3.99 \(\int \frac {a+b \sinh ^{-1}(c x)}{x^4 (\pi +c^2 \pi x^2)^{3/2}} \, dx\)

Optimal. Leaf size=153 \[ \frac {4 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi x \sqrt {\pi c^2 x^2+\pi }}-\frac {a+b \sinh ^{-1}(c x)}{3 \pi x^3 \sqrt {\pi c^2 x^2+\pi }}+\frac {8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi \sqrt {\pi c^2 x^2+\pi }}-\frac {5 b c^3 \log (x)}{3 \pi ^{3/2}}-\frac {b c^3 \log \left (c^2 x^2+1\right )}{2 \pi ^{3/2}}-\frac {b c}{6 \pi ^{3/2} x^2} \]

[Out]

-1/6*b*c/Pi^(3/2)/x^2-5/3*b*c^3*ln(x)/Pi^(3/2)-1/2*b*c^3*ln(c^2*x^2+1)/Pi^(3/2)+1/3*(-a-b*arcsinh(c*x))/Pi/x^3
/(Pi*c^2*x^2+Pi)^(1/2)+4/3*c^2*(a+b*arcsinh(c*x))/Pi/x/(Pi*c^2*x^2+Pi)^(1/2)+8/3*c^4*x*(a+b*arcsinh(c*x))/Pi/(
Pi*c^2*x^2+Pi)^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 156, normalized size of antiderivative = 1.02, 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 = {271, 191, 5732, 12, 1251, 893} \[ \frac {8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{3/2} \sqrt {c^2 x^2+1}}+\frac {4 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{3/2} x \sqrt {c^2 x^2+1}}-\frac {a+b \sinh ^{-1}(c x)}{3 \pi ^{3/2} x^3 \sqrt {c^2 x^2+1}}-\frac {b c^3 \log \left (c^2 x^2+1\right )}{2 \pi ^{3/2}}-\frac {5 b c^3 \log (x)}{3 \pi ^{3/2}}-\frac {b c}{6 \pi ^{3/2} x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(x^4*(Pi + c^2*Pi*x^2)^(3/2)),x]

[Out]

-(b*c)/(6*Pi^(3/2)*x^2) - (a + b*ArcSinh[c*x])/(3*Pi^(3/2)*x^3*Sqrt[1 + c^2*x^2]) + (4*c^2*(a + b*ArcSinh[c*x]
))/(3*Pi^(3/2)*x*Sqrt[1 + c^2*x^2]) + (8*c^4*x*(a + b*ArcSinh[c*x]))/(3*Pi^(3/2)*Sqrt[1 + c^2*x^2]) - (5*b*c^3
*Log[x])/(3*Pi^(3/2)) - (b*c^3*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 191

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 271

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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 5732

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

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{x^4 \left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{3 \pi ^{3/2} x^3 \sqrt {1+c^2 x^2}}+\frac {4 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{3/2} x \sqrt {1+c^2 x^2}}+\frac {8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{3/2} \sqrt {1+c^2 x^2}}-\frac {(b c) \int \frac {-1+4 c^2 x^2+8 c^4 x^4}{3 x^3 \left (1+c^2 x^2\right )} \, dx}{\pi ^{3/2}}\\ &=-\frac {a+b \sinh ^{-1}(c x)}{3 \pi ^{3/2} x^3 \sqrt {1+c^2 x^2}}+\frac {4 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{3/2} x \sqrt {1+c^2 x^2}}+\frac {8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{3/2} \sqrt {1+c^2 x^2}}-\frac {(b c) \int \frac {-1+4 c^2 x^2+8 c^4 x^4}{x^3 \left (1+c^2 x^2\right )} \, dx}{3 \pi ^{3/2}}\\ &=-\frac {a+b \sinh ^{-1}(c x)}{3 \pi ^{3/2} x^3 \sqrt {1+c^2 x^2}}+\frac {4 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{3/2} x \sqrt {1+c^2 x^2}}+\frac {8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{3/2} \sqrt {1+c^2 x^2}}-\frac {(b c) \operatorname {Subst}\left (\int \frac {-1+4 c^2 x+8 c^4 x^2}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )}{6 \pi ^{3/2}}\\ &=-\frac {a+b \sinh ^{-1}(c x)}{3 \pi ^{3/2} x^3 \sqrt {1+c^2 x^2}}+\frac {4 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{3/2} x \sqrt {1+c^2 x^2}}+\frac {8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{3/2} \sqrt {1+c^2 x^2}}-\frac {(b c) \operatorname {Subst}\left (\int \left (-\frac {1}{x^2}+\frac {5 c^2}{x}+\frac {3 c^4}{1+c^2 x}\right ) \, dx,x,x^2\right )}{6 \pi ^{3/2}}\\ &=-\frac {b c}{6 \pi ^{3/2} x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 \pi ^{3/2} x^3 \sqrt {1+c^2 x^2}}+\frac {4 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{3/2} x \sqrt {1+c^2 x^2}}+\frac {8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{3/2} \sqrt {1+c^2 x^2}}-\frac {5 b c^3 \log (x)}{3 \pi ^{3/2}}-\frac {b c^3 \log \left (1+c^2 x^2\right )}{2 \pi ^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 127, normalized size = 0.83 \[ \frac {2 a \left (8 c^4 x^4+4 c^2 x^2-1\right )-b c x \sqrt {c^2 x^2+1}+2 b \left (8 c^4 x^4+4 c^2 x^2-1\right ) \sinh ^{-1}(c x)}{6 \pi ^{3/2} x^3 \sqrt {c^2 x^2+1}}+\frac {-5 b c^3 \log (x)-\frac {3}{2} b c^3 \log \left (c^2 x^2+1\right )}{3 \pi ^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])/(x^4*(Pi + c^2*Pi*x^2)^(3/2)),x]

[Out]

(-(b*c*x*Sqrt[1 + c^2*x^2]) + 2*a*(-1 + 4*c^2*x^2 + 8*c^4*x^4) + 2*b*(-1 + 4*c^2*x^2 + 8*c^4*x^4)*ArcSinh[c*x]
)/(6*Pi^(3/2)*x^3*Sqrt[1 + c^2*x^2]) + (-5*b*c^3*Log[x] - (3*b*c^3*Log[1 + c^2*x^2])/2)/(3*Pi^(3/2))

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\pi + \pi c^{2} x^{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{\pi ^{2} c^{4} x^{8} + 2 \, \pi ^{2} c^{2} x^{6} + \pi ^{2} x^{4}}, x\right ) \]

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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(b*arcsinh(c*x) + a)/(pi^2*c^4*x^8 + 2*pi^2*c^2*x^6 + pi^2*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x^{4}}\,{d x} \]

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)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/((pi + pi*c^2*x^2)^(3/2)*x^4), x)

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maple [B]  time = 0.32, size = 601, normalized size = 3.93 \[ -\frac {a}{3 \pi \,x^{3} \sqrt {\pi \,c^{2} x^{2}+\pi }}+\frac {4 a \,c^{2}}{3 \pi x \sqrt {\pi \,c^{2} x^{2}+\pi }}+\frac {8 a \,c^{4} x}{3 \pi \sqrt {\pi \,c^{2} x^{2}+\pi }}+\frac {16 b \,c^{3} \arcsinh \left (c x \right )}{3 \pi ^{\frac {3}{2}}}-\frac {32 b \,x^{8} c^{11}}{3 \pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right ) \left (c^{2} x^{2}+1\right )}+\frac {32 b \,x^{6} c^{9}}{3 \pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right )}-\frac {64 b \,x^{6} c^{9}}{3 \pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right ) \left (c^{2} x^{2}+1\right )}+\frac {32 b \,x^{4} c^{7}}{3 \pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right )}-\frac {64 b \,x^{4} \arcsinh \left (c x \right ) c^{7}}{3 \pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right ) \left (c^{2} x^{2}+1\right )}+\frac {64 b \,x^{3} \arcsinh \left (c x \right ) c^{6}}{3 \pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right ) \sqrt {c^{2} x^{2}+1}}-\frac {32 b \,x^{4} c^{7}}{3 \pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right ) \left (c^{2} x^{2}+1\right )}-\frac {56 b \,x^{2} \arcsinh \left (c x \right ) c^{5}}{3 \pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right ) \left (c^{2} x^{2}+1\right )}+\frac {8 b x \arcsinh \left (c x \right ) c^{4}}{\pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right ) \sqrt {c^{2} x^{2}+1}}-\frac {4 b \,c^{3}}{3 \pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right )}+\frac {8 b \arcsinh \left (c x \right ) c^{3}}{3 \pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right ) \left (c^{2} x^{2}+1\right )}-\frac {4 b \arcsinh \left (c x \right ) c^{2}}{\pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right ) x \sqrt {c^{2} x^{2}+1}}+\frac {b c}{6 \pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right ) x^{2}}+\frac {b \arcsinh \left (c x \right )}{3 \pi ^{\frac {3}{2}} \left (8 c^{2} x^{2}-1\right ) x^{3} \sqrt {c^{2} x^{2}+1}}-\frac {b \,c^{3} \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{\pi ^{\frac {3}{2}}}-\frac {5 b \,c^{3} \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right )}{3 \pi ^{\frac {3}{2}}} \]

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)^(3/2),x)

[Out]

-1/3*a/Pi/x^3/(Pi*c^2*x^2+Pi)^(1/2)+4/3*a*c^2/Pi/x/(Pi*c^2*x^2+Pi)^(1/2)+8/3*a*c^4/Pi*x/(Pi*c^2*x^2+Pi)^(1/2)+
16/3*b*c^3/Pi^(3/2)*arcsinh(c*x)-32/3*b/Pi^(3/2)/(8*c^2*x^2-1)*x^8/(c^2*x^2+1)*c^11+32/3*b/Pi^(3/2)/(8*c^2*x^2
-1)*x^6*c^9-64/3*b/Pi^(3/2)/(8*c^2*x^2-1)*x^6/(c^2*x^2+1)*c^9+32/3*b/Pi^(3/2)/(8*c^2*x^2-1)*x^4*c^7-64/3*b/Pi^
(3/2)/(8*c^2*x^2-1)*x^4/(c^2*x^2+1)*arcsinh(c*x)*c^7+64/3*b/Pi^(3/2)/(8*c^2*x^2-1)*x^3/(c^2*x^2+1)^(1/2)*arcsi
nh(c*x)*c^6-32/3*b/Pi^(3/2)/(8*c^2*x^2-1)*x^4/(c^2*x^2+1)*c^7-56/3*b/Pi^(3/2)/(8*c^2*x^2-1)*x^2/(c^2*x^2+1)*ar
csinh(c*x)*c^5+8*b/Pi^(3/2)/(8*c^2*x^2-1)*x/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*c^4-4/3*b/Pi^(3/2)/(8*c^2*x^2-1)*c^
3+8/3*b/Pi^(3/2)/(8*c^2*x^2-1)/(c^2*x^2+1)*arcsinh(c*x)*c^3-4*b/Pi^(3/2)/(8*c^2*x^2-1)/x/(c^2*x^2+1)^(1/2)*arc
sinh(c*x)*c^2+1/6*b/Pi^(3/2)/(8*c^2*x^2-1)/x^2*c+1/3*b/Pi^(3/2)/(8*c^2*x^2-1)/x^3/(c^2*x^2+1)^(1/2)*arcsinh(c*
x)-b*c^3/Pi^(3/2)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)-5/3*b*c^3/Pi^(3/2)*ln((c*x+(c^2*x^2+1)^(1/2))^2-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, {\left (\frac {8 \, c^{4} x}{\pi \sqrt {\pi + \pi c^{2} x^{2}}} + \frac {4 \, c^{2}}{\pi \sqrt {\pi + \pi c^{2} x^{2}} x} - \frac {1}{\pi \sqrt {\pi + \pi c^{2} x^{2}} x^{3}}\right )} a + b \int \frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x^{4}}\,{d x} \]

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)^(3/2),x, algorithm="maxima")

[Out]

1/3*(8*c^4*x/(pi*sqrt(pi + pi*c^2*x^2)) + 4*c^2/(pi*sqrt(pi + pi*c^2*x^2)*x) - 1/(pi*sqrt(pi + pi*c^2*x^2)*x^3
))*a + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/((pi + pi*c^2*x^2)^(3/2)*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{x^4\,{\left (\Pi \,c^2\,x^2+\Pi \right )}^{3/2}} \,d x \]

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)^(3/2)),x)

[Out]

int((a + b*asinh(c*x))/(x^4*(Pi + Pi*c^2*x^2)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{2} x^{6} \sqrt {c^{2} x^{2} + 1} + x^{4} \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{6} \sqrt {c^{2} x^{2} + 1} + x^{4} \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {3}{2}}} \]

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)**(3/2),x)

[Out]

(Integral(a/(c**2*x**6*sqrt(c**2*x**2 + 1) + x**4*sqrt(c**2*x**2 + 1)), x) + Integral(b*asinh(c*x)/(c**2*x**6*
sqrt(c**2*x**2 + 1) + x**4*sqrt(c**2*x**2 + 1)), x))/pi**(3/2)

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