3.168 \(\int \frac{x^2}{(a+b \cos ^{-1}(c x))^3} \, dx\)

Optimal. Leaf size=197 \[ -\frac{\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \cos ^{-1}(c x)}{b}\right )}{8 b^3 c^3}-\frac{9 \sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 \left (a+b \cos ^{-1}(c x)\right )}{b}\right )}{8 b^3 c^3}+\frac{\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}(c x)}{b}\right )}{8 b^3 c^3}+\frac{9 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 \left (a+b \cos ^{-1}(c x)\right )}{b}\right )}{8 b^3 c^3}-\frac{x}{b^2 c^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{3 x^3}{2 b^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{x^2 \sqrt{1-c^2 x^2}}{2 b c \left (a+b \cos ^{-1}(c x)\right )^2} \]

[Out]

(x^2*Sqrt[1 - c^2*x^2])/(2*b*c*(a + b*ArcCos[c*x])^2) - x/(b^2*c^2*(a + b*ArcCos[c*x])) + (3*x^3)/(2*b^2*(a +
b*ArcCos[c*x])) - (CosIntegral[(a + b*ArcCos[c*x])/b]*Sin[a/b])/(8*b^3*c^3) - (9*CosIntegral[(3*(a + b*ArcCos[
c*x]))/b]*Sin[(3*a)/b])/(8*b^3*c^3) + (Cos[a/b]*SinIntegral[(a + b*ArcCos[c*x])/b])/(8*b^3*c^3) + (9*Cos[(3*a)
/b]*SinIntegral[(3*(a + b*ArcCos[c*x]))/b])/(8*b^3*c^3)

________________________________________________________________________________________

Rubi [A]  time = 0.534519, antiderivative size = 246, normalized size of antiderivative = 1.25, number of steps used = 16, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {4634, 4720, 4636, 4406, 3303, 3299, 3302, 4624} \[ -\frac{9 \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\cos ^{-1}(c x)\right )}{8 b^3 c^3}+\frac{\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \cos ^{-1}(c x)}{b}\right )}{b^3 c^3}-\frac{9 \sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 a}{b}+3 \cos ^{-1}(c x)\right )}{8 b^3 c^3}+\frac{9 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\cos ^{-1}(c x)\right )}{8 b^3 c^3}+\frac{9 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \cos ^{-1}(c x)\right )}{8 b^3 c^3}-\frac{\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}(c x)}{b}\right )}{b^3 c^3}-\frac{x}{b^2 c^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{3 x^3}{2 b^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{x^2 \sqrt{1-c^2 x^2}}{2 b c \left (a+b \cos ^{-1}(c x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*ArcCos[c*x])^3,x]

[Out]

(x^2*Sqrt[1 - c^2*x^2])/(2*b*c*(a + b*ArcCos[c*x])^2) - x/(b^2*c^2*(a + b*ArcCos[c*x])) + (3*x^3)/(2*b^2*(a +
b*ArcCos[c*x])) - (9*CosIntegral[a/b + ArcCos[c*x]]*Sin[a/b])/(8*b^3*c^3) + (CosIntegral[(a + b*ArcCos[c*x])/b
]*Sin[a/b])/(b^3*c^3) - (9*CosIntegral[(3*a)/b + 3*ArcCos[c*x]]*Sin[(3*a)/b])/(8*b^3*c^3) + (9*Cos[a/b]*SinInt
egral[a/b + ArcCos[c*x]])/(8*b^3*c^3) + (9*Cos[(3*a)/b]*SinIntegral[(3*a)/b + 3*ArcCos[c*x]])/(8*b^3*c^3) - (C
os[a/b]*SinIntegral[(a + b*ArcCos[c*x])/b])/(b^3*c^3)

Rule 4634

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Simp[(x^m*Sqrt[1 - c^2*x^2]*(a + b*ArcCo
s[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcCos[c*x])^(n + 1
))/Sqrt[1 - c^2*x^2], x], x] + Dist[m/(b*c*(n + 1)), Int[(x^(m - 1)*(a + b*ArcCos[c*x])^(n + 1))/Sqrt[1 - c^2*
x^2], x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 4720

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp
[((f*x)^m*(a + b*ArcCos[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] + Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x)
^(m - 1)*(a + b*ArcCos[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n,
-1] && GtQ[d, 0]

Rule 4636

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b*
x)^n*Cos[x]^m*Sin[x], x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 4624

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[1/(b*c), Subst[Int[x^n*Sin[a/b - x/b], x], x, a
+ b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, n}, x]

Rubi steps

\begin{align*} \int \frac{x^2}{\left (a+b \cos ^{-1}(c x)\right )^3} \, dx &=\frac{x^2 \sqrt{1-c^2 x^2}}{2 b c \left (a+b \cos ^{-1}(c x)\right )^2}-\frac{\int \frac{x}{\sqrt{1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^2} \, dx}{b c}+\frac{(3 c) \int \frac{x^3}{\sqrt{1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^2} \, dx}{2 b}\\ &=\frac{x^2 \sqrt{1-c^2 x^2}}{2 b c \left (a+b \cos ^{-1}(c x)\right )^2}-\frac{x}{b^2 c^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{3 x^3}{2 b^2 \left (a+b \cos ^{-1}(c x)\right )}-\frac{9 \int \frac{x^2}{a+b \cos ^{-1}(c x)} \, dx}{2 b^2}+\frac{\int \frac{1}{a+b \cos ^{-1}(c x)} \, dx}{b^2 c^2}\\ &=\frac{x^2 \sqrt{1-c^2 x^2}}{2 b c \left (a+b \cos ^{-1}(c x)\right )^2}-\frac{x}{b^2 c^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{3 x^3}{2 b^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}-\frac{x}{b}\right )}{x} \, dx,x,a+b \cos ^{-1}(c x)\right )}{b^3 c^3}+\frac{9 \operatorname{Subst}\left (\int \frac{\cos ^2(x) \sin (x)}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{2 b^2 c^3}\\ &=\frac{x^2 \sqrt{1-c^2 x^2}}{2 b c \left (a+b \cos ^{-1}(c x)\right )^2}-\frac{x}{b^2 c^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{3 x^3}{2 b^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{9 \operatorname{Subst}\left (\int \left (\frac{\sin (x)}{4 (a+b x)}+\frac{\sin (3 x)}{4 (a+b x)}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{2 b^2 c^3}-\frac{\cos \left (\frac{a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{x}{b}\right )}{x} \, dx,x,a+b \cos ^{-1}(c x)\right )}{b^3 c^3}+\frac{\sin \left (\frac{a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{x}{b}\right )}{x} \, dx,x,a+b \cos ^{-1}(c x)\right )}{b^3 c^3}\\ &=\frac{x^2 \sqrt{1-c^2 x^2}}{2 b c \left (a+b \cos ^{-1}(c x)\right )^2}-\frac{x}{b^2 c^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{3 x^3}{2 b^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{\text{Ci}\left (\frac{a+b \cos ^{-1}(c x)}{b}\right ) \sin \left (\frac{a}{b}\right )}{b^3 c^3}-\frac{\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}(c x)}{b}\right )}{b^3 c^3}+\frac{9 \operatorname{Subst}\left (\int \frac{\sin (x)}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{8 b^2 c^3}+\frac{9 \operatorname{Subst}\left (\int \frac{\sin (3 x)}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{8 b^2 c^3}\\ &=\frac{x^2 \sqrt{1-c^2 x^2}}{2 b c \left (a+b \cos ^{-1}(c x)\right )^2}-\frac{x}{b^2 c^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{3 x^3}{2 b^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{\text{Ci}\left (\frac{a+b \cos ^{-1}(c x)}{b}\right ) \sin \left (\frac{a}{b}\right )}{b^3 c^3}-\frac{\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}(c x)}{b}\right )}{b^3 c^3}+\frac{\left (9 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{8 b^2 c^3}+\frac{\left (9 \cos \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{8 b^2 c^3}-\frac{\left (9 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{8 b^2 c^3}-\frac{\left (9 \sin \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{8 b^2 c^3}\\ &=\frac{x^2 \sqrt{1-c^2 x^2}}{2 b c \left (a+b \cos ^{-1}(c x)\right )^2}-\frac{x}{b^2 c^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{3 x^3}{2 b^2 \left (a+b \cos ^{-1}(c x)\right )}-\frac{9 \text{Ci}\left (\frac{a}{b}+\cos ^{-1}(c x)\right ) \sin \left (\frac{a}{b}\right )}{8 b^3 c^3}+\frac{\text{Ci}\left (\frac{a+b \cos ^{-1}(c x)}{b}\right ) \sin \left (\frac{a}{b}\right )}{b^3 c^3}-\frac{9 \text{Ci}\left (\frac{3 a}{b}+3 \cos ^{-1}(c x)\right ) \sin \left (\frac{3 a}{b}\right )}{8 b^3 c^3}+\frac{9 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\cos ^{-1}(c x)\right )}{8 b^3 c^3}+\frac{9 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \cos ^{-1}(c x)\right )}{8 b^3 c^3}-\frac{\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}(c x)}{b}\right )}{b^3 c^3}\\ \end{align*}

Mathematica [A]  time = 0.455213, size = 169, normalized size = 0.86 \[ \frac{\frac{4 b^2 x^2 \sqrt{1-c^2 x^2}}{c \left (a+b \cos ^{-1}(c x)\right )^2}-\frac{\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\cos ^{-1}(c x)\right )}{c^3}-\frac{9 \sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (3 \left (\frac{a}{b}+\cos ^{-1}(c x)\right )\right )}{c^3}+\frac{\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\cos ^{-1}(c x)\right )}{c^3}+\frac{9 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (3 \left (\frac{a}{b}+\cos ^{-1}(c x)\right )\right )}{c^3}-\frac{8 b x}{c^2 \left (a+b \cos ^{-1}(c x)\right )}+\frac{12 b x^3}{a+b \cos ^{-1}(c x)}}{8 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*b^2*x^2*Sqrt[1 - c^2*x^2])/(c*(a + b*ArcCos[c*x])^2) - (8*b*x)/(c^2*(a + b*ArcCos[c*x])) + (12*b*x^3)/(a +
b*ArcCos[c*x]) - (CosIntegral[a/b + ArcCos[c*x]]*Sin[a/b])/c^3 - (9*CosIntegral[3*(a/b + ArcCos[c*x])]*Sin[(3
*a)/b])/c^3 + (Cos[a/b]*SinIntegral[a/b + ArcCos[c*x]])/c^3 + (9*Cos[(3*a)/b]*SinIntegral[3*(a/b + ArcCos[c*x]
)])/c^3)/(8*b^3)

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Maple [A]  time = 0.06, size = 291, normalized size = 1.5 \begin{align*}{\frac{1}{{c}^{3}} \left ({\frac{\sin \left ( 3\,\arccos \left ( cx \right ) \right ) }{8\, \left ( a+b\arccos \left ( cx \right ) \right ) ^{2}b}}-{\frac{3}{ \left ( 8\,a+8\,b\arccos \left ( cx \right ) \right ){b}^{3}} \left ( 3\,\arccos \left ( cx \right ) \sin \left ( 3\,{\frac{a}{b}} \right ){\it Ci} \left ( 3\,\arccos \left ( cx \right ) +3\,{\frac{a}{b}} \right ) b-3\,\arccos \left ( cx \right ){\it Si} \left ( 3\,\arccos \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) b+3\,\sin \left ( 3\,{\frac{a}{b}} \right ){\it Ci} \left ( 3\,\arccos \left ( cx \right ) +3\,{\frac{a}{b}} \right ) a-3\,{\it Si} \left ( 3\,\arccos \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) a-\cos \left ( 3\,\arccos \left ( cx \right ) \right ) b \right ) }+{\frac{1}{8\, \left ( a+b\arccos \left ( cx \right ) \right ) ^{2}b}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{1}{ \left ( 8\,a+8\,b\arccos \left ( cx \right ) \right ){b}^{3}} \left ( \arccos \left ( cx \right ){\it Si} \left ( \arccos \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) b-\arccos \left ( cx \right ){\it Ci} \left ( \arccos \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) b+{\it Si} \left ( \arccos \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) a-{\it Ci} \left ( \arccos \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) a+xbc \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a+b*arccos(c*x))^3,x)

[Out]

1/c^3*(1/8*sin(3*arccos(c*x))/(a+b*arccos(c*x))^2/b-3/8*(3*arccos(c*x)*sin(3*a/b)*Ci(3*arccos(c*x)+3*a/b)*b-3*
arccos(c*x)*Si(3*arccos(c*x)+3*a/b)*cos(3*a/b)*b+3*sin(3*a/b)*Ci(3*arccos(c*x)+3*a/b)*a-3*Si(3*arccos(c*x)+3*a
/b)*cos(3*a/b)*a-cos(3*arccos(c*x))*b)/(a+b*arccos(c*x))/b^3+1/8*(-c^2*x^2+1)^(1/2)/(a+b*arccos(c*x))^2/b+1/8*
(arccos(c*x)*Si(arccos(c*x)+a/b)*cos(a/b)*b-arccos(c*x)*Ci(arccos(c*x)+a/b)*sin(a/b)*b+Si(arccos(c*x)+a/b)*cos
(a/b)*a-Ci(arccos(c*x)+a/b)*sin(a/b)*a+x*b*c)/(a+b*arccos(c*x))/b^3)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3 \, a c^{2} x^{3} + \sqrt{c x + 1} \sqrt{-c x + 1} b c x^{2} - 2 \, a x +{\left (3 \, b c^{2} x^{3} - 2 \, b x\right )} \arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right ) -{\left (b^{4} c^{2} \arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right )^{2} + 2 \, a b^{3} c^{2} \arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right ) + a^{2} b^{2} c^{2}\right )} \int \frac{9 \, c^{2} x^{2} - 2}{b^{3} c^{2} \arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right ) + a b^{2} c^{2}}\,{d x}}{2 \,{\left (b^{4} c^{2} \arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right )^{2} + 2 \, a b^{3} c^{2} \arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right ) + a^{2} b^{2} c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b*arccos(c*x))^3,x, algorithm="maxima")

[Out]

1/2*(3*a*c^2*x^3 + sqrt(c*x + 1)*sqrt(-c*x + 1)*b*c*x^2 - 2*a*x + (3*b*c^2*x^3 - 2*b*x)*arctan2(sqrt(c*x + 1)*
sqrt(-c*x + 1), c*x) - 2*(b^4*c^2*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)^2 + 2*a*b^3*c^2*arctan2(sqrt(c*x
+ 1)*sqrt(-c*x + 1), c*x) + a^2*b^2*c^2)*integrate(1/2*(9*c^2*x^2 - 2)/(b^3*c^2*arctan2(sqrt(c*x + 1)*sqrt(-c*
x + 1), c*x) + a*b^2*c^2), x))/(b^4*c^2*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)^2 + 2*a*b^3*c^2*arctan2(sqr
t(c*x + 1)*sqrt(-c*x + 1), c*x) + a^2*b^2*c^2)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{2}}{b^{3} \arccos \left (c x\right )^{3} + 3 \, a b^{2} \arccos \left (c x\right )^{2} + 3 \, a^{2} b \arccos \left (c x\right ) + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2/(b^3*arccos(c*x)^3 + 3*a*b^2*arccos(c*x)^2 + 3*a^2*b*arccos(c*x) + a^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/(a + b*acos(c*x))**3, x)

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Giac [B]  time = 1.44992, size = 1997, normalized size = 10.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*b^2*c^3*x^3*arccos(c*x)/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) + 3/2*a*b*c^3*x^3/
(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) - 9/2*b^2*arccos(c*x)^2*cos(a/b)^2*cos_integra
l(3*a/b + 3*arccos(c*x))*sin(a/b)/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) + 9/2*b^2*ar
ccos(c*x)^2*cos(a/b)^3*sin_integral(3*a/b + 3*arccos(c*x))/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) +
a^2*b^3*c^3) - 9*a*b*arccos(c*x)*cos(a/b)^2*cos_integral(3*a/b + 3*arccos(c*x))*sin(a/b)/(b^5*c^3*arccos(c*x)^
2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) + 9*a*b*arccos(c*x)*cos(a/b)^3*sin_integral(3*a/b + 3*arccos(c*x))/
(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) + 1/2*sqrt(-c^2*x^2 + 1)*b^2*c^2*x^2/(b^5*c^3*
arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) + 9/8*b^2*arccos(c*x)^2*cos_integral(3*a/b + 3*arccos(c
*x))*sin(a/b)/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) - 9/2*a^2*cos(a/b)^2*cos_integra
l(3*a/b + 3*arccos(c*x))*sin(a/b)/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) - 1/8*b^2*ar
ccos(c*x)^2*cos_integral(a/b + arccos(c*x))*sin(a/b)/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^
3*c^3) - 27/8*b^2*arccos(c*x)^2*cos(a/b)*sin_integral(3*a/b + 3*arccos(c*x))/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*
c^3*arccos(c*x) + a^2*b^3*c^3) + 9/2*a^2*cos(a/b)^3*sin_integral(3*a/b + 3*arccos(c*x))/(b^5*c^3*arccos(c*x)^2
+ 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) + 1/8*b^2*arccos(c*x)^2*cos(a/b)*sin_integral(a/b + arccos(c*x))/(b^
5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) - b^2*c*x*arccos(c*x)/(b^5*c^3*arccos(c*x)^2 + 2*
a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) + 9/4*a*b*arccos(c*x)*cos_integral(3*a/b + 3*arccos(c*x))*sin(a/b)/(b^5*c
^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) - 1/4*a*b*arccos(c*x)*cos_integral(a/b + arccos(c*x)
)*sin(a/b)/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) - 27/4*a*b*arccos(c*x)*cos(a/b)*sin
_integral(3*a/b + 3*arccos(c*x))/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) + 1/4*a*b*arc
cos(c*x)*cos(a/b)*sin_integral(a/b + arccos(c*x))/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c
^3) - a*b*c*x/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) + 9/8*a^2*cos_integral(3*a/b + 3
*arccos(c*x))*sin(a/b)/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) - 1/8*a^2*cos_integral(
a/b + arccos(c*x))*sin(a/b)/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) - 27/8*a^2*cos(a/b
)*sin_integral(3*a/b + 3*arccos(c*x))/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3) + 1/8*a^
2*cos(a/b)*sin_integral(a/b + arccos(c*x))/(b^5*c^3*arccos(c*x)^2 + 2*a*b^4*c^3*arccos(c*x) + a^2*b^3*c^3)