3.2.61 \(\int \frac {x (a+b \text {sech}^{-1}(c x))}{(d+e x^2)^{3/2}} \, dx\) [161]

Optimal. Leaf size=87 \[ -\frac {a+b \text {sech}^{-1}(c x)}{e \sqrt {d+e x^2}}+\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{\sqrt {d} e} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {6434, 531, 457, 95, 213} \begin {gather*} \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{\sqrt {d} e}-\frac {a+b \text {sech}^{-1}(c x)}{e \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSech[c*x])/(e*Sqrt[d + e*x^2])) + (b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[d + e*x^2]/(S
qrt[d]*Sqrt[1 - c^2*x^2])])/(Sqrt[d]*e)

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 531

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^
(p_.), x_Symbol] :> Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n)^q, x] /; FreeQ[{a1, b1, a2, b2, c, d, n, p, q}, x]
&& EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rule 6434

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^(p +
1)*((a + b*ArcSech[c*x])/(2*e*(p + 1))), x] + Dist[b*(Sqrt[1 + c*x]/(2*e*(p + 1)))*Sqrt[1/(1 + c*x)], Int[(d +
e*x^2)^(p + 1)/(x*Sqrt[1 - c*x]*Sqrt[1 + c*x]), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]

Rubi steps

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

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Mathematica [A]
time = 20.41, size = 135, normalized size = 1.55 \begin {gather*} -\frac {a+b \text {sech}^{-1}(c x)}{e \sqrt {d+e x^2}}-\frac {b \sqrt {\frac {1-c x}{1+c x}} \sqrt {1-c^2 x^2} \sqrt {-d-e x^2} \text {ArcTan}\left (\frac {\sqrt {d} \sqrt {1-c^2 x^2}}{\sqrt {-d-e x^2}}\right )}{\sqrt {d} e (-1+c x) \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSech[c*x])/(e*Sqrt[d + e*x^2])) - (b*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*Sqrt[-d - e*x^2]*
ArcTan[(Sqrt[d]*Sqrt[1 - c^2*x^2])/Sqrt[-d - e*x^2]])/(Sqrt[d]*e*(-1 + c*x)*Sqrt[d + e*x^2])

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Maple [F]
time = 0.50, size = 0, normalized size = 0.00 \[\int \frac {x \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (57) = 114\).
time = 0.57, size = 575, normalized size = 6.61 \begin {gather*} \left [-\frac {4 \, \sqrt {x^{2} \cosh \left (1\right ) + x^{2} \sinh \left (1\right ) + d} b d \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) + 4 \, \sqrt {x^{2} \cosh \left (1\right ) + x^{2} \sinh \left (1\right ) + d} a d - {\left (b x^{2} \cosh \left (1\right ) + b x^{2} \sinh \left (1\right ) + b d\right )} \sqrt {d} \log \left (\frac {c^{4} d^{2} x^{4} - 8 \, c^{2} d^{2} x^{2} + x^{4} \cosh \left (1\right )^{2} + x^{4} \sinh \left (1\right )^{2} - 4 \, {\left (c^{3} d x^{3} - c x^{3} \cosh \left (1\right ) - c x^{3} \sinh \left (1\right ) - 2 \, c d x\right )} \sqrt {x^{2} \cosh \left (1\right ) + x^{2} \sinh \left (1\right ) + d} \sqrt {d} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 8 \, d^{2} - 2 \, {\left (3 \, c^{2} d x^{4} - 4 \, d x^{2}\right )} \cosh \left (1\right ) - 2 \, {\left (3 \, c^{2} d x^{4} - x^{4} \cosh \left (1\right ) - 4 \, d x^{2}\right )} \sinh \left (1\right )}{x^{4}}\right )}{4 \, {\left (d x^{2} \cosh \left (1\right )^{2} + d x^{2} \sinh \left (1\right )^{2} + d^{2} \cosh \left (1\right ) + {\left (2 \, d x^{2} \cosh \left (1\right ) + d^{2}\right )} \sinh \left (1\right )\right )}}, -\frac {2 \, \sqrt {x^{2} \cosh \left (1\right ) + x^{2} \sinh \left (1\right ) + d} b d \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) + 2 \, \sqrt {x^{2} \cosh \left (1\right ) + x^{2} \sinh \left (1\right ) + d} a d - {\left (b x^{2} \cosh \left (1\right ) + b x^{2} \sinh \left (1\right ) + b d\right )} \sqrt {-d} \arctan \left (-\frac {{\left (c^{3} d x^{3} - c x^{3} \cosh \left (1\right ) - c x^{3} \sinh \left (1\right ) - 2 \, c d x\right )} \sqrt {x^{2} \cosh \left (1\right ) + x^{2} \sinh \left (1\right ) + d} \sqrt {-d} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{2 \, {\left (c^{2} d^{2} x^{2} - d^{2} + {\left (c^{2} d x^{4} - d x^{2}\right )} \cosh \left (1\right ) + {\left (c^{2} d x^{4} - d x^{2}\right )} \sinh \left (1\right )\right )}}\right )}{2 \, {\left (d x^{2} \cosh \left (1\right )^{2} + d x^{2} \sinh \left (1\right )^{2} + d^{2} \cosh \left (1\right ) + {\left (2 \, d x^{2} \cosh \left (1\right ) + d^{2}\right )} \sinh \left (1\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(4*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*b*d*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) + 4*sqrt(
x^2*cosh(1) + x^2*sinh(1) + d)*a*d - (b*x^2*cosh(1) + b*x^2*sinh(1) + b*d)*sqrt(d)*log((c^4*d^2*x^4 - 8*c^2*d^
2*x^2 + x^4*cosh(1)^2 + x^4*sinh(1)^2 - 4*(c^3*d*x^3 - c*x^3*cosh(1) - c*x^3*sinh(1) - 2*c*d*x)*sqrt(x^2*cosh(
1) + x^2*sinh(1) + d)*sqrt(d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 8*d^2 - 2*(3*c^2*d*x^4 - 4*d*x^2)*cosh(1) - 2*(
3*c^2*d*x^4 - x^4*cosh(1) - 4*d*x^2)*sinh(1))/x^4))/(d*x^2*cosh(1)^2 + d*x^2*sinh(1)^2 + d^2*cosh(1) + (2*d*x^
2*cosh(1) + d^2)*sinh(1)), -1/2*(2*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*b*d*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x
^2)) + 1)/(c*x)) + 2*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*a*d - (b*x^2*cosh(1) + b*x^2*sinh(1) + b*d)*sqrt(-d)*
arctan(-1/2*(c^3*d*x^3 - c*x^3*cosh(1) - c*x^3*sinh(1) - 2*c*d*x)*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*sqrt(-d)
*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))/(c^2*d^2*x^2 - d^2 + (c^2*d*x^4 - d*x^2)*cosh(1) + (c^2*d*x^4 - d*x^2)*sinh(1)
)))/(d*x^2*cosh(1)^2 + d*x^2*sinh(1)^2 + d^2*cosh(1) + (2*d*x^2*cosh(1) + d^2)*sinh(1))]

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

[Out]

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

[Out]

integrate((b*arcsech(c*x) + a)*x/(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 {x\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\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((x*(a + b*acosh(1/(c*x))))/(d + e*x^2)^(3/2),x)

[Out]

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

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