\(\int \frac {x^3}{a+b \csc (c+d x^2)} \, dx\) [18]

Optimal result

Integrand size = 18, antiderivative size = 271 \[ \int \frac {x^3}{a+b \csc \left (c+d x^2\right )} \, dx=\frac {x^4}{4 a}+\frac {i b x^2 \log \left (1-\frac {i a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^2 \log \left (1-\frac {i a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {b \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d^2}-\frac {b \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d^2} \] Output:

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(987\) vs. \(2(271)=542\).

Time = 3.67 (sec) , antiderivative size = 987, normalized size of antiderivative = 3.64 \[ \int \frac {x^3}{a+b \csc \left (c+d x^2\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

(Csc[c + d*x^2]*(x^4 - (2*b*((Pi*ArcTan[(a + b*Tan[(c + d*x^2)/2])/Sqrt[-a 
^2 + b^2]])/Sqrt[-a^2 + b^2] + (2*(c - ArcCos[-(b/a)])*ArcTanh[((a - b)*Co 
t[(2*c + Pi + 2*d*x^2)/4])/Sqrt[a^2 - b^2]] + (-2*c + Pi - 2*d*x^2)*ArcTan 
h[((a + b)*Tan[(2*c + Pi + 2*d*x^2)/4])/Sqrt[a^2 - b^2]] - (ArcCos[-(b/a)] 
 - (2*I)*ArcTanh[((a - b)*Cot[(2*c + Pi + 2*d*x^2)/4])/Sqrt[a^2 - b^2]])*L 
og[((a + b)*(a - b - I*Sqrt[a^2 - b^2])*(1 + I*Cot[(2*c + Pi + 2*d*x^2)/4] 
))/(a*(a + b + Sqrt[a^2 - b^2]*Cot[(2*c + Pi + 2*d*x^2)/4]))] + (ArcCos[-( 
b/a)] + (2*I)*(-ArcTanh[((a - b)*Cot[(2*c + Pi + 2*d*x^2)/4])/Sqrt[a^2 - b 
^2]] + ArcTanh[((a + b)*Tan[(2*c + Pi + 2*d*x^2)/4])/Sqrt[a^2 - b^2]]))*Lo 
g[((-1)^(1/4)*Sqrt[a^2 - b^2])/(Sqrt[2]*Sqrt[a]*E^((I/2)*(c + d*x^2))*Sqrt 
[b + a*Sin[c + d*x^2]])] + (ArcCos[-(b/a)] + (2*I)*ArcTanh[((a - b)*Cot[(2 
*c + Pi + 2*d*x^2)/4])/Sqrt[a^2 - b^2]] - (2*I)*ArcTanh[((a + b)*Tan[(2*c 
+ Pi + 2*d*x^2)/4])/Sqrt[a^2 - b^2]])*Log[-(((-1)^(3/4)*Sqrt[a^2 - b^2]*E^ 
((I/2)*(c + d*x^2)))/(Sqrt[2]*Sqrt[a]*Sqrt[b + a*Sin[c + d*x^2]]))] - (Arc 
Cos[-(b/a)] + (2*I)*ArcTanh[((a - b)*Cot[(2*c + Pi + 2*d*x^2)/4])/Sqrt[a^2 
 - b^2]])*Log[1 + (I*(I*b + Sqrt[a^2 - b^2])*(a + b + Sqrt[a^2 - b^2]*Tan[ 
(2*c - Pi + 2*d*x^2)/4]))/(a*(a + b + Sqrt[a^2 - b^2]*Cot[(2*c + Pi + 2*d* 
x^2)/4]))] + I*(PolyLog[2, ((b - I*Sqrt[a^2 - b^2])*(a + b + Sqrt[a^2 - b^ 
2]*Tan[(2*c - Pi + 2*d*x^2)/4]))/(a*(a + b + Sqrt[a^2 - b^2]*Cot[(2*c + Pi 
 + 2*d*x^2)/4]))] - PolyLog[2, ((b + I*Sqrt[a^2 - b^2])*(a + b + Sqrt[a...
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4693, 3042, 4679, 2009}

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.

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

\(\Big \downarrow \) 4693

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4679

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si 
n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt 
Q[m, 0]
 

rule 4693

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Csc[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]

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

Input:

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

Output:

int(x^3/(a+b*csc(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. 1050 vs. \(2 (223) = 446\).

Time = 0.20 (sec) , antiderivative size = 1050, normalized size of antiderivative = 3.87 \[ \int \frac {x^3}{a+b \csc \left (c+d x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*((a^2 - b^2)*d^2*x^4 - a*b*c*sqrt((a^2 - b^2)/a^2)*log(2*a*cos(d*x^2 + 
 c) + 2*I*a*sin(d*x^2 + c) + 2*a*sqrt((a^2 - b^2)/a^2) + 2*I*b) - a*b*c*sq 
rt((a^2 - b^2)/a^2)*log(2*a*cos(d*x^2 + c) - 2*I*a*sin(d*x^2 + c) + 2*a*sq 
rt((a^2 - b^2)/a^2) - 2*I*b) + a*b*c*sqrt((a^2 - b^2)/a^2)*log(-2*a*cos(d* 
x^2 + c) + 2*I*a*sin(d*x^2 + c) + 2*a*sqrt((a^2 - b^2)/a^2) + 2*I*b) + a*b 
*c*sqrt((a^2 - b^2)/a^2)*log(-2*a*cos(d*x^2 + c) - 2*I*a*sin(d*x^2 + c) + 
2*a*sqrt((a^2 - b^2)/a^2) - 2*I*b) + I*a*b*sqrt((a^2 - b^2)/a^2)*dilog((I* 
b*cos(d*x^2 + c) - b*sin(d*x^2 + c) + (a*cos(d*x^2 + c) + I*a*sin(d*x^2 + 
c))*sqrt((a^2 - b^2)/a^2) - a)/a + 1) - I*a*b*sqrt((a^2 - b^2)/a^2)*dilog( 
(I*b*cos(d*x^2 + c) - b*sin(d*x^2 + c) - (a*cos(d*x^2 + c) + I*a*sin(d*x^2 
 + c))*sqrt((a^2 - b^2)/a^2) - a)/a + 1) - I*a*b*sqrt((a^2 - b^2)/a^2)*dil 
og((-I*b*cos(d*x^2 + c) - b*sin(d*x^2 + c) + (a*cos(d*x^2 + c) - I*a*sin(d 
*x^2 + c))*sqrt((a^2 - b^2)/a^2) - a)/a + 1) + I*a*b*sqrt((a^2 - b^2)/a^2) 
*dilog((-I*b*cos(d*x^2 + c) - b*sin(d*x^2 + c) - (a*cos(d*x^2 + c) - I*a*s 
in(d*x^2 + c))*sqrt((a^2 - b^2)/a^2) - a)/a + 1) - (a*b*d*x^2 + a*b*c)*sqr 
t((a^2 - b^2)/a^2)*log(-(I*b*cos(d*x^2 + c) - b*sin(d*x^2 + c) + (a*cos(d* 
x^2 + c) + I*a*sin(d*x^2 + c))*sqrt((a^2 - b^2)/a^2) - a)/a) + (a*b*d*x^2 
+ a*b*c)*sqrt((a^2 - b^2)/a^2)*log(-(I*b*cos(d*x^2 + c) - b*sin(d*x^2 + c) 
 - (a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt((a^2 - b^2)/a^2) - a)/a) - 
 (a*b*d*x^2 + a*b*c)*sqrt((a^2 - b^2)/a^2)*log(-(-I*b*cos(d*x^2 + c) - ...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/4*(x^4 - 8*a*b*integrate((2*b*x^3*cos(d*x^2 + c)^2 + a*x^3*cos(d*x^2 + c 
)*sin(2*d*x^2 + 2*c) - a*x^3*cos(2*d*x^2 + 2*c)*sin(d*x^2 + c) + 2*b*x^3*s 
in(d*x^2 + c)^2 + a*x^3*sin(d*x^2 + c))/(a^3*cos(2*d*x^2 + 2*c)^2 + 4*a*b^ 
2*cos(d*x^2 + c)^2 + 4*a^2*b*cos(d*x^2 + c)*sin(2*d*x^2 + 2*c) + a^3*sin(2 
*d*x^2 + 2*c)^2 + 4*a*b^2*sin(d*x^2 + c)^2 + 4*a^2*b*sin(d*x^2 + c) + a^3 
- 2*(2*a^2*b*sin(d*x^2 + c) + a^3)*cos(2*d*x^2 + 2*c)), x))/a
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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