3.1.89 \(\int \frac {a+b \sinh ^{-1}(c x)}{x^4 \sqrt {\pi +c^2 \pi x^2}} \, dx\) [89]

Optimal. Leaf size=97 \[ -\frac {b c}{6 \sqrt {\pi } x^2}-\frac {\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi x^3}+\frac {2 c^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi x}-\frac {2 b c^3 \log (x)}{3 \sqrt {\pi }} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 97, normalized size of antiderivative = 1.00, 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 = {5809, 5800, 29, 30} \begin {gather*} \frac {2 c^2 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi x}-\frac {\sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi x^3}-\frac {2 b c^3 \log (x)}{3 \sqrt {\pi }}-\frac {b c}{6 \sqrt {\pi } x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(x^4*Sqrt[Pi + c^2*Pi*x^2]),x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 30

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

Rule 5800

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(d*f*(m + 1))), x] - Dist[b*c*(n/(f*(m + 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, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]

Rule 5809

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(d*f*(m + 1))), x] + (-Dist[c^2*((m + 2*p + 3)/(f^2*
(m + 1))), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*c*(n/(f*(m + 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] && ILtQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 99, normalized size = 1.02 \begin {gather*} \frac {2 a \sqrt {1+c^2 x^2} \left (-1+2 c^2 x^2\right )+b c x \left (-1+6 c^2 x^2\right )+2 b \sqrt {1+c^2 x^2} \left (-1+2 c^2 x^2\right ) \sinh ^{-1}(c x)}{6 \sqrt {\pi } x^3}-\frac {2 b c^3 \log (x)}{3 \sqrt {\pi }} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])/(x^4*Sqrt[Pi + c^2*Pi*x^2]),x]

[Out]

(2*a*Sqrt[1 + c^2*x^2]*(-1 + 2*c^2*x^2) + b*c*x*(-1 + 6*c^2*x^2) + 2*b*Sqrt[1 + c^2*x^2]*(-1 + 2*c^2*x^2)*ArcS
inh[c*x])/(6*Sqrt[Pi]*x^3) - (2*b*c^3*Log[x])/(3*Sqrt[Pi])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(372\) vs. \(2(81)=162\).
time = 3.88, size = 373, normalized size = 3.85

method	result	size
default	\(a \left (-\frac {\sqrt {\pi \,c^{2} x^{2}+\pi }}{3 \pi \,x^{3}}+\frac {2 c^{2} \sqrt {\pi \,c^{2} x^{2}+\pi }}{3 \pi x}\right )+\frac {4 b \,c^{3} \arcsinh \left (c x \right )}{3 \sqrt {\pi }}-\frac {2 b \,x^{4} c^{7}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}+\frac {2 b \,x^{2} \left (c^{2} x^{2}+1\right ) c^{5}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}-\frac {2 b \,x^{2} \arcsinh \left (c x \right ) c^{5}}{\sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}+\frac {2 b x \sqrt {c^{2} x^{2}+1}\, \arcsinh \left (c x \right ) c^{4}}{\sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}-\frac {2 b \left (c^{2} x^{2}+1\right ) c^{3}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}+\frac {2 b \arcsinh \left (c x \right ) c^{3}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}-\frac {5 b \sqrt {c^{2} x^{2}+1}\, \arcsinh \left (c x \right ) c^{2}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right ) x}+\frac {b \left (c^{2} x^{2}+1\right ) c}{6 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right ) x^{2}}+\frac {b \sqrt {c^{2} x^{2}+1}\, \arcsinh \left (c x \right )}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right ) x^{3}}-\frac {2 b \,c^{3} \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right )}{3 \sqrt {\pi }}\)	\(373\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/x^4/(Pi*c^2*x^2+Pi)^(1/2),x,method=_RETURNVERBOSE)

[Out]

a*(-1/3/Pi/x^3*(Pi*c^2*x^2+Pi)^(1/2)+2/3/Pi*c^2/x*(Pi*c^2*x^2+Pi)^(1/2))+4/3*b*c^3/Pi^(1/2)*arcsinh(c*x)-2/3*b
/Pi^(1/2)/(3*c^2*x^2-1)*x^4*c^7+2/3*b/Pi^(1/2)/(3*c^2*x^2-1)*x^2*(c^2*x^2+1)*c^5-2*b/Pi^(1/2)/(3*c^2*x^2-1)*x^
2*arcsinh(c*x)*c^5+2*b/Pi^(1/2)/(3*c^2*x^2-1)*x*(c^2*x^2+1)^(1/2)*arcsinh(c*x)*c^4-2/3*b/Pi^(1/2)/(3*c^2*x^2-1
)*(c^2*x^2+1)*c^3+2/3*b/Pi^(1/2)/(3*c^2*x^2-1)*arcsinh(c*x)*c^3-5/3*b/Pi^(1/2)/(3*c^2*x^2-1)/x*(c^2*x^2+1)^(1/
2)*arcsinh(c*x)*c^2+1/6*b/Pi^(1/2)/(3*c^2*x^2-1)/x^2*(c^2*x^2+1)*c+1/3*b/Pi^(1/2)/(3*c^2*x^2-1)/x^3*(c^2*x^2+1
)^(1/2)*arcsinh(c*x)-2/3*b*c^3/Pi^(1/2)*ln((c*x+(c^2*x^2+1)^(1/2))^2-1)

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Maxima [A]
time = 0.30, size = 121, normalized size = 1.25 \begin {gather*} -\frac {1}{6} \, {\left (\frac {4 \, c^{2} \log \left (x\right )}{\sqrt {\pi }} + \frac {1}{\sqrt {\pi } x^{2}}\right )} b c + \frac {1}{3} \, b {\left (\frac {2 \, \sqrt {\pi + \pi c^{2} x^{2}} c^{2}}{\pi x} - \frac {\sqrt {\pi + \pi c^{2} x^{2}}}{\pi x^{3}}\right )} \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, a {\left (\frac {2 \, \sqrt {\pi + \pi c^{2} x^{2}} c^{2}}{\pi x} - \frac {\sqrt {\pi + \pi c^{2} x^{2}}}{\pi x^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(pi*c^2*x^2+pi)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (81) = 162\).
time = 0.40, size = 222, normalized size = 2.29 \begin {gather*} \frac {2 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (2 \, b c^{4} x^{4} + b c^{2} x^{2} - b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, \sqrt {\pi } {\left (b c^{5} x^{5} + b c^{3} x^{3}\right )} \log \left (\frac {\pi + \pi c^{2} x^{6} + \pi c^{2} x^{2} + \pi x^{4} - \sqrt {\pi } \sqrt {\pi + \pi c^{2} x^{2}} \sqrt {c^{2} x^{2} + 1} {\left (x^{4} - 1\right )}}{c^{2} x^{4} + x^{2}}\right ) + \sqrt {\pi + \pi c^{2} x^{2}} {\left (4 \, a c^{4} x^{4} + 2 \, a c^{2} x^{2} + {\left (b c x^{3} - b c x\right )} \sqrt {c^{2} x^{2} + 1} - 2 \, a\right )}}{6 \, {\left (\pi c^{2} x^{5} + \pi x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(pi*c^2*x^2+pi)^(1/2),x, algorithm="fricas")

[Out]

1/6*(2*sqrt(pi + pi*c^2*x^2)*(2*b*c^4*x^4 + b*c^2*x^2 - b)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*sqrt(pi)*(b*c^5*x^
5 + b*c^3*x^3)*log((pi + pi*c^2*x^6 + pi*c^2*x^2 + pi*x^4 - sqrt(pi)*sqrt(pi + pi*c^2*x^2)*sqrt(c^2*x^2 + 1)*(
x^4 - 1))/(c^2*x^4 + x^2)) + sqrt(pi + pi*c^2*x^2)*(4*a*c^4*x^4 + 2*a*c^2*x^2 + (b*c*x^3 - b*c*x)*sqrt(c^2*x^2
+ 1) - 2*a))/(pi*c^2*x^5 + pi*x^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/x**4/(pi*c**2*x**2+pi)**(1/2),x)

[Out]

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

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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^4/(pi*c^2*x^2+pi)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/(sqrt(pi + pi*c^2*x^2)*x^4), 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^4\,\sqrt {\Pi \,c^2\,x^2+\Pi }} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))/(x^4*(Pi + Pi*c^2*x^2)^(1/2)),x)

[Out]

int((a + b*asinh(c*x))/(x^4*(Pi + Pi*c^2*x^2)^(1/2)), x)

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