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

Optimal. Leaf size=296 \[ -\frac{2 b x \sqrt{1-c^2 x^2} \sqrt{\frac{e x^2}{d}+1} \text{EllipticF}\left (\sin ^{-1}(c x),-\frac{e}{c^2 d}\right )}{3 d^2 \sqrt{c^2 x^2} \sqrt{c^2 x^2-1} \sqrt{d+e x^2}}+\frac{2 x \left (a+b \sec ^{-1}(c x)\right )}{3 d^2 \sqrt{d+e x^2}}+\frac{x \left (a+b \sec ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac{b c e x^2 \sqrt{c^2 x^2-1}}{3 d^2 \sqrt{c^2 x^2} \left (c^2 d+e\right ) \sqrt{d+e x^2}}-\frac{b c^2 x \sqrt{1-c^2 x^2} \sqrt{d+e x^2} E\left (\sin ^{-1}(c x)|-\frac{e}{c^2 d}\right )}{3 d^2 \sqrt{c^2 x^2} \sqrt{c^2 x^2-1} \left (c^2 d+e\right ) \sqrt{\frac{e x^2}{d}+1}} \]

[Out]

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

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Rubi [A]  time = 0.23508, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 11, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.55, Rules used = {192, 191, 5228, 12, 527, 524, 427, 426, 424, 421, 419} \[ \frac{2 x \left (a+b \sec ^{-1}(c x)\right )}{3 d^2 \sqrt{d+e x^2}}+\frac{x \left (a+b \sec ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac{b c e x^2 \sqrt{c^2 x^2-1}}{3 d^2 \sqrt{c^2 x^2} \left (c^2 d+e\right ) \sqrt{d+e x^2}}-\frac{2 b x \sqrt{1-c^2 x^2} \sqrt{\frac{e x^2}{d}+1} F\left (\sin ^{-1}(c x)|-\frac{e}{c^2 d}\right )}{3 d^2 \sqrt{c^2 x^2} \sqrt{c^2 x^2-1} \sqrt{d+e x^2}}-\frac{b c^2 x \sqrt{1-c^2 x^2} \sqrt{d+e x^2} E\left (\sin ^{-1}(c x)|-\frac{e}{c^2 d}\right )}{3 d^2 \sqrt{c^2 x^2} \sqrt{c^2 x^2-1} \left (c^2 d+e\right ) \sqrt{\frac{e x^2}{d}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 5228

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

Rule 12

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

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 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 427

Int[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[Sqrt[a + b*x^2]/Sqrt[1 + (d*x^2)/c], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] &&  !GtQ[c, 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 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 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 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)])

Rubi steps

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

Mathematica [C]  time = 0.556276, size = 248, normalized size = 0.84 \[ \frac{x \left (a \left (c^2 d+e\right ) \left (3 d+2 e x^2\right )+b c e x \sqrt{1-\frac{1}{c^2 x^2}} \left (d+e x^2\right )+b \left (c^2 d+e\right ) \sec ^{-1}(c x) \left (3 d+2 e x^2\right )\right )}{3 d^2 \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}-\frac{i b c x \sqrt{1-\frac{1}{c^2 x^2}} \sqrt{\frac{e x^2}{d}+1} \left (2 \left (c^2 d+e\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-c^2} x\right ),-\frac{e}{c^2 d}\right )+c^2 d E\left (i \sinh ^{-1}\left (\sqrt{-c^2} x\right )|-\frac{e}{c^2 d}\right )\right )}{3 \sqrt{-c^2} d^2 \sqrt{1-c^2 x^2} \left (c^2 d+e\right ) \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.08, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\rm arcsec} \left (cx\right )) \left ( e{x}^{2}+d \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{\arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )}{{\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt{e x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{e x^{2} + d}{\left (b \operatorname{arcsec}\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(e*x^2 + d)*(b*arcsec(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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arcsec}\left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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