3.76 \(\int \frac{(\pi +c^2 \pi x^2)^{5/2} (a+b \sinh ^{-1}(c x))}{x^2} \, dx\)

Optimal. Leaf size=157 \[ \frac{5}{4} \pi c^2 x \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{15}{8} \pi ^2 c^2 x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}+\frac{15 \pi ^{5/2} c \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b}-\frac{1}{16} \pi ^{5/2} b c^5 x^4-\frac{9}{16} \pi ^{5/2} b c^3 x^2+\pi ^{5/2} b c \log (x) \]

[Out]

(-9*b*c^3*Pi^(5/2)*x^2)/16 - (b*c^5*Pi^(5/2)*x^4)/16 + (15*c^2*Pi^2*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x
]))/8 + (5*c^2*Pi*x*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/4 - ((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[
c*x]))/x + (15*c*Pi^(5/2)*(a + b*ArcSinh[c*x])^2)/(16*b) + b*c*Pi^(5/2)*Log[x]

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Rubi [A]  time = 0.235875, antiderivative size = 257, normalized size of antiderivative = 1.64, number of steps used = 10, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {5739, 5684, 5682, 5675, 30, 14, 266, 43} \[ \frac{5}{4} \pi c^2 x \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{15}{8} \pi ^2 c^2 x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac{15 \pi ^2 c \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b \sqrt{c^2 x^2+1}}-\frac{\left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{\pi ^2 b c^5 x^4 \sqrt{\pi c^2 x^2+\pi }}{16 \sqrt{c^2 x^2+1}}-\frac{9 \pi ^2 b c^3 x^2 \sqrt{\pi c^2 x^2+\pi }}{16 \sqrt{c^2 x^2+1}}+\frac{\pi ^2 b c \sqrt{\pi c^2 x^2+\pi } \log (x)}{\sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/x^2,x]

[Out]

(-9*b*c^3*Pi^2*x^2*Sqrt[Pi + c^2*Pi*x^2])/(16*Sqrt[1 + c^2*x^2]) - (b*c^5*Pi^2*x^4*Sqrt[Pi + c^2*Pi*x^2])/(16*
Sqrt[1 + c^2*x^2]) + (15*c^2*Pi^2*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/8 + (5*c^2*Pi*x*(Pi + c^2*Pi*x
^2)^(3/2)*(a + b*ArcSinh[c*x]))/4 - ((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/x + (15*c*Pi^2*Sqrt[Pi + c^
2*Pi*x^2]*(a + b*ArcSinh[c*x])^2)/(16*b*Sqrt[1 + c^2*x^2]) + (b*c*Pi^2*Sqrt[Pi + c^2*Pi*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 5684

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

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]]

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 \frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x^2} \, dx &=-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}+\left (5 c^2 \pi \right ) \int \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx+\frac{\left (b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \frac{\left (1+c^2 x^2\right )^2}{x} \, dx}{\sqrt{1+c^2 x^2}}\\ &=\frac{5}{4} c^2 \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}+\frac{1}{4} \left (15 c^2 \pi ^2\right ) \int \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx+\frac{\left (b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \operatorname{Subst}\left (\int \frac{\left (1+c^2 x\right )^2}{x} \, dx,x,x^2\right )}{2 \sqrt{1+c^2 x^2}}-\frac{\left (5 b c^3 \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int x \left (1+c^2 x^2\right ) \, dx}{4 \sqrt{1+c^2 x^2}}\\ &=\frac{15}{8} c^2 \pi ^2 x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{4} c^2 \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}+\frac{\left (b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \operatorname{Subst}\left (\int \left (2 c^2+\frac{1}{x}+c^4 x\right ) \, dx,x,x^2\right )}{2 \sqrt{1+c^2 x^2}}+\frac{\left (15 c^2 \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{8 \sqrt{1+c^2 x^2}}-\frac{\left (5 b c^3 \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \left (x+c^2 x^3\right ) \, dx}{4 \sqrt{1+c^2 x^2}}-\frac{\left (15 b c^3 \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int x \, dx}{8 \sqrt{1+c^2 x^2}}\\ &=-\frac{9 b c^3 \pi ^2 x^2 \sqrt{\pi +c^2 \pi x^2}}{16 \sqrt{1+c^2 x^2}}-\frac{b c^5 \pi ^2 x^4 \sqrt{\pi +c^2 \pi x^2}}{16 \sqrt{1+c^2 x^2}}+\frac{15}{8} c^2 \pi ^2 x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{4} c^2 \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x}+\frac{15 c \pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b \sqrt{1+c^2 x^2}}+\frac{b c \pi ^2 \sqrt{\pi +c^2 \pi x^2} \log (x)}{\sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.394832, size = 168, normalized size = 1.07 \[ \frac{\pi ^{5/2} \left (4 \sinh ^{-1}(c x) \left (60 a c x-32 b \sqrt{c^2 x^2+1}+16 b c x \sinh \left (2 \sinh ^{-1}(c x)\right )+b c x \sinh \left (4 \sinh ^{-1}(c x)\right )\right )+32 a c^4 x^4 \sqrt{c^2 x^2+1}+144 a c^2 x^2 \sqrt{c^2 x^2+1}-128 a \sqrt{c^2 x^2+1}+128 b c x \log (c x)+120 b c x \sinh ^{-1}(c x)^2-32 b c x \cosh \left (2 \sinh ^{-1}(c x)\right )-b c x \cosh \left (4 \sinh ^{-1}(c x)\right )\right )}{128 x} \]

Antiderivative was successfully verified.

[In]

Integrate[((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/x^2,x]

[Out]

(Pi^(5/2)*(-128*a*Sqrt[1 + c^2*x^2] + 144*a*c^2*x^2*Sqrt[1 + c^2*x^2] + 32*a*c^4*x^4*Sqrt[1 + c^2*x^2] + 120*b
*c*x*ArcSinh[c*x]^2 - 32*b*c*x*Cosh[2*ArcSinh[c*x]] - b*c*x*Cosh[4*ArcSinh[c*x]] + 128*b*c*x*Log[c*x] + 4*ArcS
inh[c*x]*(60*a*c*x - 32*b*Sqrt[1 + c^2*x^2] + 16*b*c*x*Sinh[2*ArcSinh[c*x]] + b*c*x*Sinh[4*ArcSinh[c*x]])))/(1
28*x)

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Maple [B]  time = 0.191, size = 283, normalized size = 1.8 \begin{align*} -{\frac{a}{\pi \,x} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{7}{2}}}}+a{c}^{2}x \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{5}{2}}}+{\frac{5\,a{c}^{2}\pi \,x}{4} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{3}{2}}}}+{\frac{15\,a{c}^{2}{\pi }^{2}x}{8}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}+{\frac{15\,a{c}^{2}{\pi }^{3}}{8}\ln \left ({\pi \,{c}^{2}x{\frac{1}{\sqrt{\pi \,{c}^{2}}}}}+\sqrt{\pi \,{c}^{2}{x}^{2}+\pi } \right ){\frac{1}{\sqrt{\pi \,{c}^{2}}}}}-{\frac{b{c}^{5}{\pi }^{{\frac{5}{2}}}{x}^{4}}{16}}-{\frac{9\,b{c}^{3}{\pi }^{5/2}{x}^{2}}{16}}+bc{\pi }^{{\frac{5}{2}}}\ln \left ( \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2}-1 \right ) -bc{\pi }^{{\frac{5}{2}}}{\it Arcsinh} \left ( cx \right ) -{\frac{33\,b{\pi }^{5/2}c}{128}}+{\frac{15\,bc{\pi }^{5/2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{16}}+{\frac{b{\it Arcsinh} \left ( cx \right ){\pi }^{{\frac{5}{2}}}{x}^{3}{c}^{4}}{4}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{9\,b{\it Arcsinh} \left ( cx \right ){\pi }^{5/2}x{c}^{2}}{8}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{b{\it Arcsinh} \left ( cx \right ){\pi }^{{\frac{5}{2}}}}{x}\sqrt{{c}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((Pi*c^2*x^2+Pi)^(5/2)*(a+b*arcsinh(c*x))/x^2,x)

[Out]

-a/Pi/x*(Pi*c^2*x^2+Pi)^(7/2)+a*c^2*x*(Pi*c^2*x^2+Pi)^(5/2)+5/4*a*c^2*Pi*x*(Pi*c^2*x^2+Pi)^(3/2)+15/8*a*c^2*Pi
^2*x*(Pi*c^2*x^2+Pi)^(1/2)+15/8*a*c^2*Pi^3*ln(Pi*x*c^2/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)-1/
16*b*c^5*Pi^(5/2)*x^4-9/16*b*c^3*Pi^(5/2)*x^2+b*c*Pi^(5/2)*ln((c*x+(c^2*x^2+1)^(1/2))^2-1)-b*c*Pi^(5/2)*arcsin
h(c*x)-33/128*b*Pi^(5/2)*c+15/16*b*c*Pi^(5/2)*arcsinh(c*x)^2+1/4*b*arcsinh(c*x)*Pi^(5/2)*(c^2*x^2+1)^(1/2)*x^3
*c^4+9/8*b*arcsinh(c*x)*Pi^(5/2)*(c^2*x^2+1)^(1/2)*x*c^2-b*Pi^(5/2)*arcsinh(c*x)/x*(c^2*x^2+1)^(1/2)

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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)^(5/2)*(a+b*arcsinh(c*x))/x^2,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 ^{2} a c^{4} x^{4} + 2 \, \pi ^{2} a c^{2} x^{2} + \pi ^{2} a +{\left (\pi ^{2} b c^{4} x^{4} + 2 \, \pi ^{2} b c^{2} x^{2} + \pi ^{2} b\right )} \operatorname{arsinh}\left (c x\right )\right )}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((pi*c^2*x^2+pi)^(5/2)*(a+b*arcsinh(c*x))/x^2,x, algorithm="fricas")

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(pi^2*a*c^4*x^4 + 2*pi^2*a*c^2*x^2 + pi^2*a + (pi^2*b*c^4*x^4 + 2*pi^2*b*c^2*x^
2 + pi^2*b)*arcsinh(c*x))/x^2, 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((pi*c**2*x**2+pi)**(5/2)*(a+b*asinh(c*x))/x**2,x)

[Out]

Timed out

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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{5}{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((pi*c^2*x^2+pi)^(5/2)*(a+b*arcsinh(c*x))/x^2,x, algorithm="giac")

[Out]

integrate((pi + pi*c^2*x^2)^(5/2)*(b*arcsinh(c*x) + a)/x^2, x)