3.2.70 \(\int \frac {x (a+b \text {sech}^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\) [170]
Optimal. Leaf size=154 \[ -\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {d+e x^2}}-\frac {a+b \text {sech}^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}+\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{3 d^{3/2} e} \]
[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 = 0.19, 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 = {6434, 531, 457,
98, 95, 213} \begin {gather*} -\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}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(x*(a + b*ArcSech[c*x]))/(d + e*x^2)^(5/2),x]
[Out]
-1/3*(b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(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 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 98
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 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 )^{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 ) \text {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 ) \text {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 ) \text {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 = 0.20, size = 204, normalized size = 1.32 \begin {gather*} \frac {-a d \left (c^2 d+e\right )-b e \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (d+e x^2\right )-b d \left (c^2 d+e\right ) \text {sech}^{-1}(c x)}{3 d e \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}-\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 )}{3 d^{3/2} 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)^(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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Maple [F]
time = 0.49, 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 \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)^(5/2),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)^(5/2), x) - 1/3*a*e^(-1)/(x^2*e +
d)^(3/2)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 709 vs.
\(2 (96) = 192\).
time = 0.58, size = 1456, normalized size = 9.45 \begin {gather*} \text {Too large to display} \end {gather*}
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*cosh(1) + b*d^2*sinh(1))*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*log((c*x*sqrt(-(c^2*
x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - (b*x^4*cosh(1)^3 + b*x^4*sinh(1)^3 + b*c^2*d^3 + (b*c^2*d*x^4 + 2*b*d*x^2)*c
osh(1)^2 + (b*c^2*d*x^4 + 3*b*x^4*cosh(1) + 2*b*d*x^2)*sinh(1)^2 + (2*b*c^2*d^2*x^2 + b*d^2)*cosh(1) + (2*b*c^
2*d^2*x^2 + 3*b*x^4*cosh(1)^2 + b*d^2 + 2*(b*c^2*d*x^4 + 2*b*d*x^2)*cosh(1))*sinh(1))*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)*sqr
t(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)*co
sh(1) - 2*(3*c^2*d*x^4 - x^4*cosh(1) - 4*d*x^2)*sinh(1))/x^4) + 4*(a*c^2*d^3 + a*d^2*cosh(1) + a*d^2*sinh(1) +
(b*c*d*x^3*cosh(1)^2 + b*c*d*x^3*sinh(1)^2 + b*c*d^2*x*cosh(1) + (2*b*c*d*x^3*cosh(1) + b*c*d^2*x)*sinh(1))*s
qrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqrt(x^2*cosh(1) + x^2*sinh(1) + d))/(d^2*x^4*cosh(1)^4 + d^2*x^4*sinh(1)^4 + c
^2*d^5*cosh(1) + (c^2*d^3*x^4 + 2*d^3*x^2)*cosh(1)^3 + (c^2*d^3*x^4 + 4*d^2*x^4*cosh(1) + 2*d^3*x^2)*sinh(1)^3
+ (2*c^2*d^4*x^2 + d^4)*cosh(1)^2 + (2*c^2*d^4*x^2 + 6*d^2*x^4*cosh(1)^2 + d^4 + 3*(c^2*d^3*x^4 + 2*d^3*x^2)*
cosh(1))*sinh(1)^2 + (4*d^2*x^4*cosh(1)^3 + c^2*d^5 + 3*(c^2*d^3*x^4 + 2*d^3*x^2)*cosh(1)^2 + 2*(2*c^2*d^4*x^2
+ d^4)*cosh(1))*sinh(1)), 1/6*((b*x^4*cosh(1)^3 + b*x^4*sinh(1)^3 + b*c^2*d^3 + (b*c^2*d*x^4 + 2*b*d*x^2)*cos
h(1)^2 + (b*c^2*d*x^4 + 3*b*x^4*cosh(1) + 2*b*d*x^2)*sinh(1)^2 + (2*b*c^2*d^2*x^2 + b*d^2)*cosh(1) + (2*b*c^2*
d^2*x^2 + 3*b*x^4*cosh(1)^2 + b*d^2 + 2*(b*c^2*d*x^4 + 2*b*d*x^2)*cosh(1))*sinh(1))*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))) - 2*(b*c^2*d^3
+ b*d^2*cosh(1) + b*d^2*sinh(1))*sqrt(x^2*cosh(1) + x^2*sinh(1) + 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*cosh(1) + a*d^2*sinh(1) + (b*c*d*x^3*cosh(1)^2 + b*c*d*x^3*sinh(1)^2 + b*c*
d^2*x*cosh(1) + (2*b*c*d*x^3*cosh(1) + b*c*d^2*x)*sinh(1))*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqrt(x^2*cosh(1) +
x^2*sinh(1) + d))/(d^2*x^4*cosh(1)^4 + d^2*x^4*sinh(1)^4 + c^2*d^5*cosh(1) + (c^2*d^3*x^4 + 2*d^3*x^2)*cosh(1)
^3 + (c^2*d^3*x^4 + 4*d^2*x^4*cosh(1) + 2*d^3*x^2)*sinh(1)^3 + (2*c^2*d^4*x^2 + d^4)*cosh(1)^2 + (2*c^2*d^4*x^
2 + 6*d^2*x^4*cosh(1)^2 + d^4 + 3*(c^2*d^3*x^4 + 2*d^3*x^2)*cosh(1))*sinh(1)^2 + (4*d^2*x^4*cosh(1)^3 + c^2*d^
5 + 3*(c^2*d^3*x^4 + 2*d^3*x^2)*cosh(1)^2 + 2*(2*c^2*d^4*x^2 + d^4)*cosh(1))*sinh(1))]
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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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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)^(5/2),x, algorithm="giac")
[Out]
integrate((b*arcsech(c*x) + a)*x/(e*x^2 + d)^(5/2), 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 )}^{5/2}} \,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)^(5/2),x)
[Out]
int((x*(a + b*acosh(1/(c*x))))/(d + e*x^2)^(5/2), x)
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