3.158 \(\int \frac {x (a+b \text {csch}^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=144 \[ -\frac {a+b \text {csch}^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}-\frac {b c x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-c^2 x^2-1}}\right )}{3 d^{3/2} e \sqrt {-c^2 x^2}}+\frac {b c x \sqrt {-c^2 x^2-1}}{3 d \sqrt {-c^2 x^2} \left (c^2 d-e\right ) \sqrt {d+e x^2}} \]

[Out]

1/3*(-a-b*arccsch(c*x))/e/(e*x^2+d)^(3/2)-1/3*b*c*x*arctan((e*x^2+d)^(1/2)/d^(1/2)/(-c^2*x^2-1)^(1/2))/d^(3/2)
/e/(-c^2*x^2)^(1/2)+1/3*b*c*x*(-c^2*x^2-1)^(1/2)/d/(c^2*d-e)/(-c^2*x^2)^(1/2)/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 144, 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 = {6300, 446, 96, 93, 204} \[ -\frac {a+b \text {csch}^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}-\frac {b c x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-c^2 x^2-1}}\right )}{3 d^{3/2} e \sqrt {-c^2 x^2}}+\frac {b c x \sqrt {-c^2 x^2-1}}{3 d \sqrt {-c^2 x^2} \left (c^2 d-e\right ) \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*ArcCsch[c*x]))/(d + e*x^2)^(5/2),x]

[Out]

(b*c*x*Sqrt[-1 - c^2*x^2])/(3*d*(c^2*d - e)*Sqrt[-(c^2*x^2)]*Sqrt[d + e*x^2]) - (a + b*ArcCsch[c*x])/(3*e*(d +
 e*x^2)^(3/2)) - (b*c*x*ArcTan[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[-1 - c^2*x^2])])/(3*d^(3/2)*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 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
 b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
 + b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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 )^{5/2}} \, dx &=-\frac {a+b \text {csch}^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}+\frac {(b c x) \int \frac {1}{x \sqrt {-1-c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx}{3 e \sqrt {-c^2 x^2}}\\ &=-\frac {a+b \text {csch}^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}+\frac {(b c x) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {-1-c^2 x} (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 e \sqrt {-c^2 x^2}}\\ &=\frac {b c x \sqrt {-1-c^2 x^2}}{3 d \left (c^2 d-e\right ) \sqrt {-c^2 x^2} \sqrt {d+e x^2}}-\frac {a+b \text {csch}^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}+\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 )}{6 d e \sqrt {-c^2 x^2}}\\ &=\frac {b c x \sqrt {-1-c^2 x^2}}{3 d \left (c^2 d-e\right ) \sqrt {-c^2 x^2} \sqrt {d+e x^2}}-\frac {a+b \text {csch}^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}+\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 )}{3 d e \sqrt {-c^2 x^2}}\\ &=\frac {b c x \sqrt {-1-c^2 x^2}}{3 d \left (c^2 d-e\right ) \sqrt {-c^2 x^2} \sqrt {d+e x^2}}-\frac {a+b \text {csch}^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}-\frac {b c x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1-c^2 x^2}}\right )}{3 d^{3/2} e \sqrt {-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 185, normalized size = 1.28 \[ \frac {a d \left (e-c^2 d\right )+b c e x \sqrt {\frac {1}{c^2 x^2}+1} \left (d+e x^2\right )+b d \left (e-c^2 d\right ) \text {csch}^{-1}(c x)}{3 d e \left (c^2 d-e\right ) \left (d+e x^2\right )^{3/2}}+\frac {b c x \sqrt {\frac {1}{c^2 x^2}+1} \sqrt {-d-e x^2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {c^2 x^2+1}}{\sqrt {-d-e x^2}}\right )}{3 d^{3/2} e \sqrt {c^2 x^2+1} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(a + b*ArcCsch[c*x]))/(d + e*x^2)^(5/2),x]

[Out]

(a*d*(-(c^2*d) + e) + b*c*e*Sqrt[1 + 1/(c^2*x^2)]*x*(d + e*x^2) + b*d*(-(c^2*d) + e)*ArcCsch[c*x])/(3*d*(c^2*d
 - e)*e*(d + e*x^2)^(3/2)) + (b*c*Sqrt[1 + 1/(c^2*x^2)]*x*Sqrt[-d - e*x^2]*ArcTan[(Sqrt[d]*Sqrt[1 + c^2*x^2])/
Sqrt[-d - e*x^2]])/(3*d^(3/2)*e*Sqrt[1 + c^2*x^2]*Sqrt[d + e*x^2])

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fricas [B]  time = 0.66, size = 698, normalized size = 4.85 \[ \left [-\frac {4 \, {\left (b c^{2} d^{3} - b d^{2} e\right )} \sqrt {e x^{2} + d} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (b c^{2} d^{3} + {\left (b c^{2} d e^{2} - b e^{3}\right )} x^{4} - b d^{2} e + 2 \, {\left (b c^{2} d^{2} e - b d e^{2}\right )} x^{2}\right )} \sqrt {d} \log \left (\frac {{\left (c^{4} d^{2} + 6 \, c^{2} d e + e^{2}\right )} x^{4} + 8 \, {\left (c^{2} d^{2} + d e\right )} x^{2} + 4 \, {\left ({\left (c^{3} d + c e\right )} x^{3} + 2 \, c d x\right )} \sqrt {e x^{2} + d} \sqrt {d} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 8 \, d^{2}}{x^{4}}\right ) + 4 \, {\left (a c^{2} d^{3} - a d^{2} e - {\left (b c d e^{2} x^{3} + b c d^{2} e x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}\right )} \sqrt {e x^{2} + d}}{12 \, {\left (c^{2} d^{5} e - d^{4} e^{2} + {\left (c^{2} d^{3} e^{3} - d^{2} e^{4}\right )} x^{4} + 2 \, {\left (c^{2} d^{4} e^{2} - d^{3} e^{3}\right )} x^{2}\right )}}, -\frac {{\left (b c^{2} d^{3} + {\left (b c^{2} d e^{2} - b e^{3}\right )} x^{4} - b d^{2} e + 2 \, {\left (b c^{2} d^{2} e - b d e^{2}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {{\left ({\left (c^{3} d + c e\right )} x^{3} + 2 \, c d x\right )} \sqrt {e x^{2} + d} \sqrt {-d} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}}{2 \, {\left (c^{2} d e x^{4} + {\left (c^{2} d^{2} + d e\right )} x^{2} + d^{2}\right )}}\right ) + 2 \, {\left (b c^{2} d^{3} - b d^{2} e\right )} \sqrt {e x^{2} + d} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) + 2 \, {\left (a c^{2} d^{3} - a d^{2} e - {\left (b c d e^{2} x^{3} + b c d^{2} e x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}\right )} \sqrt {e x^{2} + d}}{6 \, {\left (c^{2} d^{5} e - d^{4} e^{2} + {\left (c^{2} d^{3} e^{3} - d^{2} e^{4}\right )} x^{4} + 2 \, {\left (c^{2} d^{4} e^{2} - d^{3} e^{3}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(4*(b*c^2*d^3 - b*d^2*e)*sqrt(e*x^2 + d)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) - (b*c^2*d^
3 + (b*c^2*d*e^2 - b*e^3)*x^4 - b*d^2*e + 2*(b*c^2*d^2*e - b*d*e^2)*x^2)*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) + 4*(a*c^2*d^3 - a*d^2*e - (b*c*d*e^2*x^3 + b*c*d^2*e*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2))
)*sqrt(e*x^2 + d))/(c^2*d^5*e - d^4*e^2 + (c^2*d^3*e^3 - d^2*e^4)*x^4 + 2*(c^2*d^4*e^2 - d^3*e^3)*x^2), -1/6*(
(b*c^2*d^3 + (b*c^2*d*e^2 - b*e^3)*x^4 - b*d^2*e + 2*(b*c^2*d^2*e - b*d*e^2)*x^2)*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)) + 2*(b*c^2*d^3 - b*d^2*e)*sqrt(e*x^2 + d)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + 2*(a*
c^2*d^3 - a*d^2*e - (b*c*d*e^2*x^3 + b*c*d^2*e*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))*sqrt(e*x^2 + d))/(c^2*d^5*e -
 d^4*e^2 + (c^2*d^3*e^3 - d^2*e^4)*x^4 + 2*(c^2*d^4*e^2 - d^3*e^3)*x^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcsch}\left (c x\right ) + a\right )} x}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccsch(c*x) + a)*x/(e*x^2 + d)^(5/2), x)

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maple [F]  time = 0.43, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a +b \,\mathrm {arccsch}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x)

[Out]

int(x*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {x \log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} + \frac {1}{c x}\right )}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} - \frac {a}{3 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

b*integrate(x*log(sqrt(1/(c^2*x^2) + 1) + 1/(c*x))/(e*x^2 + d)^(5/2), x) - 1/3*a/((e*x^2 + d)^(3/2)*e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\left (a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*asinh(1/(c*x))))/(d + e*x^2)^(5/2),x)

[Out]

int((x*(a + b*asinh(1/(c*x))))/(d + e*x^2)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*acsch(c*x))/(e*x**2+d)**(5/2),x)

[Out]

Timed out

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