3.1.25 \(\int \frac {x^3}{(a+b \csc (c+d x^2))^2} \, dx\) [25]

3.1.25.1 Optimal result

Integrand size = 18, antiderivative size = 616 \[ \int \frac {x^3}{\left (a+b \csc \left (c+d x^2\right )\right )^2} \, dx=\frac {x^4}{4 a^2}-\frac {i b^3 x^2 \log \left (1-\frac {i a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a^2 \left (-a^2+b^2\right )^{3/2} d}+\frac {i b x^2 \log \left (1-\frac {i a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d}+\frac {i b^3 x^2 \log \left (1-\frac {i a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a^2 \left (-a^2+b^2\right )^{3/2} d}-\frac {i b x^2 \log \left (1-\frac {i a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d}+\frac {b^2 \log \left (b+a \sin \left (c+d x^2\right )\right )}{2 a^2 \left (a^2-b^2\right ) d^2}-\frac {b^3 \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a^2 \left (-a^2+b^2\right )^{3/2} d^2}+\frac {b \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d^2}+\frac {b^3 \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a^2 \left (-a^2+b^2\right )^{3/2} d^2}-\frac {b \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d^2}-\frac {b^2 x^2 \cos \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (b+a \sin \left (c+d x^2\right )\right )} \]

output

1/4*x^4/a^2+1/2*b^2*ln(b+a*sin(d*x^2+c))/a^2/(a^2-b^2)/d^2-1/2*I*b^3*x^2*l 
n(1-I*a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d+1/2* 
I*b^3*x^2*ln(1-I*a*exp(I*(d*x^2+c))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^( 
3/2)/d-1/2*b^3*polylog(2,I*a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1/2)))/a^2/(- 
a^2+b^2)^(3/2)/d^2+1/2*b^3*polylog(2,I*a*exp(I*(d*x^2+c))/(b+(-a^2+b^2)^(1 
/2)))/a^2/(-a^2+b^2)^(3/2)/d^2-1/2*b^2*x^2*cos(d*x^2+c)/a/(a^2-b^2)/d/(b+a 
*sin(d*x^2+c))+I*b*x^2*ln(1-I*a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1/2)))/a^2 
/d/(-a^2+b^2)^(1/2)-I*b*x^2*ln(1-I*a*exp(I*(d*x^2+c))/(b+(-a^2+b^2)^(1/2)) 
)/a^2/d/(-a^2+b^2)^(1/2)+b*polylog(2,I*a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1 
/2)))/a^2/d^2/(-a^2+b^2)^(1/2)-b*polylog(2,I*a*exp(I*(d*x^2+c))/(b+(-a^2+b 
^2)^(1/2)))/a^2/d^2/(-a^2+b^2)^(1/2)
 

3.1.25.2 Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2566\) vs. \(2(616)=1232\).

Time = 16.93 (sec) , antiderivative size = 2566, normalized size of antiderivative = 4.17 \[ \int \frac {x^3}{\left (a+b \csc \left (c+d x^2\right )\right )^2} \, dx=\text {Result too large to show} \]

input

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

output

((-(b^2*c*Cos[c + d*x^2]) + b^2*(c + d*x^2)*Cos[c + d*x^2])*Csc[c + d*x^2] 
^2*(b + a*Sin[c + d*x^2]))/(2*a*(-a + b)*(a + b)*d^2*(a + b*Csc[c + d*x^2] 
)^2) + ((-c + d*x^2)*(c + d*x^2)*Csc[c + d*x^2]^2*(b + a*Sin[c + d*x^2])^2 
)/(4*a^2*d^2*(a + b*Csc[c + d*x^2])^2) + (Csc[c + d*x^2]^2*(-2*a*b*ArcTanh 
[(a + b*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2]] + 2*(a*b + 2*a^2*c - b^2*c)*A 
rcTanh[(a + b*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2]] + b*Sqrt[a^2 - b^2]*Log 
[Sec[(c + d*x^2)/2]^2] - b*Sqrt[a^2 - b^2]*Log[Sec[(c + d*x^2)/2]^2*(b + a 
*Sin[c + d*x^2])] + I*(2*a^2 - b^2)*Log[1 - I*Tan[(c + d*x^2)/2]]*Log[(a - 
 Sqrt[a^2 - b^2] + b*Tan[(c + d*x^2)/2])/(a - I*b - Sqrt[a^2 - b^2])] - I* 
(2*a^2 - b^2)*Log[1 + I*Tan[(c + d*x^2)/2]]*Log[(a - Sqrt[a^2 - b^2] + b*T 
an[(c + d*x^2)/2])/(a + I*b - Sqrt[a^2 - b^2])] - I*(2*a^2 - b^2)*Log[1 - 
I*Tan[(c + d*x^2)/2]]*Log[(a + Sqrt[a^2 - b^2] + b*Tan[(c + d*x^2)/2])/(a 
- I*b + Sqrt[a^2 - b^2])] + I*(2*a^2 - b^2)*Log[1 + I*Tan[(c + d*x^2)/2]]* 
Log[(a + Sqrt[a^2 - b^2] + b*Tan[(c + d*x^2)/2])/(a + I*b + Sqrt[a^2 - b^2 
])] - I*(2*a^2 - b^2)*PolyLog[2, (b*(1 + I*Tan[(c + d*x^2)/2]))/((-I)*a + 
b + I*Sqrt[a^2 - b^2])] + I*(2*a^2 - b^2)*PolyLog[2, (b*(1 + I*Tan[(c + d* 
x^2)/2]))/(b - I*(a + Sqrt[a^2 - b^2]))] - I*(2*a^2 - b^2)*PolyLog[2, -((b 
*(I + Tan[(c + d*x^2)/2]))/(a - I*b + Sqrt[a^2 - b^2]))] + I*(2*a^2 - b^2) 
*PolyLog[2, (b*(I + Tan[(c + d*x^2)/2]))/(-a + I*b + Sqrt[a^2 - b^2])])*(b 
 + a*Sin[c + d*x^2])^2*((2*b*c)/((a^2 - b^2)*d*(b + a*Sin[c + d*x^2])) ...
 

3.1.25.3 Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 607, normalized size of antiderivative = 0.99, 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.

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

\(\Big \downarrow \) 4693

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {2 b \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (d x^2+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{a^2 d^2 \sqrt {b^2-a^2}}-\frac {2 b \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (d x^2+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{a^2 d^2 \sqrt {b^2-a^2}}+\frac {b^2 \log \left (a \sin \left (c+d x^2\right )+b\right )}{a^2 d^2 \left (a^2-b^2\right )}+\frac {2 i b x^2 \log \left (1-\frac {i a e^{i \left (c+d x^2\right )}}{b-\sqrt {b^2-a^2}}\right )}{a^2 d \sqrt {b^2-a^2}}-\frac {2 i b x^2 \log \left (1-\frac {i a e^{i \left (c+d x^2\right )}}{\sqrt {b^2-a^2}+b}\right )}{a^2 d \sqrt {b^2-a^2}}-\frac {b^2 x^2 \cos \left (c+d x^2\right )}{a d \left (a^2-b^2\right ) \left (a \sin \left (c+d x^2\right )+b\right )}-\frac {b^3 \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (d x^2+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{a^2 d^2 \left (b^2-a^2\right )^{3/2}}+\frac {b^3 \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (d x^2+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{a^2 d^2 \left (b^2-a^2\right )^{3/2}}-\frac {i b^3 x^2 \log \left (1-\frac {i a e^{i \left (c+d x^2\right )}}{b-\sqrt {b^2-a^2}}\right )}{a^2 d \left (b^2-a^2\right )^{3/2}}+\frac {i b^3 x^2 \log \left (1-\frac {i a e^{i \left (c+d x^2\right )}}{\sqrt {b^2-a^2}+b}\right )}{a^2 d \left (b^2-a^2\right )^{3/2}}+\frac {x^4}{2 a^2}\right )\)

input

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

output

(x^4/(2*a^2) - (I*b^3*x^2*Log[1 - (I*a*E^(I*(c + d*x^2)))/(b - Sqrt[-a^2 + 
 b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d) + ((2*I)*b*x^2*Log[1 - (I*a*E^(I*(c + 
d*x^2)))/(b - Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*d) + (I*b^3*x^2*Lo 
g[1 - (I*a*E^(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^( 
3/2)*d) - ((2*I)*b*x^2*Log[1 - (I*a*E^(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b^ 
2])])/(a^2*Sqrt[-a^2 + b^2]*d) + (b^2*Log[b + a*Sin[c + d*x^2]])/(a^2*(a^2 
 - b^2)*d^2) - (b^3*PolyLog[2, (I*a*E^(I*(c + d*x^2)))/(b - Sqrt[-a^2 + b^ 
2])])/(a^2*(-a^2 + b^2)^(3/2)*d^2) + (2*b*PolyLog[2, (I*a*E^(I*(c + d*x^2) 
))/(b - Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*d^2) + (b^3*PolyLog[2, ( 
I*a*E^(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d^ 
2) - (2*b*PolyLog[2, (I*a*E^(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b^2])])/(a^2 
*Sqrt[-a^2 + b^2]*d^2) - (b^2*x^2*Cos[c + d*x^2])/(a*(a^2 - b^2)*d*(b + a* 
Sin[c + d*x^2])))/2
 

3.1.25.3.1 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]
 

3.1.25.4 Maple [F]

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

input

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

output

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

3.1.25.5 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. 1906 vs. \(2 (526) = 1052\).

Time = 0.42 (sec) , antiderivative size = 1906, normalized size of antiderivative = 3.09 \[ \int \frac {x^3}{\left (a+b \csc \left (c+d x^2\right )\right )^2} \, dx=\text {Too large to display} \]

input

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

output

1/4*((a^5 - 2*a^3*b^2 + a*b^4)*d^2*x^4*sin(d*x^2 + c) + (a^4*b - 2*a^2*b^3 
 + b^5)*d^2*x^4 - 2*(a^3*b^2 - a*b^4)*d*x^2*cos(d*x^2 + c) + (2*I*a^3*b^2 
- I*a*b^4 + (2*I*a^4*b - I*a^2*b^3)*sin(d*x^2 + c))*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) + (-2*I*a^3*b^2 + I*a*b^4 + 
 (-2*I*a^4*b + I*a^2*b^3)*sin(d*x^2 + c))*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) + (-2*I*a^3*b^2 + I*a*b^4 + (-2*I*a^4 
*b + I*a^2*b^3)*sin(d*x^2 + c))*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) + (2*I*a^3*b^2 - I*a*b^4 + (2*I*a^4*b - I*a^2* 
b^3)*sin(d*x^2 + c))*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) - ((2*a^3*b^2 - a*b^4)*d*x^2 + (2*a^3*b^2 - a*b^4)*c + (( 
2*a^4*b - a^2*b^3)*d*x^2 + (2*a^4*b - a^2*b^3)*c)*sin(d*x^2 + 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) + ((2*a^3*b^2 - a* 
b^4)*d*x^2 + (2*a^3*b^2 - a*b^4)*c + ((2*a^4*b - a^2*b^3)*d*x^2 + (2*a^4*b 
 - a^2*b^3)*c)*sin(d*x^2 + 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((...
 

3.1.25.6 Sympy [F]

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

input

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

output

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

3.1.25.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3}{\left (a+b \csc \left (c+d x^2\right )\right )^2} \, dx=\text {Exception raised: ValueError} \]

input

integrate(x^3/(a+b*csc(d*x^2+c))^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(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 

3.1.25.8 Giac [F]

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

input

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

output

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

3.1.25.9 Mupad [F(-1)]

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

input

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

output

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