3.2.54 \(\int \frac {a+b \text {csch}^{-1}(c x)}{(d+e x^2)^{3/2}} \, dx\) [154]

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

[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 = 0.05, 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 = {197, 6427, 12, 429} \begin {gather*} \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 (\text {ArcTan}(c x)\left |1-\frac {e}{c^2 d}\right .\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 )}}} \end {gather*}

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 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 429

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

Rule 6427

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 = 0.67, size = 113, normalized size = 1.02 \begin {gather*} \frac {x \left (a+b \text {csch}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}+\frac {b c \sqrt {1+\frac {1}{c^2 x^2}} x \sqrt {1+\frac {e x^2}{d}} F\left (\text {ArcSin}\left (\sqrt {-c^2} x\right )|\frac {e}{c^2 d}\right )}{\sqrt {-c^2} d \sqrt {1+c^2 x^2} \sqrt {d+e x^2}} \end {gather*}

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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Maple [F]
time = 0.14, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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))/(x^2*e + d)^(3/2), x) + a*x/(sqrt(x^2*e + d)*d)

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Fricas [A]
time = 0.11, size = 165, normalized size = 1.49 \begin {gather*} \frac {\sqrt {x^{2} \cosh \left (1\right ) + x^{2} \sinh \left (1\right ) + d} b c^{2} d x \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) + \sqrt {x^{2} \cosh \left (1\right ) + x^{2} \sinh \left (1\right ) + d} a c^{2} d x - {\left (b x^{2} \cosh \left (1\right ) + b x^{2} \sinh \left (1\right ) + b d\right )} \sqrt {-c^{2}} \sqrt {d} {\rm ellipticF}\left (\sqrt {-c^{2}} x, \frac {\cosh \left (1\right ) + \sinh \left (1\right )}{c^{2} d}\right )}{c^{2} d^{2} x^{2} \cosh \left (1\right ) + c^{2} d^{2} x^{2} \sinh \left (1\right ) + c^{2} d^{3}} \end {gather*}

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]

(sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*b*c^2*d*x*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + sqrt(x^2*c
osh(1) + x^2*sinh(1) + d)*a*c^2*d*x - (b*x^2*cosh(1) + b*x^2*sinh(1) + b*d)*sqrt(-c^2)*sqrt(d)*ellipticF(sqrt(
-c^2)*x, (cosh(1) + sinh(1))/(c^2*d)))/(c^2*d^2*x^2*cosh(1) + c^2*d^2*x^2*sinh(1) + c^2*d^3)

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

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

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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