3.32 \(\int \frac{a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx\)

Optimal. Leaf size=70 \[ -\frac{i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c d}+\frac{i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c d}+\frac{2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d} \]

[Out]

(2*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(c*d) - (I*b*PolyLog[2, (-I)*E^ArcSinh[c*x]])/(c*d) + (I*b*Pol
yLog[2, I*E^ArcSinh[c*x]])/(c*d)

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Rubi [A]  time = 0.0641422, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {5693, 4180, 2279, 2391} \[ -\frac{i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c d}+\frac{i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c d}+\frac{2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(d + c^2*d*x^2),x]

[Out]

(2*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(c*d) - (I*b*PolyLog[2, (-I)*E^ArcSinh[c*x]])/(c*d) + (I*b*Pol
yLog[2, I*E^ArcSinh[c*x]])/(c*d)

Rule 5693

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/(c*d), Subst[Int[(a +
b*x)^n*Sech[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0]

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && 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{a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx &=\frac{\operatorname{Subst}\left (\int (a+b x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{c d}\\ &=\frac{2 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c d}-\frac{(i b) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c d}+\frac{(i b) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c d}\\ &=\frac{2 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c d}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c d}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c d}\\ &=\frac{2 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c d}-\frac{i b \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c d}+\frac{i b \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c d}\\ \end{align*}

Mathematica [A]  time = 0.103496, size = 135, normalized size = 1.93 \[ -\frac{c \left (b c \text{PolyLog}\left (2,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right )-b c \text{PolyLog}\left (2,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+a \sqrt{-c^2} \tan ^{-1}(c x)-b c \sinh ^{-1}(c x) \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )+b c \sinh ^{-1}(c x) \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )\right )}{\left (-c^2\right )^{3/2} d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])/(d + c^2*d*x^2),x]

[Out]

-((c*(a*Sqrt[-c^2]*ArcTan[c*x] - b*c*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + b*c*ArcSinh[c*x]*Lo
g[1 + (Sqrt[-c^2]*E^ArcSinh[c*x])/c] + b*c*PolyLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] - b*c*PolyLog[2, (Sqrt[-c
^2]*E^ArcSinh[c*x])/c]))/((-c^2)^(3/2)*d))

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Maple [A]  time = 0.036, size = 171, normalized size = 2.4 \begin{align*}{\frac{a\arctan \left ( cx \right ) }{cd}}+{\frac{b{\it Arcsinh} \left ( cx \right ) \arctan \left ( cx \right ) }{cd}}-{\frac{b\arctan \left ( cx \right ) }{cd}\ln \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{b\arctan \left ( cx \right ) }{cd}\ln \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{ib}{cd}{\it dilog} \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{ib}{cd}{\it dilog} \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/(c^2*d*x^2+d),x)

[Out]

1/c*a/d*arctan(c*x)+1/c*b/d*arcsinh(c*x)*arctan(c*x)-1/c*b/d*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+1
/c*b/d*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I/c*b/d*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-I/c*b/d*
dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(c^2*d*x^2+d),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{arsinh}\left (c x\right ) + a}{c^{2} d x^{2} + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(c^2*d*x^2+d),x, algorithm="fricas")

[Out]

integral((b*arcsinh(c*x) + a)/(c^2*d*x^2 + d), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c^{2} x^{2} + 1}\, dx + \int \frac{b \operatorname{asinh}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/(c**2*d*x**2+d),x)

[Out]

(Integral(a/(c**2*x**2 + 1), x) + Integral(b*asinh(c*x)/(c**2*x**2 + 1), x))/d

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arsinh}\left (c x\right ) + a}{c^{2} d x^{2} + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(c^2*d*x^2+d),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/(c^2*d*x^2 + d), x)