\(\int \frac {x^3}{(a+b \tan (c+d x^2))^2} \, dx\) [19]

Optimal result

Integrand size = 18, antiderivative size = 202 \[ \int \frac {x^3}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx=-\frac {x^4}{4 \left (a^2+b^2\right )}+\frac {\left (b+2 a d x^2\right )^2}{8 a (a+i b) \left (a^2+b^2\right ) d^2}+\frac {b \left (b+2 a d x^2\right ) \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right )^2 d^2}-\frac {i a b \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right )^2 d^2}-\frac {b x^2}{2 \left (a^2+b^2\right ) d \left (a+b \tan \left (c+d x^2\right )\right )} \] Output:

-1/4*x^4/(a^2+b^2)+1/8*(2*a*d*x^2+b)^2/a/(a+I*b)/(a^2+b^2)/d^2+1/2*b*(2*a* 
d*x^2+b)*ln(1+(a^2+b^2)*exp(2*I*(d*x^2+c))/(a+I*b)^2)/(a^2+b^2)^2/d^2-1/2* 
I*a*b*polylog(2,-(a^2+b^2)*exp(2*I*(d*x^2+c))/(a+I*b)^2)/(a^2+b^2)^2/d^2-1 
/2*b*x^2/(a^2+b^2)/d/(a+b*tan(d*x^2+c))
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(460\) vs. \(2(202)=404\).

Time = 5.10 (sec) , antiderivative size = 460, normalized size of antiderivative = 2.28 \[ \int \frac {x^3}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx=\frac {\sec ^2\left (c+d x^2\right ) \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right ) \left (2 b^2 \left (a^2+b^2\right ) d x^2 \sin \left (c+d x^2\right )-a \left (a^2+b^2\right ) \left (c-d x^2\right ) \left (c+d x^2\right ) \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right )-2 b^2 \left (b \left (c+d x^2\right )-a \log \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right )\right ) \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right )+4 a b c \left (b \left (c+d x^2\right )-a \log \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right )\right ) \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right )-2 a b \left (\sqrt {1+\frac {a^2}{b^2}} b e^{i \arctan \left (\frac {a}{b}\right )} \left (c+d x^2\right )^2+a \left (-i \left (c+d x^2\right ) \left (\pi -2 \arctan \left (\frac {a}{b}\right )\right )-\pi \log \left (1+e^{-2 i \left (c+d x^2\right )}\right )-2 \left (c+d x^2+\arctan \left (\frac {a}{b}\right )\right ) \log \left (1-e^{2 i \left (c+d x^2+\arctan \left (\frac {a}{b}\right )\right )}\right )+\pi \log \left (\cos \left (c+d x^2\right )\right )+2 \arctan \left (\frac {a}{b}\right ) \log \left (\sin \left (c+d x^2+\arctan \left (\frac {a}{b}\right )\right )\right )+i \operatorname {PolyLog}\left (2,e^{2 i \left (c+d x^2+\arctan \left (\frac {a}{b}\right )\right )}\right )\right )\right ) \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right )\right )}{4 a \left (a^2+b^2\right )^2 d^2 \left (a+b \tan \left (c+d x^2\right )\right )^2} \] Input:

Integrate[x^3/(a + b*Tan[c + d*x^2])^2,x]
 

Output:

(Sec[c + d*x^2]^2*(a*Cos[c + d*x^2] + b*Sin[c + d*x^2])*(2*b^2*(a^2 + b^2) 
*d*x^2*Sin[c + d*x^2] - a*(a^2 + b^2)*(c - d*x^2)*(c + d*x^2)*(a*Cos[c + d 
*x^2] + b*Sin[c + d*x^2]) - 2*b^2*(b*(c + d*x^2) - a*Log[a*Cos[c + d*x^2] 
+ b*Sin[c + d*x^2]])*(a*Cos[c + d*x^2] + b*Sin[c + d*x^2]) + 4*a*b*c*(b*(c 
 + d*x^2) - a*Log[a*Cos[c + d*x^2] + b*Sin[c + d*x^2]])*(a*Cos[c + d*x^2] 
+ b*Sin[c + d*x^2]) - 2*a*b*(Sqrt[1 + a^2/b^2]*b*E^(I*ArcTan[a/b])*(c + d* 
x^2)^2 + a*((-I)*(c + d*x^2)*(Pi - 2*ArcTan[a/b]) - Pi*Log[1 + E^((-2*I)*( 
c + d*x^2))] - 2*(c + d*x^2 + ArcTan[a/b])*Log[1 - E^((2*I)*(c + d*x^2 + A 
rcTan[a/b]))] + Pi*Log[Cos[c + d*x^2]] + 2*ArcTan[a/b]*Log[Sin[c + d*x^2 + 
 ArcTan[a/b]]] + I*PolyLog[2, E^((2*I)*(c + d*x^2 + ArcTan[a/b]))]))*(a*Co 
s[c + d*x^2] + b*Sin[c + d*x^2])))/(4*a*(a^2 + b^2)^2*d^2*(a + b*Tan[c + d 
*x^2])^2)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {4234, 3042, 4216, 3042, 4215, 2620, 2715, 2838}

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.

\(\displaystyle \int \frac {x^3}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx\)

\(\Big \downarrow \) 4234

\(\displaystyle \frac {1}{2} \int \frac {x^2}{\left (a+b \tan \left (d x^2+c\right )\right )^2}dx^2\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \frac {x^2}{\left (a+b \tan \left (d x^2+c\right )\right )^2}dx^2\)

\(\Big \downarrow \) 4216

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {2 a d x^2+b}{a+b \tan \left (d x^2+c\right )}dx^2}{d \left (a^2+b^2\right )}-\frac {b x^2}{d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d x^2\right )\right )}-\frac {x^4}{2 \left (a^2+b^2\right )}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {2 a d x^2+b}{a+b \tan \left (d x^2+c\right )}dx^2}{d \left (a^2+b^2\right )}-\frac {b x^2}{d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d x^2\right )\right )}-\frac {x^4}{2 \left (a^2+b^2\right )}\right )\)

\(\Big \downarrow \) 4215

\(\displaystyle \frac {1}{2} \left (\frac {2 i b \int \frac {e^{2 i \left (d x^2+c\right )} \left (2 a d x^2+b\right )}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i \left (d x^2+c\right )}}dx^2+\frac {\left (2 a d x^2+b\right )^2}{4 a d (a+i b)}}{d \left (a^2+b^2\right )}-\frac {b x^2}{d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d x^2\right )\right )}-\frac {x^4}{2 \left (a^2+b^2\right )}\right )\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {1}{2} \left (\frac {2 i b \left (\frac {i a \int \log \left (\frac {e^{2 i \left (d x^2+c\right )} \left (a^2+b^2\right )}{(a+i b)^2}+1\right )dx^2}{a^2+b^2}-\frac {i \left (2 a d x^2+b\right ) \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {\left (2 a d x^2+b\right )^2}{4 a d (a+i b)}}{d \left (a^2+b^2\right )}-\frac {b x^2}{d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d x^2\right )\right )}-\frac {x^4}{2 \left (a^2+b^2\right )}\right )\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {1}{2} \left (\frac {2 i b \left (\frac {a \int \frac {\log \left (\frac {e^{2 i \left (d x^2+c\right )} \left (a^2+b^2\right )}{(a+i b)^2}+1\right )}{x^2}de^{2 i \left (d x^2+c\right )}}{2 d \left (a^2+b^2\right )}-\frac {i \left (2 a d x^2+b\right ) \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {\left (2 a d x^2+b\right )^2}{4 a d (a+i b)}}{d \left (a^2+b^2\right )}-\frac {b x^2}{d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d x^2\right )\right )}-\frac {x^4}{2 \left (a^2+b^2\right )}\right )\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {1}{2} \left (\frac {2 i b \left (-\frac {a \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (d x^2+c\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}-\frac {i \left (2 a d x^2+b\right ) \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {\left (2 a d x^2+b\right )^2}{4 a d (a+i b)}}{d \left (a^2+b^2\right )}-\frac {b x^2}{d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d x^2\right )\right )}-\frac {x^4}{2 \left (a^2+b^2\right )}\right )\)

Input:

Int[x^3/(a + b*Tan[c + d*x^2])^2,x]
 

Output:

(-1/2*x^4/(a^2 + b^2) + ((b + 2*a*d*x^2)^2/(4*a*(a + I*b)*d) + (2*I)*b*((( 
-1/2*I)*(b + 2*a*d*x^2)*Log[1 + ((a^2 + b^2)*E^((2*I)*(c + d*x^2)))/(a + I 
*b)^2])/((a^2 + b^2)*d) - (a*PolyLog[2, -(((a^2 + b^2)*E^((2*I)*(c + d*x^2 
)))/(a + I*b)^2)])/(2*(a^2 + b^2)*d)))/((a^2 + b^2)*d) - (b*x^2)/((a^2 + b 
^2)*d*(a + b*Tan[c + d*x^2])))/2
 

Defintions of rubi rules used

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

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

rule 4215

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[(c + d*x)^(m + 1)/(d*(m + 1)*(a + I*b)), x] + Simp[2*I*b   In 
t[(c + d*x)^m*(E^Simp[2*I*(e + f*x), x]/((a + I*b)^2 + (a^2 + b^2)*E^Simp[2 
*I*(e + f*x), x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2 
, 0] && IGtQ[m, 0]
 

rule 4216

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

rule 4234

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Tan[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 
 1)/n], 0] && IntegerQ[p]
 

Maple [F]

\[\int \frac {x^{3}}{{\left (a +b \tan \left (d \,x^{2}+c \right )\right )}^{2}}d x\]

Input:

int(x^3/(a+b*tan(d*x^2+c))^2,x)
 

Output:

int(x^3/(a+b*tan(d*x^2+c))^2,x)
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 800 vs. \(2 (179) = 358\).

Time = 0.11 (sec) , antiderivative size = 800, normalized size of antiderivative = 3.96 \[ \int \frac {x^3}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx =\text {Too large to display} \] Input:

integrate(x^3/(a+b*tan(d*x^2+c))^2,x, algorithm="fricas")
 

Output:

1/4*((a^3 - a*b^2)*d^2*x^4 - 2*b^3*d*x^2 + (I*a*b^2*tan(d*x^2 + c) + I*a^2 
*b)*dilog(2*((I*a*b - b^2)*tan(d*x^2 + c)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b 
 - I*b^2)*tan(d*x^2 + c))/((a^2 + b^2)*tan(d*x^2 + c)^2 + a^2 + b^2) + 1) 
+ (-I*a*b^2*tan(d*x^2 + c) - I*a^2*b)*dilog(2*((-I*a*b - b^2)*tan(d*x^2 + 
c)^2 - a^2 + I*a*b + (-I*a^2 - 2*a*b + I*b^2)*tan(d*x^2 + c))/((a^2 + b^2) 
*tan(d*x^2 + c)^2 + a^2 + b^2) + 1) + 2*(a^2*b*d*x^2 + a^2*b*c + (a*b^2*d* 
x^2 + a*b^2*c)*tan(d*x^2 + c))*log(-2*((I*a*b - b^2)*tan(d*x^2 + c)^2 - a^ 
2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(d*x^2 + c))/((a^2 + b^2)*tan(d*x^2 
 + c)^2 + a^2 + b^2)) + 2*(a^2*b*d*x^2 + a^2*b*c + (a*b^2*d*x^2 + a*b^2*c) 
*tan(d*x^2 + c))*log(-2*((-I*a*b - b^2)*tan(d*x^2 + c)^2 - a^2 + I*a*b + ( 
-I*a^2 - 2*a*b + I*b^2)*tan(d*x^2 + c))/((a^2 + b^2)*tan(d*x^2 + c)^2 + a^ 
2 + b^2)) - (2*a^2*b*c - a*b^2 + (2*a*b^2*c - b^3)*tan(d*x^2 + c))*log(((I 
*a*b + b^2)*tan(d*x^2 + c)^2 - a^2 + I*a*b + (I*a^2 + I*b^2)*tan(d*x^2 + c 
))/(tan(d*x^2 + c)^2 + 1)) - (2*a^2*b*c - a*b^2 + (2*a*b^2*c - b^3)*tan(d* 
x^2 + c))*log(((I*a*b - b^2)*tan(d*x^2 + c)^2 + a^2 + I*a*b + (I*a^2 + I*b 
^2)*tan(d*x^2 + c))/(tan(d*x^2 + c)^2 + 1)) + ((a^2*b - b^3)*d^2*x^4 + 2*a 
*b^2*d*x^2)*tan(d*x^2 + c))/((a^4*b + 2*a^2*b^3 + b^5)*d^2*tan(d*x^2 + c) 
+ (a^5 + 2*a^3*b^2 + a*b^4)*d^2)
 

Sympy [F]

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

integrate(x**3/(a+b*tan(d*x**2+c))**2,x)
 

Output:

Integral(x**3/(a + b*tan(c + d*x**2))**2, x)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1001 vs. \(2 (179) = 358\).

Time = 0.17 (sec) , antiderivative size = 1001, normalized size of antiderivative = 4.96 \[ \int \frac {x^3}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(x^3/(a+b*tan(d*x^2+c))^2,x, algorithm="maxima")
 

Output:

1/4*((a^3 - I*a^2*b + a*b^2 - I*b^3)*d^2*x^4 - 2*(-I*a*b^2 + b^3 + (-I*a*b 
^2 - b^3)*cos(2*d*x^2 + 2*c) + (a*b^2 - I*b^3)*sin(2*d*x^2 + 2*c))*arctan2 
(-b*cos(2*d*x^2 + 2*c) + a*sin(2*d*x^2 + 2*c) + b, a*cos(2*d*x^2 + 2*c) + 
b*sin(2*d*x^2 + 2*c) + a) - 4*((I*a^2*b + a*b^2)*d*x^2*cos(2*d*x^2 + 2*c) 
- (a^2*b - I*a*b^2)*d*x^2*sin(2*d*x^2 + 2*c) + (I*a^2*b - a*b^2)*d*x^2)*ar 
ctan2((2*a*b*cos(2*d*x^2 + 2*c) - (a^2 - b^2)*sin(2*d*x^2 + 2*c))/(a^2 + b 
^2), (2*a*b*sin(2*d*x^2 + 2*c) + a^2 + b^2 + (a^2 - b^2)*cos(2*d*x^2 + 2*c 
))/(a^2 + b^2)) + ((a^3 - 3*I*a^2*b - 3*a*b^2 + I*b^3)*d^2*x^4 - 4*(I*a*b^ 
2 + b^3)*d*x^2)*cos(2*d*x^2 + 2*c) - 2*(I*a^2*b - a*b^2 + (I*a^2*b + a*b^2 
)*cos(2*d*x^2 + 2*c) - (a^2*b - I*a*b^2)*sin(2*d*x^2 + 2*c))*dilog((I*a + 
b)*e^(2*I*d*x^2 + 2*I*c)/(-I*a + b)) + (a*b^2 + I*b^3 + (a*b^2 - I*b^3)*co 
s(2*d*x^2 + 2*c) + (I*a*b^2 + b^3)*sin(2*d*x^2 + 2*c))*log((a^2 + b^2)*cos 
(2*d*x^2 + 2*c)^2 + 4*a*b*sin(2*d*x^2 + 2*c) + (a^2 + b^2)*sin(2*d*x^2 + 2 
*c)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*d*x^2 + 2*c)) + 2*((a^2*b - I*a*b^ 
2)*d*x^2*cos(2*d*x^2 + 2*c) - (-I*a^2*b - a*b^2)*d*x^2*sin(2*d*x^2 + 2*c) 
+ (a^2*b + I*a*b^2)*d*x^2)*log(((a^2 + b^2)*cos(2*d*x^2 + 2*c)^2 + 4*a*b*s 
in(2*d*x^2 + 2*c) + (a^2 + b^2)*sin(2*d*x^2 + 2*c)^2 + a^2 + b^2 + 2*(a^2 
- b^2)*cos(2*d*x^2 + 2*c))/(a^2 + b^2)) + ((I*a^3 + 3*a^2*b - 3*I*a*b^2 - 
b^3)*d^2*x^4 + 4*(a*b^2 - I*b^3)*d*x^2)*sin(2*d*x^2 + 2*c))/((a^5 - I*a^4* 
b + 2*a^3*b^2 - 2*I*a^2*b^3 + a*b^4 - I*b^5)*d^2*cos(2*d*x^2 + 2*c) - (...
 

Giac [F]

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

integrate(x^3/(a+b*tan(d*x^2+c))^2,x, algorithm="giac")
 

Output:

integrate(x^3/(b*tan(d*x^2 + c) + a)^2, x)
 

Mupad [F(-1)]

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

int(x^3/(a + b*tan(c + d*x^2))^2,x)
 

Output:

int(x^3/(a + b*tan(c + d*x^2))^2, x)
 

Reduce [F]

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

int(x^3/(a+b*tan(d*x^2+c))^2,x)
 

Output:

int(x**3/(tan(c + d*x**2)**2*b**2 + 2*tan(c + d*x**2)*a*b + a**2),x)