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

Optimal. Leaf size=177 \[ \frac{3}{2} c^2 d x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{3 c d \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b \sqrt{c^2 x^2+1}}-\frac{\left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{b c^3 d x^2 \sqrt{c^2 d x^2+d}}{4 \sqrt{c^2 x^2+1}}+\frac{b c d \log (x) \sqrt{c^2 d x^2+d}}{\sqrt{c^2 x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.170462, antiderivative size = 177, 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, 5682, 5675, 30, 14} \[ \frac{3}{2} c^2 d x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{3 c d \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b \sqrt{c^2 x^2+1}}-\frac{\left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{b c^3 d x^2 \sqrt{c^2 d x^2+d}}{4 \sqrt{c^2 x^2+1}}+\frac{b c d \log (x) \sqrt{c^2 d x^2+d}}{\sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c^3*d*x^2*Sqrt[d + c^2*d*x^2])/(4*Sqrt[1 + c^2*x^2]) + (3*c^2*d*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])
)/2 - ((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/x + (3*c*d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(4*b
*Sqrt[1 + c^2*x^2]) + (b*c*d*Sqrt[d + c^2*d*x^2]*Log[x])/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 5682

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

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

Mathematica [A]  time = 0.777871, size = 200, normalized size = 1.13 \[ \frac{1}{8} \left (12 a c d^{3/2} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )+\frac{4 a d \left (c^2 x^2-2\right ) \sqrt{c^2 d x^2+d}}{x}+\frac{4 b d \sqrt{c^2 d x^2+d} \left (-2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)+2 c x \log (c x)+c x \sinh ^{-1}(c x)^2\right )}{x \sqrt{c^2 x^2+1}}+\frac{b c d \sqrt{c^2 d x^2+d} \left (2 \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+\sinh \left (2 \sinh ^{-1}(c x)\right )\right )-\cosh \left (2 \sinh ^{-1}(c x)\right )\right )}{\sqrt{c^2 x^2+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a*d*(-2 + c^2*x^2)*Sqrt[d + c^2*d*x^2])/x + (4*b*d*Sqrt[d + c^2*d*x^2]*(-2*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]
+ c*x*ArcSinh[c*x]^2 + 2*c*x*Log[c*x]))/(x*Sqrt[1 + c^2*x^2]) + 12*a*c*d^(3/2)*Log[c*d*x + Sqrt[d]*Sqrt[d + c^
2*d*x^2]] + (b*c*d*Sqrt[d + c^2*d*x^2]*(-Cosh[2*ArcSinh[c*x]] + 2*ArcSinh[c*x]*(ArcSinh[c*x] + Sinh[2*ArcSinh[
c*x]])))/Sqrt[1 + c^2*x^2])/8

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Maple [B]  time = 0.146, size = 392, normalized size = 2.2 \begin{align*} -{\frac{a}{dx} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{5}{2}}}}+a{c}^{2}x \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}+{\frac{3\,a{c}^{2}dx}{2}\sqrt{{c}^{2}d{x}^{2}+d}}+{\frac{3\,a{c}^{2}{d}^{2}}{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{3\,b \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}cd}{4}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{b{c}^{4}d{\it Arcsinh} \left ( cx \right ){x}^{3}}{2\,{c}^{2}{x}^{2}+2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{b{c}^{3}d{x}^{2}}{4}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{{c}^{2}db{\it Arcsinh} \left ( cx \right ) x}{2\,{c}^{2}{x}^{2}+2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{bcd{\it Arcsinh} \left ( cx \right ) \sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{bcd}{8}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{b{\it Arcsinh} \left ( cx \right ) d}{x \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{bcd\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\ln \left ( \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2}-1 \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \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^2,x)

[Out]

-a/d/x*(c^2*d*x^2+d)^(5/2)+a*c^2*x*(c^2*d*x^2+d)^(3/2)+3/2*a*c^2*d*x*(c^2*d*x^2+d)^(1/2)+3/2*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)+3/4*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x
)^2*c*d+1/2*b*(d*(c^2*x^2+1))^(1/2)*c^4*d/(c^2*x^2+1)*arcsinh(c*x)*x^3-1/4*b*(d*(c^2*x^2+1))^(1/2)*c^3*d/(c^2*
x^2+1)^(1/2)*x^2-1/2*b*(d*(c^2*x^2+1))^(1/2)*c^2*d/(c^2*x^2+1)*arcsinh(c*x)*x-b*(d*(c^2*x^2+1))^(1/2)*c*d/(c^2
*x^2+1)^(1/2)*arcsinh(c*x)-1/8*b*(d*(c^2*x^2+1))^(1/2)*c*d/(c^2*x^2+1)^(1/2)-b*(d*(c^2*x^2+1))^(1/2)*arcsinh(c
*x)*d/x/(c^2*x^2+1)+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*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^2,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^{2}}, 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^2,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^2, 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^{2}}\, 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**2,x)

[Out]

Integral((d*(c**2*x**2 + 1))**(3/2)*(a + b*asinh(c*x))/x**2, 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^{2}}\,{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^2,x, algorithm="giac")

[Out]

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