3.184 \(\int x (a+b \cos ^{-1}(c x))^{5/2} \, dx\)

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

[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)

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Rubi [A]  time = 0.702873, antiderivative size = 216, normalized size of antiderivative = 1., 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 = {4630, 4708, 4642, 4724, 3312, 3306, 3305, 3351, 3304, 3352} \[ \frac{15 \sqrt{\pi } b^{5/2} \cos \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi } \sqrt{b}}\right )}{128 c^2}+\frac{15 \sqrt{\pi } b^{5/2} \sin \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b} \sqrt{\pi }}\right )}{128 c^2}+\frac{15 b^2 \sqrt{a+b \cos ^{-1}(c x)}}{64 c^2}-\frac{15}{32} b^2 x^2 \sqrt{a+b \cos ^{-1}(c x)}-\frac{5 b x \sqrt{1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^{3/2}}{8 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^{5/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{5/2} \]

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 4630

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 4708

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

Rule 4642

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

Rule 4724

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

Rule 3312

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 3306

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 3305

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 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3304

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 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 1.93229, size = 201, normalized size = 0.93 \[ \frac{2 \sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}(c x)} \left (\left (16 a^2-15 b^2\right ) \cos \left (2 \cos ^{-1}(c x)\right )+4 b \cos ^{-1}(c x) \left (8 a \cos \left (2 \cos ^{-1}(c x)\right )-5 b \sin \left (2 \cos ^{-1}(c x)\right )\right )-20 a b \sin \left (2 \cos ^{-1}(c x)\right )+16 b^2 \cos \left (2 \cos ^{-1}(c x)\right ) \cos ^{-1}(c x)^2\right )+15 \sqrt{\pi } b^2 \cos \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi }}\right )+15 \sqrt{\pi } b^2 \sin \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi }}\right )}{128 \sqrt{\frac{1}{b}} c^2} \]

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]  time = 0.132, size = 366, normalized size = 1.7 \begin{align*}{\frac{b}{128\,{c}^{2}\sqrt{\pi }} \left ( 32\, \left ( \arccos \left ( cx \right ) \right ) ^{2}\sqrt{a+b\arccos \left ( cx \right ) }\cos \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ) \sqrt{\pi }\sqrt{{b}^{-1}}{b}^{2}+64\,\arccos \left ( cx \right ) \sqrt{a+b\arccos \left ( cx \right ) }\cos \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ) \sqrt{\pi }\sqrt{{b}^{-1}}ab-40\,\arccos \left ( cx \right ) \sqrt{a+b\arccos \left ( cx \right ) }\sin \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ) \sqrt{\pi }\sqrt{{b}^{-1}}{b}^{2}+32\,\sqrt{a+b\arccos \left ( cx \right ) }\cos \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ) \sqrt{\pi }\sqrt{{b}^{-1}}{a}^{2}-30\,\sqrt{a+b\arccos \left ( cx \right ) }\cos \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ) \sqrt{\pi }\sqrt{{b}^{-1}}{b}^{2}-40\,\sqrt{a+b\arccos \left ( cx \right ) }\sin \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ) \sqrt{\pi }\sqrt{{b}^{-1}}ab+15\,\pi \,{b}^{2}{\it FresnelC} \left ( 2\,{\frac{\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) +15\,\pi \,{b}^{2}\sin \left ( 2\,{\frac{a}{b}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \right ) \sqrt{{b}^{-1}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128/c^2*b*(32*arccos(c*x)^2*(a+b*arccos(c*x))^(1/2)*cos(2*(a+b*arccos(c*x))/b-2*a/b)*Pi^(1/2)*(1/b)^(1/2)*b^
2+64*arccos(c*x)*(a+b*arccos(c*x))^(1/2)*cos(2*(a+b*arccos(c*x))/b-2*a/b)*Pi^(1/2)*(1/b)^(1/2)*a*b-40*arccos(c
*x)*(a+b*arccos(c*x))^(1/2)*sin(2*(a+b*arccos(c*x))/b-2*a/b)*Pi^(1/2)*(1/b)^(1/2)*b^2+32*(a+b*arccos(c*x))^(1/
2)*cos(2*(a+b*arccos(c*x))/b-2*a/b)*Pi^(1/2)*(1/b)^(1/2)*a^2-30*(a+b*arccos(c*x))^(1/2)*cos(2*(a+b*arccos(c*x)
)/b-2*a/b)*Pi^(1/2)*(1/b)^(1/2)*b^2-40*(a+b*arccos(c*x))^(1/2)*sin(2*(a+b*arccos(c*x))/b-2*a/b)*Pi^(1/2)*(1/b)
^(1/2)*a*b+15*Pi*b^2*FresnelC(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*cos(2*a/b)+15*Pi*b^2*sin(2*a/b
)*FresnelS(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b))/Pi^(1/2)*(1/b)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arccos \left (c x\right ) + a\right )}^{\frac{5}{2}} x\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

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: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 2.40081, size = 1638, normalized size = 7.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/32*sqrt(pi)*a*b^(7/2)*i*erf(-sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sqrt(b*arccos(c*x) + a)/sqrt(b))*e^
(2*a*i/b)/((b^3*i/abs(b) + b^2)*c^2) - 3/32*sqrt(pi)*a*b^(7/2)*i*erf(sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b)
- sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(-2*a*i/b)/((b^3*i/abs(b) - b^2)*c^2) - 1/8*sqrt(pi)*a^2*b^(5/2)*erf(-sqr
t(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(2*a*i/b)/((b^3*i/abs(b) + b^2)*c^2
) + 3/32*sqrt(pi)*a*b^(5/2)*i*erf(-sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sqrt(b*arccos(c*x) + a)/sqrt(b))
*e^(2*a*i/b)/((b^2*i/abs(b) + b)*c^2) + 1/8*sqrt(pi)*a^2*b^(5/2)*erf(sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b)
- sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(-2*a*i/b)/((b^3*i/abs(b) - b^2)*c^2) + 3/32*sqrt(pi)*a*b^(5/2)*i*erf(sqr
t(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(-2*a*i/b)/((b^2*i/abs(b) - b)*c^2)
+ 5/32*sqrt(b*arccos(c*x) + a)*b^2*i*arccos(c*x)*e^(2*i*arccos(c*x))/c^2 + 1/8*sqrt(b*arccos(c*x) + a)*b^2*ar
ccos(c*x)^2*e^(2*i*arccos(c*x))/c^2 - 5/32*sqrt(b*arccos(c*x) + a)*b^2*i*arccos(c*x)*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 + 1/16*sqrt(pi)*a^2*b^2*erf(-sqrt(b*a
rccos(c*x) + a)*sqrt(b)*i/abs(b) - sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(2*a*i/b)/((b^(5/2)*i/abs(b) + b^(3/2))*
c^2) - 1/16*sqrt(pi)*a^2*b^2*erf(sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sqrt(b*arccos(c*x) + a)/sqrt(b))*e
^(-2*a*i/b)/((b^(5/2)*i/abs(b) - b^(3/2))*c^2) + 1/16*sqrt(pi)*a^2*b^(3/2)*erf(-sqrt(b*arccos(c*x) + a)*sqrt(b
)*i/abs(b) - sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(2*a*i/b)/((b^2*i/abs(b) + b)*c^2) - 15/256*sqrt(pi)*b^(7/2)*e
rf(-sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(2*a*i/b)/((b^2*i/abs(b) + b
)*c^2) - 1/16*sqrt(pi)*a^2*b^(3/2)*erf(sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sqrt(b*arccos(c*x) + a)/sqrt
(b))*e^(-2*a*i/b)/((b^2*i/abs(b) - b)*c^2) + 15/256*sqrt(pi)*b^(7/2)*erf(sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs
(b) - sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(-2*a*i/b)/((b^2*i/abs(b) - b)*c^2) + 5/32*sqrt(b*arccos(c*x) + a)*a*
b*i*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*sqrt(
b*arccos(c*x) + a)*a*b*i*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 + 1/8*sqrt(b*arccos(c*x) + a)*a^2*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 + 1/8*sqrt(b*arccos(c*x) + a)*a^2*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