3.4.6 \(\int \frac {1}{(a+b \sin (c+\frac {d}{x}))^2} \, dx\) [306]

Optimal. Leaf size=17 \[ \text {Int}\left (\frac {1}{\left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2},x\right ) \]

[Out]

Unintegrable(1/(a+b*sin(c+d/x))^2,x)

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Rubi [A]
time = 0.00, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1}{\left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(a + b*Sin[c + d/x])^(-2),x]

[Out]

Defer[Int][(a + b*Sin[c + d/x])^(-2), x]

Rubi steps

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

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Mathematica [A]
time = 2.21, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(a + b*Sin[c + d/x])^(-2),x]

[Out]

Integrate[(a + b*Sin[c + d/x])^(-2), x]

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Maple [A]
time = 0.12, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \sin \left (c +\frac {d}{x}\right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(c+d/x))^2,x, algorithm="maxima")

[Out]

-(2*a*b*x^2*cos(2*(c*x + d)/x)*cos((c*x + d)/x) + 2*a*b*x^2*cos((c*x + d)/x) + ((a^2*b^2 - b^4)*d*cos(2*(c*x +
d)/x)^2 + 4*(a^4 - a^2*b^2)*d*cos((c*x + d)/x)^2 + 4*(a^3*b - a*b^3)*d*cos((c*x + d)/x)*sin(2*(c*x + d)/x) +
(a^2*b^2 - b^4)*d*sin(2*(c*x + d)/x)^2 + 4*(a^4 - a^2*b^2)*d*sin((c*x + d)/x)^2 + 4*(a^3*b - a*b^3)*d*sin((c*x
+ d)/x) + (a^2*b^2 - b^4)*d - 2*(2*(a^3*b - a*b^3)*d*sin((c*x + d)/x) + (a^2*b^2 - b^4)*d)*cos(2*(c*x + d)/x)
)*integrate(-2*(2*a^2*d*cos((c*x + d)/x)^2 + 2*a^2*d*sin((c*x + d)/x)^2 + 2*a*b*x*cos((c*x + d)/x) + a*b*d*sin
((c*x + d)/x) + (2*a*b*x*cos((c*x + d)/x) - a*b*d*sin((c*x + d)/x))*cos(2*(c*x + d)/x) + (a*b*d*cos((c*x + d)/
x) + 2*a*b*x*sin((c*x + d)/x) + 2*b^2*x)*sin(2*(c*x + d)/x))/((a^2*b^2 - b^4)*d*cos(2*(c*x + d)/x)^2 + 4*(a^4
- a^2*b^2)*d*cos((c*x + d)/x)^2 + 4*(a^3*b - a*b^3)*d*cos((c*x + d)/x)*sin(2*(c*x + d)/x) + (a^2*b^2 - b^4)*d*
sin(2*(c*x + d)/x)^2 + 4*(a^4 - a^2*b^2)*d*sin((c*x + d)/x)^2 + 4*(a^3*b - a*b^3)*d*sin((c*x + d)/x) + (a^2*b^
2 - b^4)*d - 2*(2*(a^3*b - a*b^3)*d*sin((c*x + d)/x) + (a^2*b^2 - b^4)*d)*cos(2*(c*x + d)/x)), x) + 2*(a*b*x^2
*sin((c*x + d)/x) + b^2*x^2)*sin(2*(c*x + d)/x))/((a^2*b^2 - b^4)*d*cos(2*(c*x + d)/x)^2 + 4*(a^4 - a^2*b^2)*d
*cos((c*x + d)/x)^2 + 4*(a^3*b - a*b^3)*d*cos((c*x + d)/x)*sin(2*(c*x + d)/x) + (a^2*b^2 - b^4)*d*sin(2*(c*x +
d)/x)^2 + 4*(a^4 - a^2*b^2)*d*sin((c*x + d)/x)^2 + 4*(a^3*b - a*b^3)*d*sin((c*x + d)/x) + (a^2*b^2 - b^4)*d -
2*(2*(a^3*b - a*b^3)*d*sin((c*x + d)/x) + (a^2*b^2 - b^4)*d)*cos(2*(c*x + d)/x))

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Fricas [A]
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*sin(c+d/x))^2,x, algorithm="fricas")

[Out]

integral(-1/(b^2*cos((c*x + d)/x)^2 - 2*a*b*sin((c*x + d)/x) - a^2 - b^2), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
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*sin(c+d/x))^2,x, algorithm="giac")

[Out]

integrate((b*sin(c + d/x) + a)^(-2), x)

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Mupad [A]
time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int \frac {1}{{\left (a+b\,\sin \left (c+\frac {d}{x}\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(c + d/x))^2,x)

[Out]

int(1/(a + b*sin(c + d/x))^2, x)

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