3.1.95 \(\int \frac {a+b \sinh ^{-1}(c x)}{(\pi +c^2 \pi x^2)^{3/2}} \, dx\) [95]

Optimal. Leaf size=51 \[ \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {b \log \left (1+c^2 x^2\right )}{2 c \pi ^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {5787, 266} \begin {gather*} \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\pi \sqrt {\pi c^2 x^2+\pi }}-\frac {b \log \left (c^2 x^2+1\right )}{2 \pi ^{3/2} c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 266

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

Rule 5787

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[x*((a + b*ArcSinh
[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Dist[b*c*(n/d)*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Int[x*((a + b*ArcS
inh[c*x])^(n - 1)/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 66, normalized size = 1.29 \begin {gather*} \frac {2 a c x+2 b c x \sinh ^{-1}(c x)-b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{2 c \pi ^{3/2} \sqrt {1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(131\) vs. \(2(45)=90\).
time = 1.60, size = 132, normalized size = 2.59

method	result	size
default	\(\frac {a x}{\pi \sqrt {\pi \,c^{2} x^{2}+\pi }}+\frac {2 b \arcsinh \left (c x \right )}{c \,\pi ^{\frac {3}{2}}}-\frac {b \arcsinh \left (c x \right ) c \,x^{2}}{\pi ^{\frac {3}{2}} \left (c^{2} x^{2}+1\right )}+\frac {b \arcsinh \left (c x \right ) x}{\pi ^{\frac {3}{2}} \sqrt {c^{2} x^{2}+1}}-\frac {b \arcsinh \left (c x \right )}{\pi ^{\frac {3}{2}} c \left (c^{2} x^{2}+1\right )}-\frac {b \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{c \,\pi ^{\frac {3}{2}}}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 58, normalized size = 1.14 \begin {gather*} \frac {b x \operatorname {arsinh}\left (c x\right )}{\pi \sqrt {\pi + \pi c^{2} x^{2}}} + \frac {a x}{\pi \sqrt {\pi + \pi c^{2} x^{2}}} - \frac {b \log \left (x^{2} + \frac {1}{c^{2}}\right )}{2 \, \pi ^{\frac {3}{2}} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(b*arcsinh(c*x) + a)/(pi^2*c^4*x^4 + 2*pi^2*c^2*x^2 + pi^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a}{c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsinh(c*x) + a)/(pi + pi*c^2*x^2)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{{\left (\Pi \,c^2\,x^2+\Pi \right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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