3.314 \(\int \frac{x (a+b \sinh ^{-1}(c x))^2}{(d+c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=270 \[ -\frac{i b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt{c^2 d x^2+d}}+\frac{i b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt{c^2 d x^2+d}}+\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}+\frac{2 b \sqrt{c^2 x^2+1} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d^2 \sqrt{c^2 d x^2+d}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}+\frac{b^2}{3 c^2 d^2 \sqrt{c^2 d x^2+d}} \]

[Out]

b^2/(3*c^2*d^2*Sqrt[d + c^2*d*x^2]) + (b*x*(a + b*ArcSinh[c*x]))/(3*c*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2
]) - (a + b*ArcSinh[c*x])^2/(3*c^2*d*(d + c^2*d*x^2)^(3/2)) + (2*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*ArcT
an[E^ArcSinh[c*x]])/(3*c^2*d^2*Sqrt[d + c^2*d*x^2]) - ((I/3)*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, (-I)*E^ArcSinh[c
*x]])/(c^2*d^2*Sqrt[d + c^2*d*x^2]) + ((I/3)*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, I*E^ArcSinh[c*x]])/(c^2*d^2*Sqrt
[d + c^2*d*x^2])

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Rubi [A]  time = 0.21746, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {5717, 5690, 5693, 4180, 2279, 2391, 261} \[ -\frac{i b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt{c^2 d x^2+d}}+\frac{i b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt{c^2 d x^2+d}}+\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}+\frac{2 b \sqrt{c^2 x^2+1} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d^2 \sqrt{c^2 d x^2+d}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}+\frac{b^2}{3 c^2 d^2 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2)^(5/2),x]

[Out]

b^2/(3*c^2*d^2*Sqrt[d + c^2*d*x^2]) + (b*x*(a + b*ArcSinh[c*x]))/(3*c*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2
]) - (a + b*ArcSinh[c*x])^2/(3*c^2*d*(d + c^2*d*x^2)^(3/2)) + (2*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*ArcT
an[E^ArcSinh[c*x]])/(3*c^2*d^2*Sqrt[d + c^2*d*x^2]) - ((I/3)*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, (-I)*E^ArcSinh[c
*x]])/(c^2*d^2*Sqrt[d + c^2*d*x^2]) + ((I/3)*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, I*E^ArcSinh[c*x]])/(c^2*d^2*Sqrt
[d + c^2*d*x^2])

Rule 5717

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

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{x \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^{5/2}} \, dx &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{\left (2 b \sqrt{1+c^2 x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt{d+c^2 d x^2}}\\ &=\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac{\left (b^2 \sqrt{1+c^2 x^2}\right ) \int \frac{x}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (b \sqrt{1+c^2 x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{1+c^2 x^2} \, dx}{3 c d^2 \sqrt{d+c^2 d x^2}}\\ &=\frac{b^2}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}+\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{\left (b \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}\\ &=\frac{b^2}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}+\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (i b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (i b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}\\ &=\frac{b^2}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}+\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (i b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (i b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}\\ &=\frac{b^2}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}+\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}-\frac{i b^2 \sqrt{1+c^2 x^2} \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}+\frac{i b^2 \sqrt{1+c^2 x^2} \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt{d+c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 1.02615, size = 254, normalized size = 0.94 \[ \frac{b^2 \left (-i \left (c^2 x^2+1\right )^{3/2} \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )+i \left (c^2 x^2+1\right )^{3/2} \text{PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )+c^2 x^2+c x \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)-i \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+i \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )-\sinh ^{-1}(c x)^2+1\right )-a^2+a b \left (\sqrt{c^2 x^2+1} \left (2 \left (c^2 x^2+1\right ) \tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )+c x\right )-2 \sinh ^{-1}(c x)\right )}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2)^(5/2),x]

[Out]

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

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Maple [B]  time = 0.184, size = 591, normalized size = 2.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(5/2),x)

[Out]

-1/3*a^2/c^2/d/(c^2*d*x^2+d)^(3/2)+1/3*b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^(3/2)/c*arcsinh(c*x)*x+1/3*b^
2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2*x^2-1/3*b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^2*arcsinh(c*x)
^2+1/3*b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^2-1/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d
^3*arcsinh(c*x)*ln(1+I*(c*x+(c^2*x^2+1)^(1/2)))+1/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d^3*arcs
inh(c*x)*ln(1-I*(c*x+(c^2*x^2+1)^(1/2)))-1/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d^3*dilog(1+I*(
c*x+(c^2*x^2+1)^(1/2)))+1/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d^3*dilog(1-I*(c*x+(c^2*x^2+1)^(
1/2)))+1/3*a*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^(3/2)/c*x-2/3*a*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2
/c^2*arcsinh(c*x)+1/3*I*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d^3*ln(c*x+(c^2*x^2+1)^(1/2)+I)-1/3*I*
a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d^3*ln(c*x+(c^2*x^2+1)^(1/2)-I)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{a^{2}}{3 \,{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{2} d} + \int \frac{b^{2} x \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac{5}{2}}} + \frac{2 \, a b x \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{{\left (c^{2} d x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

-1/3*a^2/((c^2*d*x^2 + d)^(3/2)*c^2*d) + integrate(b^2*x*log(c*x + sqrt(c^2*x^2 + 1))^2/(c^2*d*x^2 + d)^(5/2)
+ 2*a*b*x*log(c*x + sqrt(c^2*x^2 + 1))/(c^2*d*x^2 + d)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c^{2} d x^{2} + d}{\left (b^{2} x \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b x \operatorname{arsinh}\left (c x\right ) + a^{2} x\right )}}{c^{6} d^{3} x^{6} + 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} + d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b^2*x*arcsinh(c*x)^2 + 2*a*b*x*arcsinh(c*x) + a^2*x)/(c^6*d^3*x^6 + 3*c^4*d^3*x^
4 + 3*c^2*d^3*x^2 + d^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (a + b \operatorname{asinh}{\left (c x \right )}\right )^{2}}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*asinh(c*x))**2/(c**2*d*x**2+d)**(5/2),x)

[Out]

Integral(x*(a + b*asinh(c*x))**2/(d*(c**2*x**2 + 1))**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2} x}{{\left (c^{2} d x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2*x/(c^2*d*x^2 + d)^(5/2), x)