3.1.37 \(\int \text {ArcCos}(a+b x)^{5/2} \, dx\) [37]

Optimal. Leaf size=111 \[ -\frac {15 (a+b x) \sqrt {\text {ArcCos}(a+b x)}}{4 b}-\frac {5 \sqrt {1-(a+b x)^2} \text {ArcCos}(a+b x)^{3/2}}{2 b}+\frac {(a+b x) \text {ArcCos}(a+b x)^{5/2}}{b}+\frac {15 \sqrt {\frac {\pi }{2}} \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\text {ArcCos}(a+b x)}\right )}{4 b} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4888, 4716, 4768, 4810, 3385, 3433} \begin {gather*} \frac {15 \sqrt {\frac {\pi }{2}} \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\text {ArcCos}(a+b x)}\right )}{4 b}+\frac {(a+b x) \text {ArcCos}(a+b x)^{5/2}}{b}-\frac {5 \sqrt {1-(a+b x)^2} \text {ArcCos}(a+b x)^{3/2}}{2 b}-\frac {15 (a+b x) \sqrt {\text {ArcCos}(a+b x)}}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]^(5/2),x]

[Out]

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

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]

Rule 4888

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCos[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.04, size = 90, normalized size = 0.81 \begin {gather*} -\frac {\frac {\sqrt {\text {ArcCos}(a+b x)} \text {Gamma}\left (\frac {7}{2},-i \text {ArcCos}(a+b x)\right )}{2 \sqrt {-i \text {ArcCos}(a+b x)}}+\frac {\sqrt {\text {ArcCos}(a+b x)} \text {Gamma}\left (\frac {7}{2},i \text {ArcCos}(a+b x)\right )}{2 \sqrt {i \text {ArcCos}(a+b x)}}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a + b*x]^(5/2),x]

[Out]

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

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Maple [A]
time = 0.39, size = 140, normalized size = 1.26

method	result	size
default	\(-\frac {\sqrt {2}\, \left (-4 \arccos \left (b x +a \right )^{\frac {5}{2}} \sqrt {2}\, \sqrt {\pi }\, b x -4 \arccos \left (b x +a \right )^{\frac {5}{2}} \sqrt {2}\, \sqrt {\pi }\, a +10 \arccos \left (b x +a \right )^{\frac {3}{2}} \sqrt {2}\, \sqrt {\pi }\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}+15 \sqrt {2}\, \sqrt {\arccos \left (b x +a \right )}\, \sqrt {\pi }\, b x +15 \sqrt {2}\, \sqrt {\arccos \left (b x +a \right )}\, \sqrt {\pi }\, a -15 \pi \FresnelC \left (\frac {\sqrt {2}\, \sqrt {\arccos \left (b x +a \right )}}{\sqrt {\pi }}\right )\right )}{8 b \sqrt {\pi }}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acos(a + b*x)**(5/2), x)

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Giac [C] Result contains complex when optimal does not.
time = 0.53, size = 183, normalized size = 1.65 \begin {gather*} \frac {\arccos \left (b x + a\right )^{\frac {5}{2}} e^{\left (i \, \arccos \left (b x + a\right )\right )}}{2 \, b} + \frac {\arccos \left (b x + a\right )^{\frac {5}{2}} e^{\left (-i \, \arccos \left (b x + a\right )\right )}}{2 \, b} + \frac {5 i \, \arccos \left (b x + a\right )^{\frac {3}{2}} e^{\left (i \, \arccos \left (b x + a\right )\right )}}{4 \, b} - \frac {5 i \, \arccos \left (b x + a\right )^{\frac {3}{2}} e^{\left (-i \, \arccos \left (b x + a\right )\right )}}{4 \, b} - \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 (b x + a\right )}\right )}{32 \, b} + \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 (b x + a\right )}\right )}{32 \, b} - \frac {15 \, \sqrt {\arccos \left (b x + a\right )} e^{\left (i \, \arccos \left (b x + a\right )\right )}}{8 \, b} - \frac {15 \, \sqrt {\arccos \left (b x + a\right )} e^{\left (-i \, \arccos \left (b x + a\right )\right )}}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {acos}\left (a+b\,x\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(a + b*x)^(5/2),x)

[Out]

int(acos(a + b*x)^(5/2), x)

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