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

Optimal. Leaf size=154 \[ -\frac {a+b \text {sech}^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{3 d^{3/2} e}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {d+e x^2}} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6299, 517, 446, 96, 93, 207} \[ -\frac {a+b \text {sech}^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{3 d^{3/2} e}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[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 517

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 6299

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

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Mathematica [A]  time = 0.32, size = 204, normalized size = 1.32 \[ \frac {-a d \left (c^2 d+e\right )-b d \left (c^2 d+e\right ) \text {sech}^{-1}(c x)-b e \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (d+e x^2\right )}{3 d e \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}-\frac {b \sqrt {\frac {1-c x}{c x+1}} \sqrt {1-c^2 x^2} \sqrt {-d-e x^2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {1-c^2 x^2}}{\sqrt {-d-e x^2}}\right )}{3 d^{3/2} e (c x-1) \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.41, size = 692, normalized size = 4.49 \[ \left [-\frac {4 \, {\left (b c^{2} d^{3} + b d^{2} e\right )} \sqrt {e x^{2} + d} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (b c^{2} d^{3} + {\left (b c^{2} d e^{2} + b e^{3}\right )} x^{4} + b d^{2} e + 2 \, {\left (b c^{2} d^{2} e + b d e^{2}\right )} x^{2}\right )} \sqrt {d} \log \left (\frac {{\left (c^{4} d^{2} - 6 \, c^{2} d e + e^{2}\right )} x^{4} - 8 \, {\left (c^{2} d^{2} - d e\right )} x^{2} - 4 \, {\left ({\left (c^{3} d - c e\right )} x^{3} - 2 \, c d x\right )} \sqrt {e x^{2} + d} \sqrt {d} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 8 \, d^{2}}{x^{4}}\right ) + 4 \, {\left (a c^{2} d^{3} + a d^{2} e + {\left (b c d e^{2} x^{3} + b c d^{2} e x\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}\right )} \sqrt {e x^{2} + d}}{12 \, {\left (c^{2} d^{5} e + d^{4} e^{2} + {\left (c^{2} d^{3} e^{3} + d^{2} e^{4}\right )} x^{4} + 2 \, {\left (c^{2} d^{4} e^{2} + d^{3} e^{3}\right )} x^{2}\right )}}, \frac {{\left (b c^{2} d^{3} + {\left (b c^{2} d e^{2} + b e^{3}\right )} x^{4} + b d^{2} e + 2 \, {\left (b c^{2} d^{2} e + b d e^{2}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (-\frac {{\left ({\left (c^{3} d - c e\right )} x^{3} - 2 \, c d x\right )} \sqrt {e x^{2} + d} \sqrt {-d} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{2 \, {\left (c^{2} d e x^{4} + {\left (c^{2} d^{2} - d e\right )} x^{2} - d^{2}\right )}}\right ) - 2 \, {\left (b c^{2} d^{3} + b d^{2} e\right )} \sqrt {e x^{2} + d} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - 2 \, {\left (a c^{2} d^{3} + a d^{2} e + {\left (b c d e^{2} x^{3} + b c d^{2} e x\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}\right )} \sqrt {e x^{2} + d}}{6 \, {\left (c^{2} d^{5} e + d^{4} e^{2} + {\left (c^{2} d^{3} e^{3} + d^{2} e^{4}\right )} x^{4} + 2 \, {\left (c^{2} d^{4} e^{2} + d^{3} e^{3}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsech(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 \[ \int \frac {{\left (b \operatorname {arsech}\left (c x\right ) + a\right )} x}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.58, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(x*log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 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 \[ \int \frac {x\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x*(a + b*acosh(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*asech(c*x))/(e*x**2+d)**(5/2),x)

[Out]

Timed out

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