\(\int \frac {x}{(a+b \arccos (c x))^{5/2}} \, dx\) [199]

Optimal result

Integrand size = 14, antiderivative size = 180 \[ \int \frac {x}{(a+b \arccos (c x))^{5/2}} \, dx=\frac {2 x \sqrt {1-c^2 x^2}}{3 b c (a+b \arccos (c x))^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \arccos (c x)}}+\frac {8 x^2}{3 b^2 \sqrt {a+b \arccos (c x)}}+\frac {8 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arccos (c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{3 b^{5/2} c^2}-\frac {8 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arccos (c x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{3 b^{5/2} c^2} \]

[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)

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {4730, 4808, 4732, 4491, 12, 3387, 3386, 3432, 3385, 3433, 4738} \[ \int \frac {x}{(a+b \arccos (c x))^{5/2}} \, dx=-\frac {8 \sqrt {\pi } \sin \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arccos (c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{3 b^{5/2} c^2}+\frac {8 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arccos (c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{3 b^{5/2} c^2}-\frac {4}{3 b^2 c^2 \sqrt {a+b \arccos (c x)}}+\frac {8 x^2}{3 b^2 \sqrt {a+b \arccos (c x)}}+\frac {2 x \sqrt {1-c^2 x^2}}{3 b c (a+b \arccos (c x))^{3/2}} \]

[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 3385

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 3386

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 3387

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 3432

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

Rule 3433

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

Rule 4491

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 4730

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(-x^m)*Sqrt[1 - c^2*x^2]*((a + b*Arc
Cos[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 4732

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

Rule 4738

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

Rule 4808

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

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

Mathematica [F]

\[ \int \frac {x}{(a+b \arccos (c x))^{5/2}} \, dx=\int \frac {x}{(a+b \arccos (c x))^{5/2}} \, dx \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(339\) vs. \(2(142)=284\).

Time = 2.08 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.89

method	result	size
default	\(\frac {-8 \arccos \left (c x \right ) \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) b -8 \arccos \left (c x \right ) \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) b -8 \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) a -8 \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) a +4 \arccos \left (c x \right ) \cos \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) b -\sin \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) b +4 \cos \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a}{3 c^{2} b^{2} \left (a +b \arccos \left (c x \right )\right )^{\frac {3}{2}}}\)	\(340\)

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {x}{(a+b \arccos (c x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[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)

Sympy [F]

\[ \int \frac {x}{(a+b \arccos (c x))^{5/2}} \, dx=\int \frac {x}{\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{(a+b \arccos (c x))^{5/2}} \, dx=\int \frac {x}{{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(x/(a + b*acos(c*x))^(5/2),x)

[Out]

int(x/(a + b*acos(c*x))^(5/2), x)