3.152 \(\int \frac{x^3 (a+b \sec ^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=163 \[ -\frac{a+b \sec ^{-1}(c x)}{e^2 \sqrt{d+e x^2}}+\frac{d \left (a+b \sec ^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac{2 b c x \tan ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{c^2 x^2-1}}\right )}{3 \sqrt{d} e^2 \sqrt{c^2 x^2}}+\frac{b c x \sqrt{c^2 x^2-1}}{3 e \sqrt{c^2 x^2} \left (c^2 d+e\right ) \sqrt{d+e x^2}} \]

[Out]

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

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Rubi [A]  time = 0.242974, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {266, 43, 5238, 12, 573, 152, 93, 204} \[ -\frac{a+b \sec ^{-1}(c x)}{e^2 \sqrt{d+e x^2}}+\frac{d \left (a+b \sec ^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac{2 b c x \tan ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{c^2 x^2-1}}\right )}{3 \sqrt{d} e^2 \sqrt{c^2 x^2}}+\frac{b c x \sqrt{c^2 x^2-1}}{3 e \sqrt{c^2 x^2} \left (c^2 d+e\right ) \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcSec[c*x]))/(d + e*x^2)^(5/2),x]

[Out]

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

Rule 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 5238

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSec[c*x], u, x] - Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&  !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 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 573

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n],
x] /; FreeQ[{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.256064, size = 172, normalized size = 1.06 \[ \frac{-a \left (c^2 d+e\right ) \left (2 d+3 e x^2\right )+b c e x \sqrt{1-\frac{1}{c^2 x^2}} \left (d+e x^2\right )-b \left (c^2 d+e\right ) \sec ^{-1}(c x) \left (2 d+3 e x^2\right )}{3 e^2 \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}+\frac{2 b c x \sqrt{1-\frac{1}{c^2 x^2}} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{c^2 x^2-1}}{\sqrt{d+e x^2}}\right )}{3 \sqrt{d} e^2 \sqrt{c^2 x^2-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcSec[c*x]))/(d + e*x^2)^(5/2),x]

[Out]

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

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Maple [F]  time = 1.631, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3} \left ( a+b{\rm arcsec} \left (cx\right ) \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(x^3*(a+b*arcsec(c*x))/(e*x^2+d)^(5/2),x)

[Out]

int(x^3*(a+b*arcsec(c*x))/(e*x^2+d)^(5/2),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsec(c*x))/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.34775, size = 1380, normalized size = 8.47 \begin{align*} \left [-\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} \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 \, \sqrt{c^{2} x^{2} - 1}{\left ({\left (c^{2} d - e\right )} x^{2} - 2 \, d\right )} \sqrt{e x^{2} + d} \sqrt{-d} + 8 \, d^{2}}{x^{4}}\right ) + 2 \,{\left (2 \, a c^{2} d^{3} + 2 \, a d^{2} e + 3 \,{\left (a c^{2} d^{2} e + a d e^{2}\right )} x^{2} +{\left (2 \, b c^{2} d^{3} + 2 \, b d^{2} e + 3 \,{\left (b c^{2} d^{2} e + b d e^{2}\right )} x^{2}\right )} \operatorname{arcsec}\left (c x\right ) -{\left (b d e^{2} x^{2} + b d^{2} e\right )} \sqrt{c^{2} x^{2} - 1}\right )} \sqrt{e x^{2} + d}}{6 \,{\left (c^{2} d^{4} e^{2} + d^{3} e^{3} +{\left (c^{2} d^{2} e^{4} + d e^{5}\right )} x^{4} + 2 \,{\left (c^{2} d^{3} e^{3} + d^{2} e^{4}\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{\sqrt{c^{2} x^{2} - 1}{\left ({\left (c^{2} d - e\right )} x^{2} - 2 \, d\right )} \sqrt{e x^{2} + d} \sqrt{d}}{2 \,{\left (c^{2} d e x^{4} +{\left (c^{2} d^{2} - d e\right )} x^{2} - d^{2}\right )}}\right ) +{\left (2 \, a c^{2} d^{3} + 2 \, a d^{2} e + 3 \,{\left (a c^{2} d^{2} e + a d e^{2}\right )} x^{2} +{\left (2 \, b c^{2} d^{3} + 2 \, b d^{2} e + 3 \,{\left (b c^{2} d^{2} e + b d e^{2}\right )} x^{2}\right )} \operatorname{arcsec}\left (c x\right ) -{\left (b d e^{2} x^{2} + b d^{2} e\right )} \sqrt{c^{2} x^{2} - 1}\right )} \sqrt{e x^{2} + d}}{3 \,{\left (c^{2} d^{4} e^{2} + d^{3} e^{3} +{\left (c^{2} d^{2} e^{4} + d e^{5}\right )} x^{4} + 2 \,{\left (c^{2} d^{3} e^{3} + d^{2} e^{4}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsec(c*x))/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

[-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)*log(((c^4*d^
2 - 6*c^2*d*e + e^2)*x^4 - 8*(c^2*d^2 - d*e)*x^2 - 4*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)
*sqrt(-d) + 8*d^2)/x^4) + 2*(2*a*c^2*d^3 + 2*a*d^2*e + 3*(a*c^2*d^2*e + a*d*e^2)*x^2 + (2*b*c^2*d^3 + 2*b*d^2*
e + 3*(b*c^2*d^2*e + b*d*e^2)*x^2)*arcsec(c*x) - (b*d*e^2*x^2 + b*d^2*e)*sqrt(c^2*x^2 - 1))*sqrt(e*x^2 + d))/(
c^2*d^4*e^2 + d^3*e^3 + (c^2*d^2*e^4 + d*e^5)*x^4 + 2*(c^2*d^3*e^3 + d^2*e^4)*x^2), -1/3*((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*sqrt(c^2*x^2 - 1)*((c^2*d -
e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(d)/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) + (2*a*c^2*d^3 + 2*a*d^2*e +
3*(a*c^2*d^2*e + a*d*e^2)*x^2 + (2*b*c^2*d^3 + 2*b*d^2*e + 3*(b*c^2*d^2*e + b*d*e^2)*x^2)*arcsec(c*x) - (b*d*e
^2*x^2 + b*d^2*e)*sqrt(c^2*x^2 - 1))*sqrt(e*x^2 + d))/(c^2*d^4*e^2 + d^3*e^3 + (c^2*d^2*e^4 + d*e^5)*x^4 + 2*(
c^2*d^3*e^3 + d^2*e^4)*x^2)]

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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(x**3*(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{{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )} x^{3}}{{\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(x^3*(a+b*arcsec(c*x))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsec(c*x) + a)*x^3/(e*x^2 + d)^(5/2), x)