\(\int \arccos (a x)^{5/2} \, dx\) [90]

Optimal result

Integrand size = 8, antiderivative size = 88 \[ \int \arccos (a x)^{5/2} \, dx=-\frac {15}{4} x \sqrt {\arccos (a x)}-\frac {5 \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a}+x \arccos (a x)^{5/2}+\frac {15 \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arccos (a x)}\right )}{4 a} \]

[Out]

x*arccos(a*x)^(5/2)+15/8*FresnelC(2^(1/2)/Pi^(1/2)*arccos(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a-5/2*arccos(a*x)^(3/2)
*(-a^2*x^2+1)^(1/2)/a-15/4*x*arccos(a*x)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4716, 4768, 4810, 3385, 3433} \[ \int \arccos (a x)^{5/2} \, dx=-\frac {5 \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a}+\frac {15 \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arccos (a x)}\right )}{4 a}+x \arccos (a x)^{5/2}-\frac {15}{4} x \sqrt {\arccos (a x)} \]

[In]

Int[ArcCos[a*x]^(5/2),x]

[Out]

(-15*x*Sqrt[ArcCos[a*x]])/4 - (5*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^(3/2))/(2*a) + x*ArcCos[a*x]^(5/2) + (15*Sqrt[P
i/2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcCos[a*x]]])/(4*a)

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 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 4716

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

Rule 4768

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

Rule 4810

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

Rubi steps \begin{align*} \text {integral}& = x \arccos (a x)^{5/2}+\frac {1}{2} (5 a) \int \frac {x \arccos (a x)^{3/2}}{\sqrt {1-a^2 x^2}} \, dx \\ & = -\frac {5 \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a}+x \arccos (a x)^{5/2}-\frac {15}{4} \int \sqrt {\arccos (a x)} \, dx \\ & = -\frac {15}{4} x \sqrt {\arccos (a x)}-\frac {5 \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a}+x \arccos (a x)^{5/2}-\frac {1}{8} (15 a) \int \frac {x}{\sqrt {1-a^2 x^2} \sqrt {\arccos (a x)}} \, dx \\ & = -\frac {15}{4} x \sqrt {\arccos (a x)}-\frac {5 \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a}+x \arccos (a x)^{5/2}+\frac {15 \text {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\arccos (a x)\right )}{8 a} \\ & = -\frac {15}{4} x \sqrt {\arccos (a x)}-\frac {5 \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a}+x \arccos (a x)^{5/2}+\frac {15 \text {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\arccos (a x)}\right )}{4 a} \\ & = -\frac {15}{4} x \sqrt {\arccos (a x)}-\frac {5 \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{2 a}+x \arccos (a x)^{5/2}+\frac {15 \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arccos (a x)}\right )}{4 a} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \arccos (a x)^{5/2} \, dx=-\frac {i \left (\sqrt {-i \arccos (a x)} \Gamma \left (\frac {7}{2},-i \arccos (a x)\right )-\sqrt {i \arccos (a x)} \Gamma \left (\frac {7}{2},i \arccos (a x)\right )\right )}{2 a \sqrt {\arccos (a x)}} \]

[In]

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

[Out]

((-1/2*I)*(Sqrt[(-I)*ArcCos[a*x]]*Gamma[7/2, (-I)*ArcCos[a*x]] - Sqrt[I*ArcCos[a*x]]*Gamma[7/2, I*ArcCos[a*x]]
))/(a*Sqrt[ArcCos[a*x]])

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {\sqrt {2}\, \left (-4 \arccos \left (a x \right )^{\frac {5}{2}} \sqrt {2}\, \sqrt {\pi }\, a x +10 \arccos \left (a x \right )^{\frac {3}{2}} \sqrt {2}\, \sqrt {\pi }\, \sqrt {-a^{2} x^{2}+1}+15 \sqrt {2}\, \sqrt {\arccos \left (a x \right )}\, \sqrt {\pi }\, a x -15 \pi \,\operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right )\right )}{8 a \sqrt {\pi }}\)	\(88\)

[In]

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

[Out]

-1/8/a*2^(1/2)*(-4*arccos(a*x)^(5/2)*2^(1/2)*Pi^(1/2)*a*x+10*arccos(a*x)^(3/2)*2^(1/2)*Pi^(1/2)*(-a^2*x^2+1)^(
1/2)+15*2^(1/2)*arccos(a*x)^(1/2)*Pi^(1/2)*a*x-15*Pi*FresnelC(2^(1/2)/Pi^(1/2)*arccos(a*x)^(1/2)))/Pi^(1/2)

Fricas [F(-2)]

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

[In]

integrate(arccos(a*x)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(acos(a*x)**(5/2),x)

[Out]

Integral(acos(a*x)**(5/2), x)

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.76 \[ \int \arccos (a x)^{5/2} \, dx=\frac {\arccos \left (a x\right )^{\frac {5}{2}} e^{\left (i \, \arccos \left (a x\right )\right )}}{2 \, a} + \frac {\arccos \left (a x\right )^{\frac {5}{2}} e^{\left (-i \, \arccos \left (a x\right )\right )}}{2 \, a} + \frac {5 i \, \arccos \left (a x\right )^{\frac {3}{2}} e^{\left (i \, \arccos \left (a x\right )\right )}}{4 \, a} - \frac {5 i \, \arccos \left (a x\right )^{\frac {3}{2}} e^{\left (-i \, \arccos \left (a x\right )\right )}}{4 \, a} - \frac {\left (15 i + 15\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\arccos \left (a x\right )}\right )}{32 \, a} + \frac {\left (15 i - 15\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\arccos \left (a x\right )}\right )}{32 \, a} - \frac {15 \, \sqrt {\arccos \left (a x\right )} e^{\left (i \, \arccos \left (a x\right )\right )}}{8 \, a} - \frac {15 \, \sqrt {\arccos \left (a x\right )} e^{\left (-i \, \arccos \left (a x\right )\right )}}{8 \, a} \]

[In]

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

[Out]

1/2*arccos(a*x)^(5/2)*e^(I*arccos(a*x))/a + 1/2*arccos(a*x)^(5/2)*e^(-I*arccos(a*x))/a + 5/4*I*arccos(a*x)^(3/
2)*e^(I*arccos(a*x))/a - 5/4*I*arccos(a*x)^(3/2)*e^(-I*arccos(a*x))/a - (15/32*I + 15/32)*sqrt(2)*sqrt(pi)*erf
((1/2*I - 1/2)*sqrt(2)*sqrt(arccos(a*x)))/a + (15/32*I - 15/32)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sq
rt(arccos(a*x)))/a - 15/8*sqrt(arccos(a*x))*e^(I*arccos(a*x))/a - 15/8*sqrt(arccos(a*x))*e^(-I*arccos(a*x))/a

Mupad [F(-1)]

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

[In]

int(acos(a*x)^(5/2),x)

[Out]

int(acos(a*x)^(5/2), x)