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

Optimal. Leaf size=90 \[ \frac {4 \sqrt {2 \pi } S\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a+b x)}\right )}{3 b}+\frac {4 (a+b x)}{3 b \sqrt {\cos ^{-1}(a+b x)}}+\frac {2 \sqrt {1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4804, 4622, 4720, 4624, 3305, 3351} \[ \frac {4 \sqrt {2 \pi } S\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a+b x)}\right )}{3 b}+\frac {4 (a+b x)}{3 b \sqrt {\cos ^{-1}(a+b x)}}+\frac {2 \sqrt {1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]^(-5/2),x]

[Out]

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

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 4622

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(Sqrt[1 - c^2*x^2]*(a + b*ArcCos[c*x])^(n + 1)
)/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcCos[c*x])^(n + 1))/Sqrt[1 - c^2*x^2], x], x] /; Fre
eQ[{a, b, c}, x] && LtQ[n, -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 4720

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

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]

Rubi steps

\begin {align*} \int \frac {1}{\cos ^{-1}(a+b x)^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\cos ^{-1}(x)^{5/2}} \, dx,x,a+b x\right )}{b}\\ &=\frac {2 \sqrt {1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}}+\frac {2 \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \cos ^{-1}(x)^{3/2}} \, dx,x,a+b x\right )}{3 b}\\ &=\frac {2 \sqrt {1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}}+\frac {4 (a+b x)}{3 b \sqrt {\cos ^{-1}(a+b x)}}-\frac {4 \operatorname {Subst}\left (\int \frac {1}{\sqrt {\cos ^{-1}(x)}} \, dx,x,a+b x\right )}{3 b}\\ &=\frac {2 \sqrt {1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}}+\frac {4 (a+b x)}{3 b \sqrt {\cos ^{-1}(a+b x)}}+\frac {4 \operatorname {Subst}\left (\int \frac {\sin (x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a+b x)\right )}{3 b}\\ &=\frac {2 \sqrt {1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}}+\frac {4 (a+b x)}{3 b \sqrt {\cos ^{-1}(a+b x)}}+\frac {8 \operatorname {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a+b x)}\right )}{3 b}\\ &=\frac {2 \sqrt {1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}}+\frac {4 (a+b x)}{3 b \sqrt {\cos ^{-1}(a+b x)}}+\frac {4 \sqrt {2 \pi } S\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a+b x)}\right )}{3 b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.31, size = 139, normalized size = 1.54 \[ -\frac {2 \left (-\sqrt {1-(a+b x)^2}-e^{-i \cos ^{-1}(a+b x)} \cos ^{-1}(a+b x)-e^{i \cos ^{-1}(a+b x)} \cos ^{-1}(a+b x)+i \left (-i \cos ^{-1}(a+b x)\right )^{3/2} \Gamma \left (\frac {1}{2},-i \cos ^{-1}(a+b x)\right )-i \left (i \cos ^{-1}(a+b x)\right )^{3/2} \Gamma \left (\frac {1}{2},i \cos ^{-1}(a+b x)\right )\right )}{3 b \cos ^{-1}(a+b x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCos[a + b*x]^(-5/2),x]

[Out]

(-2*(-Sqrt[1 - (a + b*x)^2] - ArcCos[a + b*x]/E^(I*ArcCos[a + b*x]) - E^(I*ArcCos[a + b*x])*ArcCos[a + b*x] +
I*((-I)*ArcCos[a + b*x])^(3/2)*Gamma[1/2, (-I)*ArcCos[a + b*x]] - I*(I*ArcCos[a + b*x])^(3/2)*Gamma[1/2, I*Arc
Cos[a + b*x]]))/(3*b*ArcCos[a + b*x]^(3/2))

________________________________________________________________________________________

fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\arccos \left (b x + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate(arccos(b*x + a)^(-5/2), x)

________________________________________________________________________________________

maple [A]  time = 0.15, size = 120, normalized size = 1.33 \[ \frac {\sqrt {2}\, \left (4 \pi \arccos \left (b x +a \right )^{2} \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {\arccos \left (b x +a \right )}}{\sqrt {\pi }}\right )+2 \arccos \left (b x +a \right )^{\frac {3}{2}} \sqrt {2}\, \sqrt {\pi }\, x b +2 \arccos \left (b x +a \right )^{\frac {3}{2}} \sqrt {2}\, \sqrt {\pi }\, a +\sqrt {2}\, \sqrt {\pi }\, \sqrt {\arccos \left (b x +a \right )}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\right )}{3 b \sqrt {\pi }\, \arccos \left (b x +a \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arccos(b*x+a)^(5/2),x)

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {acos}\left (a+b\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/acos(a + b*x)^(5/2),x)

[Out]

int(1/acos(a + b*x)^(5/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\operatorname {acos}^{\frac {5}{2}}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/acos(b*x+a)**(5/2),x)

[Out]

Integral(acos(a + b*x)**(-5/2), x)

________________________________________________________________________________________