3.154 \(\int (e x)^m \tan ^p(a+b \log (x)) \, dx\)

Optimal. Leaf size=162 \[ \frac {(e x)^{m+1} \left (1-e^{2 i a} x^{2 i b}\right )^{-p} \left (\frac {i \left (1-e^{2 i a} x^{2 i b}\right )}{1+e^{2 i a} x^{2 i b}}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^p F_1\left (-\frac {i (m+1)}{2 b};-p,p;1-\frac {i (m+1)}{2 b};e^{2 i a} x^{2 i b},-e^{2 i a} x^{2 i b}\right )}{e (m+1)} \]

[Out]

(e*x)^(1+m)*(I*(1-exp(2*I*a)*x^(2*I*b))/(1+exp(2*I*a)*x^(2*I*b)))^p*(1+exp(2*I*a)*x^(2*I*b))^p*AppellF1(-1/2*I
*(1+m)/b,-p,p,1-1/2*I*(1+m)/b,exp(2*I*a)*x^(2*I*b),-exp(2*I*a)*x^(2*I*b))/e/(1+m)/((1-exp(2*I*a)*x^(2*I*b))^p)

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Rubi [F]  time = 0.13, 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 = {} \[ \int (e x)^m \tan ^p(a+b \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[(e*x)^m*Tan[a + b*Log[x]]^p,x]

[Out]

Defer[Int][(e*x)^m*Tan[a + b*Log[x]]^p, x]

Rubi steps

\begin {align*} \int (e x)^m \tan ^p(a+b \log (x)) \, dx &=\int (e x)^m \tan ^p(a+b \log (x)) \, dx\\ \end {align*}

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Mathematica [A]  time = 0.67, size = 157, normalized size = 0.97 \[ \frac {x (e x)^m \left (1-e^{2 i a} x^{2 i b}\right )^{-p} \left (-\frac {i \left (-1+e^{2 i a} x^{2 i b}\right )}{1+e^{2 i a} x^{2 i b}}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^p F_1\left (-\frac {i (m+1)}{2 b};-p,p;1-\frac {i (m+1)}{2 b};e^{2 i a} x^{2 i b},-e^{2 i a} x^{2 i b}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Tan[a + b*Log[x]]^p,x]

[Out]

(x*(e*x)^m*(((-I)*(-1 + E^((2*I)*a)*x^((2*I)*b)))/(1 + E^((2*I)*a)*x^((2*I)*b)))^p*(1 + E^((2*I)*a)*x^((2*I)*b
))^p*AppellF1[((-1/2*I)*(1 + m))/b, -p, p, 1 - ((I/2)*(1 + m))/b, E^((2*I)*a)*x^((2*I)*b), -(E^((2*I)*a)*x^((2
*I)*b))])/((1 + m)*(1 - E^((2*I)*a)*x^((2*I)*b))^p)

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (e x\right )^{m} \tan \left (b \log \relax (x) + a\right )^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*tan(a+b*log(x))^p,x, algorithm="fricas")

[Out]

integral((e*x)^m*tan(b*log(x) + a)^p, x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*tan(a+b*log(x))^p,x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{m} \left (\tan ^{p}\left (a +b \ln \relax (x )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*tan(a+b*ln(x))^p,x)

[Out]

int((e*x)^m*tan(a+b*ln(x))^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{m} \tan \left (b \log \relax (x) + a\right )^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*tan(a+b*log(x))^p,x, algorithm="maxima")

[Out]

integrate((e*x)^m*tan(b*log(x) + a)^p, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (a+b\,\ln \relax (x)\right )}^p\,{\left (e\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(a + b*log(x))^p*(e*x)^m,x)

[Out]

int(tan(a + b*log(x))^p*(e*x)^m, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{m} \tan ^{p}{\left (a + b \log {\relax (x )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*tan(a+b*ln(x))**p,x)

[Out]

Integral((e*x)**m*tan(a + b*log(x))**p, x)

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