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

Optimal. Leaf size=124 \[ -\frac{i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{2 c d^2}+\frac{i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{2 c d^2}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (c^2 x^2+1\right )}+\frac{\tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d^2}+\frac{b}{2 c d^2 \sqrt{c^2 x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.0978521, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5690, 5693, 4180, 2279, 2391, 261} \[ -\frac{i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{2 c d^2}+\frac{i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{2 c d^2}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (c^2 x^2+1\right )}+\frac{\tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d^2}+\frac{b}{2 c d^2 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5690

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSinh[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a +
b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 + c^2*x^2)^FracPar
t[p]), Int[x*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ
[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

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]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{a+b \sinh ^{-1}(c x)}{\left (d+c^2 d x^2\right )^2} \, dx &=\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}-\frac{(b c) \int \frac{x}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{2 d^2}+\frac{\int \frac{a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx}{2 d}\\ &=\frac{b}{2 c d^2 \sqrt{1+c^2 x^2}}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac{\operatorname{Subst}\left (\int (a+b x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 c d^2}\\ &=\frac{b}{2 c d^2 \sqrt{1+c^2 x^2}}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac{\left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c d^2}-\frac{(i b) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 c d^2}+\frac{(i b) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 c d^2}\\ &=\frac{b}{2 c d^2 \sqrt{1+c^2 x^2}}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac{\left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c d^2}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 c d^2}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 c d^2}\\ &=\frac{b}{2 c d^2 \sqrt{1+c^2 x^2}}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac{\left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c d^2}-\frac{i b \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{2 c d^2}+\frac{i b \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{2 c d^2}\\ \end{align*}

Mathematica [A]  time = 0.105299, size = 216, normalized size = 1.74 \[ \frac{-i b \left (c^2 x^2+1\right ) \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )+i b \left (c^2 x^2+1\right ) \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )+a c^2 x^2 \tan ^{-1}(c x)+a c x+a \tan ^{-1}(c x)+b \sqrt{c^2 x^2+1}+i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+b c x \sinh ^{-1}(c x)+i b \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-i b \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \left (c^3 x^2+c\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*c*x + b*Sqrt[1 + c^2*x^2] + b*c*x*ArcSinh[c*x] + a*ArcTan[c*x] + a*c^2*x^2*ArcTan[c*x] + I*b*ArcSinh[c*x]*L
og[1 - I*E^ArcSinh[c*x]] + I*b*c^2*x^2*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] - I*b*ArcSinh[c*x]*Log[1 + I*E^A
rcSinh[c*x]] - I*b*c^2*x^2*ArcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] - I*b*(1 + c^2*x^2)*PolyLog[2, (-I)*E^ArcSin
h[c*x]] + I*b*(1 + c^2*x^2)*PolyLog[2, I*E^ArcSinh[c*x]])/(2*d^2*(c + c^3*x^2))

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Maple [A]  time = 0.01, size = 234, normalized size = 1.9 \begin{align*}{\frac{ax}{2\,{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{a\arctan \left ( cx \right ) }{2\,c{d}^{2}}}+{\frac{b{\it Arcsinh} \left ( cx \right ) x}{2\,{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{b{\it Arcsinh} \left ( cx \right ) \arctan \left ( cx \right ) }{2\,c{d}^{2}}}+{\frac{b}{2\,c{d}^{2}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{b\arctan \left ( cx \right ) }{2\,c{d}^{2}}\ln \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{b\arctan \left ( cx \right ) }{2\,c{d}^{2}}\ln \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{\frac{i}{2}}b}{c{d}^{2}}{\it dilog} \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{i}{2}}b}{c{d}^{2}}{\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)^2,x)

[Out]

1/2*a/d^2*x/(c^2*x^2+1)+1/2/c*a/d^2*arctan(c*x)+1/2*b/d^2*arcsinh(c*x)*x/(c^2*x^2+1)+1/2/c*b/d^2*arcsinh(c*x)*
arctan(c*x)+1/2*b/c/d^2/(c^2*x^2+1)^(1/2)+1/2/c*b/d^2*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2/c*b/
d^2*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2*I/c*b/d^2*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/2*I
/c*b/d^2*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*} \frac{1}{2} \, a{\left (\frac{x}{c^{2} d^{2} x^{2} + d^{2}} + \frac{\arctan \left (c x\right )}{c d^{2}}\right )} + b \int \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{4} d^{2} x^{4} + 2 \, c^{2} d^{2} x^{2} + d^{2}}\,{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)^2,x, algorithm="maxima")

[Out]

1/2*a*(x/(c^2*d^2*x^2 + d^2) + arctan(c*x)/(c*d^2)) + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/(c^4*d^2*x^4 +
2*c^2*d^2*x^2 + d^2), x)

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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^{4} d^{2} x^{4} + 2 \, c^{2} d^{2} x^{2} + d^{2}}, 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)^2,x, algorithm="fricas")

[Out]

integral((b*arcsinh(c*x) + a)/(c^4*d^2*x^4 + 2*c^2*d^2*x^2 + d^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac{b \operatorname{asinh}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \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)**2,x)

[Out]

(Integral(a/(c**4*x**4 + 2*c**2*x**2 + 1), x) + Integral(b*asinh(c*x)/(c**4*x**4 + 2*c**2*x**2 + 1), x))/d**2

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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}{{\left (c^{2} d x^{2} + d\right )}^{2}}\,{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)^2,x, algorithm="giac")

[Out]

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