3.1.35 \(\int \frac {x^5}{a+b \sin (c+d x^2)} \, dx\) [35]

3.1.35.1 Optimal result

Integrand size = 18, antiderivative size = 362 \[ \int \frac {x^5}{a+b \sin \left (c+d x^2\right )} \, dx=-\frac {i x^4 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2} d}+\frac {i x^4 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2} d}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}+\frac {x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}-\frac {i \operatorname {PolyLog}\left (3,\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^3}+\frac {i \operatorname {PolyLog}\left (3,\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^3} \]

output

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

3.1.35.2 Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.80 \[ \int \frac {x^5}{a+b \sin \left (c+d x^2\right )} \, dx=\frac {-2 d x^2 \operatorname {PolyLog}\left (2,-\frac {i b e^{i \left (c+d x^2\right )}}{-a+\sqrt {a^2-b^2}}\right )-i \left (d^2 x^4 \log \left (1+\frac {i b e^{i \left (c+d x^2\right )}}{-a+\sqrt {a^2-b^2}}\right )-d^2 x^4 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )+2 i d x^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )+2 \operatorname {PolyLog}\left (3,\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )-2 \operatorname {PolyLog}\left (3,\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )\right )}{2 \sqrt {a^2-b^2} d^3} \]

input

Integrate[x^5/(a + b*Sin[c + d*x^2]),x]
 

output

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

3.1.35.3 Rubi [A] (verified)

Time = 1.40 (sec) , antiderivative size = 341, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3860, 3042, 3804, 2694, 27, 2620, 3011, 2720, 7143}

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.

\(\displaystyle \int \frac {x^5}{a+b \sin \left (c+d x^2\right )} \, dx\)

\(\Big \downarrow \) 3860

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3804

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

input

Int[x^5/(a + b*Sin[c + d*x^2]),x]
 

output

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

3.1.35.3.1 Defintions of rubi rules used

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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

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

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 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]))
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

3.1.35.4 Maple [F]

\[\int \frac {x^{5}}{a +b \sin \left (d \,x^{2}+c \right )}d x\]

input

int(x^5/(a+b*sin(d*x^2+c)),x)
 

output

int(x^5/(a+b*sin(d*x^2+c)),x)
 

3.1.35.5 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. 1435 vs. \(2 (300) = 600\).

Time = 0.46 (sec) , antiderivative size = 1435, normalized size of antiderivative = 3.96 \[ \int \frac {x^5}{a+b \sin \left (c+d x^2\right )} \, dx=\text {Too large to display} \]

input

integrate(x^5/(a+b*sin(d*x^2+c)),x, algorithm="fricas")
 

output

1/4*(2*I*b*d*x^2*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) - 2*I*b*d*x^2*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) - 2*I*b*d*x^2*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) + 2*I*b*d*x^2*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*si 
n(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + b*c^2*sqrt(-(a^2 - b^2) 
/b^2)*log(2*b*cos(d*x^2 + c) + 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-(a^2 - b^2 
)/b^2) + 2*I*a) + b*c^2*sqrt(-(a^2 - b^2)/b^2)*log(2*b*cos(d*x^2 + c) - 2* 
I*b*sin(d*x^2 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) - b*c^2*sqrt(-(a^ 
2 - b^2)/b^2)*log(-2*b*cos(d*x^2 + c) + 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-( 
a^2 - b^2)/b^2) + 2*I*a) - b*c^2*sqrt(-(a^2 - b^2)/b^2)*log(-2*b*cos(d*x^2 
 + c) - 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) - (b*d^ 
2*x^4 - b*c^2)*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) + (b*d^2*x^4 - b*c^2)*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) - (b*d^2*x^4 - b*c^2)*sqrt(-(a^2 - b^2)/b^2)*log(-...
 

3.1.35.6 Sympy [F]

\[ \int \frac {x^5}{a+b \sin \left (c+d x^2\right )} \, dx=\int \frac {x^{5}}{a + b \sin {\left (c + d x^{2} \right )}}\, dx \]

input

integrate(x**5/(a+b*sin(d*x**2+c)),x)
 

output

Integral(x**5/(a + b*sin(c + d*x**2)), x)
 

3.1.35.7 Maxima [F]

\[ \int \frac {x^5}{a+b \sin \left (c+d x^2\right )} \, dx=\int { \frac {x^{5}}{b \sin \left (d x^{2} + c\right ) + a} \,d x } \]

input

integrate(x^5/(a+b*sin(d*x^2+c)),x, algorithm="maxima")
 

output

integrate(x^5/(b*sin(d*x^2 + c) + a), x)
 

3.1.35.8 Giac [F]

\[ \int \frac {x^5}{a+b \sin \left (c+d x^2\right )} \, dx=\int { \frac {x^{5}}{b \sin \left (d x^{2} + c\right ) + a} \,d x } \]

input

integrate(x^5/(a+b*sin(d*x^2+c)),x, algorithm="giac")
 

output

integrate(x^5/(b*sin(d*x^2 + c) + a), x)
 

3.1.35.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{a+b \sin \left (c+d x^2\right )} \, dx=\int \frac {x^5}{a+b\,\sin \left (d\,x^2+c\right )} \,d x \]

input

int(x^5/(a + b*sin(c + d*x^2)),x)
 

output

int(x^5/(a + b*sin(c + d*x^2)), x)