\(\int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx\) [605]

Optimal result

Integrand size = 25, antiderivative size = 178 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=-\frac {2 b \left (3 a^2-b^2\right )}{f \sqrt {d \sec (e+f x)}}+\frac {2 b^3 \sec ^2(e+f x)}{3 f \sqrt {d \sec (e+f x)}}+\frac {2 a \left (a^2-6 b^2\right ) E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{f \sqrt {d \sec (e+f x)}}-\frac {2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt {d \sec (e+f x)}}+\frac {2 a \left (a^2-3 b^2\right ) \tan (e+f x)}{f \sqrt {d \sec (e+f x)}} \] Output:

-2*b*(3*a^2-b^2)/f/(d*sec(f*x+e))^(1/2)+2/3*b^3*sec(f*x+e)^2/f/(d*sec(f*x+ 
e))^(1/2)+2*a*(a^2-6*b^2)*EllipticE(sin(1/2*arctan(tan(f*x+e))),2^(1/2))*( 
sec(f*x+e)^2)^(1/4)/f/(d*sec(f*x+e))^(1/2)-2*a*(a^2-6*b^2)*tan(f*x+e)/f/(d 
*sec(f*x+e))^(1/2)+2*a*(a^2-3*b^2)*tan(f*x+e)/f/(d*sec(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 2.85 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.73 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=\frac {d \left (6 a \left (a^2-6 b^2\right ) \cos ^{\frac {3}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+b \left (-9 a^2+5 b^2+\left (-9 a^2+3 b^2\right ) \cos (2 (e+f x))+9 a b \sin (2 (e+f x))\right )\right ) (a+b \tan (e+f x))^3}{3 f (d \sec (e+f x))^{3/2} (a \cos (e+f x)+b \sin (e+f x))^3} \] Input:

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

Output:

(d*(6*a*(a^2 - 6*b^2)*Cos[e + f*x]^(3/2)*EllipticE[(e + f*x)/2, 2] + b*(-9 
*a^2 + 5*b^2 + (-9*a^2 + 3*b^2)*Cos[2*(e + f*x)] + 9*a*b*Sin[2*(e + f*x)]) 
)*(a + b*Tan[e + f*x])^3)/(3*f*(d*Sec[e + f*x])^(3/2)*(a*Cos[e + f*x] + b* 
Sin[e + f*x])^3)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3994, 495, 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.

\(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}}dx\)

\(\Big \downarrow \) 3994

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

\(\Big \downarrow \) 495

\(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (2 b^2 \int \frac {(a+b \tan (e+f x)) \left (\left (4-\frac {a^2}{b^2}\right ) b^2-5 a b \tan (e+f x)\right )}{2 b^2 \sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{b f \sqrt {d \sec (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (\int -\frac {(a+b \tan (e+f x)) \left (a^2+5 b \tan (e+f x) a-4 b^2\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{b f \sqrt {d \sec (e+f x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (-\int \frac {(a+b \tan (e+f x)) \left (a^2+5 b \tan (e+f x) a-4 b^2\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))-\frac {2 \left (b^2-a b \tan (e+f x)\right ) (a+b \tan (e+f x))^2}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{b f \sqrt {d \sec (e+f x)}}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (-a \left (a^2-6 b^2\right ) \int \frac {1}{\sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))-\frac {4}{3} b^2 \left (3 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4}-2 a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{3/4}-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{b f \sqrt {d \sec (e+f x)}}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (-a \left (a^2-6 b^2\right ) \left (\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}-\int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{5/4}}d(b \tan (e+f x))\right )-\frac {4}{3} b^2 \left (3 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4}-2 a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{3/4}-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{b f \sqrt {d \sec (e+f x)}}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (-a \left (a^2-6 b^2\right ) \left (\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )\right )-\frac {4}{3} b^2 \left (3 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4}-2 a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{3/4}-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{b f \sqrt {d \sec (e+f x)}}\)

Input:

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

Output:

((Sec[e + f*x]^2)^(1/4)*((-2*(a + b*Tan[e + f*x])^2*(b^2 - a*b*Tan[e + f*x 
]))/(1 + Tan[e + f*x]^2)^(1/4) - (4*b^2*(3*a^2 - 2*b^2)*(1 + Tan[e + f*x]^ 
2)^(3/4))/3 - 2*a*b^3*Tan[e + f*x]*(1 + Tan[e + f*x]^2)^(3/4) - a*(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))))/(b*f*Sqrt[d*Sec[e + f*x]])
 

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 212

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]
 

rule 225

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]
 

rule 495

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(a*d - b*c*x)*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - 
 Simp[1/(2*a*b*(p + 1))   Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1)*Simp[a* 
d^2*(n - 1) - b*c^2*(2*p + 3) - b*c*d*(n + 2*p + 2)*x, x], x], x] /; FreeQ[ 
{a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 1] && IntQuadraticQ[a, 0, b, c, d, 
 n, p, x]
 

rule 676

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]
 

rule 3042

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

rule 3994

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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 21.91 (sec) , antiderivative size = 627, normalized size of antiderivative = 3.52

method	result	size
parts	\(\frac {2 a^{3} \left (i \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 ) \left (-\cos \left (f x +e \right )-2-\sec \left (f x +e \right )\right )+i \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 ) \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right )+\sin \left (f x +e \right )\right )}{f \left (1+\cos \left (f x +e \right )\right ) \sqrt {d \sec \left (f x +e \right )}}-\frac {b^{3} \left (3 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-\cos \left (f x +e \right )+1}{1+\cos \left (f x +e \right )}\right )-3 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-2 \cos \left (f x +e \right )+2}{1+\cos \left (f x +e \right )}\right )+\sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (-12 \cos \left (f x +e \right )-12-4 \sec \left (f x +e \right )-4 \sec \left (f x +e \right )^{2}\right )\right )}{6 f \left (1+\cos \left (f x +e \right )\right ) \sqrt {d \sec \left (f x +e \right )}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}-\frac {6 a \,b^{2} \left (\sin \left (f x +e \right )-\tan \left (f x +e \right )-2 i \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 )}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+2 i \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 ) \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right )\right )}{f \left (1+\cos \left (f x +e \right )\right ) \sqrt {d \sec \left (f x +e \right )}}-\frac {6 a^{2} b}{\sqrt {d \sec \left (f x +e \right )}\, f}\)	\(627\)
default	\(\text {Expression too large to display}\)	\(894\)

Input:

int((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2*a^3/f/(1+cos(f*x+e))/(d*sec(f*x+e))^(1/2)*(I*(1/(1+cos(f*x+e)))^(1/2)*(c 
os(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(-c 
os(f*x+e)-2-sec(f*x+e))+I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+ 
e)))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(2+cos(f*x+e)+sec(f*x+e) 
)+sin(f*x+e))-1/6*b^3/f/(1+cos(f*x+e))/(d*sec(f*x+e))^(1/2)/(-cos(f*x+e)/( 
1+cos(f*x+e))^2)^(1/2)*(3*ln((2*cos(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^ 
(1/2)+2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)-cos(f*x+e)+1)/(1+cos(f*x+e))) 
-3*ln(2*(2*cos(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*(-cos(f*x+e)/ 
(1+cos(f*x+e))^2)^(1/2)-cos(f*x+e)+1)/(1+cos(f*x+e)))+(-cos(f*x+e)/(1+cos( 
f*x+e))^2)^(1/2)*(-12*cos(f*x+e)-12-4*sec(f*x+e)-4*sec(f*x+e)^2))-6*a*b^2/ 
f/(1+cos(f*x+e))/(d*sec(f*x+e))^(1/2)*(sin(f*x+e)-tan(f*x+e)-2*I*(1/(1+cos 
(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(2+cos(f*x+e)+sec(f*x+e) 
)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)+2*I*(1/(1+cos(f*x+e)))^(1/2)*(cos 
(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(2+co 
s(f*x+e)+sec(f*x+e)))-6*a^2*b/(d*sec(f*x+e))^(1/2)/f
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.95 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=-\frac {3 \, \sqrt {2} {\left (-i \, a^{3} + 6 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, a^{3} - 6 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 2 \, {\left (9 \, a b^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b^{3} - 3 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, d f \cos \left (f x + e\right )} \] Input:

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(1/2),x, algorithm="fricas")
 

Output:

-1/3*(3*sqrt(2)*(-I*a^3 + 6*I*a*b^2)*sqrt(d)*cos(f*x + e)*weierstrassZeta( 
-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + 3*sqrt 
(2)*(I*a^3 - 6*I*a*b^2)*sqrt(d)*cos(f*x + e)*weierstrassZeta(-4, 0, weiers 
trassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - 2*(9*a*b^2*cos(f*x 
+ e)*sin(f*x + e) + b^3 - 3*(3*a^2*b - b^3)*cos(f*x + e)^2)*sqrt(d/cos(f*x 
 + e)))/(d*f*cos(f*x + e))
 

Sympy [F]

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

integrate((a+b*tan(f*x+e))**3/(d*sec(f*x+e))**(1/2),x)
 

Output:

Integral((a + b*tan(e + f*x))**3/sqrt(d*sec(e + f*x)), x)
 

Maxima [F]

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

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(d)*(int(sqrt(sec(e + f*x))/sec(e + f*x),x)*a**3 + int((sqrt(sec(e + 
f*x))*tan(e + f*x)**3)/sec(e + f*x),x)*b**3 + 3*int((sqrt(sec(e + f*x))*ta 
n(e + f*x)**2)/sec(e + f*x),x)*a*b**2 + 3*int((sqrt(sec(e + f*x))*tan(e + 
f*x))/sec(e + f*x),x)*a**2*b))/d