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

Optimal. Leaf size=254 \[ \frac{1}{6} x^3 \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{8} d x^3 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{d x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^2}-\frac{d \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c^3 \sqrt{c^2 x^2+1}}-\frac{b c^3 d x^6 \sqrt{c^2 d x^2+d}}{36 \sqrt{c^2 x^2+1}}-\frac{7 b c d x^4 \sqrt{c^2 d x^2+d}}{96 \sqrt{c^2 x^2+1}}-\frac{b d x^2 \sqrt{c^2 d x^2+d}}{32 c \sqrt{c^2 x^2+1}} \]

[Out]

-(b*d*x^2*Sqrt[d + c^2*d*x^2])/(32*c*Sqrt[1 + c^2*x^2]) - (7*b*c*d*x^4*Sqrt[d + c^2*d*x^2])/(96*Sqrt[1 + c^2*x
^2]) - (b*c^3*d*x^6*Sqrt[d + c^2*d*x^2])/(36*Sqrt[1 + c^2*x^2]) + (d*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]
))/(16*c^2) + (d*x^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/8 + (x^3*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c
*x]))/6 - (d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(32*b*c^3*Sqrt[1 + c^2*x^2])

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Rubi [A]  time = 0.310098, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {5744, 5742, 5758, 5675, 30, 14} \[ \frac{1}{6} x^3 \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{8} d x^3 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{d x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^2}-\frac{d \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c^3 \sqrt{c^2 x^2+1}}-\frac{b c^3 d x^6 \sqrt{c^2 d x^2+d}}{36 \sqrt{c^2 x^2+1}}-\frac{7 b c d x^4 \sqrt{c^2 d x^2+d}}{96 \sqrt{c^2 x^2+1}}-\frac{b d x^2 \sqrt{c^2 d x^2+d}}{32 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d*x^2*Sqrt[d + c^2*d*x^2])/(32*c*Sqrt[1 + c^2*x^2]) - (7*b*c*d*x^4*Sqrt[d + c^2*d*x^2])/(96*Sqrt[1 + c^2*x
^2]) - (b*c^3*d*x^6*Sqrt[d + c^2*d*x^2])/(36*Sqrt[1 + c^2*x^2]) + (d*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]
))/(16*c^2) + (d*x^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/8 + (x^3*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c
*x]))/6 - (d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(32*b*c^3*Sqrt[1 + c^2*x^2])

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 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 5675

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

Rule 30

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int x^2 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{1}{6} x^3 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{2} d \int x^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx-\frac{\left (b c d \sqrt{d+c^2 d x^2}\right ) \int x^3 \left (1+c^2 x^2\right ) \, dx}{6 \sqrt{1+c^2 x^2}}\\ &=\frac{1}{8} d x^3 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} x^3 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{\left (d \sqrt{d+c^2 d x^2}\right ) \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{8 \sqrt{1+c^2 x^2}}-\frac{\left (b c d \sqrt{d+c^2 d x^2}\right ) \int x^3 \, dx}{8 \sqrt{1+c^2 x^2}}-\frac{\left (b c d \sqrt{d+c^2 d x^2}\right ) \int \left (x^3+c^2 x^5\right ) \, dx}{6 \sqrt{1+c^2 x^2}}\\ &=-\frac{7 b c d x^4 \sqrt{d+c^2 d x^2}}{96 \sqrt{1+c^2 x^2}}-\frac{b c^3 d x^6 \sqrt{d+c^2 d x^2}}{36 \sqrt{1+c^2 x^2}}+\frac{d x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^2}+\frac{1}{8} d x^3 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} x^3 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (d \sqrt{d+c^2 d x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{16 c^2 \sqrt{1+c^2 x^2}}-\frac{\left (b d \sqrt{d+c^2 d x^2}\right ) \int x \, dx}{16 c \sqrt{1+c^2 x^2}}\\ &=-\frac{b d x^2 \sqrt{d+c^2 d x^2}}{32 c \sqrt{1+c^2 x^2}}-\frac{7 b c d x^4 \sqrt{d+c^2 d x^2}}{96 \sqrt{1+c^2 x^2}}-\frac{b c^3 d x^6 \sqrt{d+c^2 d x^2}}{36 \sqrt{1+c^2 x^2}}+\frac{d x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^2}+\frac{1}{8} d x^3 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} x^3 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{d \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c^3 \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.712962, size = 251, normalized size = 0.99 \[ \frac{-144 a d^{3/2} \sqrt{c^2 x^2+1} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )+48 a c d x \sqrt{c^2 x^2+1} \left (8 c^4 x^4+14 c^2 x^2+3\right ) \sqrt{c^2 d x^2+d}-18 b d \sqrt{c^2 d x^2+d} \left (8 \sinh ^{-1}(c x)^2-4 \sinh \left (4 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)+\cosh \left (4 \sinh ^{-1}(c x)\right )\right )+b d \sqrt{c^2 d x^2+d} \left (72 \sinh ^{-1}(c x)^2+12 \left (-3 \sinh \left (2 \sinh ^{-1}(c x)\right )-3 \sinh \left (4 \sinh ^{-1}(c x)\right )+\sinh \left (6 \sinh ^{-1}(c x)\right )\right ) \sinh ^{-1}(c x)+18 \cosh \left (2 \sinh ^{-1}(c x)\right )+9 \cosh \left (4 \sinh ^{-1}(c x)\right )-2 \cosh \left (6 \sinh ^{-1}(c x)\right )\right )}{2304 c^3 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*a*c*d*x*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]*(3 + 14*c^2*x^2 + 8*c^4*x^4) - 144*a*d^(3/2)*Sqrt[1 + c^2*x^
2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]] - 18*b*d*Sqrt[d + c^2*d*x^2]*(8*ArcSinh[c*x]^2 + Cosh[4*ArcSinh[c*
x]] - 4*ArcSinh[c*x]*Sinh[4*ArcSinh[c*x]]) + b*d*Sqrt[d + c^2*d*x^2]*(72*ArcSinh[c*x]^2 + 18*Cosh[2*ArcSinh[c*
x]] + 9*Cosh[4*ArcSinh[c*x]] - 2*Cosh[6*ArcSinh[c*x]] + 12*ArcSinh[c*x]*(-3*Sinh[2*ArcSinh[c*x]] - 3*Sinh[4*Ar
cSinh[c*x]] + Sinh[6*ArcSinh[c*x]])))/(2304*c^3*Sqrt[1 + c^2*x^2])

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Maple [A]  time = 0.214, size = 421, normalized size = 1.7 \begin{align*}{\frac{ax}{6\,{c}^{2}d} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{5}{2}}}}-{\frac{ax}{24\,{c}^{2}} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{adx}{16\,{c}^{2}}\sqrt{{c}^{2}d{x}^{2}+d}}-{\frac{a{d}^{2}}{16\,{c}^{2}}\ln \left ({{c}^{2}dx{\frac{1}{\sqrt{{c}^{2}d}}}}+\sqrt{{c}^{2}d{x}^{2}+d} \right ){\frac{1}{\sqrt{{c}^{2}d}}}}-{\frac{b \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}d}{32\,{c}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{7\,bd}{2304\,{c}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{bd{x}^{2}}{32\,c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{bd{\it Arcsinh} \left ( cx \right ) x}{16\,{c}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{bd{c}^{4}{\it Arcsinh} \left ( cx \right ){x}^{7}}{6\,{c}^{2}{x}^{2}+6}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{bd{c}^{3}{x}^{6}}{36}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{11\,{c}^{2}db{\it Arcsinh} \left ( cx \right ){x}^{5}}{24\,{c}^{2}{x}^{2}+24}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{7\,bdc{x}^{4}}{96}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{17\,bd{\it Arcsinh} \left ( cx \right ){x}^{3}}{48\,{c}^{2}{x}^{2}+48}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*x*(c^2*d*x^2+d)^(5/2)/c^2/d-1/24*a/c^2*x*(c^2*d*x^2+d)^(3/2)-1/16*a/c^2*d*x*(c^2*d*x^2+d)^(1/2)-1/16*a/c
^2*d^2*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)-1/32*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1
/2)/c^3*arcsinh(c*x)^2*d+7/2304*b*(d*(c^2*x^2+1))^(1/2)*d/c^3/(c^2*x^2+1)^(1/2)-1/32*b*(d*(c^2*x^2+1))^(1/2)*d
/c/(c^2*x^2+1)^(1/2)*x^2+1/16*b*(d*(c^2*x^2+1))^(1/2)*d/c^2/(c^2*x^2+1)*arcsinh(c*x)*x+1/6*b*(d*(c^2*x^2+1))^(
1/2)*d*c^4/(c^2*x^2+1)*arcsinh(c*x)*x^7-1/36*b*(d*(c^2*x^2+1))^(1/2)*d*c^3/(c^2*x^2+1)^(1/2)*x^6+11/24*b*(d*(c
^2*x^2+1))^(1/2)*d*c^2/(c^2*x^2+1)*arcsinh(c*x)*x^5-7/96*b*(d*(c^2*x^2+1))^(1/2)*d*c/(c^2*x^2+1)^(1/2)*x^4+17/
48*b*(d*(c^2*x^2+1))^(1/2)*d/(c^2*x^2+1)*arcsinh(c*x)*x^3

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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^2*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(c**2*d*x**2+d)**(3/2)*(a+b*asinh(c*x)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="giac")

[Out]

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