3.258 \(\int x^3 \sqrt{d+c^2 d x^2} (a+b \sinh ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=358 \[ \frac{4 a b x \sqrt{c^2 d x^2+d}}{15 c^3 \sqrt{c^2 x^2+1}}-\frac{2 b c x^5 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt{c^2 x^2+1}}+\frac{1}{5} x^4 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{2 b x^3 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{45 c \sqrt{c^2 x^2+1}}+\frac{x^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{15 c^2}-\frac{2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{15 c^4}+\frac{2 b^2 \left (c^2 x^2+1\right )^2 \sqrt{c^2 d x^2+d}}{125 c^4}-\frac{26 b^2 \left (c^2 x^2+1\right ) \sqrt{c^2 d x^2+d}}{675 c^4}-\frac{52 b^2 \sqrt{c^2 d x^2+d}}{225 c^4}+\frac{4 b^2 x \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x)}{15 c^3 \sqrt{c^2 x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.478806, antiderivative size = 358, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5742, 5758, 5717, 5653, 261, 5661, 266, 43} \[ \frac{4 a b x \sqrt{c^2 d x^2+d}}{15 c^3 \sqrt{c^2 x^2+1}}-\frac{2 b c x^5 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt{c^2 x^2+1}}+\frac{1}{5} x^4 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{2 b x^3 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{45 c \sqrt{c^2 x^2+1}}+\frac{x^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{15 c^2}-\frac{2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{15 c^4}+\frac{2 b^2 \left (c^2 x^2+1\right )^2 \sqrt{c^2 d x^2+d}}{125 c^4}-\frac{26 b^2 \left (c^2 x^2+1\right ) \sqrt{c^2 d x^2+d}}{675 c^4}-\frac{52 b^2 \sqrt{c^2 d x^2+d}}{225 c^4}+\frac{4 b^2 x \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x)}{15 c^3 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5742

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(
(f*x)^(m + 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(f*(m + 2)), x] + (Dist[Sqrt[d + e*x^2]/((m + 2)*Sqrt[1
+ c^2*x^2]), Int[((f*x)^m*(a + b*ArcSinh[c*x])^n)/Sqrt[1 + c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(f*
(m + 2)*Sqrt[1 + c^2*x^2]), Int[(f*x)^(m + 1)*(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] &&  !LtQ[m, -1] && (RationalQ[m] || EqQ[n, 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 5653

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[(x*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

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]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.268427, size = 222, normalized size = 0.62 \[ \frac{\sqrt{c^2 d x^2+d} \left (-30 a b c x \sqrt{c^2 x^2+1} \left (9 c^4 x^4+5 c^2 x^2-30\right )-30 b \sinh ^{-1}(c x) \left (b c x \sqrt{c^2 x^2+1} \left (9 c^4 x^4+5 c^2 x^2-30\right )-15 a \left (c^2 x^2+1\right )^2 \left (3 c^2 x^2-2\right )\right )+225 \left (3 c^2 x^2-2\right ) \left (a c^2 x^2+a\right )^2+2 b^2 \left (27 c^6 x^6+16 c^4 x^4-439 c^2 x^2-428\right )+225 \left (3 c^2 x^2-2\right ) \left (b c^2 x^2+b\right )^2 \sinh ^{-1}(c x)^2\right )}{3375 c^4 \left (c^2 x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + c^2*d*x^2]*(225*(-2 + 3*c^2*x^2)*(a + a*c^2*x^2)^2 - 30*a*b*c*x*Sqrt[1 + c^2*x^2]*(-30 + 5*c^2*x^2 +
9*c^4*x^4) + 2*b^2*(-428 - 439*c^2*x^2 + 16*c^4*x^4 + 27*c^6*x^6) - 30*b*(-15*a*(1 + c^2*x^2)^2*(-2 + 3*c^2*x
^2) + b*c*x*Sqrt[1 + c^2*x^2]*(-30 + 5*c^2*x^2 + 9*c^4*x^4))*ArcSinh[c*x] + 225*(-2 + 3*c^2*x^2)*(b + b*c^2*x^
2)^2*ArcSinh[c*x]^2))/(3375*c^4*(1 + c^2*x^2))

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Maple [B]  time = 0.349, size = 1162, normalized size = 3.3 \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)^(1/2),x)

[Out]

a^2*(1/5*x^2*(c^2*d*x^2+d)^(3/2)/c^2/d-2/15/d/c^4*(c^2*d*x^2+d)^(3/2))+b^2*(1/4000*(d*(c^2*x^2+1))^(1/2)*(16*c
^6*x^6+16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4+20*c^3*x^3*(c^2*x^2+1)^(1/2)+13*c^2*x^2+5*c*x*(c^2*x^2+1)^(1/2)
+1)*(25*arcsinh(c*x)^2-10*arcsinh(c*x)+2)/c^4/(c^2*x^2+1)-1/864*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4+4*c^3*x^3*(c^
2*x^2+1)^(1/2)+5*c^2*x^2+3*c*x*(c^2*x^2+1)^(1/2)+1)*(9*arcsinh(c*x)^2-6*arcsinh(c*x)+2)/c^4/(c^2*x^2+1)-1/16*(
d*(c^2*x^2+1))^(1/2)*(c^2*x^2+c*x*(c^2*x^2+1)^(1/2)+1)*(arcsinh(c*x)^2-2*arcsinh(c*x)+2)/c^4/(c^2*x^2+1)-1/16*
(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-c*x*(c^2*x^2+1)^(1/2)+1)*(arcsinh(c*x)^2+2*arcsinh(c*x)+2)/c^4/(c^2*x^2+1)-1/86
4*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4-4*c^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*x^2-3*c*x*(c^2*x^2+1)^(1/2)+1)*(9*arcsinh
(c*x)^2+6*arcsinh(c*x)+2)/c^4/(c^2*x^2+1)+1/4000*(d*(c^2*x^2+1))^(1/2)*(16*c^6*x^6-16*c^5*x^5*(c^2*x^2+1)^(1/2
)+28*c^4*x^4-20*c^3*x^3*(c^2*x^2+1)^(1/2)+13*c^2*x^2-5*c*x*(c^2*x^2+1)^(1/2)+1)*(25*arcsinh(c*x)^2+10*arcsinh(
c*x)+2)/c^4/(c^2*x^2+1))+2*a*b*(1/800*(d*(c^2*x^2+1))^(1/2)*(16*c^6*x^6+16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^
4+20*c^3*x^3*(c^2*x^2+1)^(1/2)+13*c^2*x^2+5*c*x*(c^2*x^2+1)^(1/2)+1)*(-1+5*arcsinh(c*x))/c^4/(c^2*x^2+1)-1/288
*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4+4*c^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*x^2+3*c*x*(c^2*x^2+1)^(1/2)+1)*(-1+3*arcsi
nh(c*x))/c^4/(c^2*x^2+1)-1/16*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2+c*x*(c^2*x^2+1)^(1/2)+1)*(-1+arcsinh(c*x))/c^4/(c
^2*x^2+1)-1/16*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-c*x*(c^2*x^2+1)^(1/2)+1)*(1+arcsinh(c*x))/c^4/(c^2*x^2+1)-1/288*
(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4-4*c^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*x^2-3*c*x*(c^2*x^2+1)^(1/2)+1)*(1+3*arcsinh
(c*x))/c^4/(c^2*x^2+1)+1/800*(d*(c^2*x^2+1))^(1/2)*(16*c^6*x^6-16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4-20*c^3*
x^3*(c^2*x^2+1)^(1/2)+13*c^2*x^2-5*c*x*(c^2*x^2+1)^(1/2)+1)*(1+5*arcsinh(c*x))/c^4/(c^2*x^2+1))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.90919, size = 694, normalized size = 1.94 \begin{align*} \frac{225 \,{\left (3 \, b^{2} c^{6} x^{6} + 4 \, b^{2} c^{4} x^{4} - b^{2} c^{2} x^{2} - 2 \, b^{2}\right )} \sqrt{c^{2} d x^{2} + d} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} + 30 \,{\left (45 \, a b c^{6} x^{6} + 60 \, a b c^{4} x^{4} - 15 \, a b c^{2} x^{2} - 30 \, a b -{\left (9 \, b^{2} c^{5} x^{5} + 5 \, b^{2} c^{3} x^{3} - 30 \, b^{2} c x\right )} \sqrt{c^{2} x^{2} + 1}\right )} \sqrt{c^{2} d x^{2} + d} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) +{\left (27 \,{\left (25 \, a^{2} + 2 \, b^{2}\right )} c^{6} x^{6} + 4 \,{\left (225 \, a^{2} + 8 \, b^{2}\right )} c^{4} x^{4} -{\left (225 \, a^{2} + 878 \, b^{2}\right )} c^{2} x^{2} - 450 \, a^{2} - 856 \, b^{2} - 30 \,{\left (9 \, a b c^{5} x^{5} + 5 \, a b c^{3} x^{3} - 30 \, a b c x\right )} \sqrt{c^{2} x^{2} + 1}\right )} \sqrt{c^{2} d x^{2} + d}}{3375 \,{\left (c^{6} x^{2} + c^{4}\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)^(1/2),x, algorithm="fricas")

[Out]

1/3375*(225*(3*b^2*c^6*x^6 + 4*b^2*c^4*x^4 - b^2*c^2*x^2 - 2*b^2)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 +
1))^2 + 30*(45*a*b*c^6*x^6 + 60*a*b*c^4*x^4 - 15*a*b*c^2*x^2 - 30*a*b - (9*b^2*c^5*x^5 + 5*b^2*c^3*x^3 - 30*b
^2*c*x)*sqrt(c^2*x^2 + 1))*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + (27*(25*a^2 + 2*b^2)*c^6*x^6 + 4
*(225*a^2 + 8*b^2)*c^4*x^4 - (225*a^2 + 878*b^2)*c^2*x^2 - 450*a^2 - 856*b^2 - 30*(9*a*b*c^5*x^5 + 5*a*b*c^3*x
^3 - 30*a*b*c*x)*sqrt(c^2*x^2 + 1))*sqrt(c^2*d*x^2 + d))/(c^6*x^2 + c^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \sqrt{d \left (c^{2} x^{2} + 1\right )} \left (a + b \operatorname{asinh}{\left (c x \right )}\right )^{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)**(1/2),x)

[Out]

Integral(x**3*sqrt(d*(c**2*x**2 + 1))*(a + b*asinh(c*x))**2, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \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)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError