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

Optimal. Leaf size=197 \[ -\frac {m \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \sinh ^{-1}(c x) \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}+1\right )}{c}-\frac {m \sinh ^{-1}(c x) \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}+1\right )}{c}+\frac {\sinh ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}+\frac {m \sinh ^{-1}(c x)^2}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {215, 2404, 5799, 5561, 2190, 2279, 2391} \[ -\frac {m \text {PolyLog}\left (2,-\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \text {PolyLog}\left (2,-\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}\right )}{c}-\frac {m \sinh ^{-1}(c x) \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}+1\right )}{c}-\frac {m \sinh ^{-1}(c x) \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}+1\right )}{c}+\frac {\sinh ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}+\frac {m \sinh ^{-1}(c x)^2}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 215

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

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]

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 5561

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

Rule 5799

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

Rubi steps

\begin {align*} \int \frac {\log \left (h (f+g x)^m\right )}{\sqrt {1+c^2 x^2}} \, dx &=\frac {\sinh ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}-(g m) \int \frac {\sinh ^{-1}(c x)}{c f+c g x} \, dx\\ &=\frac {\sinh ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}-(g m) \operatorname {Subst}\left (\int \frac {x \cosh (x)}{c^2 f+c g \sinh (x)} \, dx,x,\sinh ^{-1}(c x)\right )\\ &=\frac {m \sinh ^{-1}(c x)^2}{2 c}+\frac {\sinh ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}-(g m) \operatorname {Subst}\left (\int \frac {e^x x}{c^2 f+c e^x g-c \sqrt {c^2 f^2+g^2}} \, dx,x,\sinh ^{-1}(c x)\right )-(g m) \operatorname {Subst}\left (\int \frac {e^x x}{c^2 f+c e^x g+c \sqrt {c^2 f^2+g^2}} \, dx,x,\sinh ^{-1}(c x)\right )\\ &=\frac {m \sinh ^{-1}(c x)^2}{2 c}-\frac {m \sinh ^{-1}(c x) \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \sinh ^{-1}(c x) \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {\sinh ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}+\frac {m \operatorname {Subst}\left (\int \log \left (1+\frac {c e^x g}{c^2 f-c \sqrt {c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}+\frac {m \operatorname {Subst}\left (\int \log \left (1+\frac {c e^x g}{c^2 f+c \sqrt {c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}\\ &=\frac {m \sinh ^{-1}(c x)^2}{2 c}-\frac {m \sinh ^{-1}(c x) \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \sinh ^{-1}(c x) \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {\sinh ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}+\frac {m \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {c g x}{c^2 f-c \sqrt {c^2 f^2+g^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c}+\frac {m \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {c g x}{c^2 f+c \sqrt {c^2 f^2+g^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c}\\ &=\frac {m \sinh ^{-1}(c x)^2}{2 c}-\frac {m \sinh ^{-1}(c x) \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \sinh ^{-1}(c x) \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {\sinh ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}-\frac {m \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 206, normalized size = 1.05 \[ -\frac {m \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \sinh ^{-1}(c x) \log \left (\frac {c g e^{\sinh ^{-1}(c x)}}{c^2 f-c \sqrt {c^2 f^2+g^2}}+1\right )}{c}-\frac {m \sinh ^{-1}(c x) \log \left (\frac {c g e^{\sinh ^{-1}(c x)}}{c \sqrt {c^2 f^2+g^2}+c^2 f}+1\right )}{c}+\frac {\sinh ^{-1}(c x) \log \left (h (f+g x)^m\right )}{c}+\frac {m \sinh ^{-1}(c x)^2}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt {c^{2} x^{2} + 1}}, x\right ) \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt {c^{2} x^{2} + 1}}\,{d x} \]

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)

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (h \left (g x +f \right )^{m}\right )}{\sqrt {c^{2} x^{2}+1}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt {c^{2} x^{2} + 1}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (h\,{\left (f+g\,x\right )}^m\right )}{\sqrt {c^2\,x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log {\left (h \left (f + g x\right )^{m} \right )}}{\sqrt {c^{2} x^{2} + 1}}\, dx \]

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**2*x**2 + 1), x)

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