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]

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

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Rubi [A]  time = 0.24, 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 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 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])

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 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]

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]

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*}

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Mathematica [A]  time = 0.42, size = 168, normalized size = 1.07 \[ \frac {\pi ^{5/2} \left (4 \sinh ^{-1}(c x) \left (60 a c x-32 b \sqrt {c^2 x^2+1}+16 b c x \sinh \left (2 \sinh ^{-1}(c x)\right )+b c x \sinh \left (4 \sinh ^{-1}(c x)\right )\right )+144 a c^2 x^2 \sqrt {c^2 x^2+1}-128 a \sqrt {c^2 x^2+1}+32 a c^4 x^4 \sqrt {c^2 x^2+1}+128 b c x \log (c x)+120 b c x \sinh ^{-1}(c x)^2-32 b c x \cosh \left (2 \sinh ^{-1}(c x)\right )-b c x \cosh \left (4 \sinh ^{-1}(c x)\right )\right )}{128 x} \]

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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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\pi + \pi c^{2} x^{2}} {\left (\pi ^{2} a c^{4} x^{4} + 2 \, \pi ^{2} a c^{2} x^{2} + \pi ^{2} a + {\left (\pi ^{2} b c^{4} x^{4} + 2 \, \pi ^{2} b c^{2} x^{2} + \pi ^{2} b\right )} \operatorname {arsinh}\left (c x\right )\right )}}{x^{2}}, x\right ) \]

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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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^2,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 [B]  time = 0.29, size = 283, normalized size = 1.80 \[ -\frac {a \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {7}{2}}}{\pi x}+a \,c^{2} x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}+\frac {5 a \,c^{2} \pi x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{4}+\frac {15 a \,c^{2} \pi ^{2} x \sqrt {\pi \,c^{2} x^{2}+\pi }}{8}+\frac {15 a \,c^{2} \pi ^{3} \ln \left (\frac {\pi x \,c^{2}}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{8 \sqrt {\pi \,c^{2}}}+\frac {b \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{4}}{4}+\frac {9 b \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x \,c^{2}}{8}-\frac {33 b \,\pi ^{\frac {5}{2}} c}{128}-\frac {b \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}}{x}+b c \,\pi ^{\frac {5}{2}} \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right )-\frac {b \,c^{5} \pi ^{\frac {5}{2}} x^{4}}{16}-\frac {9 b \,c^{3} \pi ^{\frac {5}{2}} x^{2}}{16}+\frac {15 b c \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right )^{2}}{16}-b c \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right ) \]

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

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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^2,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 \[ \int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (\Pi \,c^2\,x^2+\Pi \right )}^{5/2}}{x^2} \,d x \]

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^2,x)

[Out]

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

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

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]

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

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