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

Optimal. Leaf size=89 \[ \frac{3 \sqrt{\frac{\pi }{2}} S\left (\sqrt{\frac{2}{\pi }} \sqrt{\cos ^{-1}(a+b x)}\right )}{2 b}+\frac{(a+b x) \cos ^{-1}(a+b x)^{3/2}}{b}-\frac{3 \sqrt{1-(a+b x)^2} \sqrt{\cos ^{-1}(a+b x)}}{2 b} \]

[Out]

(-3*Sqrt[1 - (a + b*x)^2]*Sqrt[ArcCos[a + b*x]])/(2*b) + ((a + b*x)*ArcCos[a + b*x]^(3/2))/b + (3*Sqrt[Pi/2]*F
resnelS[Sqrt[2/Pi]*Sqrt[ArcCos[a + b*x]]])/(2*b)

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Rubi [A]  time = 0.09211, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4804, 4620, 4678, 4624, 3305, 3351} \[ \frac{3 \sqrt{\frac{\pi }{2}} S\left (\sqrt{\frac{2}{\pi }} \sqrt{\cos ^{-1}(a+b x)}\right )}{2 b}+\frac{(a+b x) \cos ^{-1}(a+b x)^{3/2}}{b}-\frac{3 \sqrt{1-(a+b x)^2} \sqrt{\cos ^{-1}(a+b x)}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]^(3/2),x]

[Out]

(-3*Sqrt[1 - (a + b*x)^2]*Sqrt[ArcCos[a + b*x]])/(2*b) + ((a + b*x)*ArcCos[a + b*x]^(3/2))/b + (3*Sqrt[Pi/2]*F
resnelS[Sqrt[2/Pi]*Sqrt[ArcCos[a + b*x]]])/(2*b)

Rule 4804

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCos[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

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 4624

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

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]

Rubi steps

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

Mathematica [C]  time = 0.0353963, size = 76, normalized size = 0.85 \[ -\frac{\sqrt{-i \cos ^{-1}(a+b x)} \text{Gamma}\left (\frac{5}{2},-i \cos ^{-1}(a+b x)\right )+\sqrt{i \cos ^{-1}(a+b x)} \text{Gamma}\left (\frac{5}{2},i \cos ^{-1}(a+b x)\right )}{2 b \sqrt{\cos ^{-1}(a+b x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCos[a + b*x]^(3/2),x]

[Out]

-(Sqrt[(-I)*ArcCos[a + b*x]]*Gamma[5/2, (-I)*ArcCos[a + b*x]] + Sqrt[I*ArcCos[a + b*x]]*Gamma[5/2, I*ArcCos[a
+ b*x]])/(2*b*Sqrt[ArcCos[a + b*x]])

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Maple [A]  time = 0.082, size = 105, normalized size = 1.2 \begin{align*}{\frac{\sqrt{2}}{4\,b\sqrt{\pi }} \left ( 2\, \left ( \arccos \left ( bx+a \right ) \right ) ^{3/2}\sqrt{2}\sqrt{\pi }xb+2\, \left ( \arccos \left ( bx+a \right ) \right ) ^{3/2}\sqrt{2}\sqrt{\pi }a-3\,\sqrt{2}\sqrt{\pi }\sqrt{\arccos \left ( bx+a \right ) }\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}+3\,\pi \,{\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{\arccos \left ( bx+a \right ) }}{\sqrt{\pi }}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/b*2^(1/2)/Pi^(1/2)*(2*arccos(b*x+a)^(3/2)*2^(1/2)*Pi^(1/2)*x*b+2*arccos(b*x+a)^(3/2)*2^(1/2)*Pi^(1/2)*a-3*
2^(1/2)*Pi^(1/2)*arccos(b*x+a)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+3*Pi*FresnelS(2^(1/2)/Pi^(1/2)*arccos(b*x+
a)^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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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(arccos(b*x+a)^(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 \operatorname{acos}^{\frac{3}{2}}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acos(a + b*x)**(3/2), x)

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Giac [B]  time = 1.62439, size = 221, normalized size = 2.48 \begin{align*} \frac{3 \, i \sqrt{\arccos \left (b x + a\right )} e^{\left (i \arccos \left (b x + a\right )\right )}}{4 \, b} + \frac{\arccos \left (b x + a\right )^{\frac{3}{2}} e^{\left (i \arccos \left (b x + a\right )\right )}}{2 \, b} - \frac{3 \, i \sqrt{\arccos \left (b x + a\right )} e^{\left (-i \arccos \left (b x + a\right )\right )}}{4 \, b} + \frac{\arccos \left (b x + a\right )^{\frac{3}{2}} e^{\left (-i \arccos \left (b x + a\right )\right )}}{2 \, b} - \frac{3 \, \sqrt{2} \sqrt{\pi } i \operatorname{erf}\left (-\frac{\sqrt{2} i \sqrt{\arccos \left (b x + a\right )}}{i - 1}\right )}{8 \, b{\left (i - 1\right )}} + \frac{3 \, \sqrt{2} \sqrt{\pi } \operatorname{erf}\left (\frac{\sqrt{2} \sqrt{\arccos \left (b x + a\right )}}{i - 1}\right )}{8 \, b{\left (i - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*i*sqrt(arccos(b*x + a))*e^(i*arccos(b*x + a))/b + 1/2*arccos(b*x + a)^(3/2)*e^(i*arccos(b*x + a))/b - 3/4*
i*sqrt(arccos(b*x + a))*e^(-i*arccos(b*x + a))/b + 1/2*arccos(b*x + a)^(3/2)*e^(-i*arccos(b*x + a))/b - 3/8*sq
rt(2)*sqrt(pi)*i*erf(-sqrt(2)*i*sqrt(arccos(b*x + a))/(i - 1))/(b*(i - 1)) + 3/8*sqrt(2)*sqrt(pi)*erf(sqrt(2)*
sqrt(arccos(b*x + a))/(i - 1))/(b*(i - 1))