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

Optimal. Leaf size=143 \[ -\frac {2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {c^2 d x^2+d}}-\frac {a+b \sinh ^{-1}(c x)}{d x \sqrt {c^2 d x^2+d}}+\frac {b c \log (x) \sqrt {c^2 d x^2+d}}{d^2 \sqrt {c^2 x^2+1}}+\frac {b c \sqrt {c^2 d x^2+d} \log \left (c^2 x^2+1\right )}{2 d^2 \sqrt {c^2 x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {5747, 5687, 260, 266, 36, 29, 31} \[ -\frac {2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {c^2 d x^2+d}}-\frac {a+b \sinh ^{-1}(c x)}{d x \sqrt {c^2 d x^2+d}}+\frac {b c \sqrt {c^2 x^2+1} \log (x)}{d \sqrt {c^2 d x^2+d}}+\frac {b c \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )}{2 d \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

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

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 260

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 266

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

Rule 5687

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*Sqrt[1 + c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSinh[
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]

Rule 5747

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*d^IntPart[p]*(d + e*x^
2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 163, normalized size = 1.14 \[ -\frac {\sqrt {c^2 d x^2+d} \left (4 a c^2 x^2 \sqrt {c^2 x^2+1}+2 a \sqrt {c^2 x^2+1}+b c x \left (c^2 x^2+1\right ) \log \left (\frac {1}{c^2 x^2}+1\right )-2 b c x \log \left (c^2 x^2+1\right )+2 b \sqrt {c^2 x^2+1} \left (2 c^2 x^2+1\right ) \sinh ^{-1}(c x)-2 b c^3 x^3 \log \left (c^2 x^2+1\right )\right )}{2 d^2 x \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[d + c^2*d*x^2]*(2*a*Sqrt[1 + c^2*x^2] + 4*a*c^2*x^2*Sqrt[1 + c^2*x^2] + 2*b*Sqrt[1 + c^2*x^2]*(1 +
2*c^2*x^2)*ArcSinh[c*x] + b*c*x*(1 + c^2*x^2)*Log[1 + 1/(c^2*x^2)] - 2*b*c*x*Log[1 + c^2*x^2] - 2*b*c^3*x^3*Lo
g[1 + c^2*x^2]))/(d^2*x*(1 + c^2*x^2)^(3/2))

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c^{2} d x^{2} + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{c^{4} d^{2} x^{6} + 2 \, c^{2} d^{2} x^{4} + d^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.17, size = 205, normalized size = 1.43 \[ -\frac {a}{d x \sqrt {c^{2} d \,x^{2}+d}}-\frac {2 a \,c^{2} x}{d \sqrt {c^{2} d \,x^{2}+d}}-\frac {2 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) c}{\sqrt {c^{2} x^{2}+1}\, d^{2}}-\frac {2 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x \,c^{2}}{\left (c^{2} x^{2}+1\right ) d^{2}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{\left (c^{2} x^{2}+1\right ) d^{2} x}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) c}{\sqrt {c^{2} x^{2}+1}\, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/x^2/(c^2*d*x^2+d)^(3/2),x)

[Out]

-a/d/x/(c^2*d*x^2+d)^(1/2)-2*a*c^2/d*x/(c^2*d*x^2+d)^(1/2)-2*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^2*arc
sinh(c*x)*c-2*b*(d*(c^2*x^2+1))^(1/2)*arcsinh(c*x)/(c^2*x^2+1)/d^2*x*c^2-b*(d*(c^2*x^2+1))^(1/2)*arcsinh(c*x)/
(c^2*x^2+1)/d^2/x+b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^2*ln((c*x+(c^2*x^2+1)^(1/2))^4-1)*c

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maxima [A]  time = 0.41, size = 119, normalized size = 0.83 \[ \frac {1}{2} \, b c {\left (\frac {\log \left (c^{2} x^{2} + 1\right )}{d^{\frac {3}{2}}} + \frac {2 \, \log \relax (x)}{d^{\frac {3}{2}}}\right )} - {\left (\frac {2 \, c^{2} x}{\sqrt {c^{2} d x^{2} + d} d} + \frac {1}{\sqrt {c^{2} d x^{2} + d} d x}\right )} b \operatorname {arsinh}\left (c x\right ) - {\left (\frac {2 \, c^{2} x}{\sqrt {c^{2} d x^{2} + d} d} + \frac {1}{\sqrt {c^{2} d x^{2} + d} d x}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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