3.154 \(\int \frac {a+b \text {csch}^{-1}(c x)}{(d+e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=111 \[ \frac {x \left (a+b \text {csch}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b x \sqrt {d+e x^2} F\left (\tan ^{-1}(c x)|1-\frac {e}{c^2 d}\right )}{d^2 \sqrt {-c^2 x^2} \sqrt {-c^2 x^2-1} \sqrt {\frac {d+e x^2}{d \left (c^2 x^2+1\right )}}} \]

[Out]

x*(a+b*arccsch(c*x))/d/(e*x^2+d)^(1/2)-b*x*(1/(c^2*x^2+1))^(1/2)*(c^2*x^2+1)^(1/2)*EllipticF(c*x/(c^2*x^2+1)^(
1/2),(1-e/c^2/d)^(1/2))*(e*x^2+d)^(1/2)/d^2/(-c^2*x^2)^(1/2)/(-c^2*x^2-1)^(1/2)/((e*x^2+d)/d/(c^2*x^2+1))^(1/2
)

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Rubi [A]  time = 0.08, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {191, 6292, 12, 418} \[ \frac {x \left (a+b \text {csch}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b x \sqrt {d+e x^2} F\left (\tan ^{-1}(c x)|1-\frac {e}{c^2 d}\right )}{d^2 \sqrt {-c^2 x^2} \sqrt {-c^2 x^2-1} \sqrt {\frac {d+e x^2}{d \left (c^2 x^2+1\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*ArcCsch[c*x]))/(d*Sqrt[d + e*x^2]) - (b*x*Sqrt[d + e*x^2]*EllipticF[ArcTan[c*x], 1 - e/(c^2*d)])/(d^
2*Sqrt[-(c^2*x^2)]*Sqrt[-1 - c^2*x^2]*Sqrt[(d + e*x^2)/(d*(1 + c^2*x^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 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 6292

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

Rubi steps

\begin {align*} \int \frac {a+b \text {csch}^{-1}(c x)}{\left (d+e x^2\right )^{3/2}} \, dx &=\frac {x \left (a+b \text {csch}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {(b c x) \int \frac {1}{d \sqrt {-1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{\sqrt {-c^2 x^2}}\\ &=\frac {x \left (a+b \text {csch}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {(b c x) \int \frac {1}{\sqrt {-1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{d \sqrt {-c^2 x^2}}\\ &=\frac {x \left (a+b \text {csch}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b x \sqrt {d+e x^2} F\left (\tan ^{-1}(c x)|1-\frac {e}{c^2 d}\right )}{d^2 \sqrt {-c^2 x^2} \sqrt {-1-c^2 x^2} \sqrt {\frac {d+e x^2}{d \left (1+c^2 x^2\right )}}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 113, normalized size = 1.02 \[ \frac {x \left (a+b \text {csch}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}+\frac {b c x \sqrt {\frac {1}{c^2 x^2}+1} \sqrt {\frac {e x^2}{d}+1} F\left (\sin ^{-1}\left (\sqrt {-c^2} x\right )|\frac {e}{c^2 d}\right )}{\sqrt {-c^2} d \sqrt {c^2 x^2+1} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*ArcCsch[c*x]))/(d*Sqrt[d + e*x^2]) + (b*c*Sqrt[1 + 1/(c^2*x^2)]*x*Sqrt[1 + (e*x^2)/d]*EllipticF[ArcS
in[Sqrt[-c^2]*x], e/(c^2*d)])/(Sqrt[-c^2]*d*Sqrt[1 + c^2*x^2]*Sqrt[d + e*x^2])

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x^{2} + d} {\left (b \operatorname {arcsch}\left (c x\right ) + a\right )}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.46, size = 0, normalized size = 0.00 \[ \int \frac {a +b \,\mathrm {arccsch}\left (c x \right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(log(sqrt(1/(c^2*x^2) + 1) + 1/(c*x))/(e*x^2 + d)^(3/2), x) + a*x/(sqrt(e*x^2 + d)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(1/(c*x)))/(d + e*x^2)^(3/2),x)

[Out]

int((a + b*asinh(1/(c*x)))/(d + e*x^2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acsch(c*x))/(d + e*x**2)**(3/2), x)

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