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

Optimal. Leaf size=115 \[ -\frac{\pi c^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{\left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\frac{\pi ^{3/2} c^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+\frac{4}{3} \pi ^{3/2} b c^3 \log (x)-\frac{\pi ^{3/2} b c}{6 x^2} \]

[Out]

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

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Rubi [A]  time = 0.218711, antiderivative size = 184, normalized size of antiderivative = 1.6, 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} \[ \frac{\pi c^3 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b \sqrt{c^2 x^2+1}}-\frac{\pi c^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{\left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac{\pi b c \sqrt{\pi c^2 x^2+\pi }}{6 x^2 \sqrt{c^2 x^2+1}}+\frac{4 \pi b c^3 \sqrt{\pi c^2 x^2+\pi } \log (x)}{3 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.2272, size = 125, normalized size = 1.09 \[ \frac{\pi ^{3/2} \left (\sinh ^{-1}(c x) \left (6 a c^3 x^3-2 b \sqrt{c^2 x^2+1} \left (4 c^2 x^2+1\right )\right )-8 a c^2 x^2 \sqrt{c^2 x^2+1}-2 a \sqrt{c^2 x^2+1}+8 b c^3 x^3 \log (c x)+3 b c^3 x^3 \sinh ^{-1}(c x)^2-b c x\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Pi^(3/2)*(-(b*c*x) - 2*a*Sqrt[1 + c^2*x^2] - 8*a*c^2*x^2*Sqrt[1 + c^2*x^2] + (6*a*c^3*x^3 - 2*b*Sqrt[1 + c^2*
x^2]*(1 + 4*c^2*x^2))*ArcSinh[c*x] + 3*b*c^3*x^3*ArcSinh[c*x]^2 + 8*b*c^3*x^3*Log[c*x]))/(6*x^3)

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Maple [B]  time = 0.221, size = 622, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a/Pi/x^3*(Pi*c^2*x^2+Pi)^(5/2)-2/3*a*c^2/Pi/x*(Pi*c^2*x^2+Pi)^(5/2)+2/3*a*c^4*x*(Pi*c^2*x^2+Pi)^(3/2)+a*c
^4*Pi*x*(Pi*c^2*x^2+Pi)^(1/2)+a*c^4*Pi^2*ln(Pi*x*c^2/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)+1/2*
b*c^3*Pi^(3/2)*arcsinh(c*x)^2-8/3*b*c^3*Pi^(3/2)*arcsinh(c*x)+32*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)*x^4*arcsi
nh(c*x)*c^7-32*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)*x^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^6+8/3*b*Pi^(3/2)/(24*c
^4*x^4+9*c^2*x^2+1)*x^4*c^7-8/3*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)*x^2*(c^2*x^2+1)*c^5+12*b*Pi^(3/2)/(24*c^4*
x^4+9*c^2*x^2+1)*x^2*arcsinh(c*x)*c^5-20*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)*x*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*
c^4-4/3*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)*(c^2*x^2+1)*c^3+4/3*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)*arcsinh(c*
x)*c^3-13/3*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)/x*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^2-1/6*b*Pi^(3/2)/(24*c^4*x^
4+9*c^2*x^2+1)/x^2*(c^2*x^2+1)*c-1/3*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)/x^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)+4/
3*b*c^3*Pi^(3/2)*ln((c*x+(c^2*x^2+1)^(1/2))^2-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (\pi + \pi c^{2} x^{2}\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((pi*c^2*x^2+pi)^(3/2)*(a+b*arcsinh(c*x))/x^4,x, algorithm="giac")

[Out]

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