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

Optimal. Leaf size=190 \[ \frac {\sqrt {-d^2 x^4-2 d x^2}}{b d x \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}-\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x}+\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \cos \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4823} \[ \frac {\sqrt {-d^2 x^4-2 d x^2}}{b d x \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}-\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x}+\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \cos \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[1 + d*x^2])^(-3/2),x]

[Out]

Sqrt[-2*d*x^2 - d^2*x^4]/(b*d*x*Sqrt[a + b*ArcCos[1 + d*x^2]]) + (2*(b^(-1))^(3/2)*Sqrt[Pi]*Cos[a/(2*b)]*Fresn
elS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[ArcCos[1 + d*x^2]/2])/(d*x) - (2*(b^(-1))^(3/2)
*Sqrt[Pi]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[a/(2*b)]*Sin[ArcCos[1 + d*x^2]/2
])/(d*x)

Rule 4823

Int[((a_.) + ArcCos[1 + (d_.)*(x_)^2]*(b_.))^(-3/2), x_Symbol] :> Simp[Sqrt[-2*d*x^2 - d^2*x^4]/(b*d*x*Sqrt[a
+ b*ArcCos[1 + d*x^2]]), x] + (-Simp[(2*(1/b)^(3/2)*Sqrt[Pi]*Sin[a/(2*b)]*Sin[ArcCos[1 + d*x^2]/2]*FresnelC[Sq
rt[1/(Pi*b)]*Sqrt[a + b*ArcCos[1 + d*x^2]]])/(d*x), x] + Simp[(2*(1/b)^(3/2)*Sqrt[Pi]*Cos[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]

Rubi steps

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

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Mathematica [A]  time = 0.42, size = 177, normalized size = 0.93 \[ \frac {-2 \sqrt {\pi } \sqrt {\frac {1}{b}} \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )+2 \sqrt {\pi } \sqrt {\frac {1}{b}} \cos \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )+\frac {\sqrt {-d x^2 \left (d x^2+2\right )}}{\sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}}{b d x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCos[1 + d*x^2])^(-3/2),x]

[Out]

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

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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))^(3/2),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 \[ \int \frac {1}{{\left (b \arccos \left (d x^{2} + 1\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arccos(d*x^2+1))^(3/2),x, algorithm="giac")

[Out]

integrate((b*arccos(d*x^2 + 1) + a)^(-3/2), x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \arccos \left (d \,x^{2}+1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arccos(d*x^2+1))^(3/2),x)

[Out]

int(1/(a+b*arccos(d*x^2+1))^(3/2),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))^(3/2),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 = -0.01 \[ \int \frac {1}{{\left (a+b\,\mathrm {acos}\left (d\,x^2+1\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*acos(d*x^2 + 1))^(3/2),x)

[Out]

int(1/(a + b*acos(d*x^2 + 1))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {acos}{\left (d x^{2} + 1 \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*acos(d*x**2+1))**(3/2),x)

[Out]

Integral((a + b*acos(d*x**2 + 1))**(-3/2), x)

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