3.1.60 \(\int \frac {1}{x (a+b \tan (c+d \sqrt [3]{x}))} \, dx\) [60]

Optimal. Leaf size=23 \[ \text {Int}\left (\frac {1}{x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )},x\right ) \]

[Out]

Unintegrable(1/x/(a+b*tan(c+d*x^(1/3))),x)

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Rubi [A]
time = 0.02, 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}{x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[1/(x*(a + b*Tan[c + d*x^(1/3)])),x]

[Out]

Defer[Int][1/(x*(a + b*Tan[c + d*x^(1/3)])), x]

Rubi steps

\begin {align*} \int \frac {1}{x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )} \, dx &=\int \frac {1}{x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )} \, dx\\ \end {align*}

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Mathematica [A]
time = 11.11, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[1/(x*(a + b*Tan[c + d*x^(1/3)])),x]

[Out]

Integrate[1/(x*(a + b*Tan[c + d*x^(1/3)])), x]

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Maple [A]
time = 0.80, size = 0, normalized size = 0.00 \[\int \frac {1}{x \left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(a+b*tan(c+d*x^(1/3))),x)

[Out]

int(1/x/(a+b*tan(c+d*x^(1/3))),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/x/(a+b*tan(c+d*x^(1/3))),x, algorithm="maxima")

[Out]

-(2*(a^2*b + b^3)*integrate((a^2*sin(2*d*x^(1/3) + 2*c) - (2*a*b*cos(2*c) + b^2*sin(2*c))*cos(2*d*x^(1/3)) - (
b^2*cos(2*c) - 2*a*b*sin(2*c))*sin(2*d*x^(1/3)))/((a^4*cos(2*d*x^(1/3) + 2*c)^2 + a^4*sin(2*d*x^(1/3) + 2*c)^2
 + a^4 + 2*a^2*b^2 + b^4 + ((4*a^2*b^2 + b^4)*cos(2*c)^2 + (4*a^2*b^2 + b^4)*sin(2*c)^2)*cos(2*d*x^(1/3))^2 +
((4*a^2*b^2 + b^4)*cos(2*c)^2 + (4*a^2*b^2 + b^4)*sin(2*c)^2)*sin(2*d*x^(1/3))^2 - 2*((a^2*b^2 + b^4)*cos(2*c)
 - 2*(a^3*b + a*b^3)*sin(2*c))*cos(2*d*x^(1/3)) + 2*(a^4 + a^2*b^2 - (a^2*b^2*cos(2*c) - 2*a^3*b*sin(2*c))*cos
(2*d*x^(1/3)) + (2*a^3*b*cos(2*c) + a^2*b^2*sin(2*c))*sin(2*d*x^(1/3)))*cos(2*d*x^(1/3) + 2*c) + 2*(2*(a^3*b +
 a*b^3)*cos(2*c) + (a^2*b^2 + b^4)*sin(2*c))*sin(2*d*x^(1/3)) - 2*((2*a^3*b*cos(2*c) + a^2*b^2*sin(2*c))*cos(2
*d*x^(1/3)) + (a^2*b^2*cos(2*c) - 2*a^3*b*sin(2*c))*sin(2*d*x^(1/3)))*sin(2*d*x^(1/3) + 2*c))*x), x) - a*log(x
))/(a^2 + b^2)

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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/x/(a+b*tan(c+d*x^(1/3))),x, algorithm="fricas")

[Out]

integral(1/(b*x*tan(d*x^(1/3) + c) + a*x), x)

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Sympy [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (a + b \tan {\left (c + d \sqrt [3]{x} \right )}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*tan(c+d*x**(1/3))),x)

[Out]

Integral(1/(x*(a + b*tan(c + d*x**(1/3)))), x)

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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/x/(a+b*tan(c+d*x^(1/3))),x, algorithm="giac")

[Out]

integrate(1/((b*tan(d*x^(1/3) + c) + a)*x), x)

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Mupad [A]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {1}{x\,\left (a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*tan(c + d*x^(1/3)))),x)

[Out]

int(1/(x*(a + b*tan(c + d*x^(1/3)))), x)

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