3.1.80 \(\int \frac {x^5 (a+b \sinh ^{-1}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx\) [80]

Optimal. Leaf size=149 \[ -\frac {8 b x}{15 c^5 \sqrt {\pi }}+\frac {4 b x^3}{45 c^3 \sqrt {\pi }}-\frac {b x^5}{25 c \sqrt {\pi }}+\frac {8 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^6 \pi }-\frac {4 x^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^4 \pi }+\frac {x^4 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 \pi } \]

[Out]

-8/15*b*x/c^5/Pi^(1/2)+4/45*b*x^3/c^3/Pi^(1/2)-1/25*b*x^5/c/Pi^(1/2)+8/15*(a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi)^(
1/2)/c^6/Pi-4/15*x^2*(a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2)/c^4/Pi+1/5*x^4*(a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi
)^(1/2)/c^2/Pi

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Rubi [A]
time = 0.18, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5812, 5798, 8, 30} \begin {gather*} \frac {x^4 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{5 \pi c^2}+\frac {8 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{15 \pi c^6}-\frac {4 x^2 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{15 \pi c^4}-\frac {8 b x}{15 \sqrt {\pi } c^5}+\frac {4 b x^3}{45 \sqrt {\pi } c^3}-\frac {b x^5}{25 \sqrt {\pi } c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b*x)/(15*c^5*Sqrt[Pi]) + (4*b*x^3)/(45*c^3*Sqrt[Pi]) - (b*x^5)/(25*c*Sqrt[Pi]) + (8*Sqrt[Pi + c^2*Pi*x^2]*
(a + b*ArcSinh[c*x]))/(15*c^6*Pi) - (4*x^2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(15*c^4*Pi) + (x^4*Sqrt
[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(5*c^2*Pi)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

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

Rule 5798

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

Rule 5812

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(e*(m + 2*p + 1))), x] + (-Dist[f^2*((m - 1)/(c^2*
(m + 2*p + 1))), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*f*(n/(c*(m + 2*p + 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, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && IGtQ[m, 1] && NeQ[m + 2*p + 1, 0
]

Rubi steps

\begin {align*} \int \frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {\pi +c^2 \pi x^2}} \, dx &=\frac {x^4 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 \pi }-\frac {4 \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {\pi +c^2 \pi x^2}} \, dx}{5 c^2}-\frac {\left (b \sqrt {1+c^2 x^2}\right ) \int x^4 \, dx}{5 c \sqrt {\pi +c^2 \pi x^2}}\\ &=-\frac {b x^5 \sqrt {1+c^2 x^2}}{25 c \sqrt {\pi +c^2 \pi x^2}}-\frac {4 x^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^4 \pi }+\frac {x^4 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 \pi }+\frac {8 \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {\pi +c^2 \pi x^2}} \, dx}{15 c^4}+\frac {\left (4 b \sqrt {1+c^2 x^2}\right ) \int x^2 \, dx}{15 c^3 \sqrt {\pi +c^2 \pi x^2}}\\ &=\frac {4 b x^3 \sqrt {1+c^2 x^2}}{45 c^3 \sqrt {\pi +c^2 \pi x^2}}-\frac {b x^5 \sqrt {1+c^2 x^2}}{25 c \sqrt {\pi +c^2 \pi x^2}}+\frac {8 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^6 \pi }-\frac {4 x^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^4 \pi }+\frac {x^4 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 \pi }-\frac {\left (8 b \sqrt {1+c^2 x^2}\right ) \int 1 \, dx}{15 c^5 \sqrt {\pi +c^2 \pi x^2}}\\ &=-\frac {8 b x \sqrt {1+c^2 x^2}}{15 c^5 \sqrt {\pi +c^2 \pi x^2}}+\frac {4 b x^3 \sqrt {1+c^2 x^2}}{45 c^3 \sqrt {\pi +c^2 \pi x^2}}-\frac {b x^5 \sqrt {1+c^2 x^2}}{25 c \sqrt {\pi +c^2 \pi x^2}}+\frac {8 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^6 \pi }-\frac {4 x^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^4 \pi }+\frac {x^4 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 \pi }\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 108, normalized size = 0.72 \begin {gather*} \frac {15 a \sqrt {1+c^2 x^2} \left (8-4 c^2 x^2+3 c^4 x^4\right )+b \left (-120 c x+20 c^3 x^3-9 c^5 x^5\right )+15 b \sqrt {1+c^2 x^2} \left (8-4 c^2 x^2+3 c^4 x^4\right ) \sinh ^{-1}(c x)}{225 c^6 \sqrt {\pi }} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a*Sqrt[1 + c^2*x^2]*(8 - 4*c^2*x^2 + 3*c^4*x^4) + b*(-120*c*x + 20*c^3*x^3 - 9*c^5*x^5) + 15*b*Sqrt[1 + c^
2*x^2]*(8 - 4*c^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x])/(225*c^6*Sqrt[Pi])

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {x^{5} \left (a +b \arcsinh \left (c x \right )\right )}{\sqrt {\pi \,c^{2} x^{2}+\pi }}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(1/2),x)

[Out]

int(x^5*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(1/2),x)

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Maxima [A]
time = 0.32, size = 174, normalized size = 1.17 \begin {gather*} \frac {1}{15} \, {\left (\frac {3 \, \sqrt {\pi + \pi c^{2} x^{2}} x^{4}}{\pi c^{2}} - \frac {4 \, \sqrt {\pi + \pi c^{2} x^{2}} x^{2}}{\pi c^{4}} + \frac {8 \, \sqrt {\pi + \pi c^{2} x^{2}}}{\pi c^{6}}\right )} b \operatorname {arsinh}\left (c x\right ) + \frac {1}{15} \, {\left (\frac {3 \, \sqrt {\pi + \pi c^{2} x^{2}} x^{4}}{\pi c^{2}} - \frac {4 \, \sqrt {\pi + \pi c^{2} x^{2}} x^{2}}{\pi c^{4}} + \frac {8 \, \sqrt {\pi + \pi c^{2} x^{2}}}{\pi c^{6}}\right )} a - \frac {{\left (9 \, c^{4} x^{5} - 20 \, c^{2} x^{3} + 120 \, x\right )} b}{225 \, \sqrt {\pi } c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="maxima")

[Out]

1/15*(3*sqrt(pi + pi*c^2*x^2)*x^4/(pi*c^2) - 4*sqrt(pi + pi*c^2*x^2)*x^2/(pi*c^4) + 8*sqrt(pi + pi*c^2*x^2)/(p
i*c^6))*b*arcsinh(c*x) + 1/15*(3*sqrt(pi + pi*c^2*x^2)*x^4/(pi*c^2) - 4*sqrt(pi + pi*c^2*x^2)*x^2/(pi*c^4) + 8
*sqrt(pi + pi*c^2*x^2)/(pi*c^6))*a - 1/225*(9*c^4*x^5 - 20*c^2*x^3 + 120*x)*b/(sqrt(pi)*c^5)

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Fricas [A]
time = 0.37, size = 161, normalized size = 1.08 \begin {gather*} \frac {15 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (3 \, b c^{6} x^{6} - b c^{4} x^{4} + 4 \, b c^{2} x^{2} + 8 \, b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + \sqrt {\pi + \pi c^{2} x^{2}} {\left (45 \, a c^{6} x^{6} - 15 \, a c^{4} x^{4} + 60 \, a c^{2} x^{2} - {\left (9 \, b c^{5} x^{5} - 20 \, b c^{3} x^{3} + 120 \, b c x\right )} \sqrt {c^{2} x^{2} + 1} + 120 \, a\right )}}{225 \, {\left (\pi c^{8} x^{2} + \pi c^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="fricas")

[Out]

1/225*(15*sqrt(pi + pi*c^2*x^2)*(3*b*c^6*x^6 - b*c^4*x^4 + 4*b*c^2*x^2 + 8*b)*log(c*x + sqrt(c^2*x^2 + 1)) + s
qrt(pi + pi*c^2*x^2)*(45*a*c^6*x^6 - 15*a*c^4*x^4 + 60*a*c^2*x^2 - (9*b*c^5*x^5 - 20*b*c^3*x^3 + 120*b*c*x)*sq
rt(c^2*x^2 + 1) + 120*a))/(pi*c^8*x^2 + pi*c^6)

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Sympy [A]
time = 5.00, size = 182, normalized size = 1.22 \begin {gather*} \frac {a \left (\begin {cases} \frac {x^{4} \sqrt {c^{2} x^{2} + 1}}{5 c^{2}} - \frac {4 x^{2} \sqrt {c^{2} x^{2} + 1}}{15 c^{4}} + \frac {8 \sqrt {c^{2} x^{2} + 1}}{15 c^{6}} & \text {for}\: c \neq 0 \\\frac {x^{6}}{6} & \text {otherwise} \end {cases}\right )}{\sqrt {\pi }} + \frac {b \left (\begin {cases} - \frac {x^{5}}{25 c} + \frac {x^{4} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{5 c^{2}} + \frac {4 x^{3}}{45 c^{3}} - \frac {4 x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{15 c^{4}} - \frac {8 x}{15 c^{5}} + \frac {8 \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{15 c^{6}} & \text {for}\: c \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{\sqrt {\pi }} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(1/2),x)

[Out]

a*Piecewise((x**4*sqrt(c**2*x**2 + 1)/(5*c**2) - 4*x**2*sqrt(c**2*x**2 + 1)/(15*c**4) + 8*sqrt(c**2*x**2 + 1)/
(15*c**6), Ne(c, 0)), (x**6/6, True))/sqrt(pi) + b*Piecewise((-x**5/(25*c) + x**4*sqrt(c**2*x**2 + 1)*asinh(c*
x)/(5*c**2) + 4*x**3/(45*c**3) - 4*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/(15*c**4) - 8*x/(15*c**5) + 8*sqrt(c**2
*x**2 + 1)*asinh(c*x)/(15*c**6), Ne(c, 0)), (0, True))/sqrt(pi)

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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(x^5*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),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 {x^5\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{\sqrt {\Pi \,c^2\,x^2+\Pi }} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(a + b*asinh(c*x)))/(Pi + Pi*c^2*x^2)^(1/2),x)

[Out]

int((x^5*(a + b*asinh(c*x)))/(Pi + Pi*c^2*x^2)^(1/2), x)

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