\(\int (a+b \sec ^2(e+f x))^{3/2} \tan ^6(e+f x) \, dx\) [395]

Optimal result

Integrand size = 25, antiderivative size = 290 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^6(e+f x) \, dx=-\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}+\frac {\left (3 a^4+20 a^3 b+90 a^2 b^2-60 a b^3-5 b^4\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{128 b^{5/2} f}-\frac {\left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{128 b^2 f}+\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{192 b f}+\frac {(9 a+b) \tan ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 f}+\frac {b \tan ^7(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f} \] Output:

-a^(3/2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f+1/128*(3* 
a^4+20*a^3*b+90*a^2*b^2-60*a*b^3-5*b^4)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b* 
tan(f*x+e)^2)^(1/2))/b^(5/2)/f-1/128*(3*a^3+17*a^2*b-55*a*b^2-5*b^3)*tan(f 
*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/b^2/f+1/192*(3*a^2-50*a*b-5*b^2)*tan(f*x+ 
e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/b/f+1/48*(9*a+b)*tan(f*x+e)^5*(a+b+b*tan(f 
*x+e)^2)^(1/2)/f+1/8*b*tan(f*x+e)^7*(a+b+b*tan(f*x+e)^2)^(1/2)/f
 

Mathematica [A] (verified)

Time = 4.37 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.22 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^6(e+f x) \, dx=-\frac {\left (128 a^{3/2} b^2 \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )-\frac {\left (3 a^4+20 a^3 b+90 a^2 b^2-60 a b^3-5 b^4\right ) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {b}}\right ) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{32 \sqrt {2} b^2 f (a+2 b+a \cos (2 e+2 f x))^{3/2}}-\frac {\left (90 a^3+498 a^2 b-1594 a b^2-626 b^3+\left (135 a^3+759 a^2 b-2303 a b^2+513 b^3\right ) \cos (2 (e+f x))+2 \left (27 a^3+159 a^2 b-523 a b^2-191 b^3\right ) \cos (4 (e+f x))+9 a^3 \cos (6 (e+f x))+57 a^2 b \cos (6 (e+f x))-337 a b^2 \cos (6 (e+f x))+15 b^3 \cos (6 (e+f x))\right ) \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \tan (e+f x)}{12288 b^2 f} \] Input:

Integrate[(a + b*Sec[e + f*x]^2)^(3/2)*Tan[e + f*x]^6,x]
 

Output:

-1/32*((128*a^(3/2)*b^2*ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e 
 + f*x]^2]] - ((3*a^4 + 20*a^3*b + 90*a^2*b^2 - 60*a*b^3 - 5*b^4)*ArcTanh[ 
(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]])/Sqrt[b])*Cos[e + f 
*x]^3*(a + b*Sec[e + f*x]^2)^(3/2))/(Sqrt[2]*b^2*f*(a + 2*b + a*Cos[2*e + 
2*f*x])^(3/2)) - ((90*a^3 + 498*a^2*b - 1594*a*b^2 - 626*b^3 + (135*a^3 + 
759*a^2*b - 2303*a*b^2 + 513*b^3)*Cos[2*(e + f*x)] + 2*(27*a^3 + 159*a^2*b 
 - 523*a*b^2 - 191*b^3)*Cos[4*(e + f*x)] + 9*a^3*Cos[6*(e + f*x)] + 57*a^2 
*b*Cos[6*(e + f*x)] - 337*a*b^2*Cos[6*(e + f*x)] + 15*b^3*Cos[6*(e + f*x)] 
)*Sec[e + f*x]^6*Sqrt[a + b*Sec[e + f*x]^2]*Tan[e + f*x])/(12288*b^2*f)
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4629, 2075, 379, 444, 27, 444, 27, 444, 398, 224, 219, 291, 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.

\(\displaystyle \int \tan ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^6 \left (a+b \sec (e+f x)^2\right )^{3/2}dx\)

\(\Big \downarrow \) 4629

\(\displaystyle \frac {\int \frac {\tan ^6(e+f x) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^{3/2}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 2075

\(\displaystyle \frac {\int \frac {\tan ^6(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 379

\(\displaystyle \frac {\frac {1}{8} \int \frac {\tan ^6(e+f x) \left (b (9 a+b) \tan ^2(e+f x)+(a+b) (8 a+b)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\frac {1}{8} b \tan ^7(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 444

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} (9 a+b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {\int \frac {b \tan ^4(e+f x) \left (5 (a+b) (9 a+b)-\left (3 a^2-50 b a-5 b^2\right ) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{6 b}\right )+\frac {1}{8} b \tan ^7(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} (9 a+b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {1}{6} \int \frac {\tan ^4(e+f x) \left (5 (a+b) (9 a+b)-\left (3 a^2-50 b a-5 b^2\right ) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)\right )+\frac {1}{8} b \tan ^7(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 444

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {\int -\frac {3 \tan ^2(e+f x) \left (\left (3 a^3+17 b a^2-55 b^2 a-5 b^3\right ) \tan ^2(e+f x)+(a+b) \left (3 a^2-50 b a-5 b^2\right )\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 b}+\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}\right )+\frac {1}{6} (9 a+b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{8} b \tan ^7(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {3 \int \frac {\tan ^2(e+f x) \left (\left (3 a^3+17 b a^2-55 b^2 a-5 b^3\right ) \tan ^2(e+f x)+(a+b) \left (3 a^2-50 b a-5 b^2\right )\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 b}\right )+\frac {1}{6} (9 a+b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{8} b \tan ^7(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 444

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {3 \left (\frac {\left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\int \frac {\left (3 a^4+20 b a^3+90 b^2 a^2-60 b^3 a-5 b^4\right ) \tan ^2(e+f x)+(a+b) \left (3 a^3+17 b a^2-55 b^2 a-5 b^3\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}\right )}{4 b}\right )+\frac {1}{6} (9 a+b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{8} b \tan ^7(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 398

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {3 \left (\frac {\left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\left (3 a^4+20 a^3 b+90 a^2 b^2-60 a b^3-5 b^4\right ) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-128 a^2 b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}\right )}{4 b}\right )+\frac {1}{6} (9 a+b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{8} b \tan ^7(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {3 \left (\frac {\left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\left (3 a^4+20 a^3 b+90 a^2 b^2-60 a b^3-5 b^4\right ) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-128 a^2 b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}\right )}{4 b}\right )+\frac {1}{6} (9 a+b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{8} b \tan ^7(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {3 \left (\frac {\left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\frac {\left (3 a^4+20 a^3 b+90 a^2 b^2-60 a b^3-5 b^4\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-128 a^2 b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}\right )}{4 b}\right )+\frac {1}{6} (9 a+b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{8} b \tan ^7(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {3 \left (\frac {\left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\frac {\left (3 a^4+20 a^3 b+90 a^2 b^2-60 a b^3-5 b^4\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-128 a^2 b^2 \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{2 b}\right )}{4 b}\right )+\frac {1}{6} (9 a+b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{8} b \tan ^7(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {3 \left (\frac {\left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\frac {\left (3 a^4+20 a^3 b+90 a^2 b^2-60 a b^3-5 b^4\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-128 a^{3/2} b^2 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 b}\right )}{4 b}\right )+\frac {1}{6} (9 a+b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{8} b \tan ^7(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

Input:

Int[(a + b*Sec[e + f*x]^2)^(3/2)*Tan[e + f*x]^6,x]
 

Output:

((b*Tan[e + f*x]^7*Sqrt[a + b + b*Tan[e + f*x]^2])/8 + (((9*a + b)*Tan[e + 
 f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/6 + (((3*a^2 - 50*a*b - 5*b^2)*Tan 
[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(4*b) - (3*(-1/2*(-128*a^(3/2) 
*b^2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]] + ((3*a 
^4 + 20*a^3*b + 90*a^2*b^2 - 60*a*b^3 - 5*b^4)*ArcTanh[(Sqrt[b]*Tan[e + f* 
x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/Sqrt[b])/b + ((3*a^3 + 17*a^2*b - 55* 
a*b^2 - 5*b^3)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*b)))/(4*b)) 
/6)/8)/f
 

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 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 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 224

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

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 379

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[d*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 
1)/(b*e*(m + 2*(p + q) + 1))), x] + Simp[1/(b*(m + 2*(p + q) + 1))   Int[(e 
*x)^m*(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*((b*c - a*d)*(m + 1) + b*c*2 
*(p + q)) + (d*(b*c - a*d)*(m + 1) + d*2*(q - 1)*(b*c - a*d) + b*c*d*2*(p + 
 q))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d, 0 
] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
 

rule 398

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

rule 444

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

rule 2075

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa 
ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi 
alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  ! 
BinomialMatchQ[{u, v}, x]
 

rule 3042

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

rule 4629

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f 
_.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim 
p[ff/f   Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 
)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte 
gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1643\) vs. \(2(260)=520\).

Time = 42.32 (sec) , antiderivative size = 1644, normalized size of antiderivative = 5.67

method	result	size
default	\(\text {Expression too large to display}\)	\(1644\)

Input:

int((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^6,x,method=_RETURNVERBOSE)
 

Output:

-1/768/f/(-a)^(1/2)/b^(13/2)*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/ 
(1+cos(f*x+e))^2)^(1/2)/(a*cos(f*x+e)^3+a*cos(f*x+e)^2+cos(f*x+e)*b+b)*(-9 
*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+b^(1 
/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x 
+e)+1))*cos(f*x+e)^3*(-a)^(1/2)*a^4*b^4-60*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^ 
2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f 
*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*cos(f*x+e)^3*(-a)^(1/2)* 
a^3*b^5-270*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos( 
f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a- 
b)/(sin(f*x+e)+1))*cos(f*x+e)^3*(-a)^(1/2)*a^2*b^6+180*ln(4*(b^(1/2)*((b+a 
*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e) 
^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*cos(f*x+e)^3 
*(-a)^(1/2)*a*b^7+15*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^( 
1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f* 
x+e)*a-a-b)/(sin(f*x+e)+1))*cos(f*x+e)^3*(-a)^(1/2)*b^8-9*ln(-4*(b^(1/2)*( 
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f* 
x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*cos(f*x+ 
e)^3*(-a)^(1/2)*a^4*b^4-60*ln(-4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e 
))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2) 
-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*cos(f*x+e)^3*(-a)^(1/2)*a^3*b^5-270*...
 

Fricas [A] (verification not implemented)

Time = 13.74 (sec) , antiderivative size = 1973, normalized size of antiderivative = 6.80 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^6(e+f x) \, dx=\text {Too large to display} \] Input:

integrate((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^6,x, algorithm="fricas")
 

Output:

[1/1536*(192*sqrt(-a)*a*b^3*cos(f*x + e)^7*log(128*a^4*cos(f*x + e)^8 - 25 
6*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x 
 + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b 
 + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 
- a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - 
(a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e 
)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) - 3*(3*a^4 + 20*a^3*b + 90*a^2*b^2 
- 60*a*b^3 - 5*b^4)*sqrt(b)*cos(f*x + e)^7*log(((a^2 - 6*a*b + b^2)*cos(f* 
x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 - 4*((a - b)*cos(f*x + e)^3 + 2*b* 
cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x 
+ e) + 8*b^2)/cos(f*x + e)^4) - 4*((9*a^3*b + 57*a^2*b^2 - 337*a*b^3 + 15* 
b^4)*cos(f*x + e)^6 - 2*(3*a^2*b^2 - 122*a*b^3 + 59*b^4)*cos(f*x + e)^4 - 
48*b^4 - 8*(9*a*b^3 - 17*b^4)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/ 
cos(f*x + e)^2)*sin(f*x + e))/(b^3*f*cos(f*x + e)^7), 1/768*(96*sqrt(-a)*a 
*b^3*cos(f*x + e)^7*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x 
 + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3* 
b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*c 
os(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 
 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^ 
2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e...
 

Sympy [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^6(e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{6}{\left (e + f x \right )}\, dx \] Input:

integrate((a+b*sec(f*x+e)**2)**(3/2)*tan(f*x+e)**6,x)
 

Output:

Integral((a + b*sec(e + f*x)**2)**(3/2)*tan(e + f*x)**6, x)
 

Maxima [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^6(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{6} \,d x } \] Input:

integrate((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^6,x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*tan(f*x + e)^6, x)
 

Giac [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^6(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{6} \,d x } \] Input:

integrate((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^6,x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*tan(f*x + e)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^6(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^6\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \] Input:

int(tan(e + f*x)^6*(a + b/cos(e + f*x)^2)^(3/2),x)
 

Output:

int(tan(e + f*x)^6*(a + b/cos(e + f*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^6(e+f x) \, dx=\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{6}d x \right ) b +\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{6}d x \right ) a \] Input:

int((a+b*sec(f*x+e)^2)^(3/2)*tan(f*x+e)^6,x)
 

Output:

int(sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**2*tan(e + f*x)**6,x)*b + int 
(sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x)**6,x)*a