3.277 \(\int \frac{a+b \log (c (d+e x)^n)}{\sqrt{2-g x} \sqrt{2+g x}} \, dx\)

Optimal. Leaf size=278 \[ \frac{i b n \text{PolyLog}\left (2,-\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{-\sqrt{4 e^2-d^2 g^2}+i d g}\right )}{g}+\frac{i b n \text{PolyLog}\left (2,-\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{\sqrt{4 e^2-d^2 g^2}+i d g}\right )}{g}+\frac{\sin ^{-1}\left (\frac{g x}{2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac{b n \sin ^{-1}\left (\frac{g x}{2}\right ) \log \left (1+\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{-\sqrt{4 e^2-d^2 g^2}+i d g}\right )}{g}-\frac{b n \sin ^{-1}\left (\frac{g x}{2}\right ) \log \left (1+\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{\sqrt{4 e^2-d^2 g^2}+i d g}\right )}{g}+\frac{i b n \sin ^{-1}\left (\frac{g x}{2}\right )^2}{2 g} \]

[Out]

((I/2)*b*n*ArcSin[(g*x)/2]^2)/g - (b*n*ArcSin[(g*x)/2]*Log[1 + (2*e*E^(I*ArcSin[(g*x)/2]))/(I*d*g - Sqrt[4*e^2
- d^2*g^2])])/g - (b*n*ArcSin[(g*x)/2]*Log[1 + (2*e*E^(I*ArcSin[(g*x)/2]))/(I*d*g + Sqrt[4*e^2 - d^2*g^2])])/
g + (ArcSin[(g*x)/2]*(a + b*Log[c*(d + e*x)^n]))/g + (I*b*n*PolyLog[2, (-2*e*E^(I*ArcSin[(g*x)/2]))/(I*d*g - S
qrt[4*e^2 - d^2*g^2])])/g + (I*b*n*PolyLog[2, (-2*e*E^(I*ArcSin[(g*x)/2]))/(I*d*g + Sqrt[4*e^2 - d^2*g^2])])/g

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Rubi [A]  time = 0.474454, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.206, Rules used = {216, 2405, 4741, 4521, 2190, 2279, 2391} \[ \frac{i b n \text{PolyLog}\left (2,-\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{-\sqrt{4 e^2-d^2 g^2}+i d g}\right )}{g}+\frac{i b n \text{PolyLog}\left (2,-\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{\sqrt{4 e^2-d^2 g^2}+i d g}\right )}{g}+\frac{\sin ^{-1}\left (\frac{g x}{2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac{b n \sin ^{-1}\left (\frac{g x}{2}\right ) \log \left (1+\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{-\sqrt{4 e^2-d^2 g^2}+i d g}\right )}{g}-\frac{b n \sin ^{-1}\left (\frac{g x}{2}\right ) \log \left (1+\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{\sqrt{4 e^2-d^2 g^2}+i d g}\right )}{g}+\frac{i b n \sin ^{-1}\left (\frac{g x}{2}\right )^2}{2 g} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])/(Sqrt[2 - g*x]*Sqrt[2 + g*x]),x]

[Out]

((I/2)*b*n*ArcSin[(g*x)/2]^2)/g - (b*n*ArcSin[(g*x)/2]*Log[1 + (2*e*E^(I*ArcSin[(g*x)/2]))/(I*d*g - Sqrt[4*e^2
- d^2*g^2])])/g - (b*n*ArcSin[(g*x)/2]*Log[1 + (2*e*E^(I*ArcSin[(g*x)/2]))/(I*d*g + Sqrt[4*e^2 - d^2*g^2])])/
g + (ArcSin[(g*x)/2]*(a + b*Log[c*(d + e*x)^n]))/g + (I*b*n*PolyLog[2, (-2*e*E^(I*ArcSin[(g*x)/2]))/(I*d*g - S
qrt[4*e^2 - d^2*g^2])])/g + (I*b*n*PolyLog[2, (-2*e*E^(I*ArcSin[(g*x)/2]))/(I*d*g + Sqrt[4*e^2 - d^2*g^2])])/g

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 2405

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/(Sqrt[(f1_) + (g1_.)*(x_)]*Sqrt[(f2_) + (g2_.)*(x_)])
, x_Symbol] :> With[{u = IntHide[1/Sqrt[f1*f2 + g1*g2*x^2], x]}, Simp[u*(a + b*Log[c*(d + e*x)^n]), x] - Dist[
b*e*n, Int[SimplifyIntegrand[u/(d + e*x), x], x], x]] /; FreeQ[{a, b, c, d, e, f1, g1, f2, g2, n}, x] && EqQ[f
2*g1 + f1*g2, 0] && GtQ[f1, 0] && GtQ[f2, 0]

Rule 4741

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Subst[Int[((a + b*x)^n*Cos[x])/
(c*d + e*Sin[x]), x], x, ArcSin[c*x]] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[n, 0]

Rule 4521

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

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(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 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{a+b \log \left (c (d+e x)^n\right )}{\sqrt{2-g x} \sqrt{2+g x}} \, dx &=\frac{\sin ^{-1}\left (\frac{g x}{2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-(b e n) \int \frac{\sin ^{-1}\left (\frac{g x}{2}\right )}{d g+e g x} \, dx\\ &=\frac{\sin ^{-1}\left (\frac{g x}{2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-(b e n) \operatorname{Subst}\left (\int \frac{x \cos (x)}{\frac{d g^2}{2}+e g \sin (x)} \, dx,x,\sin ^{-1}\left (\frac{g x}{2}\right )\right )\\ &=\frac{i b n \sin ^{-1}\left (\frac{g x}{2}\right )^2}{2 g}+\frac{\sin ^{-1}\left (\frac{g x}{2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-(i b e n) \operatorname{Subst}\left (\int \frac{e^{i x} x}{e e^{i x} g+\frac{1}{2} i d g^2-\frac{1}{2} g \sqrt{4 e^2-d^2 g^2}} \, dx,x,\sin ^{-1}\left (\frac{g x}{2}\right )\right )-(i b e n) \operatorname{Subst}\left (\int \frac{e^{i x} x}{e e^{i x} g+\frac{1}{2} i d g^2+\frac{1}{2} g \sqrt{4 e^2-d^2 g^2}} \, dx,x,\sin ^{-1}\left (\frac{g x}{2}\right )\right )\\ &=\frac{i b n \sin ^{-1}\left (\frac{g x}{2}\right )^2}{2 g}-\frac{b n \sin ^{-1}\left (\frac{g x}{2}\right ) \log \left (1+\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{i d g-\sqrt{4 e^2-d^2 g^2}}\right )}{g}-\frac{b n \sin ^{-1}\left (\frac{g x}{2}\right ) \log \left (1+\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{i d g+\sqrt{4 e^2-d^2 g^2}}\right )}{g}+\frac{\sin ^{-1}\left (\frac{g x}{2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}+\frac{(b n) \operatorname{Subst}\left (\int \log \left (1+\frac{e e^{i x} g}{\frac{1}{2} i d g^2-\frac{1}{2} g \sqrt{4 e^2-d^2 g^2}}\right ) \, dx,x,\sin ^{-1}\left (\frac{g x}{2}\right )\right )}{g}+\frac{(b n) \operatorname{Subst}\left (\int \log \left (1+\frac{e e^{i x} g}{\frac{1}{2} i d g^2+\frac{1}{2} g \sqrt{4 e^2-d^2 g^2}}\right ) \, dx,x,\sin ^{-1}\left (\frac{g x}{2}\right )\right )}{g}\\ &=\frac{i b n \sin ^{-1}\left (\frac{g x}{2}\right )^2}{2 g}-\frac{b n \sin ^{-1}\left (\frac{g x}{2}\right ) \log \left (1+\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{i d g-\sqrt{4 e^2-d^2 g^2}}\right )}{g}-\frac{b n \sin ^{-1}\left (\frac{g x}{2}\right ) \log \left (1+\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{i d g+\sqrt{4 e^2-d^2 g^2}}\right )}{g}+\frac{\sin ^{-1}\left (\frac{g x}{2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac{(i b n) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{e g x}{\frac{1}{2} i d g^2-\frac{1}{2} g \sqrt{4 e^2-d^2 g^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}\right )}{g}-\frac{(i b n) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{e g x}{\frac{1}{2} i d g^2+\frac{1}{2} g \sqrt{4 e^2-d^2 g^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}\right )}{g}\\ &=\frac{i b n \sin ^{-1}\left (\frac{g x}{2}\right )^2}{2 g}-\frac{b n \sin ^{-1}\left (\frac{g x}{2}\right ) \log \left (1+\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{i d g-\sqrt{4 e^2-d^2 g^2}}\right )}{g}-\frac{b n \sin ^{-1}\left (\frac{g x}{2}\right ) \log \left (1+\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{i d g+\sqrt{4 e^2-d^2 g^2}}\right )}{g}+\frac{\sin ^{-1}\left (\frac{g x}{2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}+\frac{i b n \text{Li}_2\left (-\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{i d g-\sqrt{4 e^2-d^2 g^2}}\right )}{g}+\frac{i b n \text{Li}_2\left (-\frac{2 e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{i d g+\sqrt{4 e^2-d^2 g^2}}\right )}{g}\\ \end{align*}

Mathematica [A]  time = 0.0283868, size = 307, normalized size = 1.1 \[ \frac{i b n \text{PolyLog}\left (2,\frac{2 i e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{d g-i \sqrt{4 e^2-d^2 g^2}}\right )}{g}+\frac{i b n \text{PolyLog}\left (2,\frac{2 i e e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{d g+i \sqrt{4 e^2-d^2 g^2}}\right )}{g}+\frac{a \sin ^{-1}\left (\frac{g x}{2}\right )}{g}+\frac{b \sin ^{-1}\left (\frac{g x}{2}\right ) \log \left (c (d+e x)^n\right )}{g}-\frac{b n \sin ^{-1}\left (\frac{g x}{2}\right ) \log \left (1+\frac{e g e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{-\frac{1}{2} g \sqrt{4 e^2-d^2 g^2}+\frac{1}{2} i d g^2}\right )}{g}-\frac{b n \sin ^{-1}\left (\frac{g x}{2}\right ) \log \left (1+\frac{e g e^{i \sin ^{-1}\left (\frac{g x}{2}\right )}}{\frac{1}{2} g \sqrt{4 e^2-d^2 g^2}+\frac{1}{2} i d g^2}\right )}{g}+\frac{i b n \sin ^{-1}\left (\frac{g x}{2}\right )^2}{2 g} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])/(Sqrt[2 - g*x]*Sqrt[2 + g*x]),x]

[Out]

(a*ArcSin[(g*x)/2])/g + ((I/2)*b*n*ArcSin[(g*x)/2]^2)/g - (b*n*ArcSin[(g*x)/2]*Log[1 + (e*E^(I*ArcSin[(g*x)/2]
)*g)/((I/2)*d*g^2 - (g*Sqrt[4*e^2 - d^2*g^2])/2)])/g - (b*n*ArcSin[(g*x)/2]*Log[1 + (e*E^(I*ArcSin[(g*x)/2])*g
)/((I/2)*d*g^2 + (g*Sqrt[4*e^2 - d^2*g^2])/2)])/g + (b*ArcSin[(g*x)/2]*Log[c*(d + e*x)^n])/g + (I*b*n*PolyLog[
2, ((2*I)*e*E^(I*ArcSin[(g*x)/2]))/(d*g - I*Sqrt[4*e^2 - d^2*g^2])])/g + (I*b*n*PolyLog[2, ((2*I)*e*E^(I*ArcSi
n[(g*x)/2]))/(d*g + I*Sqrt[4*e^2 - d^2*g^2])])/g

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Maple [F]  time = 1.102, size = 0, normalized size = 0. \begin{align*} \int{(a+b\ln \left ( c \left ( ex+d \right ) ^{n} \right ) ){\frac{1}{\sqrt{-gx+2}}}{\frac{1}{\sqrt{gx+2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(e*x+d)^n))/(-g*x+2)^(1/2)/(g*x+2)^(1/2),x)

[Out]

int((a+b*ln(c*(e*x+d)^n))/(-g*x+2)^(1/2)/(g*x+2)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{\log \left ({\left (e x + d\right )}^{n}\right ) + \log \left (c\right )}{\sqrt{g x + 2} \sqrt{-g x + 2}}\,{d x} + \frac{a \arcsin \left (\frac{g^{2} x}{2 \, \sqrt{g^{2}}}\right )}{\sqrt{g^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/(-g*x+2)^(1/2)/(g*x+2)^(1/2),x, algorithm="maxima")

[Out]

b*integrate((log((e*x + d)^n) + log(c))/(sqrt(g*x + 2)*sqrt(-g*x + 2)), x) + a*arcsin(1/2*g^2*x/sqrt(g^2))/sqr
t(g^2)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{g x + 2} \sqrt{-g x + 2} b \log \left ({\left (e x + d\right )}^{n} c\right ) + \sqrt{g x + 2} \sqrt{-g x + 2} a}{g^{2} x^{2} - 4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/(-g*x+2)^(1/2)/(g*x+2)^(1/2),x, algorithm="fricas")

[Out]

integral(-(sqrt(g*x + 2)*sqrt(-g*x + 2)*b*log((e*x + d)^n*c) + sqrt(g*x + 2)*sqrt(-g*x + 2)*a)/(g^2*x^2 - 4),
x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \log{\left (c \left (d + e x\right )^{n} \right )}}{\sqrt{- g x + 2} \sqrt{g x + 2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(e*x+d)**n))/(-g*x+2)**(1/2)/(g*x+2)**(1/2),x)

[Out]

Integral((a + b*log(c*(d + e*x)**n))/(sqrt(-g*x + 2)*sqrt(g*x + 2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{\sqrt{g x + 2} \sqrt{-g x + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/(-g*x+2)^(1/2)/(g*x+2)^(1/2),x, algorithm="giac")

[Out]

integrate((b*log((e*x + d)^n*c) + a)/(sqrt(g*x + 2)*sqrt(-g*x + 2)), x)