3.82 \(\int \frac {x^3 (a+b \sinh ^{-1}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx\)

Optimal. Leaf size=98 \[ \frac {x^2 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^2}-\frac {2 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^4}+\frac {2 b x}{3 \sqrt {\pi } c^3}-\frac {b x^3}{9 \sqrt {\pi } c} \]

[Out]

2/3*b*x/c^3/Pi^(1/2)-1/9*b*x^3/c/Pi^(1/2)-2/3*(a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2)/c^4/Pi+1/3*x^2*(a+b*arc
sinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2)/c^2/Pi

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Rubi [A]  time = 0.16, antiderivative size = 142, normalized size of antiderivative = 1.45, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5758, 5717, 8, 30} \[ \frac {x^2 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^2}-\frac {2 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^4}-\frac {b x^3 \sqrt {c^2 x^2+1}}{9 c \sqrt {\pi c^2 x^2+\pi }}+\frac {2 b x \sqrt {c^2 x^2+1}}{3 c^3 \sqrt {\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcSinh[c*x]))/Sqrt[Pi + c^2*Pi*x^2],x]

[Out]

(2*b*x*Sqrt[1 + c^2*x^2])/(3*c^3*Sqrt[Pi + c^2*Pi*x^2]) - (b*x^3*Sqrt[1 + c^2*x^2])/(9*c*Sqrt[Pi + c^2*Pi*x^2]
) - (2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(3*c^4*Pi) + (x^2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]
))/(3*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 5717

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[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 5758

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(e*m), x] + (-Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)
^(m - 2)*(a + b*ArcSinh[c*x])^n)/Sqrt[d + e*x^2], x], x] - Dist[(b*f*n*Sqrt[1 + c^2*x^2])/(c*m*Sqrt[d + e*x^2]
), Int[(f*x)^(m - 1)*(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[m, 1] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {\pi +c^2 \pi x^2}} \, dx &=\frac {x^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi }-\frac {2 \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {\pi +c^2 \pi x^2}} \, dx}{3 c^2}-\frac {\left (b \sqrt {1+c^2 x^2}\right ) \int x^2 \, dx}{3 c \sqrt {\pi +c^2 \pi x^2}}\\ &=-\frac {b x^3 \sqrt {1+c^2 x^2}}{9 c \sqrt {\pi +c^2 \pi x^2}}-\frac {2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 \pi }+\frac {x^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi }+\frac {\left (2 b \sqrt {1+c^2 x^2}\right ) \int 1 \, dx}{3 c^3 \sqrt {\pi +c^2 \pi x^2}}\\ &=\frac {2 b x \sqrt {1+c^2 x^2}}{3 c^3 \sqrt {\pi +c^2 \pi x^2}}-\frac {b x^3 \sqrt {1+c^2 x^2}}{9 c \sqrt {\pi +c^2 \pi x^2}}-\frac {2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 \pi }+\frac {x^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi }\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 82, normalized size = 0.84 \[ \frac {3 a \sqrt {c^2 x^2+1} \left (c^2 x^2-2\right )+b \left (6 c x-c^3 x^3\right )+3 b \sqrt {c^2 x^2+1} \left (c^2 x^2-2\right ) \sinh ^{-1}(c x)}{9 \sqrt {\pi } c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcSinh[c*x]))/Sqrt[Pi + c^2*Pi*x^2],x]

[Out]

(3*a*(-2 + c^2*x^2)*Sqrt[1 + c^2*x^2] + b*(6*c*x - c^3*x^3) + 3*b*(-2 + c^2*x^2)*Sqrt[1 + c^2*x^2]*ArcSinh[c*x
])/(9*c^4*Sqrt[Pi])

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fricas [A]  time = 0.66, size = 132, normalized size = 1.35 \[ \frac {3 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (b c^{4} x^{4} - b c^{2} x^{2} - 2 \, b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + \sqrt {\pi + \pi c^{2} x^{2}} {\left (3 \, a c^{4} x^{4} - 3 \, a c^{2} x^{2} - {\left (b c^{3} x^{3} - 6 \, b c x\right )} \sqrt {c^{2} x^{2} + 1} - 6 \, a\right )}}{9 \, {\left (\pi c^{6} x^{2} + \pi c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="fricas")

[Out]

1/9*(3*sqrt(pi + pi*c^2*x^2)*(b*c^4*x^4 - b*c^2*x^2 - 2*b)*log(c*x + sqrt(c^2*x^2 + 1)) + sqrt(pi + pi*c^2*x^2
)*(3*a*c^4*x^4 - 3*a*c^2*x^2 - (b*c^3*x^3 - 6*b*c*x)*sqrt(c^2*x^2 + 1) - 6*a))/(pi*c^6*x^2 + pi*c^4)

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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(x^3*(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,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 = 0.11, size = 133, normalized size = 1.36 \[ a \left (\frac {x^{2} \sqrt {\pi \,c^{2} x^{2}+\pi }}{3 \pi \,c^{2}}-\frac {2 \sqrt {\pi \,c^{2} x^{2}+\pi }}{3 \pi \,c^{4}}\right )+\frac {b \left (3 \arcsinh \left (c x \right ) c^{4} x^{4}-3 \arcsinh \left (c x \right ) c^{2} x^{2}-c^{3} x^{3} \sqrt {c^{2} x^{2}+1}-6 \arcsinh \left (c x \right )+6 c x \sqrt {c^{2} x^{2}+1}\right )}{9 c^{4} \sqrt {\pi }\, \sqrt {c^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(1/2),x)

[Out]

a*(1/3*x^2/Pi/c^2*(Pi*c^2*x^2+Pi)^(1/2)-2/3/Pi/c^4*(Pi*c^2*x^2+Pi)^(1/2))+1/9*b/c^4/Pi^(1/2)/(c^2*x^2+1)^(1/2)
*(3*arcsinh(c*x)*c^4*x^4-3*arcsinh(c*x)*c^2*x^2-c^3*x^3*(c^2*x^2+1)^(1/2)-6*arcsinh(c*x)+6*c*x*(c^2*x^2+1)^(1/
2))

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maxima [A]  time = 0.38, size = 117, normalized size = 1.19 \[ \frac {1}{3} \, b {\left (\frac {\sqrt {\pi + \pi c^{2} x^{2}} x^{2}}{\pi c^{2}} - \frac {2 \, \sqrt {\pi + \pi c^{2} x^{2}}}{\pi c^{4}}\right )} \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, a {\left (\frac {\sqrt {\pi + \pi c^{2} x^{2}} x^{2}}{\pi c^{2}} - \frac {2 \, \sqrt {\pi + \pi c^{2} x^{2}}}{\pi c^{4}}\right )} - \frac {{\left (c^{2} x^{3} - 6 \, x\right )} b}{9 \, \sqrt {\pi } c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="maxima")

[Out]

1/3*b*(sqrt(pi + pi*c^2*x^2)*x^2/(pi*c^2) - 2*sqrt(pi + pi*c^2*x^2)/(pi*c^4))*arcsinh(c*x) + 1/3*a*(sqrt(pi +
pi*c^2*x^2)*x^2/(pi*c^2) - 2*sqrt(pi + pi*c^2*x^2)/(pi*c^4)) - 1/9*(c^2*x^3 - 6*x)*b/(sqrt(pi)*c^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{\sqrt {\Pi \,c^2\,x^2+\Pi }} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*asinh(c*x)))/(Pi + Pi*c^2*x^2)^(1/2),x)

[Out]

int((x^3*(a + b*asinh(c*x)))/(Pi + Pi*c^2*x^2)^(1/2), x)

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sympy [A]  time = 3.94, size = 122, normalized size = 1.24 \[ \frac {a \left (\begin {cases} \frac {x^{2} \sqrt {c^{2} x^{2} + 1}}{3 c^{2}} - \frac {2 \sqrt {c^{2} x^{2} + 1}}{3 c^{4}} & \text {for}\: c \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases}\right )}{\sqrt {\pi }} + \frac {b \left (\begin {cases} - \frac {x^{3}}{9 c} + \frac {x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{3 c^{2}} + \frac {2 x}{3 c^{3}} - \frac {2 \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{3 c^{4}} & \text {for}\: c \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{\sqrt {\pi }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(1/2),x)

[Out]

a*Piecewise((x**2*sqrt(c**2*x**2 + 1)/(3*c**2) - 2*sqrt(c**2*x**2 + 1)/(3*c**4), Ne(c, 0)), (x**4/4, True))/sq
rt(pi) + b*Piecewise((-x**3/(9*c) + x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/(3*c**2) + 2*x/(3*c**3) - 2*sqrt(c**2*
x**2 + 1)*asinh(c*x)/(3*c**4), Ne(c, 0)), (0, True))/sqrt(pi)

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