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

Optimal. Leaf size=82 \[ -\frac {a+b \text {csch}^{-1}(c x)}{e \sqrt {d+e x^2}}-\frac {b c x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-c^2 x^2-1}}\right )}{\sqrt {d} e \sqrt {-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6300, 446, 93, 204} \[ -\frac {a+b \text {csch}^{-1}(c x)}{e \sqrt {d+e x^2}}-\frac {b c x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-c^2 x^2-1}}\right )}{\sqrt {d} e \sqrt {-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 93

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 204

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

Rule 446

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 6300

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

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 122, normalized size = 1.49 \[ \frac {b c x \sqrt {\frac {1}{c^2 x^2}+1} \sqrt {-d-e x^2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {c^2 x^2+1}}{\sqrt {-d-e x^2}}\right )}{\sqrt {d} e \sqrt {c^2 x^2+1} \sqrt {d+e x^2}}-\frac {a+b \text {csch}^{-1}(c x)}{e \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.69, size = 368, normalized size = 4.49 \[ \left [-\frac {4 \, \sqrt {e x^{2} + d} b d \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) + 4 \, \sqrt {e x^{2} + d} a d - {\left (b e x^{2} + b d\right )} \sqrt {d} \log \left (\frac {{\left (c^{4} d^{2} + 6 \, c^{2} d e + e^{2}\right )} x^{4} + 8 \, {\left (c^{2} d^{2} + d e\right )} x^{2} + 4 \, {\left ({\left (c^{3} d + c e\right )} x^{3} + 2 \, c d x\right )} \sqrt {e x^{2} + d} \sqrt {d} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 8 \, d^{2}}{x^{4}}\right )}{4 \, {\left (d e^{2} x^{2} + d^{2} e\right )}}, -\frac {2 \, \sqrt {e x^{2} + d} b d \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) + 2 \, \sqrt {e x^{2} + d} a d + {\left (b e x^{2} + b d\right )} \sqrt {-d} \arctan \left (\frac {{\left ({\left (c^{3} d + c e\right )} x^{3} + 2 \, c d x\right )} \sqrt {e x^{2} + d} \sqrt {-d} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}}{2 \, {\left (c^{2} d e x^{4} + {\left (c^{2} d^{2} + d e\right )} x^{2} + d^{2}\right )}}\right )}{2 \, {\left (d e^{2} x^{2} + d^{2} e\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(4*sqrt(e*x^2 + d)*b*d*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + 4*sqrt(e*x^2 + d)*a*d - (b*e
*x^2 + b*d)*sqrt(d)*log(((c^4*d^2 + 6*c^2*d*e + e^2)*x^4 + 8*(c^2*d^2 + d*e)*x^2 + 4*((c^3*d + c*e)*x^3 + 2*c*
d*x)*sqrt(e*x^2 + d)*sqrt(d)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 8*d^2)/x^4))/(d*e^2*x^2 + d^2*e), -1/2*(2*sqrt(e*
x^2 + d)*b*d*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + 2*sqrt(e*x^2 + d)*a*d + (b*e*x^2 + b*d)*sqrt
(-d)*arctan(1/2*((c^3*d + c*e)*x^3 + 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(-d)*sqrt((c^2*x^2 + 1)/(c^2*x^2))/(c^2*d*e*
x^4 + (c^2*d^2 + d*e)*x^2 + d^2)))/(d*e^2*x^2 + d^2*e)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.44, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a +b \,\mathrm {arccsch}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*(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 \[ -{\left (c^{2} \int \frac {x}{{\left (c^{2} e x^{2} + e\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {e x^{2} + d} + {\left (c^{2} e x^{2} + e\right )} \sqrt {e x^{2} + d}}\,{d x} + \frac {\log \left (\sqrt {c^{2} x^{2} + 1} + 1\right )}{\sqrt {e x^{2} + d} e} + \int \frac {{\left (e \log \relax (c) - e\right )} c^{2} x^{3} - {\left (c^{2} d - e \log \relax (c)\right )} x + {\left (c^{2} e x^{3} + e x\right )} \log \relax (x)}{{\left (c^{2} e^{2} x^{4} + {\left (c^{2} d e + e^{2}\right )} x^{2} + d e\right )} \sqrt {e x^{2} + d}}\,{d x}\right )} b - \frac {a}{\sqrt {e x^{2} + d} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c^2*integrate(x/((c^2*e*x^2 + e)*sqrt(c^2*x^2 + 1)*sqrt(e*x^2 + d) + (c^2*e*x^2 + e)*sqrt(e*x^2 + d)), x) +
log(sqrt(c^2*x^2 + 1) + 1)/(sqrt(e*x^2 + d)*e) + integrate(((e*log(c) - e)*c^2*x^3 - (c^2*d - e*log(c))*x + (c
^2*e*x^3 + e*x)*log(x))/((c^2*e^2*x^4 + (c^2*d*e + e^2)*x^2 + d*e)*sqrt(e*x^2 + d)), x))*b - a/(sqrt(e*x^2 + d
)*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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