\(\int \frac {x^3}{(a+b \sin (c+d x^2))^2} \, dx\) [44]

Optimal result

Integrand size = 18, antiderivative size = 324 \[ \int \frac {x^3}{\left (a+b \sin \left (c+d x^2\right )\right )^2} \, dx=-\frac {i a x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2} d}+\frac {i a x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2} d}-\frac {\log \left (a+b \sin \left (c+d x^2\right )\right )}{2 \left (a^2-b^2\right ) d^2}-\frac {a \operatorname {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2} d^2}+\frac {a \operatorname {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2} d^2}+\frac {b x^2 \cos \left (c+d x^2\right )}{2 \left (a^2-b^2\right ) d \left (a+b \sin \left (c+d x^2\right )\right )} \] Output:

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

Mathematica [A] (verified)

Time = 1.43 (sec) , antiderivative size = 302, normalized size of antiderivative = 0.93 \[ \int \frac {x^3}{\left (a+b \sin \left (c+d x^2\right )\right )^2} \, dx=\frac {-\frac {i a d x^2 \log \left (1+\frac {i b e^{i \left (c+d x^2\right )}}{-a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {i a d x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {\log \left (a+b \sin \left (c+d x^2\right )\right )}{a^2-b^2}-\frac {a \operatorname {PolyLog}\left (2,-\frac {i b e^{i \left (c+d x^2\right )}}{-a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {a \operatorname {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {b d x^2 \cos \left (c+d x^2\right )}{\left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^2\right )\right )}}{2 d^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3860, 3042, 3805, 3042, 3147, 16, 3804, 2694, 27, 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.

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

\(\Big \downarrow \) 3860

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3805

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3147

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 3804

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

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 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 2694

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.) 
*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[2*(c/q)   Int[(f + g*x) 
^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[ 
v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && 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 3147

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) 
/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p 
 - 1)/2] && NeQ[a^2 - b^2, 0]
 

rule 3804

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

rule 3805

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_ 
Symbol] :> Simp[b*(c + d*x)^m*(Cos[e + f*x]/(f*(a^2 - b^2)*(a + b*Sin[e + f 
*x]))), x] + (Simp[a/(a^2 - b^2)   Int[(c + d*x)^m/(a + b*Sin[e + f*x]), x] 
, x] - Simp[b*d*(m/(f*(a^2 - b^2)))   Int[(c + d*x)^(m - 1)*(Cos[e + f*x]/( 
a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 - 
b^2, 0] && IGtQ[m, 0]
 

rule 3860

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

Maple [F]

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

Input:

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

Output:

int(x^3/(a+b*sin(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. 1509 vs. \(2 (274) = 548\).

Time = 0.22 (sec) , antiderivative size = 1509, normalized size of antiderivative = 4.66 \[ \int \frac {x^3}{\left (a+b \sin \left (c+d x^2\right )\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(2*(a^2*b - b^3)*d*x^2*cos(d*x^2 + c) + (I*a*b^2*sin(d*x^2 + c) + I*a^ 
2*b)*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x^2 + c) - a*sin(d*x^2 + c) + 
 (b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1 
) + (-I*a*b^2*sin(d*x^2 + c) - I*a^2*b)*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a* 
cos(d*x^2 + c) - a*sin(d*x^2 + c) - (b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c) 
)*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + (-I*a*b^2*sin(d*x^2 + c) - I*a^2*b) 
*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(d*x^2 + c) - a*sin(d*x^2 + c) + (b 
*cos(d*x^2 + c) - I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 
 (I*a*b^2*sin(d*x^2 + c) + I*a^2*b)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos 
(d*x^2 + c) - a*sin(d*x^2 + c) - (b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c))*s 
qrt(-(a^2 - b^2)/b^2) - b)/b + 1) - (a^2*b*d*x^2 + a^2*b*c + (a*b^2*d*x^2 
+ a*b^2*c)*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*cos(d*x^2 + c) 
 - a*sin(d*x^2 + c) + (b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c))*sqrt(-(a^2 - 
 b^2)/b^2) - b)/b) + (a^2*b*d*x^2 + a^2*b*c + (a*b^2*d*x^2 + a*b^2*c)*sin( 
d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*cos(d*x^2 + c) - a*sin(d*x^2 
+ c) - (b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) - b) 
/b) - (a^2*b*d*x^2 + a^2*b*c + (a*b^2*d*x^2 + a*b^2*c)*sin(d*x^2 + c))*sqr 
t(-(a^2 - b^2)/b^2)*log(-(-I*a*cos(d*x^2 + c) - a*sin(d*x^2 + c) + (b*cos( 
d*x^2 + c) - I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + (a^2*b*d 
*x^2 + a^2*b*c + (a*b^2*d*x^2 + a*b^2*c)*sin(d*x^2 + c))*sqrt(-(a^2 - b...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

Reduce [F]

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

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

Output:

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