3.90 \(\int \frac {1}{\sqrt {a+b \cos ^{-1}(1+d x^2)}} \, dx\)

Optimal. Leaf size=145 \[ -\frac {2 \sqrt {\pi } \sqrt {\frac {1}{b}} \cos \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x}-\frac {2 \sqrt {\pi } \sqrt {\frac {1}{b}} \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x} \]

[Out]

-2*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))*(1/b)^(1/2
)*Pi^(1/2)/d/x-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))*(1/b)^(1/2)*Pi^(1/2)/d/x

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Rubi [A]  time = 0.02, antiderivative size = 145, 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 = {4820} \[ -\frac {2 \sqrt {\pi } \sqrt {\frac {1}{b}} \cos \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x}-\frac {2 \sqrt {\pi } \sqrt {\frac {1}{b}} \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{d x} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*ArcCos[1 + d*x^2]],x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 114, normalized size = 0.79 \[ -\frac {2 \sqrt {\pi } \sqrt {\frac {1}{b}} \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \left (\cos \left (\frac {a}{2 b}\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )+\sin \left (\frac {a}{2 b}\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )\right )}{d x} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + b*ArcCos[1 + d*x^2]],x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a +b \arccos \left (d \,x^{2}+1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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