3.78 \(\int \frac{(\pi +c^2 \pi x^2)^{5/2} (a+b \sinh ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=166 \[ \frac{5}{2} \pi ^2 c^4 x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )-\frac{5 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x}-\frac{\left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\frac{5 \pi ^{5/2} c^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b}-\frac{1}{4} \pi ^{5/2} b c^5 x^2+\frac{7}{3} \pi ^{5/2} b c^3 \log (x)-\frac{\pi ^{5/2} b c}{6 x^2} \]

[Out]

-(b*c*Pi^(5/2))/(6*x^2) - (b*c^5*Pi^(5/2)*x^2)/4 + (5*c^4*Pi^2*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/2
- (5*c^2*Pi*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(3*x) - ((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x
]))/(3*x^3) + (5*c^3*Pi^(5/2)*(a + b*ArcSinh[c*x])^2)/(4*b) + (7*b*c^3*Pi^(5/2)*Log[x])/3

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Rubi [A]  time = 0.293949, antiderivative size = 266, normalized size of antiderivative = 1.6, number of steps used = 10, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {5739, 5682, 5675, 30, 14, 266, 43} \[ \frac{5}{2} \pi ^2 c^4 x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac{5 \pi ^2 c^3 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b \sqrt{c^2 x^2+1}}-\frac{5 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x}-\frac{\left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac{\pi ^2 b c^5 x^2 \sqrt{\pi c^2 x^2+\pi }}{4 \sqrt{c^2 x^2+1}}-\frac{\pi ^2 b c \sqrt{\pi c^2 x^2+\pi }}{6 x^2 \sqrt{c^2 x^2+1}}+\frac{7 \pi ^2 b c^3 \sqrt{\pi c^2 x^2+\pi } \log (x)}{3 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/x^4,x]

[Out]

-(b*c*Pi^2*Sqrt[Pi + c^2*Pi*x^2])/(6*x^2*Sqrt[1 + c^2*x^2]) - (b*c^5*Pi^2*x^2*Sqrt[Pi + c^2*Pi*x^2])/(4*Sqrt[1
+ c^2*x^2]) + (5*c^4*Pi^2*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/2 - (5*c^2*Pi*(Pi + c^2*Pi*x^2)^(3/2)
*(a + b*ArcSinh[c*x]))/(3*x) - ((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(3*x^3) + (5*c^3*Pi^2*Sqrt[Pi +
c^2*Pi*x^2]*(a + b*ArcSinh[c*x])^2)/(4*b*Sqrt[1 + c^2*x^2]) + (7*b*c^3*Pi^2*Sqrt[Pi + c^2*Pi*x^2]*Log[x])/(3*S
qrt[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]]

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^4} \, dx &=-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\frac{1}{3} \left (5 c^2 \pi \right ) \int \frac{\left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{x^2} \, 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^3} \, dx}{3 \sqrt{1+c^2 x^2}}\\ &=-\frac{5 c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x}-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\left (5 c^4 \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^2} \, dx,x,x^2\right )}{6 \sqrt{1+c^2 x^2}}+\frac{\left (5 b c^3 \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \frac{1+c^2 x^2}{x} \, dx}{3 \sqrt{1+c^2 x^2}}\\ &=\frac{5}{2} c^4 \pi ^2 x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{5 c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x}-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\frac{\left (b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \operatorname{Subst}\left (\int \left (c^4+\frac{1}{x^2}+\frac{2 c^2}{x}\right ) \, dx,x,x^2\right )}{6 \sqrt{1+c^2 x^2}}+\frac{\left (5 b c^3 \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \left (\frac{1}{x}+c^2 x\right ) \, dx}{3 \sqrt{1+c^2 x^2}}+\frac{\left (5 c^4 \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}{2 \sqrt{1+c^2 x^2}}-\frac{\left (5 b c^5 \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int x \, dx}{2 \sqrt{1+c^2 x^2}}\\ &=-\frac{b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}}{6 x^2 \sqrt{1+c^2 x^2}}-\frac{b c^5 \pi ^2 x^2 \sqrt{\pi +c^2 \pi x^2}}{4 \sqrt{1+c^2 x^2}}+\frac{5}{2} c^4 \pi ^2 x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{5 c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x}-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\frac{5 c^3 \pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b \sqrt{1+c^2 x^2}}+\frac{7 b c^3 \pi ^2 \sqrt{\pi +c^2 \pi x^2} \log (x)}{3 \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.399065, size = 179, normalized size = 1.08 \[ \frac{\pi ^{5/2} \left (\sinh ^{-1}(c x) \left (60 a c^3 x^3-8 b \sqrt{c^2 x^2+1} \left (7 c^2 x^2+1\right )+6 b c^3 x^3 \sinh \left (2 \sinh ^{-1}(c x)\right )\right )+12 a c^4 x^4 \sqrt{c^2 x^2+1}-56 a c^2 x^2 \sqrt{c^2 x^2+1}-8 a \sqrt{c^2 x^2+1}+56 b c^3 x^3 \log (c x)+30 b c^3 x^3 \sinh ^{-1}(c x)^2-3 b c^3 x^3 \cosh \left (2 \sinh ^{-1}(c x)\right )-4 b c x\right )}{24 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/x^4,x]

[Out]

(Pi^(5/2)*(-4*b*c*x - 8*a*Sqrt[1 + c^2*x^2] - 56*a*c^2*x^2*Sqrt[1 + c^2*x^2] + 12*a*c^4*x^4*Sqrt[1 + c^2*x^2]
+ 30*b*c^3*x^3*ArcSinh[c*x]^2 - 3*b*c^3*x^3*Cosh[2*ArcSinh[c*x]] + 56*b*c^3*x^3*Log[c*x] + ArcSinh[c*x]*(60*a*
c^3*x^3 - 8*b*Sqrt[1 + c^2*x^2]*(1 + 7*c^2*x^2) + 6*b*c^3*x^3*Sinh[2*ArcSinh[c*x]])))/(24*x^3)

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Maple [B]  time = 0.24, size = 692, normalized size = 4.2 \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)^(5/2)*(a+b*arcsinh(c*x))/x^4,x)

[Out]

-1/3*a/Pi/x^3*(Pi*c^2*x^2+Pi)^(7/2)-4/3*a*c^2/Pi/x*(Pi*c^2*x^2+Pi)^(7/2)+4/3*a*c^4*x*(Pi*c^2*x^2+Pi)^(5/2)+5/3
*a*c^4*Pi*x*(Pi*c^2*x^2+Pi)^(3/2)+5/2*a*c^4*Pi^2*x*(Pi*c^2*x^2+Pi)^(1/2)+5/2*a*c^4*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)+7/3*b*c^3*Pi^(5/2)*ln((c*x+(c^2*x^2+1)^(1/2))^2-1)-1/4*b*c^5*Pi^(5/
2)*x^2-14/3*b*c^3*Pi^(5/2)*arcsinh(c*x)+5/4*b*c^3*Pi^(5/2)*arcsinh(c*x)^2-1/8*b*Pi^(5/2)*c^3+49/6*b*Pi^(5/2)/(
63*c^4*x^4+15*c^2*x^2+1)*x^4*c^7-147*b*Pi^(5/2)/(63*c^4*x^4+15*c^2*x^2+1)*x^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c
^6-56*b*Pi^(5/2)/(63*c^4*x^4+15*c^2*x^2+1)*x*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^4-22/3*b*Pi^(5/2)/(63*c^4*x^4+15
*c^2*x^2+1)/x*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^2+147*b*Pi^(5/2)/(63*c^4*x^4+15*c^2*x^2+1)*x^4*arcsinh(c*x)*c^7
-49/6*b*Pi^(5/2)/(63*c^4*x^4+15*c^2*x^2+1)*x^2*(c^2*x^2+1)*c^5+35*b*Pi^(5/2)/(63*c^4*x^4+15*c^2*x^2+1)*x^2*arc
sinh(c*x)*c^5-7/3*b*Pi^(5/2)/(63*c^4*x^4+15*c^2*x^2+1)*(c^2*x^2+1)*c^3+7/3*b*Pi^(5/2)/(63*c^4*x^4+15*c^2*x^2+1
)*arcsinh(c*x)*c^3-1/6*b*Pi^(5/2)/(63*c^4*x^4+15*c^2*x^2+1)/x^2*(c^2*x^2+1)*c-1/3*b*Pi^(5/2)/(63*c^4*x^4+15*c^
2*x^2+1)/x^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)+1/2*b*arcsinh(c*x)*Pi^(5/2)*(c^2*x^2+1)^(1/2)*x*c^4

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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^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 ^{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^{4}}, 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^4,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^4, 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**4,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^{4}}\,{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^4,x, algorithm="giac")

[Out]

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