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

Optimal. Leaf size=312 \[ \frac {b x^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d \left (c^2 d x^2+d\right )^{3/2}}+\frac {\sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^3 d^2 \sqrt {c^2 d x^2+d}}-\frac {2 b \sqrt {c^2 x^2+1} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {c^2 d x^2+d}}+\frac {b^2 x}{3 c^2 d^2 \sqrt {c^2 d x^2+d}}-\frac {b^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{3 c^3 d^2 \sqrt {c^2 d x^2+d}}-\frac {b^2 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{3 c^3 d^2 \sqrt {c^2 d x^2+d}} \]

[Out]

1/3*x^3*(a+b*arcsinh(c*x))^2/d/(c^2*d*x^2+d)^(3/2)+1/3*b^2*x/c^2/d^2/(c^2*d*x^2+d)^(1/2)+1/3*b*x^2*(a+b*arcsin
h(c*x))/c/d^2/(c^2*x^2+1)^(1/2)/(c^2*d*x^2+d)^(1/2)-1/3*b^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)/c^3/d^2/(c^2*d*x^2+
d)^(1/2)+1/3*(a+b*arcsinh(c*x))^2*(c^2*x^2+1)^(1/2)/c^3/d^2/(c^2*d*x^2+d)^(1/2)-2/3*b*(a+b*arcsinh(c*x))*ln(1+
(c*x+(c^2*x^2+1)^(1/2))^2)*(c^2*x^2+1)^(1/2)/c^3/d^2/(c^2*d*x^2+d)^(1/2)-1/3*b^2*polylog(2,-(c*x+(c^2*x^2+1)^(
1/2))^2)*(c^2*x^2+1)^(1/2)/c^3/d^2/(c^2*d*x^2+d)^(1/2)

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Rubi [A]  time = 0.36, antiderivative size = 312, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {5723, 5751, 5714, 3718, 2190, 2279, 2391, 288, 215} \[ -\frac {b^2 \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{3 c^3 d^2 \sqrt {c^2 d x^2+d}}+\frac {b x^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}+\frac {\sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^3 d^2 \sqrt {c^2 d x^2+d}}-\frac {2 b \sqrt {c^2 x^2+1} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {c^2 d x^2+d}}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d \left (c^2 d x^2+d\right )^{3/2}}+\frac {b^2 x}{3 c^2 d^2 \sqrt {c^2 d x^2+d}}-\frac {b^2 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{3 c^3 d^2 \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5714

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

Rule 5723

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.96, size = 280, normalized size = 0.90 \[ \frac {a^2 c^3 x^3-a b \sqrt {c^2 x^2+1}-a b c^2 x^2 \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )-a b \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )-b \sinh ^{-1}(c x) \left (-2 a c^3 x^3+b \sqrt {c^2 x^2+1}+2 b \left (c^2 x^2+1\right )^{3/2} \log \left (e^{-2 \sinh ^{-1}(c x)}+1\right )\right )+b^2 c^3 x^3+b^2 \left (c^2 x^2+1\right )^{3/2} \text {Li}_2\left (-e^{-2 \sinh ^{-1}(c x)}\right )-b^2 \left (-c^3 x^3+c^2 x^2 \sqrt {c^2 x^2+1}+\sqrt {c^2 x^2+1}\right ) \sinh ^{-1}(c x)^2+b^2 c x}{3 c^3 d^2 \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b x^{2} \operatorname {arsinh}\left (c x\right ) + a^{2} x^{2}\right )} \sqrt {c^{2} d x^{2} + d}}{c^{6} d^{3} x^{6} + 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*x^2*arcsinh(c*x)^2 + 2*a*b*x^2*arcsinh(c*x) + a^2*x^2)*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2} x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.46, size = 3112, normalized size = 9.97 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c^3/d^3*(c^2*x^2+1)^(1/2)-1/3*b^2/
(c^2*x^2+1)^(1/2)*(d*(c^2*x^2+1))^(1/2)/c^3/d^3*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)+1/3*b^2*(d*(c^2*x^2+1))^
(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/d^3*arcsinh(c*x)^2*x^3+2/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8
*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/d^3*(c^2*x^2+1)*x^3+1/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+
10*c^4*x^4+5*c^2*x^2+1)/d^3*arcsinh(c*x)*x^3+2/3*b^2/(c^2*x^2+1)^(1/2)*(d*(c^2*x^2+1))^(1/2)/c^3/d^3*arcsinh(c
*x)^2+1/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/d^3*x^3+2/3*b^2*(d*(c^2*x^2+1
))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^4/d^3*x^7+b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x
^6+10*c^4*x^4+5*c^2*x^2+1)*c^2/d^3*x^5+1/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2
+1)*c^4/d^3*arcsinh(c*x)*x^7-b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^3/d^3*(c
^2*x^2+1)^(1/2)*x^6+b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^2/d^3*arcsinh(c*x
)^2*x^5+1/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^2/d^3*(c^2*x^2+1)*x^5+2/3
*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^2/d^3*arcsinh(c*x)*x^5-2*b^2*(d*(c^2
*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c/d^3*(c^2*x^2+1)^(1/2)*x^4-4/3*b^2*(d*(c^2*x^2+1)
)^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c/d^3*x^2*(c^2*x^2+1)^(1/2)+1/3*b^2*(d*(c^2*x^2+1))^(1/2)
/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c^2/d^3*(c^2*x^2+1)*x-1/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9
*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c^3/d^3*arcsinh(c*x)^2*(c^2*x^2+1)^(1/2)-1/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8
*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c^3/d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)+4/3*a*b/(c^2*x^2+1)^(1/2)*(d*(c^
2*x^2+1))^(1/2)/c^3/d^3*arcsinh(c*x)+1/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1
)*c^4/d^3*x^7+2/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^2/d^3*x^5-1/3*a*b*(
d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/d^3*(c^2*x^2+1)*x^3+2/3*a*b*(d*(c^2*x^2+1))^
(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/d^3*arcsinh(c*x)*x^3-1/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x
^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c^3/d^3*(c^2*x^2+1)^(1/2)-2/3*a*b/(c^2*x^2+1)^(1/2)*(d*(c^2*x^2+1))^(1/2)
/c^3/d^3*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)+1/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x
^2+1)/d^3*x^3-2*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^3/d^3*arcsinh(c*x)*(c
^2*x^2+1)^(1/2)*x^6-4*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c/d^3*arcsinh(c*x
)*(c^2*x^2+1)^(1/2)*x^4-8/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c/d^3*arcsi
nh(c*x)*(c^2*x^2+1)^(1/2)*x^2+2*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^2/d^3
*arcsinh(c*x)*x^5-a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c/d^3*(c^2*x^2+1)^(1/
2)*x^4-b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c/d^3*arcsinh(c*x)*(c^2*x^2+1)^(
1/2)*x^4-4/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c/d^3*arcsinh(c*x)^2*(c^2*
x^2+1)^(1/2)*x^2-b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c/d^3*arcsinh(c*x)*(c^
2*x^2+1)^(1/2)*x^2+2*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^4/d^3*arcsinh(c*
x)*x^7-1/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^2/d^3*(c^2*x^2+1)*x^5-1/3*
b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^2/d^3*arcsinh(c*x)*(c^2*x^2+1)*x^5-2*
b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c/d^3*arcsinh(c*x)^2*(c^2*x^2+1)^(1/2)*
x^4-b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^3/d^3*arcsinh(c*x)^2*(c^2*x^2+1)^
(1/2)*x^6-a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c/d^3*x^2*(c^2*x^2+1)^(1/2)-2
/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c^3/d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/
2)-1/3*a^2/c^2*x/d/(c^2*d*x^2+d)^(3/2)+1/3*a^2/c^2/d^2*x/(c^2*d*x^2+d)^(1/2)-2/3*b^2/(c^2*x^2+1)^(1/2)*(d*(c^2
*x^2+1))^(1/2)/c^3/d^3*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)-1/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9
*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/d^3*arcsinh(c*x)*(c^2*x^2+1)*x^3+b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x
^6+10*c^4*x^4+5*c^2*x^2+1)*c^4/d^3*arcsinh(c*x)^2*x^7

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{3} \, a b c {\left (\frac {1}{c^{6} d^{\frac {5}{2}} x^{2} + c^{4} d^{\frac {5}{2}}} + \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4} d^{\frac {5}{2}}}\right )} + \frac {2}{3} \, a b {\left (\frac {x}{\sqrt {c^{2} d x^{2} + d} c^{2} d^{2}} - \frac {x}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d}\right )} \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, a^{2} {\left (\frac {x}{\sqrt {c^{2} d x^{2} + d} c^{2} d^{2}} - \frac {x}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d}\right )} + b^{2} \int \frac {x^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a*b*c*(1/(c^6*d^(5/2)*x^2 + c^4*d^(5/2)) + log(c^2*x^2 + 1)/(c^4*d^(5/2))) + 2/3*a*b*(x/(sqrt(c^2*d*x^2 +
d)*c^2*d^2) - x/((c^2*d*x^2 + d)^(3/2)*c^2*d))*arcsinh(c*x) + 1/3*a^2*(x/(sqrt(c^2*d*x^2 + d)*c^2*d^2) - x/((
c^2*d*x^2 + d)^(3/2)*c^2*d)) + b^2*integrate(x^2*log(c*x + sqrt(c^2*x^2 + 1))^2/(c^2*d*x^2 + d)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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