3.173 \(\int x^2 \sqrt {a+b \cos ^{-1}(c x)} \, dx\)

Optimal. Leaf size=242 \[ -\frac {\sqrt {\frac {\pi }{2}} \sqrt {b} \cos \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c^3}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {b} \cos \left (\frac {3 a}{b}\right ) C\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{12 c^3}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {b} \sin \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c^3}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {b} \sin \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{12 c^3}+\frac {1}{3} x^3 \sqrt {a+b \cos ^{-1}(c x)} \]

[Out]

-1/72*cos(3*a/b)*FresnelC(6^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*b^(1/2)*6^(1/2)*Pi^(1/2)/c^3-1/72*
FresnelS(6^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*sin(3*a/b)*b^(1/2)*6^(1/2)*Pi^(1/2)/c^3-1/8*cos(a/b
)*FresnelC(2^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*b^(1/2)*2^(1/2)*Pi^(1/2)/c^3-1/8*FresnelS(2^(1/2)
/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*sin(a/b)*b^(1/2)*2^(1/2)*Pi^(1/2)/c^3+1/3*x^3*(a+b*arccos(c*x))^(1/
2)

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Rubi [A]  time = 0.70, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4630, 4724, 3312, 3306, 3305, 3351, 3304, 3352} \[ -\frac {\sqrt {\frac {\pi }{2}} \sqrt {b} \cos \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c^3}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {b} \cos \left (\frac {3 a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{12 c^3}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {b} \sin \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c^3}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {b} \sin \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{12 c^3}+\frac {1}{3} x^3 \sqrt {a+b \cos ^{-1}(c x)} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[a + b*ArcCos[c*x]],x]

[Out]

(x^3*Sqrt[a + b*ArcCos[c*x]])/3 - (Sqrt[b]*Sqrt[Pi/2]*Cos[a/b]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/S
qrt[b]])/(4*c^3) - (Sqrt[b]*Sqrt[Pi/6]*Cos[(3*a)/b]*FresnelC[(Sqrt[6/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]])/(1
2*c^3) - (Sqrt[b]*Sqrt[Pi/2]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[a/b])/(4*c^3) - (Sqrt[
b]*Sqrt[Pi/6]*FresnelS[(Sqrt[6/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[(3*a)/b])/(12*c^3)

Rule 3304

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

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(f*x^2)/d], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3306

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

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 4630

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

Rule 4724

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> -Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Cos[x]^m*Sin[x]^(2*p + 1), x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rubi steps

\begin {align*} \int x^2 \sqrt {a+b \cos ^{-1}(c x)} \, dx &=\frac {1}{3} x^3 \sqrt {a+b \cos ^{-1}(c x)}+\frac {1}{6} (b c) \int \frac {x^3}{\sqrt {1-c^2 x^2} \sqrt {a+b \cos ^{-1}(c x)}} \, dx\\ &=\frac {1}{3} x^3 \sqrt {a+b \cos ^{-1}(c x)}-\frac {b \operatorname {Subst}\left (\int \frac {\cos ^3(x)}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{6 c^3}\\ &=\frac {1}{3} x^3 \sqrt {a+b \cos ^{-1}(c x)}-\frac {b \operatorname {Subst}\left (\int \left (\frac {3 \cos (x)}{4 \sqrt {a+b x}}+\frac {\cos (3 x)}{4 \sqrt {a+b x}}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{6 c^3}\\ &=\frac {1}{3} x^3 \sqrt {a+b \cos ^{-1}(c x)}-\frac {b \operatorname {Subst}\left (\int \frac {\cos (3 x)}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{24 c^3}-\frac {b \operatorname {Subst}\left (\int \frac {\cos (x)}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{8 c^3}\\ &=\frac {1}{3} x^3 \sqrt {a+b \cos ^{-1}(c x)}-\frac {\left (b \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{8 c^3}-\frac {\left (b \cos \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {3 a}{b}+3 x\right )}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{24 c^3}-\frac {\left (b \sin \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{8 c^3}-\frac {\left (b \sin \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {3 a}{b}+3 x\right )}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{24 c^3}\\ &=\frac {1}{3} x^3 \sqrt {a+b \cos ^{-1}(c x)}-\frac {\cos \left (\frac {a}{b}\right ) \operatorname {Subst}\left (\int \cos \left (\frac {x^2}{b}\right ) \, dx,x,\sqrt {a+b \cos ^{-1}(c x)}\right )}{4 c^3}-\frac {\cos \left (\frac {3 a}{b}\right ) \operatorname {Subst}\left (\int \cos \left (\frac {3 x^2}{b}\right ) \, dx,x,\sqrt {a+b \cos ^{-1}(c x)}\right )}{12 c^3}-\frac {\sin \left (\frac {a}{b}\right ) \operatorname {Subst}\left (\int \sin \left (\frac {x^2}{b}\right ) \, dx,x,\sqrt {a+b \cos ^{-1}(c x)}\right )}{4 c^3}-\frac {\sin \left (\frac {3 a}{b}\right ) \operatorname {Subst}\left (\int \sin \left (\frac {3 x^2}{b}\right ) \, dx,x,\sqrt {a+b \cos ^{-1}(c x)}\right )}{12 c^3}\\ &=\frac {1}{3} x^3 \sqrt {a+b \cos ^{-1}(c x)}-\frac {\sqrt {b} \sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c^3}-\frac {\sqrt {b} \sqrt {\frac {\pi }{6}} \cos \left (\frac {3 a}{b}\right ) C\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{12 c^3}-\frac {\sqrt {b} \sqrt {\frac {\pi }{2}} S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{4 c^3}-\frac {\sqrt {b} \sqrt {\frac {\pi }{6}} S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right ) \sin \left (\frac {3 a}{b}\right )}{12 c^3}\\ \end {align*}

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Mathematica [C]  time = 0.53, size = 243, normalized size = 1.00 \[ \frac {e^{-\frac {3 i a}{b}} \sqrt {a+b \cos ^{-1}(c x)} \left (9 e^{\frac {2 i a}{b}} \sqrt {\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {3}{2},-\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )+9 e^{\frac {4 i a}{b}} \sqrt {-\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {3}{2},\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )+\sqrt {3} \left (\sqrt {\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {3}{2},-\frac {3 i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )+e^{\frac {6 i a}{b}} \sqrt {-\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {3}{2},\frac {3 i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )\right )\right )}{72 c^3 \sqrt {\frac {\left (a+b \cos ^{-1}(c x)\right )^2}{b^2}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2*Sqrt[a + b*ArcCos[c*x]],x]

[Out]

(Sqrt[a + b*ArcCos[c*x]]*(9*E^(((2*I)*a)/b)*Sqrt[(I*(a + b*ArcCos[c*x]))/b]*Gamma[3/2, ((-I)*(a + b*ArcCos[c*x
]))/b] + 9*E^(((4*I)*a)/b)*Sqrt[((-I)*(a + b*ArcCos[c*x]))/b]*Gamma[3/2, (I*(a + b*ArcCos[c*x]))/b] + Sqrt[3]*
(Sqrt[(I*(a + b*ArcCos[c*x]))/b]*Gamma[3/2, ((-3*I)*(a + b*ArcCos[c*x]))/b] + E^(((6*I)*a)/b)*Sqrt[((-I)*(a +
b*ArcCos[c*x]))/b]*Gamma[3/2, ((3*I)*(a + b*ArcCos[c*x]))/b])))/(72*c^3*E^(((3*I)*a)/b)*Sqrt[(a + b*ArcCos[c*x
])^2/b^2])

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arccos(c*x))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [B]  time = 4.43, size = 1095, normalized size = 4.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arccos(c*x))^(1/2),x, algorithm="giac")

[Out]

-1/8*sqrt(2)*sqrt(pi)*a*b*i*erf(-1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*i/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcco
s(c*x) + a)*sqrt(abs(b))/b)*e^(a*i/b)/((b^2*i/sqrt(abs(b)) + b*sqrt(abs(b)))*c^3) - 1/8*sqrt(2)*sqrt(pi)*a*b*i
*erf(1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*i/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*sqrt(abs(b))/b)*
e^(-a*i/b)/((b^2*i/sqrt(abs(b)) - b*sqrt(abs(b)))*c^3) - 1/4*sqrt(pi)*a*sqrt(b)*i*erf(-1/2*sqrt(6)*sqrt(b*arcc
os(c*x) + a)*sqrt(b)*i/abs(b) - 1/2*sqrt(6)*sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(3*a*i/b)/((sqrt(6)*b^2*i/abs(b
) + sqrt(6)*b)*c^3) + 1/16*sqrt(2)*sqrt(pi)*b^2*erf(-1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*i/sqrt(abs(b)) - 1/2*
sqrt(2)*sqrt(b*arccos(c*x) + a)*sqrt(abs(b))/b)*e^(a*i/b)/((b^2*i/sqrt(abs(b)) + b*sqrt(abs(b)))*c^3) - 1/16*s
qrt(2)*sqrt(pi)*b^2*erf(1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*i/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(c*x) +
a)*sqrt(abs(b))/b)*e^(-a*i/b)/((b^2*i/sqrt(abs(b)) - b*sqrt(abs(b)))*c^3) - 1/4*sqrt(pi)*a*sqrt(b)*i*erf(1/2*s
qrt(6)*sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - 1/2*sqrt(6)*sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(-3*a*i/b)/((
sqrt(6)*b^2*i/abs(b) - sqrt(6)*b)*c^3) + 1/4*sqrt(pi)*a*i*erf(-1/2*sqrt(6)*sqrt(b*arccos(c*x) + a)*sqrt(b)*i/a
bs(b) - 1/2*sqrt(6)*sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(3*a*i/b)/((sqrt(6)*b^(3/2)*i/abs(b) + sqrt(6)*sqrt(b))
*c^3) + 1/4*sqrt(pi)*a*i*erf(-1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*i/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(c
*x) + a)*sqrt(abs(b))/b)*e^(a*i/b)/((sqrt(2)*b*i/sqrt(abs(b)) + sqrt(2)*sqrt(abs(b)))*c^3) + 1/4*sqrt(pi)*a*i*
erf(1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*i/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*sqrt(abs(b))/b)*e
^(-a*i/b)/((sqrt(2)*b*i/sqrt(abs(b)) - sqrt(2)*sqrt(abs(b)))*c^3) + 1/4*sqrt(pi)*a*i*erf(1/2*sqrt(6)*sqrt(b*ar
ccos(c*x) + a)*sqrt(b)*i/abs(b) - 1/2*sqrt(6)*sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(-3*a*i/b)/((sqrt(6)*b^(3/2)*
i/abs(b) - sqrt(6)*sqrt(b))*c^3) + 1/24*sqrt(pi)*b^(3/2)*erf(-1/2*sqrt(6)*sqrt(b*arccos(c*x) + a)*sqrt(b)*i/ab
s(b) - 1/2*sqrt(6)*sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(3*a*i/b)/((sqrt(6)*b^2*i/abs(b) + sqrt(6)*b)*c^3) - 1/2
4*sqrt(pi)*b^(3/2)*erf(1/2*sqrt(6)*sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - 1/2*sqrt(6)*sqrt(b*arccos(c*x) +
a)/sqrt(b))*e^(-3*a*i/b)/((sqrt(6)*b^2*i/abs(b) - sqrt(6)*b)*c^3) + 1/24*sqrt(b*arccos(c*x) + a)*e^(3*i*arcco
s(c*x))/c^3 + 1/8*sqrt(b*arccos(c*x) + a)*e^(i*arccos(c*x))/c^3 + 1/8*sqrt(b*arccos(c*x) + a)*e^(-i*arccos(c*x
))/c^3 + 1/24*sqrt(b*arccos(c*x) + a)*e^(-3*i*arccos(c*x))/c^3

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maple [A]  time = 0.38, size = 357, normalized size = 1.48 \[ \frac {-\sqrt {3}\, \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {3 a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) b -\sqrt {3}\, \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {3 a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) b -9 \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \cos \left (\frac {a}{b}\right ) b -9 \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sin \left (\frac {a}{b}\right ) b +6 \arccos \left (c x \right ) \cos \left (\frac {3 a +3 b \arccos \left (c x \right )}{b}-\frac {3 a}{b}\right ) b +6 \cos \left (\frac {3 a +3 b \arccos \left (c x \right )}{b}-\frac {3 a}{b}\right ) a +18 \arccos \left (c x \right ) \cos \left (\frac {a +b \arccos \left (c x \right )}{b}-\frac {a}{b}\right ) b +18 \cos \left (\frac {a +b \arccos \left (c x \right )}{b}-\frac {a}{b}\right ) a}{72 c^{3} \sqrt {a +b \arccos \left (c x \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arccos(c*x))^(1/2),x)

[Out]

1/72/c^3/(a+b*arccos(c*x))^(1/2)*(-3^(1/2)*2^(1/2)*Pi^(1/2)*(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)*cos(3*a/b)*Fre
snelC(2^(1/2)/Pi^(1/2)*3^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*b-3^(1/2)*2^(1/2)*Pi^(1/2)*(1/b)^(1/2)*(
a+b*arccos(c*x))^(1/2)*sin(3*a/b)*FresnelS(2^(1/2)/Pi^(1/2)*3^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*b-9
*2^(1/2)*Pi^(1/2)*(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^
(1/2)/b)*cos(a/b)*b-9*2^(1/2)*Pi^(1/2)*(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)/(1/b)^(1/
2)*(a+b*arccos(c*x))^(1/2)/b)*sin(a/b)*b+6*arccos(c*x)*cos(3*(a+b*arccos(c*x))/b-3*a/b)*b+6*cos(3*(a+b*arccos(
c*x))/b-3*a/b)*a+18*arccos(c*x)*cos((a+b*arccos(c*x))/b-a/b)*b+18*cos((a+b*arccos(c*x))/b-a/b)*a)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \arccos \left (c x\right ) + a} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arccos(c*x))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*arccos(c*x) + a)*x^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,\sqrt {a+b\,\mathrm {acos}\left (c\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*acos(c*x))^(1/2),x)

[Out]

int(x^2*(a + b*acos(c*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sqrt {a + b \operatorname {acos}{\left (c x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*acos(c*x))**(1/2),x)

[Out]

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

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