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

Optimal. Leaf size=175 \[ \frac{4 i \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d^2 \sqrt{a \sin (c+d x)+a}}-\frac{4 i \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d^2 \sqrt{a \sin (c+d x)+a}}-\frac{4 x \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(-4*x*ArcTanh[E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d*Sqrt[a + a*Sin[c + d*x]]) + ((4*I)*P
olyLog[2, -E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^2*Sqrt[a + a*Sin[c + d*x]]) - ((4*I)*Po
lyLog[2, E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^2*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.092206, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3319, 4183, 2279, 2391} \[ \frac{4 i \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d^2 \sqrt{a \sin (c+d x)+a}}-\frac{4 i \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d^2 \sqrt{a \sin (c+d x)+a}}-\frac{4 x \sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )}{d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-4*x*ArcTanh[E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d*Sqrt[a + a*Sin[c + d*x]]) + ((4*I)*P
olyLog[2, -E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^2*Sqrt[a + a*Sin[c + d*x]]) - ((4*I)*Po
lyLog[2, E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^2*Sqrt[a + a*Sin[c + d*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 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}{\sqrt{a+a \sin (c+d x)}} \, dx &=\frac{\sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right ) \int x \csc \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right ) \, dx}{\sqrt{a+a \sin (c+d x)}}\\ &=-\frac{4 x \tanh ^{-1}\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{\left (2 \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )\right ) \int \log \left (1-e^{i \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}\right ) \, dx}{d \sqrt{a+a \sin (c+d x)}}+\frac{\left (2 \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )\right ) \int \log \left (1+e^{i \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}\right ) \, dx}{d \sqrt{a+a \sin (c+d x)}}\\ &=-\frac{4 x \tanh ^{-1}\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{\left (4 i \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{i \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}\right )}{d^2 \sqrt{a+a \sin (c+d x)}}-\frac{\left (4 i \sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}\right )}{d^2 \sqrt{a+a \sin (c+d x)}}\\ &=-\frac{4 x \tanh ^{-1}\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{4 i \text{Li}_2\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^2 \sqrt{a+a \sin (c+d x)}}-\frac{4 i \text{Li}_2\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^2 \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.52358, size = 231, normalized size = 1.32 \[ \frac{2 \left (\frac{c \sin \left (\frac{1}{4} (2 c+2 d x-\pi )\right ) \sin ^{-1}\left (\csc \left (\frac{1}{4} (2 c+2 d x+\pi )\right )\right )}{\sqrt{\frac{\sin (c+d x)-1}{\sin (c+d x)+1}}}+\frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (2 i \left (\text{PolyLog}\left (2,-e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )-\text{PolyLog}\left (2,e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )\right )+\frac{1}{2} (2 c+2 d x+\pi ) \left (\log \left (1-e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )-\log \left (1+e^{\frac{1}{4} i (2 c+2 d x+\pi )}\right )\right )-\pi \tanh ^{-1}\left (\frac{\tan \left (\frac{1}{4} (c+d x)\right )-1}{\sqrt{2}}\right )\right )}{\sqrt{2}}\right )}{d^2 \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.06, size = 0, normalized size = 0. \begin{align*} \int{x{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x}{\sqrt{a \sin \left (d x + c\right ) + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{a \left (\sin{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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