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

Optimal. Leaf size=223 \[ \frac {\sqrt {\frac {\pi }{2}} \sin \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{2 \sqrt {b} c^3}+\frac {\sqrt {\frac {\pi }{6}} \sin \left (\frac {3 a}{b}\right ) C\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{2 \sqrt {b} c^3}-\frac {\sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{2 \sqrt {b} c^3}-\frac {\sqrt {\frac {\pi }{6}} \cos \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{2 \sqrt {b} c^3} \]

[Out]

-1/12*cos(3*a/b)*FresnelS(6^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*6^(1/2)*Pi^(1/2)/c^3/b^(1/2)+1/12*
FresnelC(6^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*sin(3*a/b)*6^(1/2)*Pi^(1/2)/c^3/b^(1/2)-1/4*cos(a/b
)*FresnelS(2^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/c^3/b^(1/2)+1/4*FresnelC(2^(1/2)
/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*sin(a/b)*2^(1/2)*Pi^(1/2)/c^3/b^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4636, 4406, 3306, 3305, 3351, 3304, 3352} \[ \frac {\sqrt {\frac {\pi }{2}} \sin \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{2 \sqrt {b} c^3}+\frac {\sqrt {\frac {\pi }{6}} \sin \left (\frac {3 a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{2 \sqrt {b} c^3}-\frac {\sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{2 \sqrt {b} c^3}-\frac {\sqrt {\frac {\pi }{6}} \cos \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{2 \sqrt {b} c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a + b*ArcCos[c*x]],x]

[Out]

-(Sqrt[Pi/2]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]])/(2*Sqrt[b]*c^3) - (Sqrt[Pi/6]*Co
s[(3*a)/b]*FresnelS[(Sqrt[6/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]])/(2*Sqrt[b]*c^3) + (Sqrt[Pi/2]*FresnelC[(Sqr
t[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[a/b])/(2*Sqrt[b]*c^3) + (Sqrt[Pi/6]*FresnelC[(Sqrt[6/Pi]*Sqrt[a
+ b*ArcCos[c*x]])/Sqrt[b]]*Sin[(3*a)/b])/(2*Sqrt[b]*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 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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.46, size = 225, normalized size = 1.01 \[ \frac {e^{-\frac {3 i a}{b}} \left (3 e^{\frac {2 i a}{b}} \sqrt {-\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )+3 e^{\frac {4 i a}{b}} \sqrt {\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )+\sqrt {3} \left (\sqrt {-\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {3 i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )+e^{\frac {6 i a}{b}} \sqrt {\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {3 i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )\right )\right )}{24 c^3 \sqrt {a+b \cos ^{-1}(c x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3*E^(((2*I)*a)/b)*Sqrt[((-I)*(a + b*ArcCos[c*x]))/b]*Gamma[1/2, ((-I)*(a + b*ArcCos[c*x]))/b] + 3*E^(((4*I)*a
)/b)*Sqrt[(I*(a + b*ArcCos[c*x]))/b]*Gamma[1/2, (I*(a + b*ArcCos[c*x]))/b] + Sqrt[3]*(Sqrt[((-I)*(a + b*ArcCos
[c*x]))/b]*Gamma[1/2, ((-3*I)*(a + b*ArcCos[c*x]))/b] + E^(((6*I)*a)/b)*Sqrt[(I*(a + b*ArcCos[c*x]))/b]*Gamma[
1/2, ((3*I)*(a + b*ArcCos[c*x]))/b]))/(24*c^3*E^(((3*I)*a)/b)*Sqrt[a + b*ArcCos[c*x]])

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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 [A]  time = 1.45, size = 330, normalized size = 1.48 \[ \frac {\sqrt {\pi } i \operatorname {erf}\left (-\frac {\sqrt {6} \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b} i}{2 \, {\left | b \right |}} - \frac {\sqrt {6} \sqrt {b \arccos \left (c x\right ) + a}}{2 \, \sqrt {b}}\right ) e^{\left (\frac {3 \, a i}{b}\right )}}{4 \, {\left (\frac {\sqrt {6} b^{\frac {3}{2}} i}{{\left | b \right |}} + \sqrt {6} \sqrt {b}\right )} c^{3}} + \frac {\sqrt {\pi } i \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {b \arccos \left (c x\right ) + a} i}{2 \, \sqrt {{\left | b \right |}}} - \frac {\sqrt {2} \sqrt {b \arccos \left (c x\right ) + a} \sqrt {{\left | b \right |}}}{2 \, b}\right ) e^{\left (\frac {a i}{b}\right )}}{4 \, {\left (\frac {\sqrt {2} b i}{\sqrt {{\left | b \right |}}} + \sqrt {2} \sqrt {{\left | b \right |}}\right )} c^{3}} + \frac {\sqrt {\pi } i \operatorname {erf}\left (\frac {\sqrt {2} \sqrt {b \arccos \left (c x\right ) + a} i}{2 \, \sqrt {{\left | b \right |}}} - \frac {\sqrt {2} \sqrt {b \arccos \left (c x\right ) + a} \sqrt {{\left | b \right |}}}{2 \, b}\right ) e^{\left (-\frac {a i}{b}\right )}}{4 \, {\left (\frac {\sqrt {2} b i}{\sqrt {{\left | b \right |}}} - \sqrt {2} \sqrt {{\left | b \right |}}\right )} c^{3}} + \frac {\sqrt {\pi } i \operatorname {erf}\left (\frac {\sqrt {6} \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b} i}{2 \, {\left | b \right |}} - \frac {\sqrt {6} \sqrt {b \arccos \left (c x\right ) + a}}{2 \, \sqrt {b}}\right ) e^{\left (-\frac {3 \, a i}{b}\right )}}{4 \, {\left (\frac {\sqrt {6} b^{\frac {3}{2}} i}{{\left | b \right |}} - \sqrt {6} \sqrt {b}\right )} c^{3}} \]

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]

1/4*sqrt(pi)*i*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^(3/2)*i/abs(b) + sqrt(6)*sqrt(b))*c^3) + 1/4*sqrt(pi)*i*erf(-1/2*sqrt(2)*sqr
t(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)*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)*i*erf(1/2*sqrt(6)*sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - 1/2*sqrt(6)*sqr
t(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)

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maple [A]  time = 0.27, size = 167, normalized size = 0.75 \[ -\frac {\sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \left (\sqrt {3}\, \cos \left (\frac {3 a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )-\sqrt {3}\, \sin \left (\frac {3 a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )+3 \cos \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )-3 \sin \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )\right )}{12 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/c^3*2^(1/2)*Pi^(1/2)*(1/b)^(1/2)*(3^(1/2)*cos(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)-3^(1/2)*sin(3*a/b)*FresnelC(2^(1/2)/Pi^(1/2)*3^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)
/b)+3*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)-3*sin(a/b)*FresnelC(2^(1/2)/Pi
^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b))

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

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(x^2/sqrt(b*arccos(c*x) + a), x)

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

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 \[ \int \frac {x^{2}}{\sqrt {a + b \operatorname {acos}{\left (c x \right )}}}\, dx \]

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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