\(\int x \arccos (a x)^{5/2} \, dx\) [89]

Optimal result

Integrand size = 10, antiderivative size = 119 \[ \int x \arccos (a x)^{5/2} \, dx=\frac {15 \sqrt {\arccos (a x)}}{64 a^2}-\frac {15}{32} x^2 \sqrt {\arccos (a x)}-\frac {5 x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{8 a}-\frac {\arccos (a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \arccos (a x)^{5/2}+\frac {15 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )}{128 a^2} \]

[Out]

-1/4*arccos(a*x)^(5/2)/a^2+1/2*x^2*arccos(a*x)^(5/2)+15/128*FresnelC(2*arccos(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^
2-5/8*x*arccos(a*x)^(3/2)*(-a^2*x^2+1)^(1/2)/a+15/64*arccos(a*x)^(1/2)/a^2-15/32*x^2*arccos(a*x)^(1/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {4726, 4796, 4738, 4810, 3393, 3385, 3433} \[ \int x \arccos (a x)^{5/2} \, dx=\frac {15 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )}{128 a^2}-\frac {5 x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{8 a}-\frac {\arccos (a x)^{5/2}}{4 a^2}+\frac {15 \sqrt {\arccos (a x)}}{64 a^2}+\frac {1}{2} x^2 \arccos (a x)^{5/2}-\frac {15}{32} x^2 \sqrt {\arccos (a x)} \]

[In]

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

[Out]

(15*Sqrt[ArcCos[a*x]])/(64*a^2) - (15*x^2*Sqrt[ArcCos[a*x]])/32 - (5*x*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^(3/2))/(8
*a) - ArcCos[a*x]^(5/2)/(4*a^2) + (x^2*ArcCos[a*x]^(5/2))/2 + (15*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcCos[a*x]])/Sqrt
[Pi]])/(128*a^2)

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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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 4726

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcCos[c*x])^n/(m
+ 1)), x] + Dist[b*c*(n/(m + 1)), Int[x^(m + 1)*((a + b*ArcCos[c*x])^(n - 1)/Sqrt[1 - c^2*x^2]), x], x] /; Fre
eQ[{a, b, c}, x] && IGtQ[m, 0] && GtQ[n, 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 4796

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

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}& = \frac {1}{2} x^2 \arccos (a x)^{5/2}+\frac {1}{4} (5 a) \int \frac {x^2 \arccos (a x)^{3/2}}{\sqrt {1-a^2 x^2}} \, dx \\ & = -\frac {5 x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{8 a}+\frac {1}{2} x^2 \arccos (a x)^{5/2}-\frac {15}{16} \int x \sqrt {\arccos (a x)} \, dx+\frac {5 \int \frac {\arccos (a x)^{3/2}}{\sqrt {1-a^2 x^2}} \, dx}{8 a} \\ & = -\frac {15}{32} x^2 \sqrt {\arccos (a x)}-\frac {5 x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{8 a}-\frac {\arccos (a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \arccos (a x)^{5/2}-\frac {1}{64} (15 a) \int \frac {x^2}{\sqrt {1-a^2 x^2} \sqrt {\arccos (a x)}} \, dx \\ & = -\frac {15}{32} x^2 \sqrt {\arccos (a x)}-\frac {5 x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{8 a}-\frac {\arccos (a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \arccos (a x)^{5/2}+\frac {15 \text {Subst}\left (\int \frac {\cos ^2(x)}{\sqrt {x}} \, dx,x,\arccos (a x)\right )}{64 a^2} \\ & = -\frac {15}{32} x^2 \sqrt {\arccos (a x)}-\frac {5 x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{8 a}-\frac {\arccos (a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \arccos (a x)^{5/2}+\frac {15 \text {Subst}\left (\int \left (\frac {1}{2 \sqrt {x}}+\frac {\cos (2 x)}{2 \sqrt {x}}\right ) \, dx,x,\arccos (a x)\right )}{64 a^2} \\ & = \frac {15 \sqrt {\arccos (a x)}}{64 a^2}-\frac {15}{32} x^2 \sqrt {\arccos (a x)}-\frac {5 x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{8 a}-\frac {\arccos (a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \arccos (a x)^{5/2}+\frac {15 \text {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\arccos (a x)\right )}{128 a^2} \\ & = \frac {15 \sqrt {\arccos (a x)}}{64 a^2}-\frac {15}{32} x^2 \sqrt {\arccos (a x)}-\frac {5 x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{8 a}-\frac {\arccos (a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \arccos (a x)^{5/2}+\frac {15 \text {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\arccos (a x)}\right )}{64 a^2} \\ & = \frac {15 \sqrt {\arccos (a x)}}{64 a^2}-\frac {15}{32} x^2 \sqrt {\arccos (a x)}-\frac {5 x \sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}{8 a}-\frac {\arccos (a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \arccos (a x)^{5/2}+\frac {15 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )}{128 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.61 \[ \int x \arccos (a x)^{5/2} \, dx=\frac {15 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )-2 \sqrt {\arccos (a x)} \left (\left (15-16 \arccos (a x)^2\right ) \cos (2 \arccos (a x))+20 \arccos (a x) \sin (2 \arccos (a x))\right )}{128 a^2} \]

[In]

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

[Out]

(15*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcCos[a*x]])/Sqrt[Pi]] - 2*Sqrt[ArcCos[a*x]]*((15 - 16*ArcCos[a*x]^2)*Cos[2*Arc
Cos[a*x]] + 20*ArcCos[a*x]*Sin[2*ArcCos[a*x]]))/(128*a^2)

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {32 \arccos \left (a x \right )^{\frac {5}{2}} \cos \left (2 \arccos \left (a x \right )\right ) \sqrt {\pi }-40 \arccos \left (a x \right )^{\frac {3}{2}} \sin \left (2 \arccos \left (a x \right )\right ) \sqrt {\pi }-30 \cos \left (2 \arccos \left (a x \right )\right ) \sqrt {\arccos \left (a x \right )}\, \sqrt {\pi }+15 \pi \,\operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right )}{128 a^{2} \sqrt {\pi }}\)	\(79\)

[In]

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

[Out]

1/128/a^2/Pi^(1/2)*(32*arccos(a*x)^(5/2)*cos(2*arccos(a*x))*Pi^(1/2)-40*arccos(a*x)^(3/2)*sin(2*arccos(a*x))*P
i^(1/2)-30*cos(2*arccos(a*x))*arccos(a*x)^(1/2)*Pi^(1/2)+15*Pi*FresnelC(2*arccos(a*x)^(1/2)/Pi^(1/2)))

Fricas [F(-2)]

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

[In]

integrate(x*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]

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

[In]

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

[Out]

Integral(x*acos(a*x)**(5/2), x)

Maxima [F(-2)]

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

[In]

integrate(x*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 = 0.33 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.20 \[ \int x \arccos (a x)^{5/2} \, dx=\frac {\arccos \left (a x\right )^{\frac {5}{2}} e^{\left (2 i \, \arccos \left (a x\right )\right )}}{8 \, a^{2}} + \frac {\arccos \left (a x\right )^{\frac {5}{2}} e^{\left (-2 i \, \arccos \left (a x\right )\right )}}{8 \, a^{2}} + \frac {5 i \, \arccos \left (a x\right )^{\frac {3}{2}} e^{\left (2 i \, \arccos \left (a x\right )\right )}}{32 \, a^{2}} - \frac {5 i \, \arccos \left (a x\right )^{\frac {3}{2}} e^{\left (-2 i \, \arccos \left (a x\right )\right )}}{32 \, a^{2}} - \frac {\left (15 i + 15\right ) \, \sqrt {\pi } \operatorname {erf}\left (\left (i - 1\right ) \, \sqrt {\arccos \left (a x\right )}\right )}{512 \, a^{2}} + \frac {\left (15 i - 15\right ) \, \sqrt {\pi } \operatorname {erf}\left (-\left (i + 1\right ) \, \sqrt {\arccos \left (a x\right )}\right )}{512 \, a^{2}} - \frac {15 \, \sqrt {\arccos \left (a x\right )} e^{\left (2 i \, \arccos \left (a x\right )\right )}}{128 \, a^{2}} - \frac {15 \, \sqrt {\arccos \left (a x\right )} e^{\left (-2 i \, \arccos \left (a x\right )\right )}}{128 \, a^{2}} \]

[In]

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

[Out]

1/8*arccos(a*x)^(5/2)*e^(2*I*arccos(a*x))/a^2 + 1/8*arccos(a*x)^(5/2)*e^(-2*I*arccos(a*x))/a^2 + 5/32*I*arccos
(a*x)^(3/2)*e^(2*I*arccos(a*x))/a^2 - 5/32*I*arccos(a*x)^(3/2)*e^(-2*I*arccos(a*x))/a^2 - (15/512*I + 15/512)*
sqrt(pi)*erf((I - 1)*sqrt(arccos(a*x)))/a^2 + (15/512*I - 15/512)*sqrt(pi)*erf(-(I + 1)*sqrt(arccos(a*x)))/a^2
- 15/128*sqrt(arccos(a*x))*e^(2*I*arccos(a*x))/a^2 - 15/128*sqrt(arccos(a*x))*e^(-2*I*arccos(a*x))/a^2

Mupad [F(-1)]

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

[In]

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

[Out]

int(x*acos(a*x)^(5/2), x)