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

Optimal. Leaf size=75 \[ -\frac {a+b \sinh ^{-1}(c x)}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {b x}{6 \pi ^{5/2} c \left (c^2 x^2+1\right )}+\frac {b \tan ^{-1}(c x)}{6 \pi ^{5/2} c^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 114, normalized size of antiderivative = 1.52, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5717, 199, 203} \[ -\frac {a+b \sinh ^{-1}(c x)}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {b x}{6 \pi ^2 c \sqrt {c^2 x^2+1} \sqrt {\pi c^2 x^2+\pi }}+\frac {b \sqrt {c^2 x^2+1} \tan ^{-1}(c x)}{6 \pi ^2 c^2 \sqrt {\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 72, normalized size = 0.96 \[ \frac {-2 a+b c x \sqrt {c^2 x^2+1}+b \left (c^2 x^2+1\right )^{3/2} \tan ^{-1}(c x)-2 b \sinh ^{-1}(c x)}{6 \pi ^{5/2} c^2 \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a + b*c*x*Sqrt[1 + c^2*x^2] - 2*b*ArcSinh[c*x] + b*(1 + c^2*x^2)^(3/2)*ArcTan[c*x])/(6*c^2*Pi^(5/2)*(1 + c
^2*x^2)^(3/2))

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fricas [B]  time = 0.63, size = 165, normalized size = 2.20 \[ -\frac {\sqrt {\pi } {\left (b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b\right )} \arctan \left (-\frac {2 \, \sqrt {\pi } \sqrt {\pi + \pi c^{2} x^{2}} \sqrt {c^{2} x^{2} + 1} c x}{\pi - \pi c^{4} x^{4}}\right ) + 4 \, \sqrt {\pi + \pi c^{2} x^{2}} b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - 2 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (\sqrt {c^{2} x^{2} + 1} b c x - 2 \, a\right )}}{12 \, {\left (\pi ^{3} c^{6} x^{4} + 2 \, \pi ^{3} c^{4} x^{2} + \pi ^{3} c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(sqrt(pi)*(b*c^4*x^4 + 2*b*c^2*x^2 + b)*arctan(-2*sqrt(pi)*sqrt(pi + pi*c^2*x^2)*sqrt(c^2*x^2 + 1)*c*x/(
pi - pi*c^4*x^4)) + 4*sqrt(pi + pi*c^2*x^2)*b*log(c*x + sqrt(c^2*x^2 + 1)) - 2*sqrt(pi + pi*c^2*x^2)*(sqrt(c^2
*x^2 + 1)*b*c*x - 2*a))/(pi^3*c^6*x^4 + 2*pi^3*c^4*x^2 + pi^3*c^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsinh(c*x) + a)*x/(pi + pi*c^2*x^2)^(5/2), x)

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maple [C]  time = 0.11, size = 124, normalized size = 1.65 \[ -\frac {a}{3 \pi \,c^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}+\frac {b x}{6 c \,\pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )}-\frac {b \arcsinh \left (c x \right )}{3 \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c^{2}}+\frac {i b \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{6 c^{2} \pi ^{\frac {5}{2}}}-\frac {i b \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{6 c^{2} \pi ^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a/Pi/c^2/(Pi*c^2*x^2+Pi)^(3/2)+1/6*b*x/c/Pi^(5/2)/(c^2*x^2+1)-1/3*b/Pi^(5/2)/(c^2*x^2+1)^(3/2)/c^2*arcsin
h(c*x)+1/6*I*b/c^2/Pi^(5/2)*ln(c*x+(c^2*x^2+1)^(1/2)+I)-1/6*I*b/c^2/Pi^(5/2)*ln(c*x+(c^2*x^2+1)^(1/2)-I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {x \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}}}\,{d x} - \frac {a}{3 \, \pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(x*log(c*x + sqrt(c^2*x^2 + 1))/(pi + pi*c^2*x^2)^(5/2), x) - 1/3*a/(pi*(pi + pi*c^2*x^2)^(3/2)*c^2
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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