\(\int \frac {x^3}{a+b \tan (c+d x^2)} \, dx\) [13]

Optimal result

Integrand size = 18, antiderivative size = 122 \[ \int \frac {x^3}{a+b \tan \left (c+d x^2\right )} \, dx=\frac {x^4}{4 (a+i b)}+\frac {b x^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{4 \left (a^2+b^2\right ) d^2} \] Output:

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

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.90 \[ \int \frac {x^3}{a+b \tan \left (c+d x^2\right )} \, dx=\frac {d x^2 \left ((a+i b) d x^2+2 b \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d x^2\right )}}{a-i b}\right )\right )+i b \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d x^2\right )}}{a-i b}\right )}{4 \left (a^2+b^2\right ) d^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4234, 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.

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

\(\Big \downarrow \) 4234

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4215

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

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

Input:

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

Output:

int(x^3/(a+b*tan(d*x^2+c)),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. 536 vs. \(2 (103) = 206\).

Time = 0.13 (sec) , antiderivative size = 536, normalized size of antiderivative = 4.39 \[ \int \frac {x^3}{a+b \tan \left (c+d x^2\right )} \, dx=\frac {2 \, a d^{2} x^{4} - 2 \, b c \log \left (\frac {{\left (i \, a b + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} - a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{\tan \left (d x^{2} + c\right )^{2} + 1}\right ) - 2 \, b c \log \left (\frac {{\left (i \, a b - b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{\tan \left (d x^{2} + c\right )^{2} + 1}\right ) + i \, b {\rm Li}_2\left (\frac {2 \, {\left ({\left (i \, a b - b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} - a^{2} - i \, a b + {\left (i \, a^{2} - 2 \, a b - i \, b^{2}\right )} \tan \left (d x^{2} + c\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + b^{2}} + 1\right ) - i \, b {\rm Li}_2\left (\frac {2 \, {\left ({\left (-i \, a b - b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} - a^{2} + i \, a b + {\left (-i \, a^{2} - 2 \, a b + i \, b^{2}\right )} \tan \left (d x^{2} + c\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + b^{2}} + 1\right ) + 2 \, {\left (b d x^{2} + b c\right )} \log \left (-\frac {2 \, {\left ({\left (i \, a b - b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} - a^{2} - i \, a b + {\left (i \, a^{2} - 2 \, a b - i \, b^{2}\right )} \tan \left (d x^{2} + c\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + b^{2}}\right ) + 2 \, {\left (b d x^{2} + b c\right )} \log \left (-\frac {2 \, {\left ({\left (-i \, a b - b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} - a^{2} + i \, a b + {\left (-i \, a^{2} - 2 \, a b + i \, b^{2}\right )} \tan \left (d x^{2} + c\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + b^{2}}\right )}{8 \, {\left (a^{2} + b^{2}\right )} d^{2}} \] Input:

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

Output:

1/8*(2*a*d^2*x^4 - 2*b*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*b*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)) + I*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*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)*ta 
n(d*x^2 + c)^2 + a^2 + b^2) + 1) + 2*(b*d*x^2 + b*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*(b*d*x^2 + b*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)))/((a^2 + b^2)*d^2 
)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

Integral(x**3/(a + b*tan(c + d*x**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. 267 vs. \(2 (103) = 206\).

Time = 0.07 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.19 \[ \int \frac {x^3}{a+b \tan \left (c+d x^2\right )} \, dx=\frac {{\left (a - i \, b\right )} d^{2} x^{4} - 2 i \, b d x^{2} \arctan \left (\frac {2 \, a b \cos \left (2 \, d x^{2} + 2 \, c\right ) - {\left (a^{2} - b^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, d x^{2} + 2 \, c\right ) + a^{2} + b^{2} + {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )}{a^{2} + b^{2}}\right ) + b d x^{2} \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + 4 \, a b \sin \left (2 \, d x^{2} + 2 \, c\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )}{a^{2} + b^{2}}\right ) - i \, b {\rm Li}_2\left (\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d x^{2} + 2 i \, c\right )}}{-i \, a + b}\right )}{4 \, {\left (a^{2} + b^{2}\right )} d^{2}} \] Input:

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

Output:

1/4*((a - I*b)*d^2*x^4 - 2*I*b*d*x^2*arctan2((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)) + b*d*x^2*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))/(a^2 + b^ 
2)) - I*b*dilog((I*a + b)*e^(2*I*d*x^2 + 2*I*c)/(-I*a + b)))/((a^2 + b^2)* 
d^2)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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