3.141 \(\int \frac{x}{(a+a \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=249 \[ \frac{i \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f^2 \sqrt{a \sin (e+f x)+a}}-\frac{i \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f^2 \sqrt{a \sin (e+f x)+a}}-\frac{1}{a f^2 \sqrt{a \sin (e+f x)+a}}-\frac{x \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f \sqrt{a \sin (e+f x)+a}}-\frac{x \cot \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right )}{2 a f \sqrt{a \sin (e+f x)+a}} \]

[Out]

-(1/(a*f^2*Sqrt[a + a*Sin[e + f*x]])) - (x*Cot[e/2 + Pi/4 + (f*x)/2])/(2*a*f*Sqrt[a + a*Sin[e + f*x]]) - (x*Ar
cTanh[E^((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f*Sqrt[a + a*Sin[e + f*x]]) + (I*PolyLog[2,
-E^((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^2*Sqrt[a + a*Sin[e + f*x]]) - (I*PolyLog[2, E^(
(I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^2*Sqrt[a + a*Sin[e + f*x]])

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Rubi [A]  time = 0.127779, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3319, 4185, 4183, 2279, 2391} \[ \frac{i \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f^2 \sqrt{a \sin (e+f x)+a}}-\frac{i \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f^2 \sqrt{a \sin (e+f x)+a}}-\frac{1}{a f^2 \sqrt{a \sin (e+f x)+a}}-\frac{x \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f \sqrt{a \sin (e+f x)+a}}-\frac{x \cot \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right )}{2 a f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

-(1/(a*f^2*Sqrt[a + a*Sin[e + f*x]])) - (x*Cot[e/2 + Pi/4 + (f*x)/2])/(2*a*f*Sqrt[a + a*Sin[e + f*x]]) - (x*Ar
cTanh[E^((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f*Sqrt[a + a*Sin[e + f*x]]) + (I*PolyLog[2,
-E^((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^2*Sqrt[a + a*Sin[e + f*x]]) - (I*PolyLog[2, E^(
(I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^2*Sqrt[a + a*Sin[e + f*x]])

Rule 3319

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((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 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

Rule 4183

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] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2391

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]

Rubi steps

\begin{align*} \int \frac{x}{(a+a \sin (e+f x))^{3/2}} \, dx &=\frac{\sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \int x \csc ^3\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \, dx}{2 a \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{1}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{x \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{2 a f \sqrt{a+a \sin (e+f x)}}+\frac{\sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \int x \csc \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \, dx}{4 a \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{1}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{x \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{2 a f \sqrt{a+a \sin (e+f x)}}-\frac{x \tanh ^{-1}\left (e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f \sqrt{a+a \sin (e+f x)}}-\frac{\sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \int \log \left (1-e^{i \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}\right ) \, dx}{2 a f \sqrt{a+a \sin (e+f x)}}+\frac{\sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \int \log \left (1+e^{i \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}\right ) \, dx}{2 a f \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{1}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{x \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{2 a f \sqrt{a+a \sin (e+f x)}}-\frac{x \tanh ^{-1}\left (e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f \sqrt{a+a \sin (e+f x)}}+\frac{\left (i \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{i \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}\right )}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{\left (i \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}\right )}{a f^2 \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{1}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{x \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{2 a f \sqrt{a+a \sin (e+f x)}}-\frac{x \tanh ^{-1}\left (e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f \sqrt{a+a \sin (e+f x)}}+\frac{i \text{Li}_2\left (-e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{i \text{Li}_2\left (e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^2 \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.54506, size = 308, normalized size = 1.24 \[ \frac{\frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \left (2 i \left (\text{PolyLog}\left (2,-e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )-\text{PolyLog}\left (2,e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )\right )+\frac{1}{2} (2 e+2 f x+\pi ) \left (\log \left (1-e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )-\log \left (1+e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )\right )-\pi \tanh ^{-1}\left (\frac{\tan \left (\frac{1}{4} (e+f x)\right )-1}{\sqrt{2}}\right )\right )}{\sqrt{2}}-(f x+2) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+2 f x \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+\frac{e (\sin (e+f x)+1) \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \sin ^{-1}\left (\csc \left (\frac{1}{4} (2 e+2 f x+\pi )\right )\right )}{\sqrt{\frac{\sin (e+f x)-1}{\sin (e+f x)+1}}}}{2 f^2 (a (\sin (e+f x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

(2*f*x*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - (2 + f*x)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]
)^2 + ((-(Pi*ArcTanh[(-1 + Tan[(e + f*x)/4])/Sqrt[2]]) + ((2*e + Pi + 2*f*x)*(Log[1 - E^((I/4)*(2*e + Pi + 2*f
*x))] - Log[1 + E^((I/4)*(2*e + Pi + 2*f*x))]))/2 + (2*I)*(PolyLog[2, -E^((I/4)*(2*e + Pi + 2*f*x))] - PolyLog
[2, E^((I/4)*(2*e + Pi + 2*f*x))]))*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/Sqrt[2] + (e*ArcSin[Csc[(2*e + Pi
+ 2*f*x)/4]]*(1 + Sin[e + f*x])*Sin[(2*e - Pi + 2*f*x)/4])/Sqrt[(-1 + Sin[e + f*x])/(1 + Sin[e + f*x])])/(2*f
^2*(a*(1 + Sin[e + f*x]))^(3/2))

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Maple [F]  time = 0.038, size = 0, normalized size = 0. \begin{align*} \int{x \left ( a+a\sin \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+a*sin(f*x+e))^(3/2),x)

[Out]

int(x/(a+a*sin(f*x+e))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate(x/(a*sin(f*x + e) + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{a \sin \left (f x + e\right ) + a} x}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(a*sin(f*x + e) + a)*x/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+a*sin(f*x+e))**(3/2),x)

[Out]

Integral(x/(a*(sin(e + f*x) + 1))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(x/(a*sin(f*x + e) + a)^(3/2), x)