3.8 \(\int \frac {a+b \sinh ^{-1}(c x)}{d+e x} \, dx\)

Optimal. Leaf size=187 \[ \frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )}{e}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b e}+\frac {b \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {b \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e} \]

[Out]

-1/2*(a+b*arcsinh(c*x))^2/b/e+(a+b*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+(
a+b*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+b*polylog(2,-e*(c*x+(c^2*x^2+1)^
(1/2))/(c*d-(c^2*d^2+e^2)^(1/2)))/e+b*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 = 0.26, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5799, 5561, 2190, 2279, 2391} \[ \frac {b \text {PolyLog}\left (2,-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {b \text {PolyLog}\left (2,-\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )}{e}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(d + e*x),x]

[Out]

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

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 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 {a+b \sinh ^{-1}(c x)}{d+e x} \, dx &=\operatorname {Subst}\left (\int \frac {(a+b x) \cosh (x)}{c d+e \sinh (x)} \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b e}+\operatorname {Subst}\left (\int \frac {e^x (a+b x)}{c d-\sqrt {c^2 d^2+e^2}+e e^x} \, dx,x,\sinh ^{-1}(c x)\right )+\operatorname {Subst}\left (\int \frac {e^x (a+b x)}{c d+\sqrt {c^2 d^2+e^2}+e e^x} \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b e}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}-\frac {b \operatorname {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 {b \operatorname {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 {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b e}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}-\frac {b \operatorname {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 {b \operatorname {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 {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b e}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {b \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {b \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 = 0.05, size = 175, normalized size = 0.94 \[ \frac {-\left (a+b \sinh ^{-1}(c x)\right ) \left (a-2 b \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )-2 b \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )+b \sinh ^{-1}(c x)\right )+2 b^2 \text {Li}_2\left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}-c d}\right )+2 b^2 \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{2 b e} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])/(d + e*x),x]

[Out]

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

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arsinh}\left (c x\right ) + a}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arcsinh(c*x) + a)/(e*x + d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (c x\right ) + a}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(e*x+d),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/(e*x + d), x)

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maple [A]  time = 0.04, size = 282, normalized size = 1.51 \[ \frac {a \ln \left (c e x +c d \right )}{e}-\frac {b \arcsinh \left (c x \right )^{2}}{2 e}+\frac {b \arcsinh \left (c x \right ) \ln \left (\frac {-\left (c x +\sqrt {c^{2} x^{2}+1}\right ) e -c d +\sqrt {c^{2} d^{2}+e^{2}}}{-c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {b \arcsinh \left (c x \right ) \ln \left (\frac {\left (c x +\sqrt {c^{2} x^{2}+1}\right ) e +c d +\sqrt {c^{2} d^{2}+e^{2}}}{c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {b \dilog \left (\frac {\left (c x +\sqrt {c^{2} x^{2}+1}\right ) e +c d +\sqrt {c^{2} d^{2}+e^{2}}}{c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {b \dilog \left (\frac {-\left (c x +\sqrt {c^{2} x^{2}+1}\right ) e -c d +\sqrt {c^{2} d^{2}+e^{2}}}{-c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/(e*x+d),x)

[Out]

a*ln(c*e*x+c*d)/e-1/2*b*arcsinh(c*x)^2/e+b/e*arcsinh(c*x)*ln((-(c*x+(c^2*x^2+1)^(1/2))*e-c*d+(c^2*d^2+e^2)^(1/
2))/(-c*d+(c^2*d^2+e^2)^(1/2)))+b/e*arcsinh(c*x)*ln(((c*x+(c^2*x^2+1)^(1/2))*e+c*d+(c^2*d^2+e^2)^(1/2))/(c*d+(
c^2*d^2+e^2)^(1/2)))+b/e*dilog(((c*x+(c^2*x^2+1)^(1/2))*e+c*d+(c^2*d^2+e^2)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2)))+
b/e*dilog((-(c*x+(c^2*x^2+1)^(1/2))*e-c*d+(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 \[ b \int \frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{e x + d}\,{d x} + \frac {a \log \left (e x + d\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/(e*x + d), x) + a*log(e*x + d)/e

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))/(d + e*x),x)

[Out]

int((a + b*asinh(c*x))/(d + e*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asinh}{\left (c x \right )}}{d + e x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asinh(c*x))/(d + e*x), x)

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