3.86 \(\int \frac{\log (h (f+g x)^m)}{\sqrt{1-c^2 x^2}} \, dx\)

Optimal. Leaf size=237 \[ \frac{i m \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{i m \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}-\frac{m \sin ^{-1}(c x) \log \left (1-\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{m \sin ^{-1}(c x) \log \left (1-\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}+\frac{\sin ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}+\frac{i m \sin ^{-1}(c x)^2}{2 c} \]

[Out]

((I/2)*m*ArcSin[c*x]^2)/c - (m*ArcSin[c*x]*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/c - (
m*ArcSin[c*x]*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/c + (ArcSin[c*x]*Log[h*(f + g*x)^m
])/c + (I*m*PolyLog[2, (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/c + (I*m*PolyLog[2, (I*E^(I*ArcSi
n[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/c

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Rubi [A]  time = 0.33166, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {216, 2404, 4741, 4519, 2190, 2279, 2391} \[ \frac{i m \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{i m \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}-\frac{m \sin ^{-1}(c x) \log \left (1-\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{m \sin ^{-1}(c x) \log \left (1-\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}+\frac{\sin ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}+\frac{i m \sin ^{-1}(c x)^2}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[Log[h*(f + g*x)^m]/Sqrt[1 - c^2*x^2],x]

[Out]

((I/2)*m*ArcSin[c*x]^2)/c - (m*ArcSin[c*x]*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/c - (
m*ArcSin[c*x]*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/c + (ArcSin[c*x]*Log[h*(f + g*x)^m
])/c + (I*m*PolyLog[2, (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/c + (I*m*PolyLog[2, (I*E^(I*ArcSi
n[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/c

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 2404

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/Sqrt[(f_) + (g_.)*(x_)^2], x_Symbol] :> With[{u = Int
Hide[1/Sqrt[f + g*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, f, g, n}, x] && GtQ[f, 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 4519

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] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[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{\log \left (h (f+g x)^m\right )}{\sqrt{1-c^2 x^2}} \, dx &=\frac{\sin ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}-(g m) \int \frac{\sin ^{-1}(c x)}{c f+c g x} \, dx\\ &=\frac{\sin ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}-(g m) \operatorname{Subst}\left (\int \frac{x \cos (x)}{c^2 f+c g \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )\\ &=\frac{i m \sin ^{-1}(c x)^2}{2 c}+\frac{\sin ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}-(g m) \operatorname{Subst}\left (\int \frac{e^{i x} x}{c^2 f-i c e^{i x} g-c \sqrt{c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )-(g m) \operatorname{Subst}\left (\int \frac{e^{i x} x}{c^2 f-i c e^{i x} g+c \sqrt{c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )\\ &=\frac{i m \sin ^{-1}(c x)^2}{2 c}-\frac{m \sin ^{-1}(c x) \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{m \sin ^{-1}(c x) \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{\sin ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}+\frac{m \operatorname{Subst}\left (\int \log \left (1-\frac{i c e^{i x} g}{c^2 f-c \sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}+\frac{m \operatorname{Subst}\left (\int \log \left (1-\frac{i c e^{i x} g}{c^2 f+c \sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}\\ &=\frac{i m \sin ^{-1}(c x)^2}{2 c}-\frac{m \sin ^{-1}(c x) \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{m \sin ^{-1}(c x) \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{\sin ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}-\frac{(i m) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{i c g x}{c^2 f-c \sqrt{c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c}-\frac{(i m) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{i c g x}{c^2 f+c \sqrt{c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c}\\ &=\frac{i m \sin ^{-1}(c x)^2}{2 c}-\frac{m \sin ^{-1}(c x) \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{m \sin ^{-1}(c x) \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{\sin ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}+\frac{i m \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{i m \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}\\ \end{align*}

Mathematica [A]  time = 0.0236326, size = 246, normalized size = 1.04 \[ \frac{i m \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{i m \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}-\frac{m \sin ^{-1}(c x) \log \left (1-\frac{i c g e^{i \sin ^{-1}(c x)}}{c^2 f-c \sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{m \sin ^{-1}(c x) \log \left (1-\frac{i c g e^{i \sin ^{-1}(c x)}}{c \sqrt{c^2 f^2-g^2}+c^2 f}\right )}{c}+\frac{\sin ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}+\frac{i m \sin ^{-1}(c x)^2}{2 c} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[h*(f + g*x)^m]/Sqrt[1 - c^2*x^2],x]

[Out]

((I/2)*m*ArcSin[c*x]^2)/c - (m*ArcSin[c*x]*Log[1 - (I*c*E^(I*ArcSin[c*x])*g)/(c^2*f - c*Sqrt[c^2*f^2 - g^2])])
/c - (m*ArcSin[c*x]*Log[1 - (I*c*E^(I*ArcSin[c*x])*g)/(c^2*f + c*Sqrt[c^2*f^2 - g^2])])/c + (ArcSin[c*x]*Log[h
*(f + g*x)^m])/c + (I*m*PolyLog[2, (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/c + (I*m*PolyLog[2, (
I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/c

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Maple [F]  time = 0.165, size = 0, normalized size = 0. \begin{align*} \int{\ln \left ( h \left ( gx+f \right ) ^{m} \right ){\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x)

[Out]

int(ln(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt{-c^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(log((g*x + f)^m*h)/sqrt(-c^2*x^2 + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-c^{2} x^{2} + 1} \log \left ({\left (g x + f\right )}^{m} h\right )}{c^{2} x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*x^2 + 1)*log((g*x + f)^m*h)/(c^2*x^2 - 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log{\left (h \left (f + g x\right )^{m} \right )}}{\sqrt{- \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(h*(g*x+f)**m)/(-c**2*x**2+1)**(1/2),x)

[Out]

Integral(log(h*(f + g*x)**m)/sqrt(-(c*x - 1)*(c*x + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt{-c^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(log((g*x + f)^m*h)/sqrt(-c^2*x^2 + 1), x)