3.2.54 \(\int (e x)^m \tan ^p(a+b \log (x)) \, dx\) [154]

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

[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 [A]
time = 0.08, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4591, 1986, 525, 524} \begin {gather*} \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)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Tan[a + b*Log[x]]^p,x]

[Out]

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

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 1986

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^(r_.))^(p_), x_Symbol] :> Dist[Simp
[(e*(a + b*x^n)^q*(c + d*x^n)^r)^p/((a + b*x^n)^(p*q)*(c + d*x^n)^(p*r))], Int[u*(a + b*x^n)^(p*q)*(c + d*x^n)
^(p*r), x], x] /; FreeQ[{a, b, c, d, e, n, p, q, r}, x]

Rule 4591

Int[((e_.)*(x_))^(m_.)*Tan[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Int[(e*x)^m*((I - I*E^(2*I*a*d)*
x^(2*I*b*d))/(1 + E^(2*I*a*d)*x^(2*I*b*d)))^p, x] /; FreeQ[{a, b, d, e, m, 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 = 0.70, size = 157, normalized size = 0.97 \begin {gather*} \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 (1+m)}{2 b};-p,p;1-\frac {i (1+m)}{2 b};e^{2 i a} x^{2 i b},-e^{2 i a} x^{2 i b}\right )}{1+m} \end {gather*}

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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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (\tan ^{p}\left (a +b \ln \left (x \right )\right )\right )\, dx\]

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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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((x*e)^m*tan(b*log(x) + a)^p, x)

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

[Out]

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

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

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

[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 \begin {gather*} \int {\mathrm {tan}\left (a+b\,\ln \left (x\right )\right )}^p\,{\left (e\,x\right )}^m \,d x \end {gather*}

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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