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

Optimal. Leaf size=221 \[ \frac {x}{3 b^2 \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}+\frac {\sqrt {-d^2 x^4-2 d x^2}}{3 b d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2}}+\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{5/2} \cos \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{3 d x}+\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{5/2} \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{3 d x} \]

[Out]

2/3*(1/b)^(5/2)*cos(1/2*a/b)*FresnelC((1/b)^(1/2)*(a+b*arccos(d*x^2+1))^(1/2)/Pi^(1/2))*sin(1/2*arccos(d*x^2+1
))*Pi^(1/2)/d/x+2/3*(1/b)^(5/2)*FresnelS((1/b)^(1/2)*(a+b*arccos(d*x^2+1))^(1/2)/Pi^(1/2))*sin(1/2*a/b)*sin(1/
2*arccos(d*x^2+1))*Pi^(1/2)/d/x+1/3*(-d^2*x^4-2*d*x^2)^(1/2)/b/d/x/(a+b*arccos(d*x^2+1))^(3/2)+1/3*x/b^2/(a+b*
arccos(d*x^2+1))^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4829, 4820} \[ \frac {x}{3 b^2 \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}+\frac {\sqrt {-d^2 x^4-2 d x^2}}{3 b d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2}}+\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{5/2} \cos \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{3 d x}+\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{5/2} \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{3 d x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[1 + d*x^2])^(-5/2),x]

[Out]

Sqrt[-2*d*x^2 - d^2*x^4]/(3*b*d*x*(a + b*ArcCos[1 + d*x^2])^(3/2)) + x/(3*b^2*Sqrt[a + b*ArcCos[1 + d*x^2]]) +
(2*(b^(-1))^(5/2)*Sqrt[Pi]*Cos[a/(2*b)]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[A
rcCos[1 + d*x^2]/2])/(3*d*x) + (2*(b^(-1))^(5/2)*Sqrt[Pi]*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]]
)/Sqrt[Pi]]*Sin[a/(2*b)]*Sin[ArcCos[1 + d*x^2]/2])/(3*d*x)

Rule 4820

Int[1/Sqrt[(a_.) + ArcCos[1 + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[(-2*Sqrt[Pi/b]*Cos[a/(2*b)]*Sin[ArcCos[1
+ d*x^2]/2]*FresnelC[Sqrt[1/(Pi*b)]*Sqrt[a + b*ArcCos[1 + d*x^2]]])/(d*x), x] - Simp[(2*Sqrt[Pi/b]*Sin[a/(2*b
)]*Sin[ArcCos[1 + d*x^2]/2]*FresnelS[Sqrt[1/(Pi*b)]*Sqrt[a + b*ArcCos[1 + d*x^2]]])/(d*x), x] /; FreeQ[{a, b,
d}, x]

Rule 4829

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[(x*(a + b*ArcCos[c + d*x^2])^(n + 2))/
(4*b^2*(n + 1)*(n + 2)), x] + (-Dist[1/(4*b^2*(n + 1)*(n + 2)), Int[(a + b*ArcCos[c + d*x^2])^(n + 2), x], x]
- Simp[(Sqrt[-2*c*d*x^2 - d^2*x^4]*(a + b*ArcCos[c + d*x^2])^(n + 1))/(2*b*d*(n + 1)*x), x]) /; FreeQ[{a, b, c
, d}, x] && EqQ[c^2, 1] && LtQ[n, -1] && NeQ[n, -2]

Rubi steps

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

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Mathematica [A]  time = 0.76, size = 234, normalized size = 1.06 \[ \frac {2 \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \left (\sqrt {\pi } \sqrt {\frac {1}{b}} \cos \left (\frac {a}{2 b}\right ) \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2} C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )+\sqrt {\pi } \sqrt {\frac {1}{b}} \sin \left (\frac {a}{2 b}\right ) \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2} S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )-a \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right )+b \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right )-b \cos ^{-1}\left (d x^2+1\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right )\right )}{3 b^2 d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCos[1 + d*x^2])^(-5/2),x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arccos(d*x^2 + 1) + a)^(-5/2), x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \arccos \left (d \,x^{2}+1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found sqrt((-_SAGE_VAR_d*_SAGE_VAR_x^2)-2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acos(d*x**2 + 1))**(-5/2), x)

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