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

Optimal. Leaf size=266 \[ \frac {2 x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}+\frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {b e x \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{3 d^2 \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {2 b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {\frac {e x^2}{d}+1} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 c d^2 \sqrt {d+e x^2}}+\frac {b c \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 d^2 \left (c^2 d+e\right ) \sqrt {\frac {e x^2}{d}+1}} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {192, 191, 6291, 12, 527, 524, 426, 424, 421, 419} \[ \frac {2 x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}+\frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {b e x \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{3 d^2 \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {2 b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {\frac {e x^2}{d}+1} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 c d^2 \sqrt {d+e x^2}}+\frac {b c \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 d^2 \left (c^2 d+e\right ) \sqrt {\frac {e x^2}{d}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e*x*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(3*d^2*(c^2*d + e)*Sqrt[d + e*x^2]) + (x*(a + b*A
rcSech[c*x]))/(3*d*(d + e*x^2)^(3/2)) + (2*x*(a + b*ArcSech[c*x]))/(3*d^2*Sqrt[d + e*x^2]) + (b*c*Sqrt[(1 + c*
x)^(-1)]*Sqrt[1 + c*x]*Sqrt[d + e*x^2]*EllipticE[ArcSin[c*x], -(e/(c^2*d))])/(3*d^2*(c^2*d + e)*Sqrt[1 + (e*x^
2)/d]) + (2*b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 + (e*x^2)/d]*EllipticF[ArcSin[c*x], -(e/(c^2*d))])/(3*
c*d^2*Sqrt[d + e*x^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 421

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d*x^2)/c]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 426

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b*x^2)/a]
, Int[Sqrt[1 + (b*x^2)/a]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 6291

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^
2)^p, x]}, Dist[a + b*ArcSech[c*x], u, x] + Dist[b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)], Int[SimplifyIntegrand[u/(x
*Sqrt[1 - c*x]*Sqrt[1 + c*x]), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rubi steps

\begin {align*} \int \frac {a+b \text {sech}^{-1}(c x)}{\left (d+e x^2\right )^{5/2}} \, dx &=\frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {2 x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {3 d+2 e x^2}{3 d^2 \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx\\ &=\frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {2 x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {3 d+2 e x^2}{\sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx}{3 d^2}\\ &=\frac {b e x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 d^2 \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {2 x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-d \left (3 c^2 d+2 e\right )-c^2 d e x^2}{\sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{3 d^3 \left (c^2 d+e\right )}\\ &=\frac {b e x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 d^2 \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {2 x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}+\frac {\left (2 b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{\sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{3 d^2}+\frac {\left (b c^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\sqrt {d+e x^2}}{\sqrt {1-c^2 x^2}} \, dx}{3 d^2 \left (c^2 d+e\right )}\\ &=\frac {b e x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 d^2 \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {2 x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}+\frac {\left (b c^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {d+e x^2}\right ) \int \frac {\sqrt {1+\frac {e x^2}{d}}}{\sqrt {1-c^2 x^2}} \, dx}{3 d^2 \left (c^2 d+e\right ) \sqrt {1+\frac {e x^2}{d}}}+\frac {\left (2 b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {1}{\sqrt {1-c^2 x^2} \sqrt {1+\frac {e x^2}{d}}} \, dx}{3 d^2 \sqrt {d+e x^2}}\\ &=\frac {b e x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 d^2 \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {2 x \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}+\frac {b c \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 d^2 \left (c^2 d+e\right ) \sqrt {1+\frac {e x^2}{d}}}+\frac {2 b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1+\frac {e x^2}{d}} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 c d^2 \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C]  time = 5.50, size = 517, normalized size = 1.94 \[ \frac {a x \left (3 d+2 e x^2\right )+\frac {b \sqrt {\frac {1-c x}{c x+1}} \left (d+e x^2\right ) (e x-c d)}{c^2 d+e}-\frac {i b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (d+e x^2\right ) \sqrt {\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{(c x+1) \left (c \sqrt {d}-i \sqrt {e}\right )}} \sqrt {\frac {c \left (\sqrt {d}+i \sqrt {e} x\right )}{(c x+1) \left (c \sqrt {d}+i \sqrt {e}\right )}} \left (\left (c \sqrt {d}-i \sqrt {e}\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {\left (d c^2+e\right ) (1-c x)}{\left (\sqrt {d} c+i \sqrt {e}\right )^2 (c x+1)}}\right )|\frac {\left (\sqrt {d} c+i \sqrt {e}\right )^2}{\left (c \sqrt {d}-i \sqrt {e}\right )^2}\right )-2 \left (3 c \sqrt {d}+2 i \sqrt {e}\right ) F\left (i \sinh ^{-1}\left (\sqrt {\frac {\left (d c^2+e\right ) (1-c x)}{\left (\sqrt {d} c+i \sqrt {e}\right )^2 (c x+1)}}\right )|\frac {\left (\sqrt {d} c+i \sqrt {e}\right )^2}{\left (c \sqrt {d}-i \sqrt {e}\right )^2}\right )\right )}{c \left (c \sqrt {d}+i \sqrt {e}\right ) \sqrt {-\frac {(c x-1) \left (c \sqrt {d}-i \sqrt {e}\right )}{(c x+1) \left (c \sqrt {d}+i \sqrt {e}\right )}}}+b x \text {sech}^{-1}(c x) \left (3 d+2 e x^2\right )}{3 d^2 \left (d+e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Sqrt[(1 - c*x)/(1 + c*x)]*(-(c*d) + e*x)*(d + e*x^2))/(c^2*d + e) + a*x*(3*d + 2*e*x^2) + b*x*(3*d + 2*e*x
^2)*ArcSech[c*x] - (I*b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*Sqrt[(c*(Sqrt[d] - I*Sqrt[e]*x))/((c*Sqrt[d] - I*S
qrt[e])*(1 + c*x))]*Sqrt[(c*(Sqrt[d] + I*Sqrt[e]*x))/((c*Sqrt[d] + I*Sqrt[e])*(1 + c*x))]*(d + e*x^2)*((c*Sqrt
[d] - I*Sqrt[e])*EllipticE[I*ArcSinh[Sqrt[((c^2*d + e)*(1 - c*x))/((c*Sqrt[d] + I*Sqrt[e])^2*(1 + c*x))]], (c*
Sqrt[d] + I*Sqrt[e])^2/(c*Sqrt[d] - I*Sqrt[e])^2] - 2*(3*c*Sqrt[d] + (2*I)*Sqrt[e])*EllipticF[I*ArcSinh[Sqrt[(
(c^2*d + e)*(1 - c*x))/((c*Sqrt[d] + I*Sqrt[e])^2*(1 + c*x))]], (c*Sqrt[d] + I*Sqrt[e])^2/(c*Sqrt[d] - I*Sqrt[
e])^2]))/(c*(c*Sqrt[d] + I*Sqrt[e])*Sqrt[-(((c*Sqrt[d] - I*Sqrt[e])*(-1 + c*x))/((c*Sqrt[d] + I*Sqrt[e])*(1 +
c*x)))]))/(3*d^2*(d + e*x^2)^(3/2))

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x^{2} + d} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )}}{e^{3} x^{6} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(e*x^2 + d)*(b*arcsech(c*x) + a)/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.02, size = 0, normalized size = 0.00 \[ \int \frac {a +b \,\mathrm {arcsech}\left (c x \right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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 \[ \frac {1}{3} \, a {\left (\frac {2 \, x}{\sqrt {e x^{2} + d} d^{2}} + \frac {x}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} d}\right )} + b \int \frac {\log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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