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

Optimal. Leaf size=111 \[ \frac {15 \sqrt {\frac {\pi }{2}} C\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a+b x)}\right )}{4 b}+\frac {(a+b x) \cos ^{-1}(a+b x)^{5/2}}{b}-\frac {5 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)^{3/2}}{2 b}-\frac {15 (a+b x) \sqrt {\cos ^{-1}(a+b x)}}{4 b} \]

[Out]

(b*x+a)*arccos(b*x+a)^(5/2)/b+15/8*FresnelC(2^(1/2)/Pi^(1/2)*arccos(b*x+a)^(1/2))*2^(1/2)*Pi^(1/2)/b-5/2*arcco
s(b*x+a)^(3/2)*(1-(b*x+a)^2)^(1/2)/b-15/4*(b*x+a)*arccos(b*x+a)^(1/2)/b

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Rubi [A]  time = 0.15, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4804, 4620, 4678, 4724, 3304, 3352} \[ \frac {15 \sqrt {\frac {\pi }{2}} \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a+b x)}\right )}{4 b}+\frac {(a+b x) \cos ^{-1}(a+b x)^{5/2}}{b}-\frac {5 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)^{3/2}}{2 b}-\frac {15 (a+b x) \sqrt {\cos ^{-1}(a+b x)}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]^(5/2),x]

[Out]

(-15*(a + b*x)*Sqrt[ArcCos[a + b*x]])/(4*b) - (5*Sqrt[1 - (a + b*x)^2]*ArcCos[a + b*x]^(3/2))/(2*b) + ((a + b*
x)*ArcCos[a + b*x]^(5/2))/b + (15*Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcCos[a + b*x]]])/(4*b)

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]

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 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 \cos ^{-1}(a+b x)^{5/2} \, dx &=\frac {\operatorname {Subst}\left (\int \cos ^{-1}(x)^{5/2} \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \cos ^{-1}(a+b x)^{5/2}}{b}+\frac {5 \operatorname {Subst}\left (\int \frac {x \cos ^{-1}(x)^{3/2}}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{2 b}\\ &=-\frac {5 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)^{3/2}}{2 b}+\frac {(a+b x) \cos ^{-1}(a+b x)^{5/2}}{b}-\frac {15 \operatorname {Subst}\left (\int \sqrt {\cos ^{-1}(x)} \, dx,x,a+b x\right )}{4 b}\\ &=-\frac {15 (a+b x) \sqrt {\cos ^{-1}(a+b x)}}{4 b}-\frac {5 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)^{3/2}}{2 b}+\frac {(a+b x) \cos ^{-1}(a+b x)^{5/2}}{b}-\frac {15 \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \sqrt {\cos ^{-1}(x)}} \, dx,x,a+b x\right )}{8 b}\\ &=-\frac {15 (a+b x) \sqrt {\cos ^{-1}(a+b x)}}{4 b}-\frac {5 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)^{3/2}}{2 b}+\frac {(a+b x) \cos ^{-1}(a+b x)^{5/2}}{b}+\frac {15 \operatorname {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a+b x)\right )}{8 b}\\ &=-\frac {15 (a+b x) \sqrt {\cos ^{-1}(a+b x)}}{4 b}-\frac {5 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)^{3/2}}{2 b}+\frac {(a+b x) \cos ^{-1}(a+b x)^{5/2}}{b}+\frac {15 \operatorname {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a+b x)}\right )}{4 b}\\ &=-\frac {15 (a+b x) \sqrt {\cos ^{-1}(a+b x)}}{4 b}-\frac {5 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)^{3/2}}{2 b}+\frac {(a+b x) \cos ^{-1}(a+b x)^{5/2}}{b}+\frac {15 \sqrt {\frac {\pi }{2}} C\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a+b x)}\right )}{4 b}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 90, normalized size = 0.81 \[ -\frac {\frac {\sqrt {\cos ^{-1}(a+b x)} \Gamma \left (\frac {7}{2},-i \cos ^{-1}(a+b x)\right )}{2 \sqrt {-i \cos ^{-1}(a+b x)}}+\frac {\sqrt {\cos ^{-1}(a+b x)} \Gamma \left (\frac {7}{2},i \cos ^{-1}(a+b x)\right )}{2 \sqrt {i \cos ^{-1}(a+b x)}}}{b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCos[a + b*x]^(5/2),x]

[Out]

-(((Sqrt[ArcCos[a + b*x]]*Gamma[7/2, (-I)*ArcCos[a + b*x]])/(2*Sqrt[(-I)*ArcCos[a + b*x]]) + (Sqrt[ArcCos[a +
b*x]]*Gamma[7/2, I*ArcCos[a + b*x]])/(2*Sqrt[I*ArcCos[a + b*x]]))/b)

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

[Out]

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

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giac [B]  time = 0.46, size = 209, normalized size = 1.88 \[ \frac {5 \, i \arccos \left (b x + a\right )^{\frac {3}{2}} e^{\left (i \arccos \left (b x + a\right )\right )}}{4 \, b} + \frac {\arccos \left (b x + a\right )^{\frac {5}{2}} e^{\left (i \arccos \left (b x + a\right )\right )}}{2 \, b} - \frac {5 \, i \arccos \left (b x + a\right )^{\frac {3}{2}} e^{\left (-i \arccos \left (b x + a\right )\right )}}{4 \, b} + \frac {\arccos \left (b x + a\right )^{\frac {5}{2}} e^{\left (-i \arccos \left (b x + a\right )\right )}}{2 \, b} - \frac {15 \, \sqrt {2} \sqrt {\pi } i \operatorname {erf}\left (\frac {\sqrt {2} \sqrt {\arccos \left (b x + a\right )}}{i - 1}\right )}{16 \, b {\left (i - 1\right )}} - \frac {15 \, \sqrt {\arccos \left (b x + a\right )} e^{\left (i \arccos \left (b x + a\right )\right )}}{8 \, b} - \frac {15 \, \sqrt {\arccos \left (b x + a\right )} e^{\left (-i \arccos \left (b x + a\right )\right )}}{8 \, b} + \frac {15 \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {\sqrt {2} i \sqrt {\arccos \left (b x + a\right )}}{i - 1}\right )}{16 \, b {\left (i - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*i*arccos(b*x + a)^(3/2)*e^(i*arccos(b*x + a))/b + 1/2*arccos(b*x + a)^(5/2)*e^(i*arccos(b*x + a))/b - 5/4*
i*arccos(b*x + a)^(3/2)*e^(-i*arccos(b*x + a))/b + 1/2*arccos(b*x + a)^(5/2)*e^(-i*arccos(b*x + a))/b - 15/16*
sqrt(2)*sqrt(pi)*i*erf(sqrt(2)*sqrt(arccos(b*x + a))/(i - 1))/(b*(i - 1)) - 15/8*sqrt(arccos(b*x + a))*e^(i*ar
ccos(b*x + a))/b - 15/8*sqrt(arccos(b*x + a))*e^(-i*arccos(b*x + a))/b + 15/16*sqrt(2)*sqrt(pi)*erf(-sqrt(2)*i
*sqrt(arccos(b*x + a))/(i - 1))/(b*(i - 1))

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maple [A]  time = 0.15, size = 140, normalized size = 1.26 \[ -\frac {\sqrt {2}\, \left (-4 \arccos \left (b x +a \right )^{\frac {5}{2}} \sqrt {2}\, \sqrt {\pi }\, x b -4 \arccos \left (b x +a \right )^{\frac {5}{2}} \sqrt {2}\, \sqrt {\pi }\, a +10 \arccos \left (b x +a \right )^{\frac {3}{2}} \sqrt {2}\, \sqrt {\pi }\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}+15 \sqrt {2}\, \sqrt {\pi }\, \sqrt {\arccos \left (b x +a \right )}\, x b +15 \sqrt {2}\, \sqrt {\pi }\, \sqrt {\arccos \left (b x +a \right )}\, a -15 \pi \FresnelC \left (\frac {\sqrt {2}\, \sqrt {\arccos \left (b x +a \right )}}{\sqrt {\pi }}\right )\right )}{8 b \sqrt {\pi }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/b*2^(1/2)*(-4*arccos(b*x+a)^(5/2)*2^(1/2)*Pi^(1/2)*x*b-4*arccos(b*x+a)^(5/2)*2^(1/2)*Pi^(1/2)*a+10*arccos
(b*x+a)^(3/2)*2^(1/2)*Pi^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+15*2^(1/2)*Pi^(1/2)*arccos(b*x+a)^(1/2)*x*b+15*2
^(1/2)*Pi^(1/2)*arccos(b*x+a)^(1/2)*a-15*Pi*FresnelC(2^(1/2)/Pi^(1/2)*arccos(b*x+a)^(1/2)))/Pi^(1/2)

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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(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.

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(a + b*x)^(5/2),x)

[Out]

int(acos(a + b*x)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acos(a + b*x)**(5/2), x)

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