\(\int \frac {x^3}{\sqrt {a+a \sin (c+d x)}} \, dx\) [134]

Optimal result

Integrand size = 18, antiderivative size = 417 \[ \int \frac {x^3}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {4 x^3 \text {arctanh}\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d \sqrt {a+a \sin (c+d x)}}+\frac {12 i x^2 \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {12 i x^2 \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {48 x \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}+\frac {48 x \operatorname {PolyLog}\left (3,e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}-\frac {96 i \operatorname {PolyLog}\left (4,-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^4 \sqrt {a+a \sin (c+d x)}}+\frac {96 i \operatorname {PolyLog}\left (4,e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^4 \sqrt {a+a \sin (c+d x)}} \] Output:

-4*x^3*arctanh(exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d/(a+a 
*sin(d*x+c))^(1/2)+12*I*x^2*polylog(2,-exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2* 
c+1/4*Pi+1/2*d*x)/d^2/(a+a*sin(d*x+c))^(1/2)-12*I*x^2*polylog(2,exp(1/4*I* 
(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d^2/(a+a*sin(d*x+c))^(1/2)-48*x 
*polylog(3,-exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d^3/(a+a* 
sin(d*x+c))^(1/2)+48*x*polylog(3,exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4* 
Pi+1/2*d*x)/d^3/(a+a*sin(d*x+c))^(1/2)-96*I*polylog(4,-exp(1/4*I*(2*d*x+Pi 
+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d^4/(a+a*sin(d*x+c))^(1/2)+96*I*polylog( 
4,exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d^4/(a+a*sin(d*x+c) 
)^(1/2)
 

Mathematica [A] (verified)

Time = 1.20 (sec) , antiderivative size = 306, normalized size of antiderivative = 0.73 \[ \int \frac {x^3}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt [4]{-1} \sqrt {2} e^{-\frac {1}{2} i (c+d x)} \left (i+e^{i (c+d x)}\right ) \left (-i d^3 x^3 \log \left (1-\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )+i d^3 x^3 \log \left (1+\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )+6 d^2 x^2 \operatorname {PolyLog}\left (2,-\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )-6 d^2 x^2 \operatorname {PolyLog}\left (2,\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )+24 i d x \operatorname {PolyLog}\left (3,-\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )-24 i d x \operatorname {PolyLog}\left (3,\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )-48 \operatorname {PolyLog}\left (4,-\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )+48 \operatorname {PolyLog}\left (4,\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )\right )}{d^4 \sqrt {-i a e^{-i (c+d x)} \left (i+e^{i (c+d x)}\right )^2}} \] Input:

Integrate[x^3/Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

((-1)^(1/4)*Sqrt[2]*(I + E^(I*(c + d*x)))*((-I)*d^3*x^3*Log[1 - (-1)^(1/4) 
*E^((I/2)*(c + d*x))] + I*d^3*x^3*Log[1 + (-1)^(1/4)*E^((I/2)*(c + d*x))] 
+ 6*d^2*x^2*PolyLog[2, -((-1)^(1/4)*E^((I/2)*(c + d*x)))] - 6*d^2*x^2*Poly 
Log[2, (-1)^(1/4)*E^((I/2)*(c + d*x))] + (24*I)*d*x*PolyLog[3, -((-1)^(1/4 
)*E^((I/2)*(c + d*x)))] - (24*I)*d*x*PolyLog[3, (-1)^(1/4)*E^((I/2)*(c + d 
*x))] - 48*PolyLog[4, -((-1)^(1/4)*E^((I/2)*(c + d*x)))] + 48*PolyLog[4, ( 
-1)^(1/4)*E^((I/2)*(c + d*x))]))/(d^4*E^((I/2)*(c + d*x))*Sqrt[((-I)*a*(I 
+ E^(I*(c + d*x)))^2)/E^(I*(c + d*x))])
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.61, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 3800, 3042, 4671, 3011, 7163, 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.

\(\displaystyle \int \frac {x^3}{\sqrt {a \sin (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {x^3}{\sqrt {a \sin (c+d x)+a}}dx\)

\(\Big \downarrow \) 3800

\(\displaystyle \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \int x^3 \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )dx}{\sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \int x^3 \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )dx}{\sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 4671

\(\displaystyle \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \left (-\frac {6 \int x^2 \log \left (1-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )dx}{d}+\frac {6 \int x^2 \log \left (1+e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )dx}{d}-\frac {4 x^3 \text {arctanh}\left (e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}\right )}{\sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \left (\frac {6 \left (\frac {2 i x^2 \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}-\frac {4 i \int x \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{d}-\frac {6 \left (\frac {2 i x^2 \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}-\frac {4 i \int x \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{d}-\frac {4 x^3 \text {arctanh}\left (e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}\right )}{\sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 7163

\(\displaystyle \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \left (\frac {6 \left (\frac {2 i x^2 \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}-\frac {4 i \left (\frac {2 i \int \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )dx}{d}-\frac {2 i x \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}\right )}{d}\right )}{d}-\frac {6 \left (\frac {2 i x^2 \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}-\frac {4 i \left (\frac {2 i \int \operatorname {PolyLog}\left (3,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )dx}{d}-\frac {2 i x \operatorname {PolyLog}\left (3,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}\right )}{d}\right )}{d}-\frac {4 x^3 \text {arctanh}\left (e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}\right )}{\sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \left (\frac {6 \left (\frac {2 i x^2 \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}-\frac {4 i \left (\frac {4 \int e^{-\frac {1}{4} i (2 c+2 d x+\pi )} \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )de^{\frac {1}{4} i (2 c+2 d x+\pi )}}{d^2}-\frac {2 i x \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}\right )}{d}\right )}{d}-\frac {6 \left (\frac {2 i x^2 \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}-\frac {4 i \left (\frac {4 \int e^{-\frac {1}{4} i (2 c+2 d x+\pi )} \operatorname {PolyLog}\left (3,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )de^{\frac {1}{4} i (2 c+2 d x+\pi )}}{d^2}-\frac {2 i x \operatorname {PolyLog}\left (3,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}\right )}{d}\right )}{d}-\frac {4 x^3 \text {arctanh}\left (e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}\right )}{\sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \left (-\frac {4 x^3 \text {arctanh}\left (e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}+\frac {6 \left (\frac {2 i x^2 \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}-\frac {4 i \left (\frac {4 \operatorname {PolyLog}\left (4,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d^2}-\frac {2 i x \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}\right )}{d}\right )}{d}-\frac {6 \left (\frac {2 i x^2 \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}-\frac {4 i \left (\frac {4 \operatorname {PolyLog}\left (4,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d^2}-\frac {2 i x \operatorname {PolyLog}\left (3,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d}\right )}{d}\right )}{d}\right )}{\sqrt {a \sin (c+d x)+a}}\)

Input:

Int[x^3/Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(((-4*x^3*ArcTanh[E^((I/4)*(2*c + Pi + 2*d*x))])/d + (6*(((2*I)*x^2*PolyLo 
g[2, -E^((I/4)*(2*c + Pi + 2*d*x))])/d - ((4*I)*(((-2*I)*x*PolyLog[3, -E^( 
(I/4)*(2*c + Pi + 2*d*x))])/d + (4*PolyLog[4, -E^((I/4)*(2*c + Pi + 2*d*x) 
)])/d^2))/d))/d - (6*(((2*I)*x^2*PolyLog[2, E^((I/4)*(2*c + Pi + 2*d*x))]) 
/d - ((4*I)*(((-2*I)*x*PolyLog[3, E^((I/4)*(2*c + Pi + 2*d*x))])/d + (4*Po 
lyLog[4, E^((I/4)*(2*c + Pi + 2*d*x))])/d^2))/d))/d)*Sin[c/2 + Pi/4 + (d*x 
)/2])/Sqrt[a + a*Sin[c + d*x]]
 

Defintions of rubi rules used

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 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), 
 x_Symbol] :> Simp[(2*a)^IntPart[n]*((a + b*Sin[e + f*x])^FracPart[n]/Sin[e 
/2 + a*(Pi/(4*b)) + f*(x/2)]^(2*FracPart[n]))   Int[(c + d*x)^m*Sin[e/2 + a 
*(Pi/(4*b)) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && 
EqQ[a^2 - b^2, 0] && IntegerQ[n + 1/2] && (GtQ[n, 0] || IGtQ[m, 0])
 

rule 4671

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[- 
2*(c + d*x)^m*(ArcTanh[E^(I*(e + f*x))]/f), x] + (-Simp[d*(m/f)   Int[(c + 
d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Simp[d*(m/f)   Int[(c + d*x 
)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IG 
tQ[m, 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]
 

rule 7163

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

Maple [F]

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

Input:

int(x^3/(a+a*sin(d*x+c))^(1/2),x)
 

Output:

int(x^3/(a+a*sin(d*x+c))^(1/2),x)
 

Fricas [F]

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

integrate(x^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

integral(x^3/sqrt(a*sin(d*x + c) + a), x)
 

Sympy [F]

\[ \int \frac {x^3}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {x^{3}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \] Input:

integrate(x**3/(a+a*sin(d*x+c))**(1/2),x)
 

Output:

Integral(x**3/sqrt(a*(sin(c + d*x) + 1)), x)
 

Maxima [F]

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

integrate(x^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(x^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(x^3/(a + a*sin(c + d*x))^(1/2),x)
 

Output:

int(x^3/(a + a*sin(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {x^3}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, x^{3}}{\sin \left (d x +c \right )+1}d x \right )}{a} \] Input:

int(x^3/(a+a*sin(d*x+c))^(1/2),x)
 

Output:

(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*x**3)/(sin(c + d*x) + 1),x))/a