\(\int \frac {\tan ^4(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [353]

Optimal result

Integrand size = 25, antiderivative size = 131 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{5/2} f}-\frac {a \tan (e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(a-4 b) \tan (e+f x)}{3 (a-b)^2 b f \sqrt {a+b \tan ^2(e+f x)}} \] Output:

arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(5/2)/f-1/3* 
a*tan(f*x+e)/(a-b)/b/f/(a+b*tan(f*x+e)^2)^(3/2)+1/3*(a-4*b)*tan(f*x+e)/(a- 
b)^2/b/f/(a+b*tan(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 6.03 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.98 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\tan ^5(e+f x) \left (1+\frac {b \tan ^2(e+f x)}{a}\right ) \left (\frac {\text {arctanh}\left (\frac {\sqrt {-\tan ^2(e+f x)+\frac {b \tan ^2(e+f x)}{a}}}{\sqrt {1+\frac {b \tan ^2(e+f x)}{a}}}\right ) \sqrt {-\tan ^2(e+f x)+\frac {b \tan ^2(e+f x)}{a}}}{\sqrt {1+\frac {b \tan ^2(e+f x)}{a}}}-\frac {-\tan ^2(e+f x)+\frac {b \tan ^2(e+f x)}{a}}{1+\frac {b \tan ^2(e+f x)}{a}}-\frac {\left (-\tan ^2(e+f x)+\frac {b \tan ^2(e+f x)}{a}\right )^2}{3 \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^2}\right )}{a^2 f \sqrt {a+b \tan ^2(e+f x)} \left (-\tan ^2(e+f x)+\frac {b \tan ^2(e+f x)}{a}\right )^3} \] Input:

Integrate[Tan[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(5/2),x]
 

Output:

(Tan[e + f*x]^5*(1 + (b*Tan[e + f*x]^2)/a)*((ArcTanh[Sqrt[-Tan[e + f*x]^2 
+ (b*Tan[e + f*x]^2)/a]/Sqrt[1 + (b*Tan[e + f*x]^2)/a]]*Sqrt[-Tan[e + f*x] 
^2 + (b*Tan[e + f*x]^2)/a])/Sqrt[1 + (b*Tan[e + f*x]^2)/a] - (-Tan[e + f*x 
]^2 + (b*Tan[e + f*x]^2)/a)/(1 + (b*Tan[e + f*x]^2)/a) - (-Tan[e + f*x]^2 
+ (b*Tan[e + f*x]^2)/a)^2/(3*(1 + (b*Tan[e + f*x]^2)/a)^2)))/(a^2*f*Sqrt[a 
 + b*Tan[e + f*x]^2]*(-Tan[e + f*x]^2 + (b*Tan[e + f*x]^2)/a)^3)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4153, 372, 402, 27, 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.

\(\displaystyle \int \frac {\tan ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (e+f x)^4}{\left (a+b \tan (e+f x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {3 b \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2}}+\frac {(a-4 b) \tan (e+f x)}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}}{3 b (a-b)}-\frac {a \tan (e+f x)}{3 b (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}}{f}\)

Input:

Int[Tan[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(5/2),x]
 

Output:

(-1/3*(a*Tan[e + f*x])/((a - b)*b*(a + b*Tan[e + f*x]^2)^(3/2)) + ((3*b*Ar 
cTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(a - b)^(3/2) 
 + ((a - 4*b)*Tan[e + f*x])/((a - b)*Sqrt[a + b*Tan[e + f*x]^2]))/(3*(a - 
b)*b))/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 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 372

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a 
, b, c, d, e, m, 2, p, q, x]
 

rule 402

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

rule 3042

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

rule 4153

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], 
 x]}, Simp[c*(ff/f)   Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, 
n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio 
nalQ[n]))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(290\) vs. \(2(117)=234\).

Time = 0.20 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.22

method	result	size
derivativedivides	\(-\frac {\tan \left (f x +e \right )}{3 f b \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {\tan \left (f x +e \right )}{3 f a b \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {b \tan \left (f x +e \right )}{3 a \left (a -b \right ) f \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {2 b \tan \left (f x +e \right )}{3 f \left (a -b \right ) a^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{f \left (a -b \right )^{3} b^{2}}-\frac {b \tan \left (f x +e \right )}{f \left (a -b \right )^{2} a \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {\tan \left (f x +e \right )}{3 f a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {2 \tan \left (f x +e \right )}{3 f \,a^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\)	\(291\)
default	\(-\frac {\tan \left (f x +e \right )}{3 f b \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {\tan \left (f x +e \right )}{3 f a b \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {b \tan \left (f x +e \right )}{3 a \left (a -b \right ) f \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {2 b \tan \left (f x +e \right )}{3 f \left (a -b \right ) a^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{f \left (a -b \right )^{3} b^{2}}-\frac {b \tan \left (f x +e \right )}{f \left (a -b \right )^{2} a \sqrt {a +b \tan \left (f x +e \right )^{2}}}-\frac {\tan \left (f x +e \right )}{3 f a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {2 \tan \left (f x +e \right )}{3 f \,a^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\)	\(291\)

Input:

int(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3/f/b*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(3/2)+1/3/f/a/b*tan(f*x+e)/(a+b*tan 
(f*x+e)^2)^(1/2)-1/3*b*tan(f*x+e)/a/(a-b)/f/(a+b*tan(f*x+e)^2)^(3/2)-2/3/f 
/(a-b)*b/a^2*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2)+1/f/(a-b)^3*(b^4*(a-b))^( 
1/2)/b^2*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*tan(f*x+e)^2)^(1/2)*tan(f 
*x+e))-1/f/(a-b)^2*b*tan(f*x+e)/a/(a+b*tan(f*x+e)^2)^(1/2)-1/3/f*tan(f*x+e 
)/a/(a+b*tan(f*x+e)^2)^(3/2)-2/3/f/a^2*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 495, normalized size of antiderivative = 3.78 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left (b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, {\left ({\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )^{3} - 3 \, {\left (a^{2} - a b\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left ({\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} f\right )}}, \frac {3 \, {\left (b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {a - b} \arctan \left (\frac {\sqrt {a - b} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a}}\right ) + {\left ({\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )^{3} - 3 \, {\left (a^{2} - a b\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{3 \, {\left ({\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} f\right )}}\right ] \] Input:

integrate(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

Output:

[-1/6*(3*(b^2*tan(f*x + e)^4 + 2*a*b*tan(f*x + e)^2 + a^2)*sqrt(-a + b)*lo 
g(-((a - 2*b)*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)*t 
an(f*x + e) - a)/(tan(f*x + e)^2 + 1)) - 2*((a^2 - 5*a*b + 4*b^2)*tan(f*x 
+ e)^3 - 3*(a^2 - a*b)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/((a^3*b^2 
 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3* 
a^2*b^3 - a*b^4)*f*tan(f*x + e)^2 + (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)* 
f), 1/3*(3*(b^2*tan(f*x + e)^4 + 2*a*b*tan(f*x + e)^2 + a^2)*sqrt(a - b)*a 
rctan(sqrt(a - b)*tan(f*x + e)/sqrt(b*tan(f*x + e)^2 + a)) + ((a^2 - 5*a*b 
 + 4*b^2)*tan(f*x + e)^3 - 3*(a^2 - a*b)*tan(f*x + e))*sqrt(b*tan(f*x + e) 
^2 + a))/((a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f*tan(f*x + e)^4 + 2*(a^4* 
b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*f*tan(f*x + e)^2 + (a^5 - 3*a^4*b + 3*a 
^3*b^2 - a^2*b^3)*f)]
 

Sympy [F]

\[ \int \frac {\tan ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\tan ^{4}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(tan(f*x+e)**4/(a+b*tan(f*x+e)**2)**(5/2),x)
 

Output:

Integral(tan(e + f*x)**4/(a + b*tan(e + f*x)**2)**(5/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more 
details)Is
 

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \] Input:

int(tan(e + f*x)^4/(a + b*tan(e + f*x)^2)^(5/2),x)
 

Output:

int(tan(e + f*x)^4/(a + b*tan(e + f*x)^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{3}-3 \left (\int \frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{6} b^{3}+3 \tan \left (f x +e \right )^{4} a \,b^{2}+3 \tan \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \right ) \tan \left (f x +e \right )^{4} a \,b^{2} f -6 \left (\int \frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{6} b^{3}+3 \tan \left (f x +e \right )^{4} a \,b^{2}+3 \tan \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \right ) \tan \left (f x +e \right )^{2} a^{2} b f -3 \left (\int \frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{6} b^{3}+3 \tan \left (f x +e \right )^{4} a \,b^{2}+3 \tan \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \right ) a^{3} f}{3 a f \left (\tan \left (f x +e \right )^{4} b^{2}+2 \tan \left (f x +e \right )^{2} a b +a^{2}\right )} \] Input:

int(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x)
 

Output:

(sqrt(tan(e + f*x)**2*b + a)*tan(e + f*x)**3 - 3*int((sqrt(tan(e + f*x)**2 
*b + a)*tan(e + f*x)**2)/(tan(e + f*x)**6*b**3 + 3*tan(e + f*x)**4*a*b**2 
+ 3*tan(e + f*x)**2*a**2*b + a**3),x)*tan(e + f*x)**4*a*b**2*f - 6*int((sq 
rt(tan(e + f*x)**2*b + a)*tan(e + f*x)**2)/(tan(e + f*x)**6*b**3 + 3*tan(e 
 + f*x)**4*a*b**2 + 3*tan(e + f*x)**2*a**2*b + a**3),x)*tan(e + f*x)**2*a* 
*2*b*f - 3*int((sqrt(tan(e + f*x)**2*b + a)*tan(e + f*x)**2)/(tan(e + f*x) 
**6*b**3 + 3*tan(e + f*x)**4*a*b**2 + 3*tan(e + f*x)**2*a**2*b + a**3),x)* 
a**3*f)/(3*a*f*(tan(e + f*x)**4*b**2 + 2*tan(e + f*x)**2*a*b + a**2))