3.5.5 \(\int \frac {1}{a+b \text {ArcSin}(1+d x^2)} \, dx\) [405]

Optimal. Leaf size=159 \[ -\frac {x \text {CosIntegral}\left (\frac {a+b \text {ArcSin}\left (1+d x^2\right )}{2 b}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{2 b \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+d x^2\right )\right )\right )}-\frac {x \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a+b \text {ArcSin}\left (1+d x^2\right )}{2 b}\right )}{2 b \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+d x^2\right )\right )\right )} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 159, 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 = {4900} \begin {gather*} -\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcSin}\left (d x^2+1\right )}{2 b}\right )}{2 b \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (d x^2+1\right )\right )\right )}-\frac {x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a+b \text {ArcSin}\left (d x^2+1\right )}{2 b}\right )}{2 b \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (d x^2+1\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[1 + d*x^2])^(-1),x]

[Out]

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

Rule 4900

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-1), x_Symbol] :> Simp[(-x)*(c*Cos[a/(2*b)] - Sin[a/(2*b)])*(
CosIntegral[(c/(2*b))*(a + b*ArcSin[c + d*x^2])]/(2*b*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2]))
), x] - Simp[x*(c*Cos[a/(2*b)] + Sin[a/(2*b)])*(SinIntegral[(c/(2*b))*(a + b*ArcSin[c + d*x^2])]/(2*b*(Cos[Arc
Sin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2]))), x] /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1]

Rubi steps

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

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Mathematica [A]
time = 0.50, size = 120, normalized size = 0.75 \begin {gather*} -\frac {x \left (\text {CosIntegral}\left (\frac {1}{2} \left (\frac {a}{b}+\text {ArcSin}\left (1+d x^2\right )\right )\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )+\left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {1}{2} \left (\frac {a}{b}+\text {ArcSin}\left (1+d x^2\right )\right )\right )\right )}{2 b \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+d x^2\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {1}{a +b \arcsin \left (d \,x^{2}+1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x^2+1)),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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{a + b \operatorname {asin}{\left (d x^{2} + 1 \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{a+b\,\mathrm {asin}\left (d\,x^2+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asin(d*x^2 + 1)),x)

[Out]

int(1/(a + b*asin(d*x^2 + 1)), x)

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