3.312 \(\int \frac{x^3 (a+b \sinh ^{-1}(c x))^2}{(d+c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=307 \[ -\frac{5 i b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{c^2 d x^2+d}}+\frac{5 i b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{c^2 d x^2+d}}-\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt{c^2 d x^2+d}}+\frac{10 b \sqrt{c^2 x^2+1} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{c^2 d x^2+d}}-\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{b^2}{3 c^4 d^2 \sqrt{c^2 d x^2+d}} \]

[Out]

-b^2/(3*c^4*d^2*Sqrt[d + c^2*d*x^2]) - (b*x*(a + b*ArcSinh[c*x]))/(3*c^3*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*
x^2]) - (x^2*(a + b*ArcSinh[c*x])^2)/(3*c^2*d*(d + c^2*d*x^2)^(3/2)) - (2*(a + b*ArcSinh[c*x])^2)/(3*c^4*d^2*S
qrt[d + c^2*d*x^2]) + (10*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(3*c^4*d^2*Sqrt[d +
c^2*d*x^2]) - (((5*I)/3)*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, (-I)*E^ArcSinh[c*x]])/(c^4*d^2*Sqrt[d + c^2*d*x^2])
+ (((5*I)/3)*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, I*E^ArcSinh[c*x]])/(c^4*d^2*Sqrt[d + c^2*d*x^2])

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Rubi [A]  time = 0.504904, antiderivative size = 307, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5751, 5717, 5693, 4180, 2279, 2391, 261} \[ -\frac{5 i b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{c^2 d x^2+d}}+\frac{5 i b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{c^2 d x^2+d}}-\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt{c^2 d x^2+d}}+\frac{10 b \sqrt{c^2 x^2+1} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{c^2 d x^2+d}}-\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{b^2}{3 c^4 d^2 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2)^(5/2),x]

[Out]

-b^2/(3*c^4*d^2*Sqrt[d + c^2*d*x^2]) - (b*x*(a + b*ArcSinh[c*x]))/(3*c^3*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*
x^2]) - (x^2*(a + b*ArcSinh[c*x])^2)/(3*c^2*d*(d + c^2*d*x^2)^(3/2)) - (2*(a + b*ArcSinh[c*x])^2)/(3*c^4*d^2*S
qrt[d + c^2*d*x^2]) + (10*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(3*c^4*d^2*Sqrt[d +
c^2*d*x^2]) - (((5*I)/3)*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, (-I)*E^ArcSinh[c*x]])/(c^4*d^2*Sqrt[d + c^2*d*x^2])
+ (((5*I)/3)*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, I*E^ArcSinh[c*x]])/(c^4*d^2*Sqrt[d + c^2*d*x^2])

Rule 5751

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

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 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^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^{5/2}} \, dx &=-\frac{x^2 \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 \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^{3/2}} \, dx}{3 c^2 d}+\frac{\left (2 b \sqrt{1+c^2 x^2}\right ) \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\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^3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}-\frac{x^2 \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 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 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^3 d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (4 b \sqrt{1+c^2 x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{1+c^2 x^2} \, dx}{3 c^3 d^2 \sqrt{d+c^2 d x^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 c^2 d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b^2}{3 c^4 d^2 \sqrt{d+c^2 d x^2}}-\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}-\frac{x^2 \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 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt{d+c^2 d x^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^4 d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (4 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^4 d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b^2}{3 c^4 d^2 \sqrt{d+c^2 d x^2}}-\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}-\frac{x^2 \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 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt{d+c^2 d x^2}}+\frac{10 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^4 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^4 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^4 d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (4 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^4 d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (4 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^4 d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b^2}{3 c^4 d^2 \sqrt{d+c^2 d x^2}}-\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}-\frac{x^2 \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 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt{d+c^2 d x^2}}+\frac{10 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^4 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^4 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^4 d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (4 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^4 d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (4 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^4 d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b^2}{3 c^4 d^2 \sqrt{d+c^2 d x^2}}-\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}-\frac{x^2 \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 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt{d+c^2 d x^2}}+\frac{10 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^4 d^2 \sqrt{d+c^2 d x^2}}-\frac{5 i b^2 \sqrt{1+c^2 x^2} \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d+c^2 d x^2}}+\frac{5 i b^2 \sqrt{1+c^2 x^2} \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d+c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 1.09377, size = 301, normalized size = 0.98 \[ \frac{-b^2 \left (5 i \left (c^2 x^2+1\right )^{3/2} \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )-5 i \left (c^2 x^2+1\right )^{3/2} \text{PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )+c^2 x^2+3 c^2 x^2 \sinh ^{-1}(c x)^2+c x \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)+5 i \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-5 i \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+2 \sinh ^{-1}(c x)^2+1\right )+a^2 \left (-\left (3 c^2 x^2+2\right )\right )+a b \left (\sqrt{c^2 x^2+1} \left (10 \left (c^2 x^2+1\right ) \tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )-c x\right )-2 \left (3 c^2 x^2+2\right ) \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \left (c^2 x^2+1\right ) \sqrt{c^2 d x^2+d}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^3*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2)^(5/2),x]

[Out]

(-(a^2*(2 + 3*c^2*x^2)) + a*b*(-2*(2 + 3*c^2*x^2)*ArcSinh[c*x] + Sqrt[1 + c^2*x^2]*(-(c*x) + 10*(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] + 2*ArcSinh[c*x]^2 + 3*
c^2*x^2*ArcSinh[c*x]^2 + (5*I)*(1 + c^2*x^2)^(3/2)*ArcSinh[c*x]*Log[1 - I/E^ArcSinh[c*x]] - (5*I)*(1 + c^2*x^2
)^(3/2)*ArcSinh[c*x]*Log[1 + I/E^ArcSinh[c*x]] + (5*I)*(1 + c^2*x^2)^(3/2)*PolyLog[2, (-I)/E^ArcSinh[c*x]] - (
5*I)*(1 + c^2*x^2)^(3/2)*PolyLog[2, I/E^ArcSinh[c*x]]))/(3*c^4*d^2*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2])

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Maple [B]  time = 0.286, size = 705, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(5/2),x)

[Out]

-a^2*x^2/c^2/d/(c^2*d*x^2+d)^(3/2)-2/3*a^2/d/c^4/(c^2*d*x^2+d)^(3/2)-b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)
^2/c^2*arcsinh(c*x)^2*x^2-1/3*b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^(3/2)/c^3*arcsinh(c*x)*x-1/3*b^2*(d*(c
^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^2*x^2-2/3*b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^4*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^4-5/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^4/d^3*
dilog(1+I*(c*x+(c^2*x^2+1)^(1/2)))+5/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^4/d^3*arcsinh(c*x)*ln(1
-I*(c*x+(c^2*x^2+1)^(1/2)))+5/3*I*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^4/d^3*ln(c*x+(c^2*x^2+1)^(1/2)
+I)+5/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^4/d^3*dilog(1-I*(c*x+(c^2*x^2+1)^(1/2)))-2*a*b*(d*(c^2
*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^2*arcsinh(c*x)*x^2-1/3*a*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^(3/2)/c^3*
x-4/3*a*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^4*arcsinh(c*x)-5/3*I*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)
^(1/2)/c^4/d^3*ln(c*x+(c^2*x^2+1)^(1/2)-I)-5/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^4/d^3*arcsinh(c
*x)*ln(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}{3} \, a b c{\left (\frac{x}{c^{6} d^{\frac{5}{2}} x^{2} + c^{4} d^{\frac{5}{2}}} - \frac{5 \, \arctan \left (c x\right )}{c^{5} d^{\frac{5}{2}}}\right )} - \frac{2}{3} \, a b{\left (\frac{3 \, x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{2} d} + \frac{2}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{4} d}\right )} \operatorname{arsinh}\left (c x\right ) - \frac{1}{3} \, a^{2}{\left (\frac{3 \, x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{2} d} + \frac{2}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{4} d}\right )} + b^{2} \int \frac{x^{3} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{{\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^3*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

-1/3*a*b*c*(x/(c^6*d^(5/2)*x^2 + c^4*d^(5/2)) - 5*arctan(c*x)/(c^5*d^(5/2))) - 2/3*a*b*(3*x^2/((c^2*d*x^2 + d)
^(3/2)*c^2*d) + 2/((c^2*d*x^2 + d)^(3/2)*c^4*d))*arcsinh(c*x) - 1/3*a^2*(3*x^2/((c^2*d*x^2 + d)^(3/2)*c^2*d) +
2/((c^2*d*x^2 + d)^(3/2)*c^4*d)) + b^2*integrate(x^3*log(c*x + sqrt(c^2*x^2 + 1))^2/(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{{\left (b^{2} x^{3} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b x^{3} \operatorname{arsinh}\left (c x\right ) + a^{2} x^{3}\right )} \sqrt{c^{2} d x^{2} + d}}{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^3*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral((b^2*x^3*arcsinh(c*x)^2 + 2*a*b*x^3*arcsinh(c*x) + a^2*x^3)*sqrt(c^2*d*x^2 + d)/(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^{3} \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**3*(a+b*asinh(c*x))**2/(c**2*d*x**2+d)**(5/2),x)

[Out]

Integral(x**3*(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^{3}}{{\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^3*(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^3/(c^2*d*x^2 + d)^(5/2), x)