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

Optimal. Leaf size=206 \[ \frac{1}{5} d x^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{4 b d x^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{45 c}-\frac{2 b d \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{25 c^3}+\frac{2 b d \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^3}+\frac{8 b d \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{45 c^3}+\frac{2}{15} d x^3 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{2}{125} b^2 c^2 d x^5-\frac{52 b^2 d x}{225 c^2}+\frac{26}{675} b^2 d x^3 \]

[Out]

(-52*b^2*d*x)/(225*c^2) + (26*b^2*d*x^3)/675 + (2*b^2*c^2*d*x^5)/125 + (8*b*d*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh
[c*x]))/(45*c^3) - (4*b*d*x^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(45*c) + (2*b*d*(1 + c^2*x^2)^(3/2)*(a +
b*ArcSinh[c*x]))/(15*c^3) - (2*b*d*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(25*c^3) + (2*d*x^3*(a + b*ArcSi
nh[c*x])^2)/15 + (d*x^3*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/5

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Rubi [A]  time = 0.341996, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 10, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {5744, 5661, 5758, 5717, 8, 30, 266, 43, 5732, 12} \[ \frac{1}{5} d x^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{4 b d x^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{45 c}-\frac{2 b d \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{25 c^3}+\frac{2 b d \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^3}+\frac{8 b d \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{45 c^3}+\frac{2}{15} d x^3 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{2}{125} b^2 c^2 d x^5-\frac{52 b^2 d x}{225 c^2}+\frac{26}{675} b^2 d x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-52*b^2*d*x)/(225*c^2) + (26*b^2*d*x^3)/675 + (2*b^2*c^2*d*x^5)/125 + (8*b*d*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh
[c*x]))/(45*c^3) - (4*b*d*x^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(45*c) + (2*b*d*(1 + c^2*x^2)^(3/2)*(a +
b*ArcSinh[c*x]))/(15*c^3) - (2*b*d*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(25*c^3) + (2*d*x^3*(a + b*ArcSi
nh[c*x])^2)/15 + (d*x^3*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/5

Rule 5744

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*(a + b*ArcSinh[c*x])^n)/(f*(m + 2*p + 1)), x] + (Dist[(2*d*p)/(m + 2*p + 1), Int
[(f*x)^m*(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]
)/(f*(m + 2*p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^
(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0] &&  !LtQ[m, -1]
&& (RationalQ[m] || EqQ[n, 1])

Rule 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5758

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

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 5732

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x
^m*(1 + c^2*x^2)^p, x]}, Dist[d^p*(a + b*ArcSinh[c*x]), u, x] - Dist[b*c*d^p, Int[SimplifyIntegrand[u/Sqrt[1 +
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegerQ[p - 1/2] && (IGtQ[(m + 1)/2,
0] || ILtQ[(m + 2*p + 3)/2, 0]) && NeQ[p, -2^(-1)] && GtQ[d, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin{align*} \int x^2 \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac{1}{5} d x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{5} (2 d) \int x^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx-\frac{1}{5} (2 b c d) \int x^3 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx\\ &=\frac{2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^3}-\frac{2 b d \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{25 c^3}+\frac{2}{15} d x^3 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{5} d x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{1}{15} (4 b c d) \int \frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx+\frac{1}{5} \left (2 b^2 c^2 d\right ) \int \frac{-2+c^2 x^2+3 c^4 x^4}{15 c^4} \, dx\\ &=-\frac{4 b d x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{45 c}+\frac{2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^3}-\frac{2 b d \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{25 c^3}+\frac{2}{15} d x^3 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{5} d x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{45} \left (4 b^2 d\right ) \int x^2 \, dx+\frac{\left (2 b^2 d\right ) \int \left (-2+c^2 x^2+3 c^4 x^4\right ) \, dx}{75 c^2}+\frac{(8 b d) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{45 c}\\ &=-\frac{4 b^2 d x}{75 c^2}+\frac{26}{675} b^2 d x^3+\frac{2}{125} b^2 c^2 d x^5+\frac{8 b d \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{45 c^3}-\frac{4 b d x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{45 c}+\frac{2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^3}-\frac{2 b d \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{25 c^3}+\frac{2}{15} d x^3 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{5} d x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{\left (8 b^2 d\right ) \int 1 \, dx}{45 c^2}\\ &=-\frac{52 b^2 d x}{225 c^2}+\frac{26}{675} b^2 d x^3+\frac{2}{125} b^2 c^2 d x^5+\frac{8 b d \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{45 c^3}-\frac{4 b d x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{45 c}+\frac{2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^3}-\frac{2 b d \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{25 c^3}+\frac{2}{15} d x^3 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{5} d x^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2\\ \end{align*}

Mathematica [A]  time = 0.235485, size = 177, normalized size = 0.86 \[ \frac{d \left (225 a^2 c^3 x^3 \left (3 c^2 x^2+5\right )-30 a b \sqrt{c^2 x^2+1} \left (9 c^4 x^4+13 c^2 x^2-26\right )-30 b \sinh ^{-1}(c x) \left (b \sqrt{c^2 x^2+1} \left (9 c^4 x^4+13 c^2 x^2-26\right )-15 a c^3 x^3 \left (3 c^2 x^2+5\right )\right )+2 b^2 c x \left (27 c^4 x^4+65 c^2 x^2-390\right )+225 b^2 c^3 x^3 \left (3 c^2 x^2+5\right ) \sinh ^{-1}(c x)^2\right )}{3375 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(225*a^2*c^3*x^3*(5 + 3*c^2*x^2) - 30*a*b*Sqrt[1 + c^2*x^2]*(-26 + 13*c^2*x^2 + 9*c^4*x^4) + 2*b^2*c*x*(-39
0 + 65*c^2*x^2 + 27*c^4*x^4) - 30*b*(-15*a*c^3*x^3*(5 + 3*c^2*x^2) + b*Sqrt[1 + c^2*x^2]*(-26 + 13*c^2*x^2 + 9
*c^4*x^4))*ArcSinh[c*x] + 225*b^2*c^3*x^3*(5 + 3*c^2*x^2)*ArcSinh[c*x]^2))/(3375*c^3)

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Maple [A]  time = 0.037, size = 260, normalized size = 1.3 \begin{align*}{\frac{1}{{c}^{3}} \left ( d{a}^{2} \left ({\frac{{c}^{5}{x}^{5}}{5}}+{\frac{{c}^{3}{x}^{3}}{3}} \right ) +d{b}^{2} \left ({\frac{ \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}cx \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}{5}}-{\frac{2\, \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}cx}{15}}-{\frac{ \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}cx \left ({c}^{2}{x}^{2}+1 \right ) }{15}}-{\frac{2\,{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}}{25} \left ({c}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{8\,{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}}{225}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{52\,{\it Arcsinh} \left ( cx \right ) }{225}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{2\,cx \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}{125}}-{\frac{856\,cx}{3375}}+{\frac{22\,cx \left ({c}^{2}{x}^{2}+1 \right ) }{3375}} \right ) +2\,dab \left ( 1/5\,{\it Arcsinh} \left ( cx \right ){c}^{5}{x}^{5}+1/3\,{\it Arcsinh} \left ( cx \right ){c}^{3}{x}^{3}-1/25\,{c}^{4}{x}^{4}\sqrt{{c}^{2}{x}^{2}+1}-{\frac{13\,{c}^{2}{x}^{2}\sqrt{{c}^{2}{x}^{2}+1}}{225}}+{\frac{26\,\sqrt{{c}^{2}{x}^{2}+1}}{225}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(d*a^2*(1/5*c^5*x^5+1/3*c^3*x^3)+d*b^2*(1/5*arcsinh(c*x)^2*c*x*(c^2*x^2+1)^2-2/15*arcsinh(c*x)^2*c*x-1/1
5*arcsinh(c*x)^2*c*x*(c^2*x^2+1)-2/25*arcsinh(c*x)*c^2*x^2*(c^2*x^2+1)^(3/2)-8/225*arcsinh(c*x)*c^2*x^2*(c^2*x
^2+1)^(1/2)+52/225*arcsinh(c*x)*(c^2*x^2+1)^(1/2)+2/125*c*x*(c^2*x^2+1)^2-856/3375*c*x+22/3375*c*x*(c^2*x^2+1)
)+2*d*a*b*(1/5*arcsinh(c*x)*c^5*x^5+1/3*arcsinh(c*x)*c^3*x^3-1/25*c^4*x^4*(c^2*x^2+1)^(1/2)-13/225*c^2*x^2*(c^
2*x^2+1)^(1/2)+26/225*(c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.18843, size = 467, normalized size = 2.27 \begin{align*} \frac{1}{5} \, b^{2} c^{2} d x^{5} \operatorname{arsinh}\left (c x\right )^{2} + \frac{1}{5} \, a^{2} c^{2} d x^{5} + \frac{1}{3} \, b^{2} d x^{3} \operatorname{arsinh}\left (c x\right )^{2} + \frac{2}{75} \,{\left (15 \, x^{5} \operatorname{arsinh}\left (c x\right ) -{\left (\frac{3 \, \sqrt{c^{2} x^{2} + 1} x^{4}}{c^{2}} - \frac{4 \, \sqrt{c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac{8 \, \sqrt{c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} a b c^{2} d - \frac{2}{1125} \,{\left (15 \,{\left (\frac{3 \, \sqrt{c^{2} x^{2} + 1} x^{4}}{c^{2}} - \frac{4 \, \sqrt{c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac{8 \, \sqrt{c^{2} x^{2} + 1}}{c^{6}}\right )} c \operatorname{arsinh}\left (c x\right ) - \frac{9 \, c^{4} x^{5} - 20 \, c^{2} x^{3} + 120 \, x}{c^{4}}\right )} b^{2} c^{2} d + \frac{1}{3} \, a^{2} d x^{3} + \frac{2}{9} \,{\left (3 \, x^{3} \operatorname{arsinh}\left (c x\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac{2 \, \sqrt{c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} a b d - \frac{2}{27} \,{\left (3 \, c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac{2 \, \sqrt{c^{2} x^{2} + 1}}{c^{4}}\right )} \operatorname{arsinh}\left (c x\right ) - \frac{c^{2} x^{3} - 6 \, x}{c^{2}}\right )} b^{2} d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^2*c^2*d*x^5*arcsinh(c*x)^2 + 1/5*a^2*c^2*d*x^5 + 1/3*b^2*d*x^3*arcsinh(c*x)^2 + 2/75*(15*x^5*arcsinh(c*x
) - (3*sqrt(c^2*x^2 + 1)*x^4/c^2 - 4*sqrt(c^2*x^2 + 1)*x^2/c^4 + 8*sqrt(c^2*x^2 + 1)/c^6)*c)*a*b*c^2*d - 2/112
5*(15*(3*sqrt(c^2*x^2 + 1)*x^4/c^2 - 4*sqrt(c^2*x^2 + 1)*x^2/c^4 + 8*sqrt(c^2*x^2 + 1)/c^6)*c*arcsinh(c*x) - (
9*c^4*x^5 - 20*c^2*x^3 + 120*x)/c^4)*b^2*c^2*d + 1/3*a^2*d*x^3 + 2/9*(3*x^3*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1
)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4))*a*b*d - 2/27*(3*c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4)*
arcsinh(c*x) - (c^2*x^3 - 6*x)/c^2)*b^2*d

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Fricas [A]  time = 2.70919, size = 509, normalized size = 2.47 \begin{align*} \frac{27 \,{\left (25 \, a^{2} + 2 \, b^{2}\right )} c^{5} d x^{5} + 5 \,{\left (225 \, a^{2} + 26 \, b^{2}\right )} c^{3} d x^{3} - 780 \, b^{2} c d x + 225 \,{\left (3 \, b^{2} c^{5} d x^{5} + 5 \, b^{2} c^{3} d x^{3}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} + 30 \,{\left (45 \, a b c^{5} d x^{5} + 75 \, a b c^{3} d x^{3} -{\left (9 \, b^{2} c^{4} d x^{4} + 13 \, b^{2} c^{2} d x^{2} - 26 \, b^{2} d\right )} \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - 30 \,{\left (9 \, a b c^{4} d x^{4} + 13 \, a b c^{2} d x^{2} - 26 \, a b d\right )} \sqrt{c^{2} x^{2} + 1}}{3375 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3375*(27*(25*a^2 + 2*b^2)*c^5*d*x^5 + 5*(225*a^2 + 26*b^2)*c^3*d*x^3 - 780*b^2*c*d*x + 225*(3*b^2*c^5*d*x^5
+ 5*b^2*c^3*d*x^3)*log(c*x + sqrt(c^2*x^2 + 1))^2 + 30*(45*a*b*c^5*d*x^5 + 75*a*b*c^3*d*x^3 - (9*b^2*c^4*d*x^4
+ 13*b^2*c^2*d*x^2 - 26*b^2*d)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1)) - 30*(9*a*b*c^4*d*x^4 + 13*a*b
*c^2*d*x^2 - 26*a*b*d)*sqrt(c^2*x^2 + 1))/c^3

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Sympy [A]  time = 5.56227, size = 313, normalized size = 1.52 \begin{align*} \begin{cases} \frac{a^{2} c^{2} d x^{5}}{5} + \frac{a^{2} d x^{3}}{3} + \frac{2 a b c^{2} d x^{5} \operatorname{asinh}{\left (c x \right )}}{5} - \frac{2 a b c d x^{4} \sqrt{c^{2} x^{2} + 1}}{25} + \frac{2 a b d x^{3} \operatorname{asinh}{\left (c x \right )}}{3} - \frac{26 a b d x^{2} \sqrt{c^{2} x^{2} + 1}}{225 c} + \frac{52 a b d \sqrt{c^{2} x^{2} + 1}}{225 c^{3}} + \frac{b^{2} c^{2} d x^{5} \operatorname{asinh}^{2}{\left (c x \right )}}{5} + \frac{2 b^{2} c^{2} d x^{5}}{125} - \frac{2 b^{2} c d x^{4} \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{25} + \frac{b^{2} d x^{3} \operatorname{asinh}^{2}{\left (c x \right )}}{3} + \frac{26 b^{2} d x^{3}}{675} - \frac{26 b^{2} d x^{2} \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{225 c} - \frac{52 b^{2} d x}{225 c^{2}} + \frac{52 b^{2} d \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{225 c^{3}} & \text{for}\: c \neq 0 \\\frac{a^{2} d x^{3}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*c**2*d*x**5/5 + a**2*d*x**3/3 + 2*a*b*c**2*d*x**5*asinh(c*x)/5 - 2*a*b*c*d*x**4*sqrt(c**2*x**2
+ 1)/25 + 2*a*b*d*x**3*asinh(c*x)/3 - 26*a*b*d*x**2*sqrt(c**2*x**2 + 1)/(225*c) + 52*a*b*d*sqrt(c**2*x**2 + 1
)/(225*c**3) + b**2*c**2*d*x**5*asinh(c*x)**2/5 + 2*b**2*c**2*d*x**5/125 - 2*b**2*c*d*x**4*sqrt(c**2*x**2 + 1)
*asinh(c*x)/25 + b**2*d*x**3*asinh(c*x)**2/3 + 26*b**2*d*x**3/675 - 26*b**2*d*x**2*sqrt(c**2*x**2 + 1)*asinh(c
*x)/(225*c) - 52*b**2*d*x/(225*c**2) + 52*b**2*d*sqrt(c**2*x**2 + 1)*asinh(c*x)/(225*c**3), Ne(c, 0)), (a**2*d
*x**3/3, True))

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Giac [B]  time = 2.20009, size = 501, normalized size = 2.43 \begin{align*} \frac{1}{5} \, a^{2} c^{2} d x^{5} + \frac{2}{75} \,{\left (15 \, x^{5} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{3 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{5}{2}} - 10 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} + 15 \, \sqrt{c^{2} x^{2} + 1}}{c^{5}}\right )} a b c^{2} d + \frac{1}{1125} \,{\left (225 \, x^{5} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} + 2 \, c{\left (\frac{9 \, c^{4} x^{5} - 20 \, c^{2} x^{3} + 120 \, x}{c^{5}} - \frac{15 \,{\left (3 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{5}{2}} - 10 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} + 15 \, \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{6}}\right )}\right )} b^{2} c^{2} d + \frac{1}{3} \, a^{2} d x^{3} + \frac{2}{9} \,{\left (3 \, x^{3} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} - 3 \, \sqrt{c^{2} x^{2} + 1}}{c^{3}}\right )} a b d + \frac{1}{27} \,{\left (9 \, x^{3} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} + 2 \, c{\left (\frac{c^{2} x^{3} - 6 \, x}{c^{3}} - \frac{3 \,{\left ({\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} - 3 \, \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{4}}\right )}\right )} b^{2} d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^2*c^2*d*x^5 + 2/75*(15*x^5*log(c*x + sqrt(c^2*x^2 + 1)) - (3*(c^2*x^2 + 1)^(5/2) - 10*(c^2*x^2 + 1)^(3/2
) + 15*sqrt(c^2*x^2 + 1))/c^5)*a*b*c^2*d + 1/1125*(225*x^5*log(c*x + sqrt(c^2*x^2 + 1))^2 + 2*c*((9*c^4*x^5 -
20*c^2*x^3 + 120*x)/c^5 - 15*(3*(c^2*x^2 + 1)^(5/2) - 10*(c^2*x^2 + 1)^(3/2) + 15*sqrt(c^2*x^2 + 1))*log(c*x +
sqrt(c^2*x^2 + 1))/c^6))*b^2*c^2*d + 1/3*a^2*d*x^3 + 2/9*(3*x^3*log(c*x + sqrt(c^2*x^2 + 1)) - ((c^2*x^2 + 1)
^(3/2) - 3*sqrt(c^2*x^2 + 1))/c^3)*a*b*d + 1/27*(9*x^3*log(c*x + sqrt(c^2*x^2 + 1))^2 + 2*c*((c^2*x^3 - 6*x)/c
^3 - 3*((c^2*x^2 + 1)^(3/2) - 3*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1))/c^4))*b^2*d