Integrand size = 25, antiderivative size = 190 \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^3 \, dx=\frac {2 b \left (3 a^2-b^2\right ) (d \sec (e+f x))^{3/2}}{3 f}+\frac {2 b^3 \sec ^2(e+f x) (d \sec (e+f x))^{3/2}}{7 f}-\frac {2 a \left (5 a^2-6 b^2\right ) E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{5 f \sec ^2(e+f x)^{3/4}}+\frac {2 a \left (5 a^2-6 b^2\right ) \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{5 f}+\frac {6 a b^2 (d \sec (e+f x))^{3/2} \tan (e+f x)}{5 f} \] Output:
2/3*b*(3*a^2-b^2)*(d*sec(f*x+e))^(3/2)/f+2/7*b^3*sec(f*x+e)^2*(d*sec(f*x+e ))^(3/2)/f-2/5*a*(5*a^2-6*b^2)*EllipticE(sin(1/2*arctan(tan(f*x+e))),2^(1/ 2))*(d*sec(f*x+e))^(3/2)/f/(sec(f*x+e)^2)^(3/4)+2/5*a*(5*a^2-6*b^2)*cos(f* x+e)*(d*sec(f*x+e))^(3/2)*sin(f*x+e)/f+6/5*a*b^2*(d*sec(f*x+e))^(3/2)*tan( f*x+e)/f
Time = 2.22 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.82 \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^3 \, dx=-\frac {d \sqrt {d \sec (e+f x)} \left (70 b \left (-3 a^2+b^2\right ) \cos ^2(e+f x)+42 a \left (5 a^2-6 b^2\right ) \cos ^{\frac {7}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )-42 a \left (5 a^2-6 b^2\right ) \cos ^3(e+f x) \sin (e+f x)-3 b^2 (10 b+21 a \sin (2 (e+f x)))\right ) (a+b \tan (e+f x))^3}{105 f (a \cos (e+f x)+b \sin (e+f x))^3} \] Input:
Integrate[(d*Sec[e + f*x])^(3/2)*(a + b*Tan[e + f*x])^3,x]
Output:
-1/105*(d*Sqrt[d*Sec[e + f*x]]*(70*b*(-3*a^2 + b^2)*Cos[e + f*x]^2 + 42*a* (5*a^2 - 6*b^2)*Cos[e + f*x]^(7/2)*EllipticE[(e + f*x)/2, 2] - 42*a*(5*a^2 - 6*b^2)*Cos[e + f*x]^3*Sin[e + f*x] - 3*b^2*(10*b + 21*a*Sin[2*(e + f*x) ]))*(a + b*Tan[e + f*x])^3)/(f*(a*Cos[e + f*x] + b*Sin[e + f*x])^3)
Time = 0.39 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3994, 497, 27, 25, 676, 225, 212}
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.
| \(\displaystyle \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^3dx\) |
| \(\Big \downarrow \) 3994 |
| \(\displaystyle \frac {(d \sec (e+f x))^{3/2} \int \frac {(a+b \tan (e+f x))^3}{\sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))}{b f \sec ^2(e+f x)^{3/4}}\) |
| \(\Big \downarrow \) 497 |
| \(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {2}{7} b^2 \int -\frac {(a+b \tan (e+f x)) \left (\left (4-\frac {7 a^2}{b^2}\right ) b^2-11 a b \tan (e+f x)\right )}{2 b^2 \sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))+\frac {2}{7} b^2 \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))^2\right )}{b f \sec ^2(e+f x)^{3/4}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {2}{7} b^2 \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))^2-\frac {1}{7} \int -\frac {(a+b \tan (e+f x)) \left (7 a^2+11 b \tan (e+f x) a-4 b^2\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))\right )}{b f \sec ^2(e+f x)^{3/4}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {1}{7} \int \frac {(a+b \tan (e+f x)) \left (7 a^2+11 b \tan (e+f x) a-4 b^2\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))+\frac {2}{7} b^2 \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))^2\right )}{b f \sec ^2(e+f x)^{3/4}}\) |
| \(\Big \downarrow \) 676 |
| \(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {1}{7} \left (\frac {7}{5} a \left (5 a^2-6 b^2\right ) \int \frac {1}{\sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))+\frac {4}{3} b^2 \left (9 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4}+\frac {22}{5} a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{3/4}\right )+\frac {2}{7} b^2 \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))^2\right )}{b f \sec ^2(e+f x)^{3/4}}\) |
| \(\Big \downarrow \) 225 |
| \(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {1}{7} \left (\frac {7}{5} a \left (5 a^2-6 b^2\right ) \left (\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}-\int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{5/4}}d(b \tan (e+f x))\right )+\frac {4}{3} b^2 \left (9 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4}+\frac {22}{5} a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{3/4}\right )+\frac {2}{7} b^2 \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))^2\right )}{b f \sec ^2(e+f x)^{3/4}}\) |
| \(\Big \downarrow \) 212 |
| \(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {1}{7} \left (\frac {7}{5} a \left (5 a^2-6 b^2\right ) \left (\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )\right )+\frac {4}{3} b^2 \left (9 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4}+\frac {22}{5} a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{3/4}\right )+\frac {2}{7} b^2 \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))^2\right )}{b f \sec ^2(e+f x)^{3/4}}\) |
Input:
Int[(d*Sec[e + f*x])^(3/2)*(a + b*Tan[e + f*x])^3,x]
Output:
((d*Sec[e + f*x])^(3/2)*((2*b^2*(a + b*Tan[e + f*x])^2*(1 + Tan[e + f*x]^2 )^(3/4))/7 + ((4*b^2*(9*a^2 - 2*b^2)*(1 + Tan[e + f*x]^2)^(3/4))/3 + (22*a *b^3*Tan[e + f*x]*(1 + Tan[e + f*x]^2)^(3/4))/5 + (7*a*(5*a^2 - 6*b^2)*(-2 *b*EllipticE[ArcTan[Tan[e + f*x]]/2, 2] + (2*b*Tan[e + f*x])/(1 + Tan[e + f*x]^2)^(1/4)))/5)/7))/(b*f*(Sec[e + f*x]^2)^(3/4))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) , x] - Simp[a Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ a, 0] && PosQ[b/a]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[1/(b *(n + 2*p + 1)) Int[(c + d*x)^(n - 2)*(a + b*x^2)^p*Simp[b*c^2*(n + 2*p + 1) - a*d^2*(n - 1) + 2*b*c*d*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, n , p}, x] && If[RationalQ[n], GtQ[n, 1], SumSimplerQ[n, -2]] && NeQ[n + 2*p + 1, 0] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x _Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p + 3))/(c*(2*p + 3)) Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g , p}, x] && !LeQ[p, -1]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[d^(2*IntPart[m/2])*((d*Sec[e + f*x])^(2*FracP art[m/2])/(b*f*(Sec[e + f*x]^2)^FracPart[m/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, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] && !IntegerQ[m] && IntegerQ[n]
Result contains complex when optimal does not.
Time = 26.26 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.30
| method | result | size |
| default | \(-\frac {2 d \sqrt {d \sec \left (f x +e \right )}\, \left (105 i \left (\cos \left (f x +e \right )^{2}+2 \cos \left (f x +e \right )+1\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, a^{3} \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+126 i \left (-\cos \left (f x +e \right )^{2}-2 \cos \left (f x +e \right )-1\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, a \,b^{2} \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+105 i \left (-\cos \left (f x +e \right )^{2}-2 \cos \left (f x +e \right )-1\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, a^{3} \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+126 i \left (\cos \left (f x +e \right )^{2}+2 \cos \left (f x +e \right )+1\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, a \,b^{2} \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )-105 a^{3} \sin \left (f x +e \right )+105 b \,a^{2} \left (-1-\sec \left (f x +e \right )\right )+63 a \,b^{2} \left (2 \sin \left (f x +e \right )-\tan \left (f x +e \right )-\sec \left (f x +e \right ) \tan \left (f x +e \right )\right )+5 b^{3} \left (-3 \sec \left (f x +e \right )^{3}-3 \sec \left (f x +e \right )^{2}+7 \sec \left (f x +e \right )+7\right )\right )}{105 f \left (1+\cos \left (f x +e \right )\right )}\) | \(437\) |
| parts | \(\frac {2 a^{3} \left (i \left (\cos \left (f x +e \right )^{2}+2 \cos \left (f x +e \right )+1\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+i \left (-\cos \left (f x +e \right )^{2}-2 \cos \left (f x +e \right )-1\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+\sin \left (f x +e \right )\right ) \sqrt {d \sec \left (f x +e \right )}\, d}{f \left (1+\cos \left (f x +e \right )\right )}+\frac {2 b^{3} \left (\frac {\left (d \sec \left (f x +e \right )\right )^{\frac {7}{2}}}{7}-\frac {d^{2} \left (d \sec \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f \,d^{2}}+\frac {2 a^{2} b \left (d \sec \left (f x +e \right )\right )^{\frac {3}{2}}}{f}-\frac {6 a \,b^{2} \sqrt {d \sec \left (f x +e \right )}\, d \left (2 \sin \left (f x +e \right )-\tan \left (f x +e \right )-\sec \left (f x +e \right ) \tan \left (f x +e \right )+i \left (2 \cos \left (f x +e \right )^{2}+4 \cos \left (f x +e \right )+2\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+i \left (-2 \cos \left (f x +e \right )^{2}-4 \cos \left (f x +e \right )-2\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )\right )}{5 f \left (1+\cos \left (f x +e \right )\right )}\) | \(455\) |
Input:
int((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
Output:
-2/105*d/f*(d*sec(f*x+e))^(1/2)/(1+cos(f*x+e))*(105*I*(cos(f*x+e)^2+2*cos( f*x+e)+1)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a^3*E llipticE(I*(csc(f*x+e)-cot(f*x+e)),I)+126*I*(-cos(f*x+e)^2-2*cos(f*x+e)-1) *(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a*b^2*Elliptic E(I*(csc(f*x+e)-cot(f*x+e)),I)+105*I*(-cos(f*x+e)^2-2*cos(f*x+e)-1)*(1/(1+ cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a^3*EllipticF(I*(csc( f*x+e)-cot(f*x+e)),I)+126*I*(cos(f*x+e)^2+2*cos(f*x+e)+1)*(1/(1+cos(f*x+e) ))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a*b^2*EllipticF(I*(csc(f*x+e)-c ot(f*x+e)),I)-105*a^3*sin(f*x+e)+105*b*a^2*(-1-sec(f*x+e))+63*a*b^2*(2*sin (f*x+e)-tan(f*x+e)-sec(f*x+e)*tan(f*x+e))+5*b^3*(-3*sec(f*x+e)^3-3*sec(f*x +e)^2+7*sec(f*x+e)+7))
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.05 \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^3 \, dx=\frac {-21 i \, \sqrt {2} {\left (5 \, a^{3} - 6 \, a b^{2}\right )} d^{\frac {3}{2}} \cos \left (f x + e\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 21 i \, \sqrt {2} {\left (5 \, a^{3} - 6 \, a b^{2}\right )} d^{\frac {3}{2}} \cos \left (f x + e\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, {\left (15 \, b^{3} d + 35 \, {\left (3 \, a^{2} b - b^{3}\right )} d \cos \left (f x + e\right )^{2} + 21 \, {\left (3 \, a b^{2} d \cos \left (f x + e\right ) + {\left (5 \, a^{3} - 6 \, a b^{2}\right )} d \cos \left (f x + e\right )^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{105 \, f \cos \left (f x + e\right )^{3}} \] Input:
integrate((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e))^3,x, algorithm="fricas")
Output:
1/105*(-21*I*sqrt(2)*(5*a^3 - 6*a*b^2)*d^(3/2)*cos(f*x + e)^3*weierstrassZ eta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + 21 *I*sqrt(2)*(5*a^3 - 6*a*b^2)*d^(3/2)*cos(f*x + e)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) + 2*(15*b^3*d + 35*(3*a^2*b - b^3)*d*cos(f*x + e)^2 + 21*(3*a*b^2*d*cos(f*x + e) + (5*a^ 3 - 6*a*b^2)*d*cos(f*x + e)^3)*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(f*cos( f*x + e)^3)
\[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^3 \, dx=\int \left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (a + b \tan {\left (e + f x \right )}\right )^{3}\, dx \] Input:
integrate((d*sec(f*x+e))**(3/2)*(a+b*tan(f*x+e))**3,x)
Output:
Integral((d*sec(e + f*x))**(3/2)*(a + b*tan(e + f*x))**3, x)
Timed out. \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^3 \, dx=\text {Timed out} \] Input:
integrate((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e))^3,x, algorithm="maxima")
Output:
Timed out
\[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^3 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{3} \,d x } \] Input:
integrate((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e))^3,x, algorithm="giac")
Output:
integrate((d*sec(f*x + e))^(3/2)*(b*tan(f*x + e) + a)^3, x)
Timed out. \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^3 \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3 \,d x \] Input:
int((d/cos(e + f*x))^(3/2)*(a + b*tan(e + f*x))^3,x)
Output:
int((d/cos(e + f*x))^(3/2)*(a + b*tan(e + f*x))^3, x)
\[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^3 \, dx=\frac {\sqrt {d}\, d \left (6 \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right ) \tan \left (f x +e \right )^{2} b^{3}+42 \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right ) a^{2} b -8 \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right ) b^{3}+63 \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right ) \tan \left (f x +e \right )^{2}d x \right ) a \,b^{2} f +21 \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )d x \right ) a^{3} f \right )}{21 f} \] Input:
int((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e))^3,x)
Output:
(sqrt(d)*d*(6*sqrt(sec(e + f*x))*sec(e + f*x)*tan(e + f*x)**2*b**3 + 42*sq rt(sec(e + f*x))*sec(e + f*x)*a**2*b - 8*sqrt(sec(e + f*x))*sec(e + f*x)*b **3 + 63*int(sqrt(sec(e + f*x))*sec(e + f*x)*tan(e + f*x)**2,x)*a*b**2*f + 21*int(sqrt(sec(e + f*x))*sec(e + f*x),x)*a**3*f))/(21*f)