos(d*x + c) - a*sin( 
d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + 
 b^2)) - 15*((4*a^5 - a^3*b^2 - 3*a*b^4)*cos(d*x + c)^5 + 2*(4*a^4*b + 3*a 
^2*b^3)*cos(d*x + c)^4*sin(d*x + c) + (4*a^3*b^2 + 3*a*b^4)*cos(d*x + c)^3 
)*log(sin(d*x + c) + 1) + 15*((4*a^5 - a^3*b^2 - 3*a*b^4)*cos(d*x + c)^5 + 
 2*(4*a^4*b + 3*a^2*b^3)*cos(d*x + c)^4*sin(d*x + c) + (4*a^3*b^2 + 3*a*b^ 
4)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 10*(a*b^4*cos(d*x + c) - 6*(3* 
a^3*b^2 + 2*a*b^4)*cos(d*x + c)^3)*sin(d*x + c))/(2*a*b^7*d*cos(d*x + c)^4 
*sin(d*x + c) + b^8*d*cos(d*x + c)^3 + (a^2*b^6 - b^8)*d*cos(d*x + c)^5)
 

Sympy [F]

\[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\sec ^{7}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate(sec(d*x+c)**7/(a+b*tan(d*x+c))**3,x)
 

Output:

Integral(sec(c + d*x)**7/(a + b*tan(c + d*x))**3, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 902 vs. \(2 (237) = 474\).

Time = 0.14 (sec) , antiderivative size = 902, normalized size of antiderivative = 3.51 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^7/(a+b*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

1/6*(2*(60*a^6 + 35*a^4*b^2 - 3*a^2*b^4 + (210*a^5*b + 125*a^3*b^3 - 6*a*b 
^5)*sin(d*x + c)/(cos(d*x + c) + 1) - 2*(120*a^6 - 10*a^4*b^2 - 55*a^2*b^4 
 + 3*b^6)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*(330*a^5*b + 205*a^3*b^3 
 - 12*a*b^5)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2*(180*a^6 - 95*a^4*b^2 
 - 120*a^2*b^4 + 9*b^6)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 12*(60*a^5*b 
 + 35*a^3*b^3 - 3*a*b^5)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 6*(40*a^6 - 
 30*a^4*b^2 - 35*a^2*b^4 + 3*b^6)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 6* 
(50*a^5*b + 25*a^3*b^3 - 4*a*b^5)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 3* 
(20*a^6 - 15*a^4*b^2 - 15*a^2*b^4 + 2*b^6)*sin(d*x + c)^8/(cos(d*x + c) + 
1)^8 + 3*(10*a^5*b + 5*a^3*b^3 - 2*a*b^5)*sin(d*x + c)^9/(cos(d*x + c) + 1 
)^9)/(a^4*b^5 + 4*a^3*b^6*sin(d*x + c)/(cos(d*x + c) + 1) - 16*a^3*b^6*sin 
(d*x + c)^3/(cos(d*x + c) + 1)^3 + 24*a^3*b^6*sin(d*x + c)^5/(cos(d*x + c) 
 + 1)^5 - 16*a^3*b^6*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 4*a^3*b^6*sin(d 
*x + c)^9/(cos(d*x + c) + 1)^9 - a^4*b^5*sin(d*x + c)^10/(cos(d*x + c) + 1 
)^10 - (5*a^4*b^5 - 4*a^2*b^7)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2*(5* 
a^4*b^5 - 6*a^2*b^7)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2*(5*a^4*b^5 - 
6*a^2*b^7)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + (5*a^4*b^5 - 4*a^2*b^7)*s 
in(d*x + c)^8/(cos(d*x + c) + 1)^8) - 15*(4*a^3 + 3*a*b^2)*log(sin(d*x + c 
)/(cos(d*x + c) + 1) + 1)/b^6 + 15*(4*a^3 + 3*a*b^2)*log(sin(d*x + c)/(cos 
(d*x + c) + 1) - 1)/b^6 - 15*(4*a^4 + 5*a^2*b^2 + b^4)*log((b - a*sin(d...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 510 vs. \(2 (237) = 474\).

Time = 0.40 (sec) , antiderivative size = 510, normalized size of antiderivative = 1.98 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^7/(a+b*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

-1/6*(15*(4*a^3 + 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^6 - 15*(4* 
a^3 + 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^6 + 15*(4*a^4 + 5*a^2* 
b^2 + b^4)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs 
(2*a*tan(1/2*d*x + 1/2*c) - 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^6 
) + 2*(9*a*b*tan(1/2*d*x + 1/2*c)^5 + 36*a^2*tan(1/2*d*x + 1/2*c)^4 + 18*b 
^2*tan(1/2*d*x + 1/2*c)^4 - 72*a^2*tan(1/2*d*x + 1/2*c)^2 - 24*b^2*tan(1/2 
*d*x + 1/2*c)^2 - 9*a*b*tan(1/2*d*x + 1/2*c) + 36*a^2 + 14*b^2)/((tan(1/2* 
d*x + 1/2*c)^2 - 1)^3*b^5) + 6*(7*a^5*b*tan(1/2*d*x + 1/2*c)^3 + 5*a^3*b^3 
*tan(1/2*d*x + 1/2*c)^3 - 2*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 8*a^6*tan(1/2*d 
*x + 1/2*c)^2 - 9*a^4*b^2*tan(1/2*d*x + 1/2*c)^2 - 15*a^2*b^4*tan(1/2*d*x 
+ 1/2*c)^2 + 2*b^6*tan(1/2*d*x + 1/2*c)^2 - 25*a^5*b*tan(1/2*d*x + 1/2*c) 
- 23*a^3*b^3*tan(1/2*d*x + 1/2*c) + 2*a*b^5*tan(1/2*d*x + 1/2*c) - 8*a^6 - 
 7*a^4*b^2 + a^2*b^4)/((a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c 
) - a)^2*a^2*b^5))/d
 

Mupad [B] (verification not implemented)

Time = 4.03 (sec) , antiderivative size = 1203, normalized size of antiderivative = 4.68 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\text {Too large to display} \] Input:

int(1/(cos(c + d*x)^7*(a + b*tan(c + d*x))^3),x)
 

Output:

((60*a^4 - 3*b^4 + 35*a^2*b^2)/(3*b^5) + (tan(c/2 + (d*x)/2)*(210*a^4 - 6* 
b^4 + 125*a^2*b^2))/(3*a*b^4) + (tan(c/2 + (d*x)/2)^8*(20*a^6 + 2*b^6 - 15 
*a^2*b^4 - 15*a^4*b^2))/(a^2*b^5) - (2*tan(c/2 + (d*x)/2)^6*(40*a^6 + 3*b^ 
6 - 35*a^2*b^4 - 30*a^4*b^2))/(a^2*b^5) - (2*tan(c/2 + (d*x)/2)^2*(120*a^6 
 + 3*b^6 - 55*a^2*b^4 - 10*a^4*b^2))/(3*a^2*b^5) + (2*tan(c/2 + (d*x)/2)^4 
*(180*a^6 + 9*b^6 - 120*a^2*b^4 - 95*a^4*b^2))/(3*a^2*b^5) + (tan(c/2 + (d 
*x)/2)^9*(10*a^4 - 2*b^4 + 5*a^2*b^2))/(a*b^4) - (2*tan(c/2 + (d*x)/2)^7*( 
50*a^4 - 4*b^4 + 25*a^2*b^2))/(a*b^4) + (4*tan(c/2 + (d*x)/2)^5*(60*a^4 - 
3*b^4 + 35*a^2*b^2))/(a*b^4) - (2*tan(c/2 + (d*x)/2)^3*(330*a^4 - 12*b^4 + 
 205*a^2*b^2))/(3*a*b^4))/(d*(tan(c/2 + (d*x)/2)^8*(5*a^2 - 4*b^2) - tan(c 
/2 + (d*x)/2)^2*(5*a^2 - 4*b^2) + tan(c/2 + (d*x)/2)^4*(10*a^2 - 12*b^2) - 
 tan(c/2 + (d*x)/2)^6*(10*a^2 - 12*b^2) - a^2*tan(c/2 + (d*x)/2)^10 + a^2 
- 16*a*b*tan(c/2 + (d*x)/2)^3 + 24*a*b*tan(c/2 + (d*x)/2)^5 - 16*a*b*tan(c 
/2 + (d*x)/2)^7 + 4*a*b*tan(c/2 + (d*x)/2)^9 + 4*a*b*tan(c/2 + (d*x)/2))) 
- (atanh((3000*a^2*tan(c/2 + (d*x)/2))/(3000*a^2 + (7000*a^4)/b^2 + (4000* 
a^6)/b^4) + (7000*a^4*tan(c/2 + (d*x)/2))/(7000*a^4 + 3000*a^2*b^2 + (4000 
*a^6)/b^2) + (4000*a^6*tan(c/2 + (d*x)/2))/(4000*a^6 + 3000*a^2*b^4 + 7000 
*a^4*b^2))*(15*a*b^2 + 20*a^3))/(b^6*d) + (5*atanh((1000*a^2*(a^2 + b^2)^( 
1/2))/(1000*a^2*b + (5000*a^4)/b + (4000*a^6)/b^3 + 10000*a^3*tan(c/2 + (d 
*x)/2) + 2000*a*b^2*tan(c/2 + (d*x)/2) + (8000*a^5*tan(c/2 + (d*x)/2))/...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 10737, normalized size of antiderivative = 41.78 \[ \int \frac {\sec ^7(c+d x)}{(a+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(sec(d*x+c)^7/(a+b*tan(d*x+c))^3,x)
                                                                                    
                                                                                    
 

Output:

( - 240*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b* 
*2))*cos(c + d*x)*sin(c + d*x)**6*tan(c + d*x)**2*a**4*b**2*i + 180*sqrt(a 
**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d 
*x)*sin(c + d*x)**6*tan(c + d*x)**2*a**2*b**4*i + 60*sqrt(a**2 + b**2)*ata 
n((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin(c + d*x 
)**6*tan(c + d*x)**2*b**6*i - 480*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2) 
*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin(c + d*x)**6*tan(c + d*x)*a 
**5*b*i + 360*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a** 
2 + b**2))*cos(c + d*x)*sin(c + d*x)**6*tan(c + d*x)*a**3*b**3*i + 120*sqr 
t(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c 
+ d*x)*sin(c + d*x)**6*tan(c + d*x)*a*b**5*i - 240*sqrt(a**2 + b**2)*atan( 
(tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin(c + d*x)* 
*6*a**6*i + 180*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a 
**2 + b**2))*cos(c + d*x)*sin(c + d*x)**6*a**4*b**2*i + 60*sqrt(a**2 + b** 
2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin(c 
 + d*x)**6*a**2*b**4*i + 720*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i 
- b*i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x)**2*a** 
4*b**2*i - 300*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a* 
*2 + b**2))*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x)**2*a**2*b**4*i - 120 
*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))...