3.1.1 \(\int \frac {\sinh ^{-1}(c x)}{d+e x} \, dx\) [1]

Optimal. Leaf size=170 \[ -\frac {\sinh ^{-1}(c x)^2}{2 e}+\frac {\sinh ^{-1}(c x) \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\sinh ^{-1}(c x) \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\text {PolyLog}\left (2,-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\text {PolyLog}\left (2,-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e} \]

[Out]

-1/2*arcsinh(c*x)^2/e+arcsinh(c*x)*ln(1+e*(c*x+(c^2*x^2+1)^(1/2))/(c*d-(c^2*d^2+e^2)^(1/2)))/e+arcsinh(c*x)*ln
(1+e*(c*x+(c^2*x^2+1)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2)))/e+polylog(2,-e*(c*x+(c^2*x^2+1)^(1/2))/(c*d-(c^2*d^2+e
^2)^(1/2)))/e+polylog(2,-e*(c*x+(c^2*x^2+1)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2)))/e

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Rubi [A]
time = 0.18, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5827, 5680, 2221, 2317, 2438} \begin {gather*} \frac {\text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\sinh ^{-1}(c x) \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )}{e}+\frac {\sinh ^{-1}(c x) \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )}{e}-\frac {\sinh ^{-1}(c x)^2}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSinh[c*x]/(d + e*x),x]

[Out]

-1/2*ArcSinh[c*x]^2/e + (ArcSinh[c*x]*Log[1 + (e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2])])/e + (ArcSinh[c*
x]*Log[1 + (e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2])])/e + PolyLog[2, -((e*E^ArcSinh[c*x])/(c*d - Sqrt[c^
2*d^2 + e^2]))]/e + PolyLog[2, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))]/e

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 5680

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 5827

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

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Mathematica [A]
time = 0.01, size = 168, normalized size = 0.99 \begin {gather*} -\frac {\sinh ^{-1}(c x)^2}{2 e}+\frac {\sinh ^{-1}(c x) \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\sinh ^{-1}(c x) \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\text {PolyLog}\left (2,\frac {e e^{\sinh ^{-1}(c x)}}{-c d+\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\text {PolyLog}\left (2,-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[c*x]/(d + e*x),x]

[Out]

-1/2*ArcSinh[c*x]^2/e + (ArcSinh[c*x]*Log[1 + (e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2])])/e + (ArcSinh[c*
x]*Log[1 + (e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2])])/e + PolyLog[2, (e*E^ArcSinh[c*x])/(-(c*d) + Sqrt[c
^2*d^2 + e^2])]/e + PolyLog[2, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))]/e

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Maple [A]
time = 3.42, size = 272, normalized size = 1.60

method	result	size
derivativedivides	\(\frac {-\frac {c \arcsinh \left (c x \right )^{2}}{2 e}+\frac {c \arcsinh \left (c x \right ) \ln \left (\frac {-c d -e \left (c x +\sqrt {c^{2} x^{2}+1}\right )+\sqrt {c^{2} d^{2}+e^{2}}}{-c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {c \arcsinh \left (c x \right ) \ln \left (\frac {c d +e \left (c x +\sqrt {c^{2} x^{2}+1}\right )+\sqrt {c^{2} d^{2}+e^{2}}}{c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {c \dilog \left (\frac {c d +e \left (c x +\sqrt {c^{2} x^{2}+1}\right )+\sqrt {c^{2} d^{2}+e^{2}}}{c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {c \dilog \left (\frac {-c d -e \left (c x +\sqrt {c^{2} x^{2}+1}\right )+\sqrt {c^{2} d^{2}+e^{2}}}{-c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e}}{c}\)	\(272\)
default	\(\frac {-\frac {c \arcsinh \left (c x \right )^{2}}{2 e}+\frac {c \arcsinh \left (c x \right ) \ln \left (\frac {-c d -e \left (c x +\sqrt {c^{2} x^{2}+1}\right )+\sqrt {c^{2} d^{2}+e^{2}}}{-c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {c \arcsinh \left (c x \right ) \ln \left (\frac {c d +e \left (c x +\sqrt {c^{2} x^{2}+1}\right )+\sqrt {c^{2} d^{2}+e^{2}}}{c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {c \dilog \left (\frac {c d +e \left (c x +\sqrt {c^{2} x^{2}+1}\right )+\sqrt {c^{2} d^{2}+e^{2}}}{c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {c \dilog \left (\frac {-c d -e \left (c x +\sqrt {c^{2} x^{2}+1}\right )+\sqrt {c^{2} d^{2}+e^{2}}}{-c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e}}{c}\)	\(272\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(c*x)/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

1/c*(-1/2*c*arcsinh(c*x)^2/e+1/e*c*arcsinh(c*x)*ln((-c*d-e*(c*x+(c^2*x^2+1)^(1/2))+(c^2*d^2+e^2)^(1/2))/(-c*d+
(c^2*d^2+e^2)^(1/2)))+1/e*c*arcsinh(c*x)*ln((c*d+e*(c*x+(c^2*x^2+1)^(1/2))+(c^2*d^2+e^2)^(1/2))/(c*d+(c^2*d^2+
e^2)^(1/2)))+1/e*c*dilog((c*d+e*(c*x+(c^2*x^2+1)^(1/2))+(c^2*d^2+e^2)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2)))+1/e*c*
dilog((-c*d-e*(c*x+(c^2*x^2+1)^(1/2))+(c^2*d^2+e^2)^(1/2))/(-c*d+(c^2*d^2+e^2)^(1/2))))

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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(arcsinh(c*x)/(e*x+d),x, algorithm="maxima")

[Out]

integrate(arcsinh(c*x)/(x*e + d), x)

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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(arcsinh(c*x)/(e*x+d),x, algorithm="fricas")

[Out]

integral(arcsinh(c*x)/(x*e + d), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asinh}{\left (c x \right )}}{d + e x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(c*x)/(e*x+d),x)

[Out]

Integral(asinh(c*x)/(d + e*x), 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(arcsinh(c*x)/(e*x+d),x, algorithm="giac")

[Out]

integrate(arcsinh(c*x)/(e*x + d), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {asinh}\left (c\,x\right )}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(c*x)/(d + e*x),x)

[Out]

int(asinh(c*x)/(d + e*x), x)

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