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

Optimal. Leaf size=151 \[ \frac {x \sin \left (\frac {a}{2 b}\right ) \text {Ci}\left (\frac {a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {-d x^2}}-\frac {x \cos \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {-d x^2}}+\frac {\sqrt {-d^2 x^4-2 d x^2}}{2 b d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )} \]

[Out]

-1/4*x*cos(1/2*a/b)*Si(1/2*(a+b*arccos(d*x^2+1))/b)/b^2*2^(1/2)/(-d*x^2)^(1/2)+1/4*x*Ci(1/2*(a+b*arccos(d*x^2+
1))/b)*sin(1/2*a/b)/b^2*2^(1/2)/(-d*x^2)^(1/2)+1/2*(-d^2*x^4-2*d*x^2)^(1/2)/b/d/x/(a+b*arccos(d*x^2+1))

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Rubi [A]  time = 0.02, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {4826} \[ \frac {x \sin \left (\frac {a}{2 b}\right ) \text {CosIntegral}\left (\frac {a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {-d x^2}}-\frac {x \cos \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {-d x^2}}+\frac {\sqrt {-d^2 x^4-2 d x^2}}{2 b d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[1 + d*x^2])^(-2),x]

[Out]

Sqrt[-2*d*x^2 - d^2*x^4]/(2*b*d*x*(a + b*ArcCos[1 + d*x^2])) + (x*CosIntegral[(a + b*ArcCos[1 + d*x^2])/(2*b)]
*Sin[a/(2*b)])/(2*Sqrt[2]*b^2*Sqrt[-(d*x^2)]) - (x*Cos[a/(2*b)]*SinIntegral[(a + b*ArcCos[1 + d*x^2])/(2*b)])/
(2*Sqrt[2]*b^2*Sqrt[-(d*x^2)])

Rule 4826

Int[((a_.) + ArcCos[1 + (d_.)*(x_)^2]*(b_.))^(-2), x_Symbol] :> Simp[Sqrt[-2*d*x^2 - d^2*x^4]/(2*b*d*x*(a + b*
ArcCos[1 + d*x^2])), x] + (Simp[(x*Sin[a/(2*b)]*CosIntegral[(a + b*ArcCos[1 + d*x^2])/(2*b)])/(2*Sqrt[2]*b^2*S
qrt[(-d)*x^2]), x] - Simp[(x*Cos[a/(2*b)]*SinIntegral[(a + b*ArcCos[1 + d*x^2])/(2*b)])/(2*Sqrt[2]*b^2*Sqrt[(-
d)*x^2]), 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 )^2} \, dx &=\frac {\sqrt {-2 d x^2-d^2 x^4}}{2 b d x \left (a+b \cos ^{-1}\left (1+d x^2\right )\right )}+\frac {x \text {Ci}\left (\frac {a+b \cos ^{-1}\left (1+d x^2\right )}{2 b}\right ) \sin \left (\frac {a}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {-d x^2}}-\frac {x \cos \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \cos ^{-1}\left (1+d x^2\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {-d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 133, normalized size = 0.88 \[ \frac {\sqrt {-d x^2 \left (d x^2+2\right )} \left (\frac {b}{a+b \cos ^{-1}\left (d x^2+1\right )}-\frac {\cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \left (\sin \left (\frac {a}{2 b}\right ) \text {Ci}\left (\frac {a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )-\cos \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )\right )}{d x^2+2}\right )}{2 b^2 d x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[-(d*x^2*(2 + d*x^2))]*(b/(a + b*ArcCos[1 + d*x^2]) - (Cos[ArcCos[1 + d*x^2]/2]*(CosIntegral[(a + b*ArcCo
s[1 + d*x^2])/(2*b)]*Sin[a/(2*b)] - Cos[a/(2*b)]*SinIntegral[(a + b*ArcCos[1 + d*x^2])/(2*b)]))/(2 + d*x^2)))/
(2*b^2*d*x)

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{2} \arccos \left (d x^{2} + 1\right )^{2} + 2 \, a b \arccos \left (d x^{2} + 1\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*acos(d*x^2 + 1))^2,x)

[Out]

int(1/(a + b*acos(d*x^2 + 1))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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