3.199 \(\int \frac {x}{(a+b \cos ^{-1}(c x))^{5/2}} \, dx\)

Optimal. Leaf size=180 \[ -\frac {8 \sqrt {\pi } \sin \left (\frac {2 a}{b}\right ) C\left (\frac {2 \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{3 b^{5/2} c^2}+\frac {8 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{3 b^{5/2} c^2}-\frac {4}{3 b^2 c^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {8 x^2}{3 b^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {2 x \sqrt {1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}} \]

[Out]

8/3*cos(2*a/b)*FresnelS(2*(a+b*arccos(c*x))^(1/2)/b^(1/2)/Pi^(1/2))*Pi^(1/2)/b^(5/2)/c^2-8/3*FresnelC(2*(a+b*a
rccos(c*x))^(1/2)/b^(1/2)/Pi^(1/2))*sin(2*a/b)*Pi^(1/2)/b^(5/2)/c^2+2/3*x*(-c^2*x^2+1)^(1/2)/b/c/(a+b*arccos(c
*x))^(3/2)-4/3/b^2/c^2/(a+b*arccos(c*x))^(1/2)+8/3*x^2/b^2/(a+b*arccos(c*x))^(1/2)

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Rubi [A]  time = 0.49, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {4634, 4720, 4636, 4406, 12, 3306, 3305, 3351, 3304, 3352, 4642} \[ -\frac {8 \sqrt {\pi } \sin \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {\pi } \sqrt {b}}\right )}{3 b^{5/2} c^2}+\frac {8 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{3 b^{5/2} c^2}-\frac {4}{3 b^2 c^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {8 x^2}{3 b^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {2 x \sqrt {1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*ArcCos[c*x])^(5/2),x]

[Out]

(2*x*Sqrt[1 - c^2*x^2])/(3*b*c*(a + b*ArcCos[c*x])^(3/2)) - 4/(3*b^2*c^2*Sqrt[a + b*ArcCos[c*x]]) + (8*x^2)/(3
*b^2*Sqrt[a + b*ArcCos[c*x]]) + (8*Sqrt[Pi]*Cos[(2*a)/b]*FresnelS[(2*Sqrt[a + b*ArcCos[c*x]])/(Sqrt[b]*Sqrt[Pi
])])/(3*b^(5/2)*c^2) - (8*Sqrt[Pi]*FresnelC[(2*Sqrt[a + b*ArcCos[c*x]])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b])/(3*b
^(5/2)*c^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 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 4642

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

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]

Rubi steps

\begin {align*} \int \frac {x}{\left (a+b \cos ^{-1}(c x)\right )^{5/2}} \, dx &=\frac {2 x \sqrt {1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}-\frac {2 \int \frac {1}{\sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^{3/2}} \, dx}{3 b c}+\frac {(4 c) \int \frac {x^2}{\sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^{3/2}} \, dx}{3 b}\\ &=\frac {2 x \sqrt {1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {8 x^2}{3 b^2 \sqrt {a+b \cos ^{-1}(c x)}}-\frac {16 \int \frac {x}{\sqrt {a+b \cos ^{-1}(c x)}} \, dx}{3 b^2}\\ &=\frac {2 x \sqrt {1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {8 x^2}{3 b^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {16 \operatorname {Subst}\left (\int \frac {\cos (x) \sin (x)}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{3 b^2 c^2}\\ &=\frac {2 x \sqrt {1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {8 x^2}{3 b^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {16 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{2 \sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{3 b^2 c^2}\\ &=\frac {2 x \sqrt {1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {8 x^2}{3 b^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {8 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{3 b^2 c^2}\\ &=\frac {2 x \sqrt {1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {8 x^2}{3 b^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {\left (8 \cos \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{3 b^2 c^2}-\frac {\left (8 \sin \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{3 b^2 c^2}\\ &=\frac {2 x \sqrt {1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {8 x^2}{3 b^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {\left (16 \cos \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \cos ^{-1}(c x)}\right )}{3 b^3 c^2}-\frac {\left (16 \sin \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \cos ^{-1}(c x)}\right )}{3 b^3 c^2}\\ &=\frac {2 x \sqrt {1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {8 x^2}{3 b^2 \sqrt {a+b \cos ^{-1}(c x)}}+\frac {8 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{3 b^{5/2} c^2}-\frac {8 \sqrt {\pi } C\left (\frac {2 \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{3 b^{5/2} c^2}\\ \end {align*}

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Mathematica [A]  time = 0.76, size = 176, normalized size = 0.98 \[ \frac {-8 \sqrt {\pi } \sqrt {\frac {1}{b}} \sin \left (\frac {2 a}{b}\right ) \left (a+b \cos ^{-1}(c x)\right )^{3/2} C\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {\pi }}\right )+8 \sqrt {\pi } \sqrt {\frac {1}{b}} \cos \left (\frac {2 a}{b}\right ) \left (a+b \cos ^{-1}(c x)\right )^{3/2} S\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {\pi }}\right )+4 a \cos \left (2 \cos ^{-1}(c x)\right )+4 b \cos ^{-1}(c x) \cos \left (2 \cos ^{-1}(c x)\right )+b \sin \left (2 \cos ^{-1}(c x)\right )}{3 b^2 c^2 \left (a+b \cos ^{-1}(c x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*ArcCos[c*x])^(5/2),x]

[Out]

(4*a*Cos[2*ArcCos[c*x]] + 4*b*ArcCos[c*x]*Cos[2*ArcCos[c*x]] + 8*Sqrt[b^(-1)]*Sqrt[Pi]*(a + b*ArcCos[c*x])^(3/
2)*Cos[(2*a)/b]*FresnelS[(2*Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[Pi]] - 8*Sqrt[b^(-1)]*Sqrt[Pi]*(a + b*A
rcCos[c*x])^(3/2)*FresnelC[(2*Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[Pi]]*Sin[(2*a)/b] + b*Sin[2*ArcCos[c*
x]])/(3*b^2*c^2*(a + b*ArcCos[c*x])^(3/2))

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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/(a+b*arccos(c*x))^(5/2),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 \[ \int \frac {x}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*arccos(c*x))^(5/2),x, algorithm="giac")

[Out]

integrate(x/(b*arccos(c*x) + a)^(5/2), x)

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maple [B]  time = 0.22, size = 311, normalized size = 1.73 \[ \frac {8 \arccos \left (c x \right ) \sqrt {\pi }\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {\frac {1}{b}}\, b -8 \arccos \left (c x \right ) \sqrt {\pi }\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {\frac {1}{b}}\, b +8 \sqrt {\pi }\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {\frac {1}{b}}\, a -8 \sqrt {\pi }\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {\frac {1}{b}}\, a +4 \arccos \left (c x \right ) \cos \left (\frac {2 a +2 b \arccos \left (c x \right )}{b}-\frac {2 a}{b}\right ) b +\sin \left (\frac {2 a +2 b \arccos \left (c x \right )}{b}-\frac {2 a}{b}\right ) b +4 \cos \left (\frac {2 a +2 b \arccos \left (c x \right )}{b}-\frac {2 a}{b}\right ) a}{3 c^{2} b^{2} \left (a +b \arccos \left (c x \right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+b*arccos(c*x))^(5/2),x)

[Out]

1/3/c^2/b^2*(8*arccos(c*x)*Pi^(1/2)*(a+b*arccos(c*x))^(1/2)*cos(2*a/b)*FresnelS(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*ar
ccos(c*x))^(1/2)/b)*(1/b)^(1/2)*b-8*arccos(c*x)*Pi^(1/2)*(a+b*arccos(c*x))^(1/2)*sin(2*a/b)*FresnelC(2/Pi^(1/2
)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*(1/b)^(1/2)*b+8*Pi^(1/2)*(a+b*arccos(c*x))^(1/2)*cos(2*a/b)*FresnelS(
2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*(1/b)^(1/2)*a-8*Pi^(1/2)*(a+b*arccos(c*x))^(1/2)*sin(2*a/b)*
FresnelC(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*(1/b)^(1/2)*a+4*arccos(c*x)*cos(2*(a+b*arccos(c*x))
/b-2*a/b)*b+sin(2*(a+b*arccos(c*x))/b-2*a/b)*b+4*cos(2*(a+b*arccos(c*x))/b-2*a/b)*a)/(a+b*arccos(c*x))^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*arccos(c*x))^(5/2),x, algorithm="maxima")

[Out]

integrate(x/(b*arccos(c*x) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*acos(c*x))^(5/2),x)

[Out]

int(x/(a + b*acos(c*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*acos(c*x))**(5/2),x)

[Out]

Integral(x/(a + b*acos(c*x))**(5/2), x)

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