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

Optimal. Leaf size=149 \[ \frac{1}{4} d x^4 \left (a+b \cos ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \cos ^{-1}(c x)\right )-\frac{b x^3 \sqrt{1-c^2 x^2} \left (9 c^2 d+5 e\right )}{144 c^3}-\frac{b x \sqrt{1-c^2 x^2} \left (9 c^2 d+5 e\right )}{96 c^5}+\frac{b \left (9 c^2 d+5 e\right ) \sin ^{-1}(c x)}{96 c^6}-\frac{b e x^5 \sqrt{1-c^2 x^2}}{36 c} \]

[Out]

-(b*(9*c^2*d + 5*e)*x*Sqrt[1 - c^2*x^2])/(96*c^5) - (b*(9*c^2*d + 5*e)*x^3*Sqrt[1 - c^2*x^2])/(144*c^3) - (b*e
*x^5*Sqrt[1 - c^2*x^2])/(36*c) + (d*x^4*(a + b*ArcCos[c*x]))/4 + (e*x^6*(a + b*ArcCos[c*x]))/6 + (b*(9*c^2*d +
5*e)*ArcSin[c*x])/(96*c^6)

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Rubi [A]  time = 0.118787, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {14, 4732, 12, 459, 321, 216} \[ \frac{1}{4} d x^4 \left (a+b \cos ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \cos ^{-1}(c x)\right )-\frac{b x^3 \sqrt{1-c^2 x^2} \left (9 c^2 d+5 e\right )}{144 c^3}-\frac{b x \sqrt{1-c^2 x^2} \left (9 c^2 d+5 e\right )}{96 c^5}+\frac{b \left (9 c^2 d+5 e\right ) \sin ^{-1}(c x)}{96 c^6}-\frac{b e x^5 \sqrt{1-c^2 x^2}}{36 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(9*c^2*d + 5*e)*x*Sqrt[1 - c^2*x^2])/(96*c^5) - (b*(9*c^2*d + 5*e)*x^3*Sqrt[1 - c^2*x^2])/(144*c^3) - (b*e
*x^5*Sqrt[1 - c^2*x^2])/(36*c) + (d*x^4*(a + b*ArcCos[c*x]))/4 + (e*x^6*(a + b*ArcCos[c*x]))/6 + (b*(9*c^2*d +
5*e)*ArcSin[c*x])/(96*c^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 4732

Int[((a_.) + ArcCos[(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*ArcCos[c*x], u, x] + Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 -
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (GtQ[p, 0] ||
(IGtQ[(m - 1)/2, 0] && LeQ[m + p, 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 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 216

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

Rubi steps

\begin{align*} \int x^3 \left (d+e x^2\right ) \left (a+b \cos ^{-1}(c x)\right ) \, dx &=\frac{1}{4} d x^4 \left (a+b \cos ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \cos ^{-1}(c x)\right )+(b c) \int \frac{x^4 \left (3 d+2 e x^2\right )}{12 \sqrt{1-c^2 x^2}} \, dx\\ &=\frac{1}{4} d x^4 \left (a+b \cos ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \cos ^{-1}(c x)\right )+\frac{1}{12} (b c) \int \frac{x^4 \left (3 d+2 e x^2\right )}{\sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{b e x^5 \sqrt{1-c^2 x^2}}{36 c}+\frac{1}{4} d x^4 \left (a+b \cos ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \cos ^{-1}(c x)\right )+\frac{1}{36} \left (b c \left (9 d+\frac{5 e}{c^2}\right )\right ) \int \frac{x^4}{\sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{b \left (9 c^2 d+5 e\right ) x^3 \sqrt{1-c^2 x^2}}{144 c^3}-\frac{b e x^5 \sqrt{1-c^2 x^2}}{36 c}+\frac{1}{4} d x^4 \left (a+b \cos ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \cos ^{-1}(c x)\right )+\frac{\left (b \left (9 c^2 d+5 e\right )\right ) \int \frac{x^2}{\sqrt{1-c^2 x^2}} \, dx}{48 c^3}\\ &=-\frac{b \left (9 c^2 d+5 e\right ) x \sqrt{1-c^2 x^2}}{96 c^5}-\frac{b \left (9 c^2 d+5 e\right ) x^3 \sqrt{1-c^2 x^2}}{144 c^3}-\frac{b e x^5 \sqrt{1-c^2 x^2}}{36 c}+\frac{1}{4} d x^4 \left (a+b \cos ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \cos ^{-1}(c x)\right )+\frac{\left (b \left (9 c^2 d+5 e\right )\right ) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx}{96 c^5}\\ &=-\frac{b \left (9 c^2 d+5 e\right ) x \sqrt{1-c^2 x^2}}{96 c^5}-\frac{b \left (9 c^2 d+5 e\right ) x^3 \sqrt{1-c^2 x^2}}{144 c^3}-\frac{b e x^5 \sqrt{1-c^2 x^2}}{36 c}+\frac{1}{4} d x^4 \left (a+b \cos ^{-1}(c x)\right )+\frac{1}{6} e x^6 \left (a+b \cos ^{-1}(c x)\right )+\frac{b \left (9 c^2 d+5 e\right ) \sin ^{-1}(c x)}{96 c^6}\\ \end{align*}

Mathematica [A]  time = 0.140823, size = 153, normalized size = 1.03 \[ \frac{1}{4} a d x^4+\frac{1}{6} a e x^6+b d \sqrt{1-c^2 x^2} \left (-\frac{3 x}{32 c^3}-\frac{x^3}{16 c}\right )+\frac{3 b d \sin ^{-1}(c x)}{32 c^4}+b e \sqrt{1-c^2 x^2} \left (-\frac{5 x^3}{144 c^3}-\frac{5 x}{96 c^5}-\frac{x^5}{36 c}\right )+\frac{5 b e \sin ^{-1}(c x)}{96 c^6}+\frac{1}{4} b d x^4 \cos ^{-1}(c x)+\frac{1}{6} b e x^6 \cos ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*d*x^4)/4 + (a*e*x^6)/6 + b*d*Sqrt[1 - c^2*x^2]*((-3*x)/(32*c^3) - x^3/(16*c)) + b*e*Sqrt[1 - c^2*x^2]*((-5*
x)/(96*c^5) - (5*x^3)/(144*c^3) - x^5/(36*c)) + (b*d*x^4*ArcCos[c*x])/4 + (b*e*x^6*ArcCos[c*x])/6 + (3*b*d*Arc
Sin[c*x])/(32*c^4) + (5*b*e*ArcSin[c*x])/(96*c^6)

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Maple [A]  time = 0.022, size = 177, normalized size = 1.2 \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{\arccos \left ( cx \right ) e{c}^{6}{x}^{6}}{6}}+{\frac{\arccos \left ( cx \right ){c}^{6}{x}^{4}d}{4}}+{\frac{e}{6} \left ( -{\frac{{c}^{5}{x}^{5}}{6}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{5\,{c}^{3}{x}^{3}}{24}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{5\,cx}{16}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{5\,\arcsin \left ( cx \right ) }{16}} \right ) }+{\frac{{c}^{2}d}{4} \left ( -{\frac{{c}^{3}{x}^{3}}{4}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{3\,cx}{8}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{3\,\arcsin \left ( cx \right ) }{8}} \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x^2+d)*(a+b*arccos(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*arccos(c*x)*e*c^6*x^6+1/4*arccos(c*x)*c^6*x^4*d+1/6*e*(-
1/6*c^5*x^5*(-c^2*x^2+1)^(1/2)-5/24*c^3*x^3*(-c^2*x^2+1)^(1/2)-5/16*c*x*(-c^2*x^2+1)^(1/2)+5/16*arcsin(c*x))+1
/4*c^2*d*(-1/4*c^3*x^3*(-c^2*x^2+1)^(1/2)-3/8*c*x*(-c^2*x^2+1)^(1/2)+3/8*arcsin(c*x))))

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Maxima [A]  time = 1.81237, size = 255, normalized size = 1.71 \begin{align*} \frac{1}{6} \, a e x^{6} + \frac{1}{4} \, a d x^{4} + \frac{1}{32} \,{\left (8 \, x^{4} \arccos \left (c x\right ) -{\left (\frac{2 \, \sqrt{-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac{3 \, \sqrt{-c^{2} x^{2} + 1} x}{c^{4}} - \frac{3 \, \arcsin \left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{4}}\right )} c\right )} b d + \frac{1}{288} \,{\left (48 \, x^{6} \arccos \left (c x\right ) -{\left (\frac{8 \, \sqrt{-c^{2} x^{2} + 1} x^{5}}{c^{2}} + \frac{10 \, \sqrt{-c^{2} x^{2} + 1} x^{3}}{c^{4}} + \frac{15 \, \sqrt{-c^{2} x^{2} + 1} x}{c^{6}} - \frac{15 \, \arcsin \left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{6}}\right )} c\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*arccos(c*x)),x, algorithm="maxima")

[Out]

1/6*a*e*x^6 + 1/4*a*d*x^4 + 1/32*(8*x^4*arccos(c*x) - (2*sqrt(-c^2*x^2 + 1)*x^3/c^2 + 3*sqrt(-c^2*x^2 + 1)*x/c
^4 - 3*arcsin(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^4))*c)*b*d + 1/288*(48*x^6*arccos(c*x) - (8*sqrt(-c^2*x^2 + 1)*x^5
/c^2 + 10*sqrt(-c^2*x^2 + 1)*x^3/c^4 + 15*sqrt(-c^2*x^2 + 1)*x/c^6 - 15*arcsin(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^6
))*c)*b*e

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Fricas [A]  time = 2.14309, size = 286, normalized size = 1.92 \begin{align*} \frac{48 \, a c^{6} e x^{6} + 72 \, a c^{6} d x^{4} + 3 \,{\left (16 \, b c^{6} e x^{6} + 24 \, b c^{6} d x^{4} - 9 \, b c^{2} d - 5 \, b e\right )} \arccos \left (c x\right ) -{\left (8 \, b c^{5} e x^{5} + 2 \,{\left (9 \, b c^{5} d + 5 \, b c^{3} e\right )} x^{3} + 3 \,{\left (9 \, b c^{3} d + 5 \, b c e\right )} x\right )} \sqrt{-c^{2} x^{2} + 1}}{288 \, c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/288*(48*a*c^6*e*x^6 + 72*a*c^6*d*x^4 + 3*(16*b*c^6*e*x^6 + 24*b*c^6*d*x^4 - 9*b*c^2*d - 5*b*e)*arccos(c*x) -
(8*b*c^5*e*x^5 + 2*(9*b*c^5*d + 5*b*c^3*e)*x^3 + 3*(9*b*c^3*d + 5*b*c*e)*x)*sqrt(-c^2*x^2 + 1))/c^6

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Sympy [A]  time = 5.14358, size = 211, normalized size = 1.42 \begin{align*} \begin{cases} \frac{a d x^{4}}{4} + \frac{a e x^{6}}{6} + \frac{b d x^{4} \operatorname{acos}{\left (c x \right )}}{4} + \frac{b e x^{6} \operatorname{acos}{\left (c x \right )}}{6} - \frac{b d x^{3} \sqrt{- c^{2} x^{2} + 1}}{16 c} - \frac{b e x^{5} \sqrt{- c^{2} x^{2} + 1}}{36 c} - \frac{3 b d x \sqrt{- c^{2} x^{2} + 1}}{32 c^{3}} - \frac{5 b e x^{3} \sqrt{- c^{2} x^{2} + 1}}{144 c^{3}} - \frac{3 b d \operatorname{acos}{\left (c x \right )}}{32 c^{4}} - \frac{5 b e x \sqrt{- c^{2} x^{2} + 1}}{96 c^{5}} - \frac{5 b e \operatorname{acos}{\left (c x \right )}}{96 c^{6}} & \text{for}\: c \neq 0 \\\left (a + \frac{\pi b}{2}\right ) \left (\frac{d x^{4}}{4} + \frac{e x^{6}}{6}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d*x**4/4 + a*e*x**6/6 + b*d*x**4*acos(c*x)/4 + b*e*x**6*acos(c*x)/6 - b*d*x**3*sqrt(-c**2*x**2 +
1)/(16*c) - b*e*x**5*sqrt(-c**2*x**2 + 1)/(36*c) - 3*b*d*x*sqrt(-c**2*x**2 + 1)/(32*c**3) - 5*b*e*x**3*sqrt(-c
**2*x**2 + 1)/(144*c**3) - 3*b*d*acos(c*x)/(32*c**4) - 5*b*e*x*sqrt(-c**2*x**2 + 1)/(96*c**5) - 5*b*e*acos(c*x
)/(96*c**6), Ne(c, 0)), ((a + pi*b/2)*(d*x**4/4 + e*x**6/6), True))

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Giac [A]  time = 1.14045, size = 231, normalized size = 1.55 \begin{align*} \frac{1}{6} \, b x^{6} \arccos \left (c x\right ) e + \frac{1}{6} \, a x^{6} e + \frac{1}{4} \, b d x^{4} \arccos \left (c x\right ) - \frac{\sqrt{-c^{2} x^{2} + 1} b x^{5} e}{36 \, c} + \frac{1}{4} \, a d x^{4} - \frac{\sqrt{-c^{2} x^{2} + 1} b d x^{3}}{16 \, c} - \frac{5 \, \sqrt{-c^{2} x^{2} + 1} b x^{3} e}{144 \, c^{3}} - \frac{3 \, \sqrt{-c^{2} x^{2} + 1} b d x}{32 \, c^{3}} - \frac{3 \, b d \arccos \left (c x\right )}{32 \, c^{4}} - \frac{5 \, \sqrt{-c^{2} x^{2} + 1} b x e}{96 \, c^{5}} - \frac{5 \, b \arccos \left (c x\right ) e}{96 \, c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b*x^6*arccos(c*x)*e + 1/6*a*x^6*e + 1/4*b*d*x^4*arccos(c*x) - 1/36*sqrt(-c^2*x^2 + 1)*b*x^5*e/c + 1/4*a*d*
x^4 - 1/16*sqrt(-c^2*x^2 + 1)*b*d*x^3/c - 5/144*sqrt(-c^2*x^2 + 1)*b*x^3*e/c^3 - 3/32*sqrt(-c^2*x^2 + 1)*b*d*x
/c^3 - 3/32*b*d*arccos(c*x)/c^4 - 5/96*sqrt(-c^2*x^2 + 1)*b*x*e/c^5 - 5/96*b*arccos(c*x)*e/c^6