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

Optimal. Leaf size=153 \[ \frac{1}{4} d x^4 \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \sec ^{-1}(c x)\right )-\frac{b x \left (c^2 x^2-1\right )^{3/2} \left (3 c^2 d+4 e\right )}{36 c^5 \sqrt{c^2 x^2}}-\frac{b x \sqrt{c^2 x^2-1} \left (3 c^2 d+2 e\right )}{12 c^5 \sqrt{c^2 x^2}}-\frac{b e x \left (c^2 x^2-1\right )^{5/2}}{30 c^5 \sqrt{c^2 x^2}} \]

[Out]

-(b*(3*c^2*d + 2*e)*x*Sqrt[-1 + c^2*x^2])/(12*c^5*Sqrt[c^2*x^2]) - (b*(3*c^2*d + 4*e)*x*(-1 + c^2*x^2)^(3/2))/
(36*c^5*Sqrt[c^2*x^2]) - (b*e*x*(-1 + c^2*x^2)^(5/2))/(30*c^5*Sqrt[c^2*x^2]) + (d*x^4*(a + b*ArcSec[c*x]))/4 +
(e*x^6*(a + b*ArcSec[c*x]))/6

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Rubi [A]  time = 0.121989, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {14, 5238, 12, 446, 77} \[ \frac{1}{4} d x^4 \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \sec ^{-1}(c x)\right )-\frac{b x \left (c^2 x^2-1\right )^{3/2} \left (3 c^2 d+4 e\right )}{36 c^5 \sqrt{c^2 x^2}}-\frac{b x \sqrt{c^2 x^2-1} \left (3 c^2 d+2 e\right )}{12 c^5 \sqrt{c^2 x^2}}-\frac{b e x \left (c^2 x^2-1\right )^{5/2}}{30 c^5 \sqrt{c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(3*c^2*d + 2*e)*x*Sqrt[-1 + c^2*x^2])/(12*c^5*Sqrt[c^2*x^2]) - (b*(3*c^2*d + 4*e)*x*(-1 + c^2*x^2)^(3/2))/
(36*c^5*Sqrt[c^2*x^2]) - (b*e*x*(-1 + c^2*x^2)^(5/2))/(30*c^5*Sqrt[c^2*x^2]) + (d*x^4*(a + b*ArcSec[c*x]))/4 +
(e*x^6*(a + b*ArcSec[c*x]))/6

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int x^3 \left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right ) \, dx &=\frac{1}{4} d x^4 \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \sec ^{-1}(c x)\right )-\frac{(b c x) \int \frac{x^3 \left (3 d+2 e x^2\right )}{12 \sqrt{-1+c^2 x^2}} \, dx}{\sqrt{c^2 x^2}}\\ &=\frac{1}{4} d x^4 \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \sec ^{-1}(c x)\right )-\frac{(b c x) \int \frac{x^3 \left (3 d+2 e x^2\right )}{\sqrt{-1+c^2 x^2}} \, dx}{12 \sqrt{c^2 x^2}}\\ &=\frac{1}{4} d x^4 \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \sec ^{-1}(c x)\right )-\frac{(b c x) \operatorname{Subst}\left (\int \frac{x (3 d+2 e x)}{\sqrt{-1+c^2 x}} \, dx,x,x^2\right )}{24 \sqrt{c^2 x^2}}\\ &=\frac{1}{4} d x^4 \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \sec ^{-1}(c x)\right )-\frac{(b c x) \operatorname{Subst}\left (\int \left (\frac{3 c^2 d+2 e}{c^4 \sqrt{-1+c^2 x}}+\frac{\left (3 c^2 d+4 e\right ) \sqrt{-1+c^2 x}}{c^4}+\frac{2 e \left (-1+c^2 x\right )^{3/2}}{c^4}\right ) \, dx,x,x^2\right )}{24 \sqrt{c^2 x^2}}\\ &=-\frac{b \left (3 c^2 d+2 e\right ) x \sqrt{-1+c^2 x^2}}{12 c^5 \sqrt{c^2 x^2}}-\frac{b \left (3 c^2 d+4 e\right ) x \left (-1+c^2 x^2\right )^{3/2}}{36 c^5 \sqrt{c^2 x^2}}-\frac{b e x \left (-1+c^2 x^2\right )^{5/2}}{30 c^5 \sqrt{c^2 x^2}}+\frac{1}{4} d x^4 \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \sec ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.253271, size = 98, normalized size = 0.64 \[ \frac{1}{180} x \left (15 a x^3 \left (3 d+2 e x^2\right )-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} \left (3 c^4 \left (5 d x^2+2 e x^4\right )+c^2 \left (30 d+8 e x^2\right )+16 e\right )}{c^5}+15 b x^3 \sec ^{-1}(c x) \left (3 d+2 e x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(15*a*x^3*(3*d + 2*e*x^2) - (b*Sqrt[1 - 1/(c^2*x^2)]*(16*e + c^2*(30*d + 8*e*x^2) + 3*c^4*(5*d*x^2 + 2*e*x^
4)))/c^5 + 15*b*x^3*(3*d + 2*e*x^2)*ArcSec[c*x]))/180

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Maple [A]  time = 0.171, size = 134, normalized size = 0.9 \begin{align*}{\frac{1}{{c}^{4}} \left ({\frac{a}{{c}^{2}} \left ({\frac{e{c}^{6}{x}^{6}}{6}}+{\frac{{x}^{4}{c}^{6}d}{4}} \right ) }+{\frac{b}{{c}^{2}} \left ({\frac{{\rm arcsec} \left (cx\right )e{c}^{6}{x}^{6}}{6}}+{\frac{{\rm arcsec} \left (cx\right ){c}^{6}{x}^{4}d}{4}}-{\frac{ \left ({c}^{2}{x}^{2}-1 \right ) \left ( 6\,{c}^{4}e{x}^{4}+15\,{c}^{4}d{x}^{2}+8\,{c}^{2}e{x}^{2}+30\,{c}^{2}d+16\,e \right ) }{180\,cx}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(a/c^2*(1/6*e*c^6*x^6+1/4*x^4*c^6*d)+b/c^2*(1/6*arcsec(c*x)*e*c^6*x^6+1/4*arcsec(c*x)*c^6*x^4*d-1/180*(c
^2*x^2-1)*(6*c^4*e*x^4+15*c^4*d*x^2+8*c^2*e*x^2+30*c^2*d+16*e)/((c^2*x^2-1)/c^2/x^2)^(1/2)/c/x))

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Maxima [A]  time = 0.968066, size = 194, normalized size = 1.27 \begin{align*} \frac{1}{6} \, a e x^{6} + \frac{1}{4} \, a d x^{4} + \frac{1}{12} \,{\left (3 \, x^{4} \operatorname{arcsec}\left (c x\right ) - \frac{c^{2} x^{3}{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} + 3 \, x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} b d + \frac{1}{90} \,{\left (15 \, x^{6} \operatorname{arcsec}\left (c x\right ) - \frac{3 \, c^{4} x^{5}{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{5}{2}} + 10 \, c^{2} x^{3}{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} + 15 \, x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c^{5}}\right )} b e \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*e*x^6 + 1/4*a*d*x^4 + 1/12*(3*x^4*arcsec(c*x) - (c^2*x^3*(-1/(c^2*x^2) + 1)^(3/2) + 3*x*sqrt(-1/(c^2*x^2
) + 1))/c^3)*b*d + 1/90*(15*x^6*arcsec(c*x) - (3*c^4*x^5*(-1/(c^2*x^2) + 1)^(5/2) + 10*c^2*x^3*(-1/(c^2*x^2) +
1)^(3/2) + 15*x*sqrt(-1/(c^2*x^2) + 1))/c^5)*b*e

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Fricas [A]  time = 2.18156, size = 247, normalized size = 1.61 \begin{align*} \frac{30 \, a c^{6} e x^{6} + 45 \, a c^{6} d x^{4} + 15 \,{\left (2 \, b c^{6} e x^{6} + 3 \, b c^{6} d x^{4}\right )} \operatorname{arcsec}\left (c x\right ) -{\left (6 \, b c^{4} e x^{4} + 30 \, b c^{2} d +{\left (15 \, b c^{4} d + 8 \, b c^{2} e\right )} x^{2} + 16 \, b e\right )} \sqrt{c^{2} x^{2} - 1}}{180 \, c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/180*(30*a*c^6*e*x^6 + 45*a*c^6*d*x^4 + 15*(2*b*c^6*e*x^6 + 3*b*c^6*d*x^4)*arcsec(c*x) - (6*b*c^4*e*x^4 + 30*
b*c^2*d + (15*b*c^4*d + 8*b*c^2*e)*x^2 + 16*b*e)*sqrt(c^2*x^2 - 1))/c^6

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \left (a + b \operatorname{asec}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x**2+d)*(a+b*asec(c*x)),x)

[Out]

Integral(x**3*(a + b*asec(c*x))*(d + e*x**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )} x^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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