3.1.77 \(\int x^3 (d+e x^2) (a+b \sec ^{-1}(c x)) \, dx\) [77]

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

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 153, normalized size of antiderivative = 1.00, 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, 5346, 12, 457, 78} \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(b*(3*c^2*d + 2*e)*x*Sqrt[-1 + c^2*x^2])/(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 12

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

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 78

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]))))

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 5346

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]))

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) \text {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) \text {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*}

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Mathematica [A]
time = 0.18, size = 98, normalized size = 0.64 \begin {gather*} \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 (16 e+c^2 \left (30 d+8 e x^2\right )+3 c^4 \left (5 d x^2+2 e x^4\right )\right )}{c^5}+15 b x^3 \left (3 d+2 e x^2\right ) \sec ^{-1}(c x)\right ) \end {gather*}

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 = 0.20, size = 134, normalized size = 0.88

method	result	size
derivativedivides	\(\frac {\frac {a \left (\frac {1}{4} c^{6} d \,x^{4}+\frac {1}{6} c^{6} e \,x^{6}\right )}{c^{2}}+\frac {b \left (\frac {\mathrm {arcsec}\left (c x \right ) d \,c^{6} x^{4}}{4}+\frac {\mathrm {arcsec}\left (c x \right ) e \,c^{6} x^{6}}{6}-\frac {\left (c^{2} x^{2}-1\right ) \left (6 c^{4} e \,x^{4}+15 c^{4} d \,x^{2}+8 c^{2} e \,x^{2}+30 c^{2} d +16 e \right )}{180 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{2}}}{c^{4}}\)	\(134\)
default	\(\frac {\frac {a \left (\frac {1}{4} c^{6} d \,x^{4}+\frac {1}{6} c^{6} e \,x^{6}\right )}{c^{2}}+\frac {b \left (\frac {\mathrm {arcsec}\left (c x \right ) d \,c^{6} x^{4}}{4}+\frac {\mathrm {arcsec}\left (c x \right ) e \,c^{6} x^{6}}{6}-\frac {\left (c^{2} x^{2}-1\right ) \left (6 c^{4} e \,x^{4}+15 c^{4} d \,x^{2}+8 c^{2} e \,x^{2}+30 c^{2} d +16 e \right )}{180 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{2}}}{c^{4}}\)	\(134\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(a/c^2*(1/4*c^6*d*x^4+1/6*c^6*e*x^6)+b/c^2*(1/4*arcsec(c*x)*d*c^6*x^4+1/6*arcsec(c*x)*e*c^6*x^6-1/180*(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 = 0.29, size = 146, normalized size = 0.95 \begin {gather*} \frac {1}{6} \, a x^{6} e + \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 {gather*}

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*x^6*e + 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 = 1.90, size = 112, normalized size = 0.73 \begin {gather*} \frac {30 \, a c^{6} x^{6} e + 45 \, a c^{6} d x^{4} + 15 \, {\left (2 \, b c^{6} x^{6} e + 3 \, b c^{6} d x^{4}\right )} \operatorname {arcsec}\left (c x\right ) - {\left (15 \, b c^{4} d x^{2} + 30 \, b c^{2} d + 2 \, {\left (3 \, b c^{4} x^{4} + 4 \, b c^{2} x^{2} + 8 \, b\right )} e\right )} \sqrt {c^{2} x^{2} - 1}}{180 \, c^{6}} \end {gather*}

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

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Sympy [A]
time = 3.49, size = 272, normalized size = 1.78 \begin {gather*} \frac {a d x^{4}}{4} + \frac {a e x^{6}}{6} + \frac {b d x^{4} \operatorname {asec}{\left (c x \right )}}{4} + \frac {b e x^{6} \operatorname {asec}{\left (c x \right )}}{6} - \frac {b d \left (\begin {cases} \frac {x^{2} \sqrt {c^{2} x^{2} - 1}}{3 c} + \frac {2 \sqrt {c^{2} x^{2} - 1}}{3 c^{3}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {i x^{2} \sqrt {- c^{2} x^{2} + 1}}{3 c} + \frac {2 i \sqrt {- c^{2} x^{2} + 1}}{3 c^{3}} & \text {otherwise} \end {cases}\right )}{4 c} - \frac {b e \left (\begin {cases} \frac {x^{4} \sqrt {c^{2} x^{2} - 1}}{5 c} + \frac {4 x^{2} \sqrt {c^{2} x^{2} - 1}}{15 c^{3}} + \frac {8 \sqrt {c^{2} x^{2} - 1}}{15 c^{5}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {i x^{4} \sqrt {- c^{2} x^{2} + 1}}{5 c} + \frac {4 i x^{2} \sqrt {- c^{2} x^{2} + 1}}{15 c^{3}} + \frac {8 i \sqrt {- c^{2} x^{2} + 1}}{15 c^{5}} & \text {otherwise} \end {cases}\right )}{6 c} \end {gather*}

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]

a*d*x**4/4 + a*e*x**6/6 + b*d*x**4*asec(c*x)/4 + b*e*x**6*asec(c*x)/6 - b*d*Piecewise((x**2*sqrt(c**2*x**2 - 1
)/(3*c) + 2*sqrt(c**2*x**2 - 1)/(3*c**3), Abs(c**2*x**2) > 1), (I*x**2*sqrt(-c**2*x**2 + 1)/(3*c) + 2*I*sqrt(-
c**2*x**2 + 1)/(3*c**3), True))/(4*c) - b*e*Piecewise((x**4*sqrt(c**2*x**2 - 1)/(5*c) + 4*x**2*sqrt(c**2*x**2
- 1)/(15*c**3) + 8*sqrt(c**2*x**2 - 1)/(15*c**5), Abs(c**2*x**2) > 1), (I*x**4*sqrt(-c**2*x**2 + 1)/(5*c) + 4*
I*x**2*sqrt(-c**2*x**2 + 1)/(15*c**3) + 8*I*sqrt(-c**2*x**2 + 1)/(15*c**5), True))/(6*c)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 7820 vs. \(2 (131) = 262\).
time = 0.49, size = 7820, normalized size = 51.11 \begin {gather*} \text {Too large to display} \end {gather*}

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]

1/180*(45*b*c^2*d*arccos(1/(c*x))/(c^7 + 6*c^7*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^2) - 1)^2/
(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + 6*
c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10 + c^7*(1/(c^2*x^2) - 1)^6/(1/(c*x) + 1)^12) + 45*a*c^2*d/(c^7 + 6*c^7
*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2) - 1)^3/(
1/(c*x) + 1)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10 + c^7
*(1/(c^2*x^2) - 1)^6/(1/(c*x) + 1)^12) - 90*b*c^2*d*(1/(c^2*x^2) - 1)*arccos(1/(c*x))/((c^7 + 6*c^7*(1/(c^2*x^
2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1
)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10 + c^7*(1/(c^2*x^
2) - 1)^6/(1/(c*x) + 1)^12)*(1/(c*x) + 1)^2) - 90*b*c^2*d*sqrt(-1/(c^2*x^2) + 1)/((c^7 + 6*c^7*(1/(c^2*x^2) -
1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 +
15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10 + c^7*(1/(c^2*x^2) -
1)^6/(1/(c*x) + 1)^12)*(1/(c*x) + 1)) - 90*a*c^2*d*(1/(c^2*x^2) - 1)/((c^7 + 6*c^7*(1/(c^2*x^2) - 1)/(1/(c*x)
+ 1)^2 + 15*c^7*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 15*c^7*(1/(
c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10 + c^7*(1/(c^2*x^2) - 1)^6/(1/(c*x
) + 1)^12)*(1/(c*x) + 1)^2) + 30*b*e*arccos(1/(c*x))/(c^7 + 6*c^7*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(
1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(
1/(c*x) + 1)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10 + c^7*(1/(c^2*x^2) - 1)^6/(1/(c*x) + 1)^12) - 45*b
*c^2*d*(1/(c^2*x^2) - 1)^2*arccos(1/(c*x))/((c^7 + 6*c^7*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^
2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) +
1)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10 + c^7*(1/(c^2*x^2) - 1)^6/(1/(c*x) + 1)^12)*(1/(c*x) + 1)^4
) + 330*b*c^2*d*(-1/(c^2*x^2) + 1)^(3/2)/((c^7 + 6*c^7*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^2)
- 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1
)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10 + c^7*(1/(c^2*x^2) - 1)^6/(1/(c*x) + 1)^12)*(1/(c*x) + 1)^3)
+ 30*a*e/(c^7 + 6*c^7*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*
(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(
1/(c*x) + 1)^10 + c^7*(1/(c^2*x^2) - 1)^6/(1/(c*x) + 1)^12) - 45*a*c^2*d*(1/(c^2*x^2) - 1)^2/((c^7 + 6*c^7*(1/
(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2) - 1)^3/(1/(c
*x) + 1)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10 + c^7*(1/
(c^2*x^2) - 1)^6/(1/(c*x) + 1)^12)*(1/(c*x) + 1)^4) - 180*b*e*(1/(c^2*x^2) - 1)*arccos(1/(c*x))/((c^7 + 6*c^7*
(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2) - 1)^3/(1
/(c*x) + 1)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10 + c^7*
(1/(c^2*x^2) - 1)^6/(1/(c*x) + 1)^12)*(1/(c*x) + 1)^2) + 180*b*c^2*d*(1/(c^2*x^2) - 1)^3*arccos(1/(c*x))/((c^7
+ 6*c^7*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2)
- 1)^3/(1/(c*x) + 1)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^
10 + c^7*(1/(c^2*x^2) - 1)^6/(1/(c*x) + 1)^12)*(1/(c*x) + 1)^6) - 60*b*e*sqrt(-1/(c^2*x^2) + 1)/((c^7 + 6*c^7*
(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2) - 1)^3/(1
/(c*x) + 1)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10 + c^7*
(1/(c^2*x^2) - 1)^6/(1/(c*x) + 1)^12)*(1/(c*x) + 1)) - 540*b*c^2*d*(1/(c^2*x^2) - 1)^2*sqrt(-1/(c^2*x^2) + 1)/
((c^7 + 6*c^7*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*
x^2) - 1)^3/(1/(c*x) + 1)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x)
+ 1)^10 + c^7*(1/(c^2*x^2) - 1)^6/(1/(c*x) + 1)^12)*(1/(c*x) + 1)^5) - 180*a*e*(1/(c^2*x^2) - 1)/((c^7 + 6*c^7
*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2) - 1)^3/(
1/(c*x) + 1)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + 6*c^7*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10 + c^7
*(1/(c^2*x^2) - 1)^6/(1/(c*x) + 1)^12)*(1/(c*x) + 1)^2) + 180*a*c^2*d*(1/(c^2*x^2) - 1)^3/((c^7 + 6*c^7*(1/(c^
2*x^2) - 1)/(1/(c*x) + 1)^2 + 15*c^7*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 20*c^7*(1/(c^2*x^2) - 1)^3/(1/(c*x)
+ 1)^6 + 15*c^7*(1/(c^2*x^2) - 1)^4/(1/(c*x) +...

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(d + e*x^2)*(a + b*acos(1/(c*x))),x)

[Out]

int(x^3*(d + e*x^2)*(a + b*acos(1/(c*x))), x)

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