\(\int \frac {(d \tan (e+f x))^{7/2}}{(a+a \tan (e+f x))^3} \, dx\) [371]

Optimal result

Integrand size = 25, antiderivative size = 165 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+a \tan (e+f x))^3} \, dx=\frac {11 d^{7/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}+\frac {d^{7/2} \arctan \left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}-\frac {7 d^3 \sqrt {d \tan (e+f x)}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a+a \tan (e+f x))^2} \] Output:

11/8*d^(7/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f+1/4*d^(7/2)*arctan 
(1/2*(d^(1/2)-d^(1/2)*tan(f*x+e))*2^(1/2)/(d*tan(f*x+e))^(1/2))*2^(1/2)/a^ 
3/f-7/8*d^3*(d*tan(f*x+e))^(1/2)/a^3/f/(1+tan(f*x+e))-1/4*d^2*(d*tan(f*x+e 
))^(3/2)/a/f/(a+a*tan(f*x+e))^2
 

Mathematica [A] (verified)

Time = 3.49 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.11 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+a \tan (e+f x))^3} \, dx=\frac {(\cos (e+f x)+\sin (e+f x))^3 \left (-\frac {\csc ^5(e+f x) (7+7 \cos (2 (e+f x))+9 \sin (2 (e+f x)))}{(1+\cot (e+f x))^2}+\frac {2 \left (2 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )-2 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right )+11 \arctan \left (\sqrt {\tan (e+f x)}\right )\right ) \csc (e+f x) \sec ^2(e+f x)}{\tan ^{\frac {5}{2}}(e+f x)}\right ) (d \tan (e+f x))^{7/2}}{16 a^3 f (1+\tan (e+f x))^3} \] Input:

Integrate[(d*Tan[e + f*x])^(7/2)/(a + a*Tan[e + f*x])^3,x]
 

Output:

((Cos[e + f*x] + Sin[e + f*x])^3*(-((Csc[e + f*x]^5*(7 + 7*Cos[2*(e + f*x) 
] + 9*Sin[2*(e + f*x)]))/(1 + Cot[e + f*x])^2) + (2*(2*Sqrt[2]*ArcTan[1 - 
Sqrt[2]*Sqrt[Tan[e + f*x]]] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f* 
x]]] + 11*ArcTan[Sqrt[Tan[e + f*x]]])*Csc[e + f*x]*Sec[e + f*x]^2)/Tan[e + 
 f*x]^(5/2))*(d*Tan[e + f*x])^(7/2))/(16*a^3*f*(1 + Tan[e + f*x])^3)
 

Rubi [A] (verified)

Time = 1.23 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.11, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4048, 27, 3042, 4128, 3042, 4136, 27, 3042, 4015, 218, 4117, 27, 73, 216}

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.

\(\displaystyle \int \frac {(d \tan (e+f x))^{7/2}}{(a \tan (e+f x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \tan (e+f x))^{7/2}}{(a \tan (e+f x)+a)^3}dx\)

\(\Big \downarrow \) 4048

\(\displaystyle \frac {\int \frac {\sqrt {d \tan (e+f x)} \left (3 a^2 d^3+7 a^2 \tan ^2(e+f x) d^3-4 a^2 \tan (e+f x) d^3\right )}{2 (\tan (e+f x) a+a)^2}dx}{4 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {d \tan (e+f x)} \left (3 a^2 d^3+7 a^2 \tan ^2(e+f x) d^3-4 a^2 \tan (e+f x) d^3\right )}{(\tan (e+f x) a+a)^2}dx}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {d \tan (e+f x)} \left (3 a^2 d^3+7 a^2 \tan (e+f x)^2 d^3-4 a^2 \tan (e+f x) d^3\right )}{(\tan (e+f x) a+a)^2}dx}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4128

\(\displaystyle \frac {\frac {\int \frac {7 a^4 d^4+7 a^4 \tan ^2(e+f x) d^4-8 a^4 \tan (e+f x) d^4}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{2 a^3}-\frac {7 d^3 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {7 a^4 d^4+7 a^4 \tan (e+f x)^2 d^4-8 a^4 \tan (e+f x) d^4}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{2 a^3}-\frac {7 d^3 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4136

\(\displaystyle \frac {\frac {11 a^4 d^4 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {\int -\frac {8 \left (d^4 a^5+d^4 \tan (e+f x) a^5\right )}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}}{2 a^3}-\frac {7 d^3 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {11 a^4 d^4 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {4 \int \frac {d^4 a^5+d^4 \tan (e+f x) a^5}{\sqrt {d \tan (e+f x)}}dx}{a^2}}{2 a^3}-\frac {7 d^3 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {11 a^4 d^4 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {4 \int \frac {d^4 a^5+d^4 \tan (e+f x) a^5}{\sqrt {d \tan (e+f x)}}dx}{a^2}}{2 a^3}-\frac {7 d^3 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4015

\(\displaystyle \frac {\frac {11 a^4 d^4 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {8 a^8 d^8 \int \frac {1}{2 d^8 a^{10}+\cot (e+f x) \left (a^5 d^4-a^5 d^4 \tan (e+f x)\right )^2}d\frac {a^5 d^4-a^5 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}}{f}}{2 a^3}-\frac {7 d^3 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {11 a^4 d^4 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 \sqrt {2} a^3 d^{7/2} \arctan \left (\frac {a^5 d^4-a^5 d^4 \tan (e+f x)}{\sqrt {2} a^5 d^{7/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3}-\frac {7 d^3 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {\frac {\frac {11 a^4 d^4 \int \frac {1}{a \sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}+\frac {4 \sqrt {2} a^3 d^{7/2} \arctan \left (\frac {a^5 d^4-a^5 d^4 \tan (e+f x)}{\sqrt {2} a^5 d^{7/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3}-\frac {7 d^3 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {11 a^3 d^4 \int \frac {1}{\sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}+\frac {4 \sqrt {2} a^3 d^{7/2} \arctan \left (\frac {a^5 d^4-a^5 d^4 \tan (e+f x)}{\sqrt {2} a^5 d^{7/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3}-\frac {7 d^3 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {22 a^3 d^3 \int \frac {1}{\tan (e+f x)+1}d\sqrt {d \tan (e+f x)}}{f}+\frac {4 \sqrt {2} a^3 d^{7/2} \arctan \left (\frac {a^5 d^4-a^5 d^4 \tan (e+f x)}{\sqrt {2} a^5 d^{7/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3}-\frac {7 d^3 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {22 a^3 d^{7/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {4 \sqrt {2} a^3 d^{7/2} \arctan \left (\frac {a^5 d^4-a^5 d^4 \tan (e+f x)}{\sqrt {2} a^5 d^{7/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3}-\frac {7 d^3 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2}\)

Input:

Int[(d*Tan[e + f*x])^(7/2)/(a + a*Tan[e + f*x])^3,x]
 

Output:

-1/4*(d^2*(d*Tan[e + f*x])^(3/2))/(a*f*(a + a*Tan[e + f*x])^2) + (((22*a^3 
*d^(7/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/f + (4*Sqrt[2]*a^3*d^(7/2)* 
ArcTan[(a^5*d^4 - a^5*d^4*Tan[e + f*x])/(Sqrt[2]*a^5*d^(7/2)*Sqrt[d*Tan[e 
+ f*x]])])/f)/(2*a^3) - (7*d^3*Sqrt[d*Tan[e + f*x]])/(f*(1 + Tan[e + f*x]) 
))/(8*a^3)
 

Defintions of rubi rules used

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 216

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

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 4015

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[-2*(d^2/f)   Subst[Int[1/(2*c*d + b*x^2), x], x, (c 
- d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && 
 EqQ[c^2 - d^2, 0]
 

rule 4048

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m 
 - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 
/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + 
f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c 
*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) 
*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( 
n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 2] && LtQ 
[n, -1] && IntegerQ[2*m]
 

rule 4117

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[A/f   Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; 
FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
 

rule 4128

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Tan[e + 
 f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Sim 
p[1/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e 
 + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c*m + a*d* 
(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b 
*(d*(B*c - A*d)*(m + n + 1) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], 
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ 
[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 4136

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^ 
n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ 
(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ 
e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 
 C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & 
&  !GtQ[n, 0] &&  !LeQ[n, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(337\) vs. \(2(136)=272\).

Time = 1.78 (sec) , antiderivative size = 338, normalized size of antiderivative = 2.05

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (\frac {-\frac {9 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}-\frac {7 d \sqrt {d \tan \left (f x +e \right )}}{4}}{4 \left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {11 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{16 \sqrt {d}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f \,a^{3}}\)	\(338\)
default	\(\frac {2 d^{4} \left (\frac {-\frac {9 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}-\frac {7 d \sqrt {d \tan \left (f x +e \right )}}{4}}{4 \left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {11 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{16 \sqrt {d}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{32 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f \,a^{3}}\)	\(338\)

Input:

int((d*tan(f*x+e))^(7/2)/(a+a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
 

Output:

2/f/a^3*d^4*(1/4*(-9/4*(d*tan(f*x+e))^(3/2)-7/4*d*(d*tan(f*x+e))^(1/2))/(d 
*tan(f*x+e)+d)^2+11/16/d^(1/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))-1/32/d 
*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^ 
(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+ 
(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arcta 
n(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))-1/32/(d^2)^(1/4)*2^(1/2)*( 
ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d* 
tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan 
(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4) 
*(d*tan(f*x+e))^(1/2)+1)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.60 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+a \tan (e+f x))^3} \, dx=\left [\frac {4 \, \sqrt {\frac {1}{2}} {\left (d^{3} \tan \left (f x + e\right )^{2} + 2 \, d^{3} \tan \left (f x + e\right ) + d^{3}\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 4 \, \sqrt {\frac {1}{2}} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) - 1\right )} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 11 \, {\left (d^{3} \tan \left (f x + e\right )^{2} + 2 \, d^{3} \tan \left (f x + e\right ) + d^{3}\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) - 2 \, {\left (9 \, d^{3} \tan \left (f x + e\right ) + 7 \, d^{3}\right )} \sqrt {d \tan \left (f x + e\right )}}{16 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}, -\frac {4 \, \sqrt {\frac {1}{2}} {\left (d^{3} \tan \left (f x + e\right )^{2} + 2 \, d^{3} \tan \left (f x + e\right ) + d^{3}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {\frac {1}{2}} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) + 11 \, {\left (d^{3} \tan \left (f x + e\right )^{2} + 2 \, d^{3} \tan \left (f x + e\right ) + d^{3}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) + {\left (9 \, d^{3} \tan \left (f x + e\right ) + 7 \, d^{3}\right )} \sqrt {d \tan \left (f x + e\right )}}{8 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}\right ] \] Input:

integrate((d*tan(f*x+e))^(7/2)/(a+a*tan(f*x+e))^3,x, algorithm="fricas")
 

Output:

[1/16*(4*sqrt(1/2)*(d^3*tan(f*x + e)^2 + 2*d^3*tan(f*x + e) + d^3)*sqrt(-d 
)*log((d*tan(f*x + e)^2 - 4*sqrt(1/2)*sqrt(d*tan(f*x + e))*sqrt(-d)*(tan(f 
*x + e) - 1) - 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) + 11*(d^3*tan(f 
*x + e)^2 + 2*d^3*tan(f*x + e) + d^3)*sqrt(-d)*log((d*tan(f*x + e) + 2*sqr 
t(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1)) - 2*(9*d^3*tan(f*x + e 
) + 7*d^3)*sqrt(d*tan(f*x + e)))/(a^3*f*tan(f*x + e)^2 + 2*a^3*f*tan(f*x + 
 e) + a^3*f), -1/8*(4*sqrt(1/2)*(d^3*tan(f*x + e)^2 + 2*d^3*tan(f*x + e) + 
 d^3)*sqrt(d)*arctan(sqrt(1/2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) - 1)/(sq 
rt(d)*tan(f*x + e))) + 11*(d^3*tan(f*x + e)^2 + 2*d^3*tan(f*x + e) + d^3)* 
sqrt(d)*arctan(sqrt(d*tan(f*x + e))/(sqrt(d)*tan(f*x + e))) + (9*d^3*tan(f 
*x + e) + 7*d^3)*sqrt(d*tan(f*x + e)))/(a^3*f*tan(f*x + e)^2 + 2*a^3*f*tan 
(f*x + e) + a^3*f)]
 

Sympy [F]

\[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+a \tan (e+f x))^3} \, dx=\frac {\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}{\tan ^{3}{\left (e + f x \right )} + 3 \tan ^{2}{\left (e + f x \right )} + 3 \tan {\left (e + f x \right )} + 1}\, dx}{a^{3}} \] Input:

integrate((d*tan(f*x+e))**(7/2)/(a+a*tan(f*x+e))**3,x)
 

Output:

Integral((d*tan(e + f*x))**(7/2)/(tan(e + f*x)**3 + 3*tan(e + f*x)**2 + 3* 
tan(e + f*x) + 1), x)/a**3
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.12 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+a \tan (e+f x))^3} \, dx=-\frac {\frac {2 \, d^{5} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )}}{a^{3}} - \frac {11 \, d^{\frac {9}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3}} + \frac {9 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{5} + 7 \, \sqrt {d \tan \left (f x + e\right )} d^{6}}{a^{3} d^{2} \tan \left (f x + e\right )^{2} + 2 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}}}{8 \, d f} \] Input:

integrate((d*tan(f*x+e))^(7/2)/(a+a*tan(f*x+e))^3,x, algorithm="maxima")
 

Output:

-1/8*(2*d^5*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f* 
x + e)))/sqrt(d))/sqrt(d) + sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 
 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d))/a^3 - 11*d^(9/2)*arctan(sqrt(d* 
tan(f*x + e))/sqrt(d))/a^3 + (9*(d*tan(f*x + e))^(3/2)*d^5 + 7*sqrt(d*tan( 
f*x + e))*d^6)/(a^3*d^2*tan(f*x + e)^2 + 2*a^3*d^2*tan(f*x + e) + a^3*d^2) 
)/(d*f)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+a \tan (e+f x))^3} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((d*tan(f*x+e))^(7/2)/(a+a*tan(f*x+e))^3,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 1.86 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.08 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+a \tan (e+f x))^3} \, dx=\frac {11\,d^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {\frac {7\,d^5\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8}+\frac {9\,d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8}}{f\,a^3\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,f\,a^3\,d^2\,\mathrm {tan}\left (e+f\,x\right )+f\,a^3\,d^2}-\frac {\sqrt {2}\,d^{7/2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{8\,a^3\,f} \] Input:

int((d*tan(e + f*x))^(7/2)/(a + a*tan(e + f*x))^3,x)
 

Output:

(11*d^(7/2)*atan((d*tan(e + f*x))^(1/2)/d^(1/2)))/(8*a^3*f) - ((7*d^5*(d*t 
an(e + f*x))^(1/2))/8 + (9*d^4*(d*tan(e + f*x))^(3/2))/8)/(a^3*d^2*f + a^3 
*d^2*f*tan(e + f*x)^2 + 2*a^3*d^2*f*tan(e + f*x)) - (2^(1/2)*d^(7/2)*(2*at 
an((2^(1/2)*(d*tan(e + f*x))^(1/2))/(2*d^(1/2))) + 2*atan((2^(1/2)*(d*tan( 
e + f*x))^(1/2))/(2*d^(1/2)) + (2^(1/2)*(d*tan(e + f*x))^(3/2))/(2*d^(3/2) 
))))/(8*a^3*f)
 

Reduce [F]

\[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+a \tan (e+f x))^3} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{3}+3 \tan \left (f x +e \right )^{2}+3 \tan \left (f x +e \right )+1}d x \right ) d^{3}}{a^{3}} \] Input:

int((d*tan(f*x+e))^(7/2)/(a+a*tan(f*x+e))^3,x)
 

Output:

(sqrt(d)*int((sqrt(tan(e + f*x))*tan(e + f*x)**3)/(tan(e + f*x)**3 + 3*tan 
(e + f*x)**2 + 3*tan(e + f*x) + 1),x)*d**3)/a**3