3.179 \(\int x (a+b \cos ^{-1}(c x))^{3/2} \, dx\)

Optimal. Leaf size=172 \[ -\frac{3 \sqrt{\pi } b^{3/2} \sin \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi } \sqrt{b}}\right )}{32 c^2}+\frac{3 \sqrt{\pi } b^{3/2} \cos \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b} \sqrt{\pi }}\right )}{32 c^2}-\frac{3 b x \sqrt{1-c^2 x^2} \sqrt{a+b \cos ^{-1}(c x)}}{8 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{3/2} \]

[Out]

(-3*b*x*Sqrt[1 - c^2*x^2]*Sqrt[a + b*ArcCos[c*x]])/(8*c) - (a + b*ArcCos[c*x])^(3/2)/(4*c^2) + (x^2*(a + b*Arc
Cos[c*x])^(3/2))/2 + (3*b^(3/2)*Sqrt[Pi]*Cos[(2*a)/b]*FresnelS[(2*Sqrt[a + b*ArcCos[c*x]])/(Sqrt[b]*Sqrt[Pi])]
)/(32*c^2) - (3*b^(3/2)*Sqrt[Pi]*FresnelC[(2*Sqrt[a + b*ArcCos[c*x]])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b])/(32*c^
2)

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Rubi [A]  time = 0.461326, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 11, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.786, Rules used = {4630, 4708, 4642, 4636, 4406, 12, 3306, 3305, 3351, 3304, 3352} \[ -\frac{3 \sqrt{\pi } b^{3/2} \sin \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi } \sqrt{b}}\right )}{32 c^2}+\frac{3 \sqrt{\pi } b^{3/2} \cos \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b} \sqrt{\pi }}\right )}{32 c^2}-\frac{3 b x \sqrt{1-c^2 x^2} \sqrt{a+b \cos ^{-1}(c x)}}{8 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcCos[c*x])^(3/2),x]

[Out]

(-3*b*x*Sqrt[1 - c^2*x^2]*Sqrt[a + b*ArcCos[c*x]])/(8*c) - (a + b*ArcCos[c*x])^(3/2)/(4*c^2) + (x^2*(a + b*Arc
Cos[c*x])^(3/2))/2 + (3*b^(3/2)*Sqrt[Pi]*Cos[(2*a)/b]*FresnelS[(2*Sqrt[a + b*ArcCos[c*x]])/(Sqrt[b]*Sqrt[Pi])]
)/(32*c^2) - (3*b^(3/2)*Sqrt[Pi]*FresnelC[(2*Sqrt[a + b*ArcCos[c*x]])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b])/(32*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 4636

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b*
x)^n*Cos[x]^m*Sin[x], x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 )^{3/2} \, dx &=\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{3/2}+\frac{1}{4} (3 b c) \int \frac{x^2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{3 b x \sqrt{1-c^2 x^2} \sqrt{a+b \cos ^{-1}(c x)}}{8 c}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{3/2}-\frac{1}{16} \left (3 b^2\right ) \int \frac{x}{\sqrt{a+b \cos ^{-1}(c x)}} \, dx+\frac{(3 b) \int \frac{\sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{1-c^2 x^2}} \, dx}{8 c}\\ &=-\frac{3 b x \sqrt{1-c^2 x^2} \sqrt{a+b \cos ^{-1}(c x)}}{8 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{3/2}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{\sqrt{a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{16 c^2}\\ &=-\frac{3 b x \sqrt{1-c^2 x^2} \sqrt{a+b \cos ^{-1}(c x)}}{8 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{3/2}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 \sqrt{a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{16 c^2}\\ &=-\frac{3 b x \sqrt{1-c^2 x^2} \sqrt{a+b \cos ^{-1}(c x)}}{8 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{3/2}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\sin (2 x)}{\sqrt{a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{32 c^2}\\ &=-\frac{3 b x \sqrt{1-c^2 x^2} \sqrt{a+b \cos ^{-1}(c x)}}{8 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{3/2}+\frac{\left (3 b^2 \cos \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 )}{32 c^2}-\frac{\left (3 b^2 \sin \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 )}{32 c^2}\\ &=-\frac{3 b x \sqrt{1-c^2 x^2} \sqrt{a+b \cos ^{-1}(c x)}}{8 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{3/2}+\frac{\left (3 b \cos \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 )}{16 c^2}-\frac{\left (3 b \sin \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 )}{16 c^2}\\ &=-\frac{3 b x \sqrt{1-c^2 x^2} \sqrt{a+b \cos ^{-1}(c x)}}{8 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^{3/2}+\frac{3 b^{3/2} \sqrt{\pi } \cos \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b} \sqrt{\pi }}\right )}{32 c^2}-\frac{3 b^{3/2} \sqrt{\pi } C\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b} \sqrt{\pi }}\right ) \sin \left (\frac{2 a}{b}\right )}{32 c^2}\\ \end{align*}

Mathematica [A]  time = 0.825214, size = 155, normalized size = 0.9 \[ \frac{-3 \sqrt{\pi } b \sin \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi }}\right )+3 \sqrt{\pi } b \cos \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi }}\right )+2 \sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}(c x)} \left (4 a \cos \left (2 \cos ^{-1}(c x)\right )+4 b \cos ^{-1}(c x) \cos \left (2 \cos ^{-1}(c x)\right )-3 b \sin \left (2 \cos ^{-1}(c x)\right )\right )}{32 \sqrt{\frac{1}{b}} c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcCos[c*x])^(3/2),x]

[Out]

(3*b*Sqrt[Pi]*Cos[(2*a)/b]*FresnelS[(2*Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[Pi]] - 3*b*Sqrt[Pi]*FresnelC
[(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]]*(4*a
*Cos[2*ArcCos[c*x]] + 4*b*ArcCos[c*x]*Cos[2*ArcCos[c*x]] - 3*b*Sin[2*ArcCos[c*x]]))/(32*Sqrt[b^(-1)]*c^2)

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Maple [A]  time = 0.114, size = 267, normalized size = 1.6 \begin{align*}{\frac{1}{32\,{c}^{2}} \left ( 3\,\sqrt{{b}^{-1}}\sqrt{\pi }\cos \left ( 2\,{\frac{a}{b}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{{b}^{-1}}\sqrt{\pi }b}} \right ) \sqrt{a+b\arccos \left ( cx \right ) }{b}^{2}-3\,\sqrt{{b}^{-1}}\sqrt{\pi }\sin \left ( 2\,{\frac{a}{b}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{{b}^{-1}}\sqrt{\pi }b}} \right ) \sqrt{a+b\arccos \left ( cx \right ) }{b}^{2}+8\, \left ( \arccos \left ( cx \right ) \right ) ^{2}\cos \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ){b}^{2}+16\,\arccos \left ( cx \right ) \cos \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ) ab-6\,\arccos \left ( cx \right ) \sin \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ){b}^{2}+8\,\cos \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ){a}^{2}-6\,\sin \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ) ab \right ){\frac{1}{\sqrt{a+b\arccos \left ( cx \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arccos(c*x))^(3/2),x)

[Out]

1/32/c^2*(3*(1/b)^(1/2)*Pi^(1/2)*cos(2*a/b)*FresnelS(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*(a+b*ar
ccos(c*x))^(1/2)*b^2-3*(1/b)^(1/2)*Pi^(1/2)*sin(2*a/b)*FresnelC(2/Pi^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)
/b)*(a+b*arccos(c*x))^(1/2)*b^2+8*arccos(c*x)^2*cos(2*(a+b*arccos(c*x))/b-2*a/b)*b^2+16*arccos(c*x)*cos(2*(a+b
*arccos(c*x))/b-2*a/b)*a*b-6*arccos(c*x)*sin(2*(a+b*arccos(c*x))/b-2*a/b)*b^2+8*cos(2*(a+b*arccos(c*x))/b-2*a/
b)*a^2-6*sin(2*(a+b*arccos(c*x))/b-2*a/b)*a*b)/(a+b*arccos(c*x))^(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{3}{2}} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arccos(c*x))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*arccos(c*x) + a)^(3/2)*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))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (a + b \operatorname{acos}{\left (c x \right )}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*acos(c*x))**(3/2),x)

[Out]

Integral(x*(a + b*acos(c*x))**(3/2), x)

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Giac [B]  time = 1.97114, size = 791, normalized size = 4.6 \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))^(3/2),x, algorithm="giac")

[Out]

-3/64*sqrt(pi)*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) - 3/64*sqrt(pi)*b^(7/2)*i*erf(sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sq
rt(b*arccos(c*x) + a)/sqrt(b))*e^(-2*a*i/b)/((b^3*i/abs(b) - b^2)*c^2) - 1/16*sqrt(pi)*a*b^(5/2)*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^3*i/abs(b) + b^2)*c^2) + 1
/16*sqrt(pi)*a*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) + 1/16*sqrt(pi)*a*b^(3/2)*erf(-sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sqr
t(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*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) + 3/32*sq
rt(b*arccos(c*x) + a)*b*i*e^(2*i*arccos(c*x))/c^2 + 1/8*sqrt(b*arccos(c*x) + a)*b*arccos(c*x)*e^(2*i*arccos(c*
x))/c^2 - 3/32*sqrt(b*arccos(c*x) + a)*b*i*e^(-2*i*arccos(c*x))/c^2 + 1/8*sqrt(b*arccos(c*x) + a)*b*arccos(c*x
)*e^(-2*i*arccos(c*x))/c^2 + 1/8*sqrt(b*arccos(c*x) + a)*a*e^(2*i*arccos(c*x))/c^2 + 1/8*sqrt(b*arccos(c*x) +
a)*a*e^(-2*i*arccos(c*x))/c^2