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

Optimal. Leaf size=165 \[ \frac {1}{6} x \left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{24} \pi x \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{16} \pi ^2 x \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac {5 \pi ^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c}-\frac {5}{96} \pi ^{5/2} b c^3 x^4-\frac {\pi ^{5/2} b \left (c^2 x^2+1\right )^3}{36 c}-\frac {25}{96} \pi ^{5/2} b c x^2 \]

[Out]

-25/96*b*c*Pi^(5/2)*x^2-5/96*b*c^3*Pi^(5/2)*x^4-1/36*b*Pi^(5/2)*(c^2*x^2+1)^3/c+5/24*Pi*x*(Pi*c^2*x^2+Pi)^(3/2
)*(a+b*arcsinh(c*x))+1/6*x*(Pi*c^2*x^2+Pi)^(5/2)*(a+b*arcsinh(c*x))+5/32*Pi^(5/2)*(a+b*arcsinh(c*x))^2/b/c+5/1
6*Pi^2*x*(a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 254, normalized size of antiderivative = 1.54, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5684, 5682, 5675, 30, 14, 261} \[ \frac {1}{6} x \left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{24} \pi x \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{16} \pi ^2 x \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac {5 \pi ^2 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \sqrt {c^2 x^2+1}}-\frac {5 \pi ^2 b c^3 x^4 \sqrt {\pi c^2 x^2+\pi }}{96 \sqrt {c^2 x^2+1}}-\frac {25 \pi ^2 b c x^2 \sqrt {\pi c^2 x^2+\pi }}{96 \sqrt {c^2 x^2+1}}-\frac {\pi ^2 b \left (c^2 x^2+1\right )^{5/2} \sqrt {\pi c^2 x^2+\pi }}{36 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-25*b*c*Pi^2*x^2*Sqrt[Pi + c^2*Pi*x^2])/(96*Sqrt[1 + c^2*x^2]) - (5*b*c^3*Pi^2*x^4*Sqrt[Pi + c^2*Pi*x^2])/(96
*Sqrt[1 + c^2*x^2]) - (b*Pi^2*(1 + c^2*x^2)^(5/2)*Sqrt[Pi + c^2*Pi*x^2])/(36*c) + (5*Pi^2*x*Sqrt[Pi + c^2*Pi*x
^2]*(a + b*ArcSinh[c*x]))/16 + (5*Pi*x*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/24 + (x*(Pi + c^2*Pi*x^2)
^(5/2)*(a + b*ArcSinh[c*x]))/6 + (5*Pi^2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x])^2)/(32*b*c*Sqrt[1 + c^2*x^
2])

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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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 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 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]

Rubi steps

\begin {align*} \int \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac {1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} (5 \pi ) \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 x \left (1+c^2 x^2\right )^2 \, dx}{6 \sqrt {1+c^2 x^2}}\\ &=-\frac {b \pi ^2 \left (1+c^2 x^2\right )^{5/2} \sqrt {\pi +c^2 \pi x^2}}{36 c}+\frac {5}{24} \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} \left (5 \pi ^2\right ) \int \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx-\frac {\left (5 b c \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int x \left (1+c^2 x^2\right ) \, dx}{24 \sqrt {1+c^2 x^2}}\\ &=-\frac {b \pi ^2 \left (1+c^2 x^2\right )^{5/2} \sqrt {\pi +c^2 \pi x^2}}{36 c}+\frac {5}{16} \pi ^2 x \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{24} \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {\left (5 \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}{16 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int \left (x+c^2 x^3\right ) \, dx}{24 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int x \, dx}{16 \sqrt {1+c^2 x^2}}\\ &=-\frac {25 b c \pi ^2 x^2 \sqrt {\pi +c^2 \pi x^2}}{96 \sqrt {1+c^2 x^2}}-\frac {5 b c^3 \pi ^2 x^4 \sqrt {\pi +c^2 \pi x^2}}{96 \sqrt {1+c^2 x^2}}-\frac {b \pi ^2 \left (1+c^2 x^2\right )^{5/2} \sqrt {\pi +c^2 \pi x^2}}{36 c}+\frac {5}{16} \pi ^2 x \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{24} \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5 \pi ^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 153, normalized size = 0.93 \[ \frac {\pi ^{5/2} \left (12 \sinh ^{-1}(c x) \left (60 a+45 b \sinh \left (2 \sinh ^{-1}(c x)\right )+9 b \sinh \left (4 \sinh ^{-1}(c x)\right )+b \sinh \left (6 \sinh ^{-1}(c x)\right )\right )+1584 a c x \sqrt {c^2 x^2+1}+384 a c^5 x^5 \sqrt {c^2 x^2+1}+1248 a c^3 x^3 \sqrt {c^2 x^2+1}+360 b \sinh ^{-1}(c x)^2-270 b \cosh \left (2 \sinh ^{-1}(c x)\right )-27 b \cosh \left (4 \sinh ^{-1}(c x)\right )-2 b \cosh \left (6 \sinh ^{-1}(c x)\right )\right )}{2304 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Pi^(5/2)*(1584*a*c*x*Sqrt[1 + c^2*x^2] + 1248*a*c^3*x^3*Sqrt[1 + c^2*x^2] + 384*a*c^5*x^5*Sqrt[1 + c^2*x^2] +
360*b*ArcSinh[c*x]^2 - 270*b*Cosh[2*ArcSinh[c*x]] - 27*b*Cosh[4*ArcSinh[c*x]] - 2*b*Cosh[6*ArcSinh[c*x]] + 12
*ArcSinh[c*x]*(60*a + 45*b*Sinh[2*ArcSinh[c*x]] + 9*b*Sinh[4*ArcSinh[c*x]] + b*Sinh[6*ArcSinh[c*x]])))/(2304*c
)

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\pi + \pi c^{2} x^{2}} {\left (\pi ^{2} a c^{4} x^{4} + 2 \, \pi ^{2} a c^{2} x^{2} + \pi ^{2} a + {\left (\pi ^{2} b c^{4} x^{4} + 2 \, \pi ^{2} b c^{2} x^{2} + \pi ^{2} b\right )} \operatorname {arsinh}\left (c x\right )\right )}, x\right ) \]

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, 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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.07, size = 228, normalized size = 1.38 \[ \frac {x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}} a}{6}+\frac {5 a \pi x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{24}+\frac {5 a \,\pi ^{2} x \sqrt {\pi \,c^{2} x^{2}+\pi }}{16}+\frac {5 a \,\pi ^{3} \ln \left (\frac {\pi x \,c^{2}}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{16 \sqrt {\pi \,c^{2}}}+\frac {b \,\pi ^{\frac {5}{2}} c^{4} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{5}}{6}-\frac {b \,\pi ^{\frac {5}{2}} c^{5} x^{6}}{36}+\frac {13 b \,\pi ^{\frac {5}{2}} c^{2} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3}}{24}-\frac {13 b \,c^{3} \pi ^{\frac {5}{2}} x^{4}}{96}+\frac {11 b \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x}{16}-\frac {11 b c \,\pi ^{\frac {5}{2}} x^{2}}{32}+\frac {5 b \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right )^{2}}{32 c}-\frac {17 b \,\pi ^{\frac {5}{2}}}{72 c} \]

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)

[Out]

1/6*x*(Pi*c^2*x^2+Pi)^(5/2)*a+5/24*a*Pi*x*(Pi*c^2*x^2+Pi)^(3/2)+5/16*a*Pi^2*x*(Pi*c^2*x^2+Pi)^(1/2)+5/16*a*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)+1/6*b*Pi^(5/2)*c^4*arcsinh(c*x)*(c^2*x^2+1)
^(1/2)*x^5-1/36*b*Pi^(5/2)*c^5*x^6+13/24*b*Pi^(5/2)*c^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^3-13/96*b*c^3*Pi^(5/2
)*x^4+11/16*b*Pi^(5/2)*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x-11/32*b*c*Pi^(5/2)*x^2+5/32*b*Pi^(5/2)/c*arcsinh(c*x)^
2-17/72*b*Pi^(5/2)/c

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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

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)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)

[Out]

Timed out

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