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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.428301, antiderivative size = 179, normalized size of antiderivative = 1., 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 = {4620, 4678, 4724, 3306, 3305, 3351, 3304, 3352} \[ \frac{15 \sqrt{\frac{\pi }{2}} b^{5/2} \cos \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right )}{4 c}+\frac{15 \sqrt{\frac{\pi }{2}} b^{5/2} \sin \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right )}{4 c}-\frac{15}{4} b^2 x \sqrt{a+b \cos ^{-1}(c x)}-\frac{5 b \sqrt{1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^{3/2}}{2 c}+x \left (a+b \cos ^{-1}(c x)\right )^{5/2} \]

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 4620

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 4678

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[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 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 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 \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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [C]  time = 4.10564, size = 383, normalized size = 2.14 \[ \frac{b e^{-\frac{i a}{b}} \left (4 i a^2 \sqrt{-\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}} \text{Gamma}\left (\frac{3}{2},-\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )-4 i a^2 e^{\frac{2 i a}{b}} \sqrt{\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}} \text{Gamma}\left (\frac{3}{2},\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )+\sqrt{\frac{\pi }{2}} \sqrt{\frac{1}{b}} \left (4 a^2+15 b^2\right ) \left (1+e^{\frac{2 i a}{b}}\right ) \sqrt{a+b \cos ^{-1}(c x)} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}(c x)}\right )-i \sqrt{\frac{\pi }{2}} \sqrt{\frac{1}{b}} \left (4 a^2+15 b^2\right ) \left (-1+e^{\frac{2 i a}{b}}\right ) \sqrt{a+b \cos ^{-1}(c x)} S\left (\sqrt{\frac{1}{b}} \sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}\right )-2 e^{\frac{i a}{b}} \left (a+b \cos ^{-1}(c x)\right ) \left (5 \left (2 a \sqrt{1-c^2 x^2}+3 b c x\right )+\cos ^{-1}(c x) \left (10 b \sqrt{1-c^2 x^2}-8 a c x\right )-4 b c x \cos ^{-1}(c x)^2\right )\right )}{8 c \sqrt{a+b \cos ^{-1}(c x)}} \]

Warning: Unable to verify antiderivative.

[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]])

________________________________________________________________________________________

Maple [B]  time = 0.123, size = 393, normalized size = 2.2 \begin{align*}{\frac{1}{8\,c} \left ( 15\,\sqrt{{b}^{-1}}\sqrt{\pi }\sqrt{2}\sqrt{a+b\arccos \left ( cx \right ) }\cos \left ({\frac{a}{b}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{{b}^{-1}}\sqrt{\pi }b}} \right ){b}^{3}+15\,\sqrt{{b}^{-1}}\sqrt{\pi }\sqrt{2}\sqrt{a+b\arccos \left ( cx \right ) }\sin \left ({\frac{a}{b}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{{b}^{-1}}\sqrt{\pi }b}} \right ){b}^{3}+8\, \left ( \arccos \left ( cx \right ) \right ) ^{3}\cos \left ({\frac{a+b\arccos \left ( cx \right ) }{b}}-{\frac{a}{b}} \right ){b}^{3}+24\, \left ( \arccos \left ( cx \right ) \right ) ^{2}\cos \left ({\frac{a+b\arccos \left ( cx \right ) }{b}}-{\frac{a}{b}} \right ) a{b}^{2}-20\, \left ( \arccos \left ( cx \right ) \right ) ^{2}\sin \left ({\frac{a+b\arccos \left ( cx \right ) }{b}}-{\frac{a}{b}} \right ){b}^{3}+24\,\arccos \left ( cx \right ) \cos \left ({\frac{a+b\arccos \left ( cx \right ) }{b}}-{\frac{a}{b}} \right ){a}^{2}b-30\,\arccos \left ( cx \right ) \cos \left ({\frac{a+b\arccos \left ( cx \right ) }{b}}-{\frac{a}{b}} \right ){b}^{3}-40\,\arccos \left ( cx \right ) \sin \left ({\frac{a+b\arccos \left ( cx \right ) }{b}}-{\frac{a}{b}} \right ) a{b}^{2}+8\,\cos \left ({\frac{a+b\arccos \left ( cx \right ) }{b}}-{\frac{a}{b}} \right ){a}^{3}-30\,\cos \left ({\frac{a+b\arccos \left ( cx \right ) }{b}}-{\frac{a}{b}} \right ) a{b}^{2}-20\,\sin \left ({\frac{a+b\arccos \left ( cx \right ) }{b}}-{\frac{a}{b}} \right ){a}^{2}b \right ){\frac{1}{\sqrt{a+b\arccos \left ( cx \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/c*(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*arccos(c*x))/b-a/b)*b^3+24*arc
cos(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*ar
ccos(c*x))/b-a/b)*a*b^2+8*cos((a+b*arccos(c*x))/b-a/b)*a^3-30*cos((a+b*arccos(c*x))/b-a/b)*a*b^2-20*sin((a+b*a
rccos(c*x))/b-a/b)*a^2*b)/(a+b*arccos(c*x))^(1/2)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

[Out]

Exception raised: UnboundLocalError

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 2.90574, size = 1621, normalized size = 9.06 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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