3.135 \(\int \frac{(d+c^2 d x^2)^{3/2} (a+b \sinh ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=184 \[ \frac{c^3 d \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b \sqrt{c^2 x^2+1}}-\frac{c^2 d \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{\left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac{b c d \sqrt{c^2 d x^2+d}}{6 x^2 \sqrt{c^2 x^2+1}}+\frac{4 b c^3 d \log (x) \sqrt{c^2 d x^2+d}}{3 \sqrt{c^2 x^2+1}} \]

[Out]

-(b*c*d*Sqrt[d + c^2*d*x^2])/(6*x^2*Sqrt[1 + c^2*x^2]) - (c^2*d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/x -
((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(3*x^3) + (c^3*d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(2*b
*Sqrt[1 + c^2*x^2]) + (4*b*c^3*d*Sqrt[d + c^2*d*x^2]*Log[x])/(3*Sqrt[1 + c^2*x^2])

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Rubi [A]  time = 0.231355, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {5739, 5737, 29, 5675, 14} \[ \frac{c^3 d \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b \sqrt{c^2 x^2+1}}-\frac{c^2 d \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{\left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac{b c d \sqrt{c^2 d x^2+d}}{6 x^2 \sqrt{c^2 x^2+1}}+\frac{4 b c^3 d \log (x) \sqrt{c^2 d x^2+d}}{3 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/x^4,x]

[Out]

-(b*c*d*Sqrt[d + c^2*d*x^2])/(6*x^2*Sqrt[1 + c^2*x^2]) - (c^2*d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/x -
((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(3*x^3) + (c^3*d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(2*b
*Sqrt[1 + c^2*x^2]) + (4*b*c^3*d*Sqrt[d + c^2*d*x^2]*Log[x])/(3*Sqrt[1 + c^2*x^2])

Rule 5739

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 + 1)), x] + (-Dist[(2*e*p)/(f^2*(m + 1)), Int[(f*x
)^(m + 2)*(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 + 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}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]

Rule 5737

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 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 \frac{\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{x^4} \, dx &=-\frac{\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\left (c^2 d\right ) \int \frac{\sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x^2} \, dx+\frac{\left (b c d \sqrt{d+c^2 d x^2}\right ) \int \frac{1+c^2 x^2}{x^3} \, dx}{3 \sqrt{1+c^2 x^2}}\\ &=-\frac{c^2 d \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\frac{\left (b c d \sqrt{d+c^2 d x^2}\right ) \int \left (\frac{1}{x^3}+\frac{c^2}{x}\right ) \, dx}{3 \sqrt{1+c^2 x^2}}+\frac{\left (b c^3 d \sqrt{d+c^2 d x^2}\right ) \int \frac{1}{x} \, dx}{\sqrt{1+c^2 x^2}}+\frac{\left (c^4 d \sqrt{d+c^2 d x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{\sqrt{1+c^2 x^2}}\\ &=-\frac{b c d \sqrt{d+c^2 d x^2}}{6 x^2 \sqrt{1+c^2 x^2}}-\frac{c^2 d \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\frac{c^3 d \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b \sqrt{1+c^2 x^2}}+\frac{4 b c^3 d \sqrt{d+c^2 d x^2} \log (x)}{3 \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.721007, size = 214, normalized size = 1.16 \[ \frac{1}{6} d \left (-\frac{2 a \left (4 c^2 x^2+1\right ) \sqrt{c^2 d x^2+d}}{x^3}+6 a c^3 \sqrt{d} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )+\frac{3 b c^2 \sqrt{c^2 d x^2+d} \left (-2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)+2 c x \log (c x)+c x \sinh ^{-1}(c x)^2\right )}{x \sqrt{c^2 x^2+1}}-\frac{b \sqrt{c^2 d x^2+d} \left (-2 c^3 x^3 \log (c x)+2 \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x)+c x\right )}{x^3 \sqrt{c^2 x^2+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/x^4,x]

[Out]

(d*((-2*a*(1 + 4*c^2*x^2)*Sqrt[d + c^2*d*x^2])/x^3 + (3*b*c^2*Sqrt[d + c^2*d*x^2]*(-2*Sqrt[1 + c^2*x^2]*ArcSin
h[c*x] + c*x*ArcSinh[c*x]^2 + 2*c*x*Log[c*x]))/(x*Sqrt[1 + c^2*x^2]) - (b*Sqrt[d + c^2*d*x^2]*(c*x + 2*(1 + c^
2*x^2)^(3/2)*ArcSinh[c*x] - 2*c^3*x^3*Log[c*x]))/(x^3*Sqrt[1 + c^2*x^2]) + 6*a*c^3*Sqrt[d]*Log[c*d*x + Sqrt[d]
*Sqrt[d + c^2*d*x^2]]))/6

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Maple [B]  time = 0.176, size = 1107, normalized size = 6. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))/x^4,x)

[Out]

-1/3*a/d/x^3*(c^2*d*x^2+d)^(5/2)-2/3*a*c^2/d/x*(c^2*d*x^2+d)^(5/2)+2/3*a*c^4*x*(c^2*d*x^2+d)^(3/2)+a*c^4*d*x*(
c^2*d*x^2+d)^(1/2)+a*c^4*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)+1/2*b*(d*(c^2*x^2+1))
^(1/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)^2*c^3*d-8/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*c^3*d
-32*b*(d*(c^2*x^2+1))^(1/2)*d/(24*c^4*x^4+9*c^2*x^2+1)*x^5/(c^2*x^2+1)*arcsinh(c*x)*c^8+32*b*(d*(c^2*x^2+1))^(
1/2)*d/(24*c^4*x^4+9*c^2*x^2+1)*x^4/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*c^7-8/3*b*(d*(c^2*x^2+1))^(1/2)*d/(24*c^4*x
^4+9*c^2*x^2+1)*x^5/(c^2*x^2+1)*c^8+8/3*b*(d*(c^2*x^2+1))^(1/2)*d/(24*c^4*x^4+9*c^2*x^2+1)*x^3*c^6-52*b*(d*(c^
2*x^2+1))^(1/2)*d/(24*c^4*x^4+9*c^2*x^2+1)*x^3/(c^2*x^2+1)*arcsinh(c*x)*c^6+12*b*(d*(c^2*x^2+1))^(1/2)*d/(24*c
^4*x^4+9*c^2*x^2+1)*x^2/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*c^5-10/3*b*(d*(c^2*x^2+1))^(1/2)*d/(24*c^4*x^4+9*c^2*x^
2+1)*x^3/(c^2*x^2+1)*c^6-4*b*(d*(c^2*x^2+1))^(1/2)*d/(24*c^4*x^4+9*c^2*x^2+1)*x^2/(c^2*x^2+1)^(1/2)*c^5+2/3*b*
(d*(c^2*x^2+1))^(1/2)*d/(24*c^4*x^4+9*c^2*x^2+1)*x*c^4-73/3*b*(d*(c^2*x^2+1))^(1/2)*d/(24*c^4*x^4+9*c^2*x^2+1)
*x/(c^2*x^2+1)*arcsinh(c*x)*c^4+4/3*b*(d*(c^2*x^2+1))^(1/2)*d/(24*c^4*x^4+9*c^2*x^2+1)/(c^2*x^2+1)^(1/2)*arcsi
nh(c*x)*c^3-2/3*b*(d*(c^2*x^2+1))^(1/2)*d/(24*c^4*x^4+9*c^2*x^2+1)*x/(c^2*x^2+1)*c^4-3/2*b*(d*(c^2*x^2+1))^(1/
2)*d/(24*c^4*x^4+9*c^2*x^2+1)/(c^2*x^2+1)^(1/2)*c^3-14/3*b*(d*(c^2*x^2+1))^(1/2)*d/(24*c^4*x^4+9*c^2*x^2+1)/x/
(c^2*x^2+1)*arcsinh(c*x)*c^2-1/6*b*(d*(c^2*x^2+1))^(1/2)*d/(24*c^4*x^4+9*c^2*x^2+1)/x^2/(c^2*x^2+1)^(1/2)*c-1/
3*b*(d*(c^2*x^2+1))^(1/2)*d/(24*c^4*x^4+9*c^2*x^2+1)/x^3/(c^2*x^2+1)*arcsinh(c*x)+4/3*b*(d*(c^2*x^2+1))^(1/2)/
(c^2*x^2+1)^(1/2)*ln((c*x+(c^2*x^2+1)^(1/2))^2-1)*c^3*d

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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((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))/x^4,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 (\frac{{\left (a c^{2} d x^{2} + a d +{\left (b c^{2} d x^{2} + b d\right )} \operatorname{arsinh}\left (c x\right )\right )} \sqrt{c^{2} d x^{2} + d}}{x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac{3}{2}} \left (a + b \operatorname{asinh}{\left (c x \right )}\right )}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c**2*d*x**2+d)**(3/2)*(a+b*asinh(c*x))/x**4,x)

[Out]

Integral((d*(c**2*x**2 + 1))**(3/2)*(a + b*asinh(c*x))/x**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))/x^4,x, algorithm="giac")

[Out]

integrate((c^2*d*x^2 + d)^(3/2)*(b*arcsinh(c*x) + a)/x^4, x)