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

Optimal. Leaf size=252 \[ -\frac {\sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^3}-\frac {\sqrt {\frac {3 \pi }{2}} \cos \left (\frac {3 a}{b}\right ) C\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^3}-\frac {\sqrt {\frac {\pi }{2}} \sin \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^3}-\frac {\sqrt {\frac {3 \pi }{2}} \sin \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^3}+\frac {2 x^2 \sqrt {1-c^2 x^2}}{b c \sqrt {a+b \cos ^{-1}(c x)}} \]

[Out]

-1/2*cos(a/b)*FresnelC(2^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(3/2)/c^3-1/2*Fres
nelS(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)/b^(3/2)/c^3-1/2*cos(3*a/b)*Fr
esnelC(6^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*6^(1/2)*Pi^(1/2)/b^(3/2)/c^3-1/2*FresnelS(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)/b^(3/2)/c^3+2*x^2*(-c^2*x^2+1)^(1/2)/b/c/(a
+b*arccos(c*x))^(1/2)

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Rubi [A]  time = 0.39, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4632, 3306, 3305, 3351, 3304, 3352} \[ -\frac {\sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^3}-\frac {\sqrt {\frac {3 \pi }{2}} \cos \left (\frac {3 a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^3}-\frac {\sqrt {\frac {\pi }{2}} \sin \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^3}-\frac {\sqrt {\frac {3 \pi }{2}} \sin \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c^3}+\frac {2 x^2 \sqrt {1-c^2 x^2}}{b c \sqrt {a+b \cos ^{-1}(c x)}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*ArcCos[c*x])^(3/2),x]

[Out]

(2*x^2*Sqrt[1 - c^2*x^2])/(b*c*Sqrt[a + b*ArcCos[c*x]]) - (Sqrt[Pi/2]*Cos[a/b]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b
*ArcCos[c*x]])/Sqrt[b]])/(b^(3/2)*c^3) - (Sqrt[(3*Pi)/2]*Cos[(3*a)/b]*FresnelC[(Sqrt[6/Pi]*Sqrt[a + b*ArcCos[c
*x]])/Sqrt[b]])/(b^(3/2)*c^3) - (Sqrt[Pi/2]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[a/b])/(
b^(3/2)*c^3) - (Sqrt[(3*Pi)/2]*FresnelS[(Sqrt[6/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[(3*a)/b])/(b^(3/2)*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 4632

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[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n + 1
), Cos[x]^(m - 1)*(m - (m + 1)*Cos[x]^2), x], x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] &&
GeQ[n, -2] && LtQ[n, -1]

Rubi steps

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

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Mathematica [C]  time = 0.57, size = 273, normalized size = 1.08 \[ \frac {e^{-\frac {3 i a}{b}} \left (8 c^2 x^2 e^{\frac {3 i a}{b}} \sqrt {1-c^2 x^2}+i e^{\frac {2 i a}{b}} \sqrt {-\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )-i e^{\frac {4 i a}{b}} \sqrt {\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )+i \sqrt {3} \sqrt {-\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {3 i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )-i \sqrt {3} e^{\frac {6 i a}{b}} \sqrt {\frac {i \left (a+b \cos ^{-1}(c x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {3 i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )\right )}{4 b c^3 \sqrt {a+b \cos ^{-1}(c x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.27, size = 295, normalized size = 1.17 \[ \frac {-\sqrt {3}\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {3 a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}-\sqrt {3}\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {3 a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}-\sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}-\sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}+\sin \left (\frac {a +b \arccos \left (c x \right )}{b}-\frac {a}{b}\right )+\sin \left (\frac {3 a +3 b \arccos \left (c x \right )}{b}-\frac {3 a}{b}\right )}{2 c^{3} b \sqrt {a +b \arccos \left (c x \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2}{{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*acos(c*x))^(3/2),x)

[Out]

int(x^2/(a + b*acos(c*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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