3.3.77 \(\int \frac {a+b \log (c (d+e x)^n)}{\sqrt {2-g x} \sqrt {2+g x}} \, dx\) [277]

Optimal. Leaf size=278 \[ \frac {i b n \sin ^{-1}\left (\frac {g x}{2}\right )^2}{2 g}-\frac {b n \sin ^{-1}\left (\frac {g x}{2}\right ) \log \left (1+\frac {2 e e^{i \sin ^{-1}\left (\frac {g x}{2}\right )}}{i d g-\sqrt {4 e^2-d^2 g^2}}\right )}{g}-\frac {b n \sin ^{-1}\left (\frac {g x}{2}\right ) \log \left (1+\frac {2 e e^{i \sin ^{-1}\left (\frac {g x}{2}\right )}}{i d g+\sqrt {4 e^2-d^2 g^2}}\right )}{g}+\frac {\sin ^{-1}\left (\frac {g x}{2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}+\frac {i b n \text {Li}_2\left (-\frac {2 e e^{i \sin ^{-1}\left (\frac {g x}{2}\right )}}{i d g-\sqrt {4 e^2-d^2 g^2}}\right )}{g}+\frac {i b n \text {Li}_2\left (-\frac {2 e e^{i \sin ^{-1}\left (\frac {g x}{2}\right )}}{i d g+\sqrt {4 e^2-d^2 g^2}}\right )}{g} \]

[Out]

1/2*I*b*n*arcsin(1/2*g*x)^2/g+arcsin(1/2*g*x)*(a+b*ln(c*(e*x+d)^n))/g-b*n*arcsin(1/2*g*x)*ln(1+2*e*(1/2*I*g*x+
1/2*(-g^2*x^2+4)^(1/2))/(I*d*g-(-d^2*g^2+4*e^2)^(1/2)))/g-b*n*arcsin(1/2*g*x)*ln(1+2*e*(1/2*I*g*x+1/2*(-g^2*x^
2+4)^(1/2))/(I*d*g+(-d^2*g^2+4*e^2)^(1/2)))/g+I*b*n*polylog(2,-2*e*(1/2*I*g*x+1/2*(-g^2*x^2+4)^(1/2))/(I*d*g-(
-d^2*g^2+4*e^2)^(1/2)))/g+I*b*n*polylog(2,-2*e*(1/2*I*g*x+1/2*(-g^2*x^2+4)^(1/2))/(I*d*g+(-d^2*g^2+4*e^2)^(1/2
)))/g

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Rubi [A]
time = 0.32, antiderivative size = 278, normalized size of antiderivative = 1.00, 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 = {222, 2452, 4825, 4617, 2221, 2317, 2438} \begin {gather*} \frac {i b n \text {PolyLog}\left (2,-\frac {2 e e^{i \text {ArcSin}\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 \text {ArcSin}\left (\frac {g x}{2}\right )}}{\sqrt {4 e^2-d^2 g^2}+i d g}\right )}{g}+\frac {\text {ArcSin}\left (\frac {g x}{2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac {b n \text {ArcSin}\left (\frac {g x}{2}\right ) \log \left (1+\frac {2 e e^{i \text {ArcSin}\left (\frac {g x}{2}\right )}}{-\sqrt {4 e^2-d^2 g^2}+i d g}\right )}{g}-\frac {b n \text {ArcSin}\left (\frac {g x}{2}\right ) \log \left (1+\frac {2 e e^{i \text {ArcSin}\left (\frac {g x}{2}\right )}}{\sqrt {4 e^2-d^2 g^2}+i d g}\right )}{g}+\frac {i b n \text {ArcSin}\left (\frac {g x}{2}\right )^2}{2 g} \end {gather*}

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 222

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 2221

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]
- 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 2317

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 2438

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]

Rule 2452

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 4617

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 4825

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]

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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 0.02, size = 307, normalized size = 1.10 \begin {gather*} \frac {a \sin ^{-1}\left (\frac {g x}{2}\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 {e e^{i \sin ^{-1}\left (\frac {g x}{2}\right )} g}{\frac {1}{2} i d g^2-\frac {1}{2} 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 {e e^{i \sin ^{-1}\left (\frac {g x}{2}\right )} g}{\frac {1}{2} i d g^2+\frac {1}{2} g \sqrt {4 e^2-d^2 g^2}}\right )}{g}+\frac {b \sin ^{-1}\left (\frac {g x}{2}\right ) \log \left (c (d+e x)^n\right )}{g}+\frac {i b n \text {Li}_2\left (\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 {Li}_2\left (\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} \end {gather*}

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 = 0.32, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \left (e x +d \right )^{n}\right )}{\sqrt {-g x +2}\, \sqrt {g x +2}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((x*e + 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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \log {\left (c \left (d + e x\right )^{n} \right )}}{\sqrt {- g x + 2} \sqrt {g x + 2}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{\sqrt {2-g\,x}\,\sqrt {g\,x+2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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