3.1.97 \(\int \frac {a+b \sinh ^{-1}(c x)}{x^2 (\pi +c^2 \pi x^2)^{3/2}} \, dx\) [97]

Optimal. Leaf size=93 \[ -\frac {a+b \sinh ^{-1}(c x)}{\pi x \sqrt {\pi +c^2 \pi x^2}}-\frac {2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{\pi \sqrt {\pi +c^2 \pi x^2}}+\frac {b c \log (x)}{\pi ^{3/2}}+\frac {b c \log \left (1+c^2 x^2\right )}{2 \pi ^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {277, 197, 5804, 12, 457, 78} \begin {gather*} -\frac {2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{\pi \sqrt {\pi c^2 x^2+\pi }}-\frac {a+b \sinh ^{-1}(c x)}{\pi x \sqrt {\pi c^2 x^2+\pi }}+\frac {b c \log \left (c^2 x^2+1\right )}{2 \pi ^{3/2}}+\frac {b c \log (x)}{\pi ^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5804

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x
^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]], Int[
SimplifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegerQ[p -
1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 69, normalized size = 0.74 \begin {gather*} -\frac {\left (1+2 c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{3/2} x \sqrt {1+c^2 x^2}}+\frac {b \left (c \log (x)+\frac {1}{2} c \log \left (1+c^2 x^2\right )\right )}{\pi ^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(180\) vs. \(2(85)=170\).
time = 2.49, size = 181, normalized size = 1.95

method	result	size
default	\(a \left (-\frac {1}{\pi x \sqrt {\pi \,c^{2} x^{2}+\pi }}-\frac {2 c^{2} x}{\pi \sqrt {\pi \,c^{2} x^{2}+\pi }}\right )-\frac {4 b c \arcsinh \left (c x \right )}{\pi ^{\frac {3}{2}}}+\frac {2 b \arcsinh \left (c x \right ) x^{2} c^{3}}{\pi ^{\frac {3}{2}} \left (c^{2} x^{2}+1\right )}-\frac {2 b \arcsinh \left (c x \right ) x \,c^{2}}{\pi ^{\frac {3}{2}} \sqrt {c^{2} x^{2}+1}}+\frac {2 b \arcsinh \left (c x \right ) c}{\pi ^{\frac {3}{2}} \left (c^{2} x^{2}+1\right )}-\frac {b \arcsinh \left (c x \right )}{\pi ^{\frac {3}{2}} x \sqrt {c^{2} x^{2}+1}}+\frac {b c \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right )}{\pi ^{\frac {3}{2}}}\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(-1/Pi/x/(Pi*c^2*x^2+Pi)^(1/2)-2/Pi*c^2*x/(Pi*c^2*x^2+Pi)^(1/2))-4*b*c/Pi^(3/2)*arcsinh(c*x)+2*b/Pi^(3/2)*ar
csinh(c*x)*x^2/(c^2*x^2+1)*c^3-2*b/Pi^(3/2)*arcsinh(c*x)*x/(c^2*x^2+1)^(1/2)*c^2+2*b/Pi^(3/2)*arcsinh(c*x)/(c^
2*x^2+1)*c-b/Pi^(3/2)*arcsinh(c*x)/x/(c^2*x^2+1)^(1/2)+b*c/Pi^(3/2)*ln((c*x+(c^2*x^2+1)^(1/2))^4-1)

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Maxima [A]
time = 0.30, size = 119, normalized size = 1.28 \begin {gather*} \frac {1}{2} \, b c {\left (\frac {\log \left (c^{2} x^{2} + 1\right )}{\pi ^{\frac {3}{2}}} + \frac {2 \, \log \left (x\right )}{\pi ^{\frac {3}{2}}}\right )} - {\left (\frac {2 \, c^{2} x}{\pi \sqrt {\pi + \pi c^{2} x^{2}}} + \frac {1}{\pi \sqrt {\pi + \pi c^{2} x^{2}} x}\right )} b \operatorname {arsinh}\left (c x\right ) - {\left (\frac {2 \, c^{2} x}{\pi \sqrt {\pi + \pi c^{2} x^{2}}} + \frac {1}{\pi \sqrt {\pi + \pi c^{2} x^{2}} x}\right )} a \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))/x^2/(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^6 + 2*pi^2*c^2*x^4 + pi^2*x^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^{4} \sqrt {c^{2} x^{2} + 1} + x^{2} \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{4} \sqrt {c^{2} x^{2} + 1} + x^{2} \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))/x**2/(pi*c**2*x**2+pi)**(3/2),x)

[Out]

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

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

[Out]

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

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