3.1.78 \(\int \frac {(\pi +c^2 \pi x^2)^{5/2} (a+b \sinh ^{-1}(c x))}{x^4} \, dx\) [78]

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

[Out]

-1/6*b*c*Pi^(5/2)/x^2-1/4*b*c^5*Pi^(5/2)*x^2-5/3*c^2*Pi*(Pi*c^2*x^2+Pi)^(3/2)*(a+b*arcsinh(c*x))/x-1/3*(Pi*c^2
*x^2+Pi)^(5/2)*(a+b*arcsinh(c*x))/x^3+5/4*c^3*Pi^(5/2)*(a+b*arcsinh(c*x))^2/b+7/3*b*c^3*Pi^(5/2)*ln(x)+5/2*c^4
*Pi^2*x*(a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2)

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Rubi [A]
time = 0.19, antiderivative size = 166, normalized size of antiderivative = 1.00, 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 = {5807, 5785, 5783, 30, 14, 272, 45} \begin {gather*} \frac {5 \pi ^{5/2} c^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b}-\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}{2} \pi ^2 c^4 x \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )-\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(b*c*Pi^(5/2))/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

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 45

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])

Rule 272

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 5783

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

Rule 5785

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[(1/2)*Simp[Sqrt[d + e*x^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/2)*Simp[Sqrt[d + e*x^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 5807

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/(f*(m + 1)))*Simp[(d + e*x^2)^p/(1
+ c^2*x^2)^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]

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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.25, size = 179, normalized size = 1.08 \begin {gather*} \frac {\pi ^{5/2} \left (-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 \sinh ^{-1}(c x)^2-3 b c^3 x^3 \cosh \left (2 \sinh ^{-1}(c x)\right )+56 b c^3 x^3 \log (c x)+\sinh ^{-1}(c x) \left (60 a c^3 x^3-8 b \sqrt {1+c^2 x^2} \left (1+7 c^2 x^2\right )+6 b c^3 x^3 \sinh \left (2 \sinh ^{-1}(c x)\right )\right )\right )}{24 x^3} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(691\) vs. \(2(138)=276\).
time = 7.28, size = 692, normalized size = 4.17

method	result	size
default	\(-\frac {a \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {7}{2}}}{3 \pi \,x^{3}}-\frac {4 a \,c^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {7}{2}}}{3 \pi x}+\frac {4 a \,c^{4} x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{3}+\frac {5 a \,c^{4} \pi x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{3}+\frac {5 a \,c^{4} \pi ^{2} x \sqrt {\pi \,c^{2} x^{2}+\pi }}{2}+\frac {5 a \,c^{4} \pi ^{3} \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{2 \sqrt {\pi \,c^{2}}}+\frac {5 b \,c^{3} \pi ^{\frac {5}{2}} \arcsinh \left (c x \right )^{2}}{4}-\frac {147 b \,\pi ^{\frac {5}{2}} x^{3} \sqrt {c^{2} x^{2}+1}\, \arcsinh \left (c x \right ) c^{6}}{63 c^{4} x^{4}+15 c^{2} x^{2}+1}-\frac {56 b \,\pi ^{\frac {5}{2}} x \sqrt {c^{2} x^{2}+1}\, \arcsinh \left (c x \right ) c^{4}}{63 c^{4} x^{4}+15 c^{2} x^{2}+1}-\frac {22 b \,\pi ^{\frac {5}{2}} \sqrt {c^{2} x^{2}+1}\, \arcsinh \left (c x \right ) c^{2}}{3 \left (63 c^{4} x^{4}+15 c^{2} x^{2}+1\right ) x}+\frac {7 b \,c^{3} \pi ^{\frac {5}{2}} \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right )}{3}-\frac {14 b \,c^{3} \pi ^{\frac {5}{2}} \arcsinh \left (c x \right )}{3}+\frac {147 b \,\pi ^{\frac {5}{2}} x^{4} \arcsinh \left (c x \right ) c^{7}}{63 c^{4} x^{4}+15 c^{2} x^{2}+1}-\frac {49 b \,\pi ^{\frac {5}{2}} x^{2} \left (c^{2} x^{2}+1\right ) c^{5}}{6 \left (63 c^{4} x^{4}+15 c^{2} x^{2}+1\right )}+\frac {35 b \,\pi ^{\frac {5}{2}} x^{2} \arcsinh \left (c x \right ) c^{5}}{63 c^{4} x^{4}+15 c^{2} x^{2}+1}-\frac {7 b \,\pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right ) c^{3}}{3 \left (63 c^{4} x^{4}+15 c^{2} x^{2}+1\right )}+\frac {7 b \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right ) c^{3}}{3 \left (63 c^{4} x^{4}+15 c^{2} x^{2}+1\right )}-\frac {b \,\pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right ) c}{6 \left (63 c^{4} x^{4}+15 c^{2} x^{2}+1\right ) x^{2}}-\frac {b \,\pi ^{\frac {5}{2}} \sqrt {c^{2} x^{2}+1}\, \arcsinh \left (c x \right )}{3 \left (63 c^{4} x^{4}+15 c^{2} x^{2}+1\right ) x^{3}}+\frac {b \,\pi ^{\frac {5}{2}} \sqrt {c^{2} x^{2}+1}\, \arcsinh \left (c x \right ) x \,c^{4}}{2}-\frac {b \,c^{5} \pi ^{\frac {5}{2}} x^{2}}{4}+\frac {49 b \,\pi ^{\frac {5}{2}} x^{4} c^{7}}{6 \left (63 c^{4} x^{4}+15 c^{2} x^{2}+1\right )}-\frac {b \,\pi ^{\frac {5}{2}} c^{3}}{8}\)	\(692\)

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,method=_RETURNVERBOSE)

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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 >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \pi ^{\frac {5}{2}} \left (\int a c^{4} \sqrt {c^{2} x^{2} + 1}\, dx + \int \frac {a \sqrt {c^{2} x^{2} + 1}}{x^{4}}\, dx + \int \frac {2 a c^{2} \sqrt {c^{2} x^{2} + 1}}{x^{2}}\, dx + \int b c^{4} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}\, dx + \int \frac {b \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{x^{4}}\, dx + \int \frac {2 b c^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{x^{2}}\, dx\right ) \end {gather*}

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]

pi**(5/2)*(Integral(a*c**4*sqrt(c**2*x**2 + 1), x) + Integral(a*sqrt(c**2*x**2 + 1)/x**4, x) + Integral(2*a*c*
*2*sqrt(c**2*x**2 + 1)/x**2, x) + Integral(b*c**4*sqrt(c**2*x**2 + 1)*asinh(c*x), x) + Integral(b*sqrt(c**2*x*
*2 + 1)*asinh(c*x)/x**4, x) + Integral(2*b*c**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/x**2, x))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (\Pi \,c^2\,x^2+\Pi \right )}^{5/2}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asinh(c*x))*(Pi + Pi*c^2*x^2)^(5/2))/x^4,x)

[Out]

int(((a + b*asinh(c*x))*(Pi + Pi*c^2*x^2)^(5/2))/x^4, x)

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