3.2.85 \(\int (a+b \text {ArcCos}(c x))^{5/2} \, dx\) [185]

Optimal. Leaf size=179 \[ -\frac {15}{4} b^2 x \sqrt {a+b \text {ArcCos}(c x)}-\frac {5 b \sqrt {1-c^2 x^2} (a+b \text {ArcCos}(c x))^{3/2}}{2 c}+x (a+b \text {ArcCos}(c x))^{5/2}+\frac {15 b^{5/2} \sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {b}}\right )}{4 c}+\frac {15 b^{5/2} \sqrt {\frac {\pi }{2}} S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{4 c} \]

[Out]

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

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Rubi [A]
time = 0.28, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {4716, 4768, 4810, 3387, 3386, 3432, 3385, 3433} \begin {gather*} \frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \cos \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {b}}\right )}{4 c}+\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \sin \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {b}}\right )}{4 c}-\frac {15}{4} b^2 x \sqrt {a+b \text {ArcCos}(c x)}-\frac {5 b \sqrt {1-c^2 x^2} (a+b \text {ArcCos}(c x))^{3/2}}{2 c}+x (a+b \text {ArcCos}(c x))^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[c*x])^(5/2),x]

[Out]

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

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

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Mathematica [C] Result contains complex when optimal does not.
time = 2.96, size = 383, normalized size = 2.14 \begin {gather*} \frac {b e^{-\frac {i a}{b}} \left (-2 e^{\frac {i a}{b}} (a+b \text {ArcCos}(c x)) \left (5 \left (3 b c x+2 a \sqrt {1-c^2 x^2}\right )+\left (-8 a c x+10 b \sqrt {1-c^2 x^2}\right ) \text {ArcCos}(c x)-4 b c x \text {ArcCos}(c x)^2\right )+\sqrt {\frac {1}{b}} \left (4 a^2+15 b^2\right ) \left (1+e^{\frac {2 i a}{b}}\right ) \sqrt {\frac {\pi }{2}} \sqrt {a+b \text {ArcCos}(c x)} \text {FresnelC}\left (\sqrt {\frac {1}{b}} \sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcCos}(c x)}\right )-i \sqrt {\frac {1}{b}} \left (4 a^2+15 b^2\right ) \left (-1+e^{\frac {2 i a}{b}}\right ) \sqrt {\frac {\pi }{2}} \sqrt {a+b \text {ArcCos}(c x)} S\left (\sqrt {\frac {1}{b}} \sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcCos}(c x)}\right )+4 i a^2 \sqrt {-\frac {i (a+b \text {ArcCos}(c x))}{b}} \text {Gamma}\left (\frac {3}{2},-\frac {i (a+b \text {ArcCos}(c x))}{b}\right )-4 i a^2 e^{\frac {2 i a}{b}} \sqrt {\frac {i (a+b \text {ArcCos}(c x))}{b}} \text {Gamma}\left (\frac {3}{2},\frac {i (a+b \text {ArcCos}(c x))}{b}\right )\right )}{8 c \sqrt {a+b \text {ArcCos}(c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(-2*E^((I*a)/b)*(a + b*ArcCos[c*x])*(5*(3*b*c*x + 2*a*Sqrt[1 - c^2*x^2]) + (-8*a*c*x + 10*b*Sqrt[1 - c^2*x^
2])*ArcCos[c*x] - 4*b*c*x*ArcCos[c*x]^2) + Sqrt[b^(-1)]*(4*a^2 + 15*b^2)*(1 + E^(((2*I)*a)/b))*Sqrt[Pi/2]*Sqrt
[a + b*ArcCos[c*x]]*FresnelC[Sqrt[b^(-1)]*Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]]] - I*Sqrt[b^(-1)]*(4*a^2 + 15*b^2
)*(-1 + E^(((2*I)*a)/b))*Sqrt[Pi/2]*Sqrt[a + b*ArcCos[c*x]]*FresnelS[Sqrt[b^(-1)]*Sqrt[2/Pi]*Sqrt[a + b*ArcCos
[c*x]]] + (4*I)*a^2*Sqrt[((-I)*(a + b*ArcCos[c*x]))/b]*Gamma[3/2, ((-I)*(a + b*ArcCos[c*x]))/b] - (4*I)*a^2*E^
(((2*I)*a)/b)*Sqrt[(I*(a + b*ArcCos[c*x]))/b]*Gamma[3/2, (I*(a + b*ArcCos[c*x]))/b]))/(8*c*E^((I*a)/b)*Sqrt[a
+ b*ArcCos[c*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(400\) vs. \(2(139)=278\).
time = 0.26, size = 401, normalized size = 2.24

method	result	size
default	\(\frac {15 \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) b^{3}-15 \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) b^{3}+8 \arccos \left (c x \right )^{3} \cos \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) b^{3}+24 \arccos \left (c x \right )^{2} \cos \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) a \,b^{2}+20 \arccos \left (c x \right )^{2} \sin \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) b^{3}+24 \arccos \left (c x \right ) \cos \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) a^{2} b -30 \arccos \left (c x \right ) \cos \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) b^{3}+40 \arccos \left (c x \right ) \sin \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) a \,b^{2}+8 \cos \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) a^{3}-30 \cos \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) a \,b^{2}+20 \sin \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) a^{2} b}{8 c \sqrt {a +b \arccos \left (c x \right )}}\)	\(401\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/c/(a+b*arccos(c*x))^(1/2)*(15*(-1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arccos(c*x))^(1/2)*cos(a/b)*FresnelC(2^(1
/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*b^3-15*(-1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arccos(c*x))^(1
/2)*sin(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*b^3+8*arccos(c*x)^3*cos(-(a+b*a
rccos(c*x))/b+a/b)*b^3+24*arccos(c*x)^2*cos(-(a+b*arccos(c*x))/b+a/b)*a*b^2+20*arccos(c*x)^2*sin(-(a+b*arccos(
c*x))/b+a/b)*b^3+24*arccos(c*x)*cos(-(a+b*arccos(c*x))/b+a/b)*a^2*b-30*arccos(c*x)*cos(-(a+b*arccos(c*x))/b+a/
b)*b^3+40*arccos(c*x)*sin(-(a+b*arccos(c*x))/b+a/b)*a*b^2+8*cos(-(a+b*arccos(c*x))/b+a/b)*a^3-30*cos(-(a+b*arc
cos(c*x))/b+a/b)*a*b^2+20*sin(-(a+b*arccos(c*x))/b+a/b)*a^2*b)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acos(c*x))**(5/2), x)

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Giac [C] Result contains complex when optimal does not.
time = 1.45, size = 1177, normalized size = 6.58 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*sqrt(2)*sqrt(pi)*a^3*b^3*erf(-1/2*I*sqrt(2)*sqrt(b*arccos(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*a
rccos(c*x) + a)*sqrt(abs(b))/b)*e^(I*a/b)/((I*b^4/sqrt(abs(b)) + b^3*sqrt(abs(b)))*c) + 1/2*I*sqrt(2)*sqrt(pi)
*a^3*b^3*erf(1/2*I*sqrt(2)*sqrt(b*arccos(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*sqrt(abs
(b))/b)*e^(-I*a/b)/((-I*b^4/sqrt(abs(b)) + b^3*sqrt(abs(b)))*c) + 3/2*sqrt(2)*sqrt(pi)*a^2*b^3*erf(-1/2*I*sqrt
(2)*sqrt(b*arccos(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*sqrt(abs(b))/b)*e^(I*a/b)/((I*b
^3/sqrt(abs(b)) + b^2*sqrt(abs(b)))*c) + 3/2*sqrt(2)*sqrt(pi)*a^2*b^3*erf(1/2*I*sqrt(2)*sqrt(b*arccos(c*x) + a
)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*sqrt(abs(b))/b)*e^(-I*a/b)/((-I*b^3/sqrt(abs(b)) + b^2*sq
rt(abs(b)))*c) - 3/2*sqrt(2)*sqrt(pi)*a^2*b^2*erf(-1/2*I*sqrt(2)*sqrt(b*arccos(c*x) + a)/sqrt(abs(b)) - 1/2*sq
rt(2)*sqrt(b*arccos(c*x) + a)*sqrt(abs(b))/b)*e^(I*a/b)/((I*b^2/sqrt(abs(b)) + b*sqrt(abs(b)))*c) - 15/16*sqrt
(2)*sqrt(pi)*b^4*erf(-1/2*I*sqrt(2)*sqrt(b*arccos(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)
*sqrt(abs(b))/b)*e^(I*a/b)/((I*b^2/sqrt(abs(b)) + b*sqrt(abs(b)))*c) - 3/2*sqrt(2)*sqrt(pi)*a^2*b^2*erf(1/2*I*
sqrt(2)*sqrt(b*arccos(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*sqrt(abs(b))/b)*e^(-I*a/b)/
((-I*b^2/sqrt(abs(b)) + b*sqrt(abs(b)))*c) - 15/16*sqrt(2)*sqrt(pi)*b^4*erf(1/2*I*sqrt(2)*sqrt(b*arccos(c*x) +
a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*sqrt(abs(b))/b)*e^(-I*a/b)/((-I*b^2/sqrt(abs(b)) + b*sq
rt(abs(b)))*c) + 1/2*sqrt(b*arccos(c*x) + a)*b^2*arccos(c*x)^2*e^(I*arccos(c*x))/c + 1/2*sqrt(b*arccos(c*x) +
a)*b^2*arccos(c*x)^2*e^(-I*arccos(c*x))/c + I*sqrt(pi)*a^3*b*erf(-1/2*I*sqrt(2)*sqrt(b*arccos(c*x) + a)/sqrt(a
bs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*sqrt(abs(b))/b)*e^(I*a/b)/((I*sqrt(2)*b^2/sqrt(abs(b)) + sqrt(2)*
b*sqrt(abs(b)))*c) - I*sqrt(pi)*a^3*b*erf(1/2*I*sqrt(2)*sqrt(b*arccos(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqr
t(b*arccos(c*x) + a)*sqrt(abs(b))/b)*e^(-I*a/b)/((-I*sqrt(2)*b^2/sqrt(abs(b)) + sqrt(2)*b*sqrt(abs(b)))*c) + s
qrt(b*arccos(c*x) + a)*a*b*arccos(c*x)*e^(I*arccos(c*x))/c + 5/4*I*sqrt(b*arccos(c*x) + a)*b^2*arccos(c*x)*e^(
I*arccos(c*x))/c + sqrt(b*arccos(c*x) + a)*a*b*arccos(c*x)*e^(-I*arccos(c*x))/c - 5/4*I*sqrt(b*arccos(c*x) + a
)*b^2*arccos(c*x)*e^(-I*arccos(c*x))/c + 1/2*sqrt(b*arccos(c*x) + a)*a^2*e^(I*arccos(c*x))/c + 5/4*I*sqrt(b*ar
ccos(c*x) + a)*a*b*e^(I*arccos(c*x))/c - 15/8*sqrt(b*arccos(c*x) + a)*b^2*e^(I*arccos(c*x))/c + 1/2*sqrt(b*arc
cos(c*x) + a)*a^2*e^(-I*arccos(c*x))/c - 5/4*I*sqrt(b*arccos(c*x) + a)*a*b*e^(-I*arccos(c*x))/c - 15/8*sqrt(b*
arccos(c*x) + a)*b^2*e^(-I*arccos(c*x))/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(c*x))^(5/2),x)

[Out]

int((a + b*acos(c*x))^(5/2), x)

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