3.2.84 \(\int x (a+b \text {ArcCos}(c x))^{5/2} \, dx\) [184]

Optimal. Leaf size=216 \[ \frac {15 b^2 \sqrt {a+b \text {ArcCos}(c x)}}{64 c^2}-\frac {15}{32} b^2 x^2 \sqrt {a+b \text {ArcCos}(c x)}-\frac {5 b x \sqrt {1-c^2 x^2} (a+b \text {ArcCos}(c x))^{3/2}}{8 c}-\frac {(a+b \text {ArcCos}(c x))^{5/2}}{4 c^2}+\frac {1}{2} x^2 (a+b \text {ArcCos}(c x))^{5/2}+\frac {15 b^{5/2} \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{128 c^2}+\frac {15 b^{5/2} \sqrt {\pi } S\left (\frac {2 \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{128 c^2} \]

[Out]

-1/4*(a+b*arccos(c*x))^(5/2)/c^2+1/2*x^2*(a+b*arccos(c*x))^(5/2)+15/128*b^(5/2)*cos(2*a/b)*FresnelC(2*(a+b*arc
cos(c*x))^(1/2)/b^(1/2)/Pi^(1/2))*Pi^(1/2)/c^2+15/128*b^(5/2)*FresnelS(2*(a+b*arccos(c*x))^(1/2)/b^(1/2)/Pi^(1
/2))*sin(2*a/b)*Pi^(1/2)/c^2-5/8*b*x*(a+b*arccos(c*x))^(3/2)*(-c^2*x^2+1)^(1/2)/c+15/64*b^2*(a+b*arccos(c*x))^
(1/2)/c^2-15/32*b^2*x^2*(a+b*arccos(c*x))^(1/2)

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Rubi [A]
time = 0.43, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {4726, 4796, 4738, 4810, 3393, 3387, 3386, 3432, 3385, 3433} \begin {gather*} \frac {15 \sqrt {\pi } b^{5/2} \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {\pi } \sqrt {b}}\right )}{128 c^2}+\frac {15 \sqrt {\pi } b^{5/2} \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{128 c^2}+\frac {15 b^2 \sqrt {a+b \text {ArcCos}(c x)}}{64 c^2}-\frac {15}{32} b^2 x^2 \sqrt {a+b \text {ArcCos}(c x)}-\frac {5 b x \sqrt {1-c^2 x^2} (a+b \text {ArcCos}(c x))^{3/2}}{8 c}-\frac {(a+b \text {ArcCos}(c x))^{5/2}}{4 c^2}+\frac {1}{2} x^2 (a+b \text {ArcCos}(c x))^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcCos[c*x])^(5/2),x]

[Out]

(15*b^2*Sqrt[a + b*ArcCos[c*x]])/(64*c^2) - (15*b^2*x^2*Sqrt[a + b*ArcCos[c*x]])/32 - (5*b*x*Sqrt[1 - c^2*x^2]
*(a + b*ArcCos[c*x])^(3/2))/(8*c) - (a + b*ArcCos[c*x])^(5/2)/(4*c^2) + (x^2*(a + b*ArcCos[c*x])^(5/2))/2 + (1
5*b^(5/2)*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[a + b*ArcCos[c*x]])/(Sqrt[b]*Sqrt[Pi])])/(128*c^2) + (15*b^(5
/2)*Sqrt[Pi]*FresnelS[(2*Sqrt[a + b*ArcCos[c*x]])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b])/(128*c^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 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 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 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 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*} \int x \left (a+b \cos ^{-1}(c x)\right )^{5/2} \, dx &=\frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{5/2}+\frac {1}{4} (5 b c) \int \frac {x^2 \left (a+b \cos ^{-1}(c x)\right )^{3/2}}{\sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {5 b x \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^{3/2}}{8 c}+\frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{5/2}-\frac {1}{16} \left (15 b^2\right ) \int x \sqrt {a+b \cos ^{-1}(c x)} \, dx+\frac {(5 b) \int \frac {\left (a+b \cos ^{-1}(c x)\right )^{3/2}}{\sqrt {1-c^2 x^2}} \, dx}{8 c}\\ &=-\frac {15}{32} b^2 x^2 \sqrt {a+b \cos ^{-1}(c x)}-\frac {5 b x \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^{3/2}}{8 c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^{5/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{5/2}-\frac {1}{64} \left (15 b^3 c\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2} \sqrt {a+b \cos ^{-1}(c x)}} \, dx\\ &=-\frac {15}{32} b^2 x^2 \sqrt {a+b \cos ^{-1}(c x)}-\frac {5 b x \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^{3/2}}{8 c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^{5/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{5/2}+\frac {\left (15 b^3\right ) \text {Subst}\left (\int \frac {\cos ^2(x)}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{64 c^2}\\ &=-\frac {15}{32} b^2 x^2 \sqrt {a+b \cos ^{-1}(c x)}-\frac {5 b x \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^{3/2}}{8 c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^{5/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{5/2}+\frac {\left (15 b^3\right ) \text {Subst}\left (\int \left (\frac {1}{2 \sqrt {a+b x}}+\frac {\cos (2 x)}{2 \sqrt {a+b x}}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{64 c^2}\\ &=\frac {15 b^2 \sqrt {a+b \cos ^{-1}(c x)}}{64 c^2}-\frac {15}{32} b^2 x^2 \sqrt {a+b \cos ^{-1}(c x)}-\frac {5 b x \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^{3/2}}{8 c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^{5/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{5/2}+\frac {\left (15 b^3\right ) \text {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{128 c^2}\\ &=\frac {15 b^2 \sqrt {a+b \cos ^{-1}(c x)}}{64 c^2}-\frac {15}{32} b^2 x^2 \sqrt {a+b \cos ^{-1}(c x)}-\frac {5 b x \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^{3/2}}{8 c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^{5/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{5/2}+\frac {\left (15 b^3 \cos \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{128 c^2}+\frac {\left (15 b^3 \sin \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{\sqrt {a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{128 c^2}\\ &=\frac {15 b^2 \sqrt {a+b \cos ^{-1}(c x)}}{64 c^2}-\frac {15}{32} b^2 x^2 \sqrt {a+b \cos ^{-1}(c x)}-\frac {5 b x \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^{3/2}}{8 c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^{5/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{5/2}+\frac {\left (15 b^2 \cos \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \cos \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \cos ^{-1}(c x)}\right )}{64 c^2}+\frac {\left (15 b^2 \sin \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \sin \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \cos ^{-1}(c x)}\right )}{64 c^2}\\ &=\frac {15 b^2 \sqrt {a+b \cos ^{-1}(c x)}}{64 c^2}-\frac {15}{32} b^2 x^2 \sqrt {a+b \cos ^{-1}(c x)}-\frac {5 b x \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^{3/2}}{8 c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^{5/2}}{4 c^2}+\frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{5/2}+\frac {15 b^{5/2} \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) C\left (\frac {2 \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{128 c^2}+\frac {15 b^{5/2} \sqrt {\pi } S\left (\frac {2 \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{128 c^2}\\ \end {align*}

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Mathematica [A]
time = 1.39, size = 201, normalized size = 0.93 \begin {gather*} \frac {15 b^2 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {\pi }}\right )+15 b^2 \sqrt {\pi } S\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )+2 \sqrt {\frac {1}{b}} \sqrt {a+b \text {ArcCos}(c x)} \left (\left (16 a^2-15 b^2\right ) \cos (2 \text {ArcCos}(c x))+16 b^2 \text {ArcCos}(c x)^2 \cos (2 \text {ArcCos}(c x))-20 a b \sin (2 \text {ArcCos}(c x))+4 b \text {ArcCos}(c x) (8 a \cos (2 \text {ArcCos}(c x))-5 b \sin (2 \text {ArcCos}(c x)))\right )}{128 \sqrt {\frac {1}{b}} c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*b^2*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[Pi]] + 15*b^2*Sqrt[Pi]*Fr
esnelS[(2*Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[Pi]]*Sin[(2*a)/b] + 2*Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[c*x]
]*((16*a^2 - 15*b^2)*Cos[2*ArcCos[c*x]] + 16*b^2*ArcCos[c*x]^2*Cos[2*ArcCos[c*x]] - 20*a*b*Sin[2*ArcCos[c*x]]
+ 4*b*ArcCos[c*x]*(8*a*Cos[2*ArcCos[c*x]] - 5*b*Sin[2*ArcCos[c*x]])))/(128*Sqrt[b^(-1)]*c^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(413\) vs. \(2(170)=340\).
time = 0.33, size = 414, normalized size = 1.92

method	result	size
default	\(\frac {15 \sqrt {-\frac {2}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) b^{3}-15 \sqrt {-\frac {2}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) b^{3}+64 \arccos \left (c x \right )^{3} \cos \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) b^{3}+192 \arccos \left (c x \right )^{2} \cos \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a \,b^{2}+80 \arccos \left (c x \right )^{2} \sin \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) b^{3}+192 \arccos \left (c x \right ) \cos \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a^{2} b -60 \arccos \left (c x \right ) \cos \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) b^{3}+160 \arccos \left (c x \right ) \sin \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a \,b^{2}+64 \cos \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a^{3}-60 \cos \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a \,b^{2}+80 \sin \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a^{2} b}{256 c^{2} \sqrt {a +b \arccos \left (c x \right )}}\)	\(414\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256/c^2/(a+b*arccos(c*x))^(1/2)*(15*(-2/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arccos(c*x))^(1/2)*cos(2*a/b)*Fresnel
C(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*b^3-15*(-2/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arccos(
c*x))^(1/2)*sin(2*a/b)*FresnelS(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*b^3+64*arccos(c*x)^
3*cos(-2*(a+b*arccos(c*x))/b+2*a/b)*b^3+192*arccos(c*x)^2*cos(-2*(a+b*arccos(c*x))/b+2*a/b)*a*b^2+80*arccos(c*
x)^2*sin(-2*(a+b*arccos(c*x))/b+2*a/b)*b^3+192*arccos(c*x)*cos(-2*(a+b*arccos(c*x))/b+2*a/b)*a^2*b-60*arccos(c
*x)*cos(-2*(a+b*arccos(c*x))/b+2*a/b)*b^3+160*arccos(c*x)*sin(-2*(a+b*arccos(c*x))/b+2*a/b)*a*b^2+64*cos(-2*(a
+b*arccos(c*x))/b+2*a/b)*a^3-60*cos(-2*(a+b*arccos(c*x))/b+2*a/b)*a*b^2+80*sin(-2*(a+b*arccos(c*x))/b+2*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(x*(a+b*arccos(c*x))^(5/2),x, algorithm="maxima")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \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(x*(a+b*acos(c*x))**(5/2),x)

[Out]

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

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Giac [C] Result contains complex when optimal does not.
time = 1.20, size = 1307, normalized size = 6.05 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*sqrt(pi)*a^3*b^(3/2)*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b) - I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e
^(2*I*a/b)/((b^2 + I*b^3/abs(b))*c^2) + 3/8*sqrt(pi)*a^2*b^(5/2)*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b) - I*sqrt
(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(2*I*a/b)/((b^2 + I*b^3/abs(b))*c^2) + 1/4*I*sqrt(pi)*a^3*b^(3/2)*erf(-s
qrt(b*arccos(c*x) + a)/sqrt(b) + I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(-2*I*a/b)/((b^2 - I*b^3/abs(b))*
c^2) + 3/8*sqrt(pi)*a^2*b^(5/2)*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b) + I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b
))*e^(-2*I*a/b)/((b^2 - I*b^3/abs(b))*c^2) + 1/8*sqrt(b*arccos(c*x) + a)*b^2*arccos(c*x)^2*e^(2*I*arccos(c*x))
/c^2 + 1/8*sqrt(b*arccos(c*x) + a)*b^2*arccos(c*x)^2*e^(-2*I*arccos(c*x))/c^2 - 3/8*sqrt(pi)*a^2*b^2*erf(-sqrt
(b*arccos(c*x) + a)/sqrt(b) - I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(2*I*a/b)/((b^(3/2) + I*b^(5/2)/abs(
b))*c^2) + 9/64*I*sqrt(pi)*a*b^3*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b) - I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(
b))*e^(2*I*a/b)/((b^(3/2) + I*b^(5/2)/abs(b))*c^2) - 1/4*I*sqrt(pi)*a^3*b*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b)
+ I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(-2*I*a/b)/((b^(3/2) - I*b^(5/2)/abs(b))*c^2) - 3/8*sqrt(pi)*a^
2*b^2*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b) + I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(-2*I*a/b)/((b^(3/2)
- I*b^(5/2)/abs(b))*c^2) - 9/64*I*sqrt(pi)*a*b^3*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b) + I*sqrt(b*arccos(c*x) +
a)*sqrt(b)/abs(b))*e^(-2*I*a/b)/((b^(3/2) - I*b^(5/2)/abs(b))*c^2) + 1/4*I*sqrt(pi)*a^3*sqrt(b)*erf(-sqrt(b*a
rccos(c*x) + a)/sqrt(b) - I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(2*I*a/b)/((b + I*b^2/abs(b))*c^2) - 9/6
4*I*sqrt(pi)*a*b^(5/2)*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b) - I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(2*I
*a/b)/((b + I*b^2/abs(b))*c^2) - 15/256*sqrt(pi)*b^(7/2)*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b) - I*sqrt(b*arcco
s(c*x) + a)*sqrt(b)/abs(b))*e^(2*I*a/b)/((b + I*b^2/abs(b))*c^2) + 9/64*I*sqrt(pi)*a*b^(5/2)*erf(-sqrt(b*arcco
s(c*x) + a)/sqrt(b) + I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(-2*I*a/b)/((b - I*b^2/abs(b))*c^2) - 15/256
*sqrt(pi)*b^(7/2)*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b) + I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(-2*I*a/b
)/((b - I*b^2/abs(b))*c^2) + 1/4*sqrt(b*arccos(c*x) + a)*a*b*arccos(c*x)*e^(2*I*arccos(c*x))/c^2 + 5/32*I*sqrt
(b*arccos(c*x) + a)*b^2*arccos(c*x)*e^(2*I*arccos(c*x))/c^2 + 1/4*sqrt(b*arccos(c*x) + a)*a*b*arccos(c*x)*e^(-
2*I*arccos(c*x))/c^2 - 5/32*I*sqrt(b*arccos(c*x) + a)*b^2*arccos(c*x)*e^(-2*I*arccos(c*x))/c^2 + 1/8*sqrt(b*ar
ccos(c*x) + a)*a^2*e^(2*I*arccos(c*x))/c^2 + 5/32*I*sqrt(b*arccos(c*x) + a)*a*b*e^(2*I*arccos(c*x))/c^2 - 15/1
28*sqrt(b*arccos(c*x) + a)*b^2*e^(2*I*arccos(c*x))/c^2 + 1/8*sqrt(b*arccos(c*x) + a)*a^2*e^(-2*I*arccos(c*x))/
c^2 - 5/32*I*sqrt(b*arccos(c*x) + a)*a*b*e^(-2*I*arccos(c*x))/c^2 - 15/128*sqrt(b*arccos(c*x) + a)*b^2*e^(-2*I
*arccos(c*x))/c^2

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

[Out]

int(x*(a + b*acos(c*x))^(5/2), x)

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