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

Optimal. Leaf size=92 \[ \frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {\frac {e x^2}{d}+1} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{c d \sqrt {d+e x^2}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {191, 6291, 12, 421, 419} \[ \frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {\frac {e x^2}{d}+1} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{c d \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*ArcSech[c*x]))/(d*Sqrt[d + e*x^2]) + (b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 + (e*x^2)/d]*Ellip
ticF[ArcSin[c*x], -(e/(c^2*d))])/(c*d*Sqrt[d + e*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 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 421

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d*x^2)/c]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 6291

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^
2)^p, x]}, Dist[a + b*ArcSech[c*x], u, x] + Dist[b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)], Int[SimplifyIntegrand[u/(x
*Sqrt[1 - c*x]*Sqrt[1 + c*x]), 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 {sech}^{-1}(c x)}{\left (d+e x^2\right )^{3/2}} \, dx &=\frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{d \sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx\\ &=\frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{\sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{d}\\ &=\frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {1}{\sqrt {1-c^2 x^2} \sqrt {1+\frac {e x^2}{d}}} \, dx}{d \sqrt {d+e x^2}}\\ &=\frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}+\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1+\frac {e x^2}{d}} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{c d \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C]  time = 1.39, size = 334, normalized size = 3.63 \[ \frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{d \sqrt {d+e x^2}}+\frac {2 i b \sqrt {\frac {1-c x}{c x+1}} \left (\sqrt {e} x-i \sqrt {d}\right ) \sqrt {\frac {(c x+1) \left (c \sqrt {d}+i \sqrt {e}\right )}{(c x-1) \left (c \sqrt {d}-i \sqrt {e}\right )}} \sqrt {-\frac {c \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right )+\frac {i \sqrt {e} x}{\sqrt {d}}-1}{1-c x}} F\left (\sin ^{-1}\left (\sqrt {\frac {-x c+\frac {i \sqrt {d} c}{\sqrt {e}}+\frac {i \sqrt {e} x}{\sqrt {d}}+1}{2-2 c x}}\right )|-\frac {4 i c \sqrt {d} \sqrt {e}}{\left (c \sqrt {d}-i \sqrt {e}\right )^2}\right )}{d \left (c \sqrt {d}+i \sqrt {e}\right ) \sqrt {d+e x^2} \sqrt {\frac {\frac {i c \sqrt {d}}{\sqrt {e}}+c (-x)+\frac {i \sqrt {e} x}{\sqrt {d}}+1}{1-c x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*ArcSech[c*x]))/(d*Sqrt[d + e*x^2]) + ((2*I)*b*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[((c*Sqrt[d] + I*Sqrt[e]
)*(1 + c*x))/((c*Sqrt[d] - I*Sqrt[e])*(-1 + c*x))]*((-I)*Sqrt[d] + Sqrt[e]*x)*Sqrt[-((-1 + (I*Sqrt[e]*x)/Sqrt[
d] + c*((I*Sqrt[d])/Sqrt[e] + x))/(1 - c*x))]*EllipticF[ArcSin[Sqrt[(1 + (I*c*Sqrt[d])/Sqrt[e] - c*x + (I*Sqrt
[e]*x)/Sqrt[d])/(2 - 2*c*x)]], ((-4*I)*c*Sqrt[d]*Sqrt[e])/(c*Sqrt[d] - I*Sqrt[e])^2])/(d*(c*Sqrt[d] + I*Sqrt[e
])*Sqrt[(1 + (I*c*Sqrt[d])/Sqrt[e] - c*x + (I*Sqrt[e]*x)/Sqrt[d])/(1 - c*x)]*Sqrt[d + e*x^2])

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x^{2} + d} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(e*x^2 + d)*(b*arcsech(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 \[ \int \frac {b \operatorname {arsech}\left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.17, size = 0, normalized size = 0.00 \[ \int \frac {a +b \,\mathrm {arcsech}\left (c x \right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 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 \[ \int \frac {a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a + b*acosh(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 \[ \int \frac {a + b \operatorname {asech}{\left (c x \right )}}{\left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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