Integrand size = 9, antiderivative size = 142 \[ \int \tan ^p(a+b \log (x)) \, dx=x \left (1-e^{2 i a} x^{2 i b}\right )^{-p} \left (\frac {i \left (1-e^{2 i a} x^{2 i b}\right )}{1+e^{2 i a} x^{2 i b}}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^p \operatorname {AppellF1}\left (-\frac {i}{2 b},-p,p,1-\frac {i}{2 b},e^{2 i a} x^{2 i b},-e^{2 i a} x^{2 i b}\right ) \] Output:
x*(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/b,-p,p,1-1/2*I/b,exp(2*I*a)*x^(2*I*b),-exp(2*I* a)*x^(2*I*b))/((1-exp(2*I*a)*x^(2*I*b))^p)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(330\) vs. \(2(142)=284\).
Time = 0.55 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.32 \[ \int \tan ^p(a+b \log (x)) \, dx=\frac {(-i+2 b) x \left (-\frac {i \left (-1+e^{2 i a} x^{2 i b}\right )}{1+e^{2 i a} x^{2 i b}}\right )^p \operatorname {AppellF1}\left (-\frac {i}{2 b},-p,p,1-\frac {i}{2 b},e^{2 i a} x^{2 i b},-e^{2 i a} x^{2 i b}\right )}{-2 b e^{2 i a} p x^{2 i b} \operatorname {AppellF1}\left (1-\frac {i}{2 b},1-p,p,2-\frac {i}{2 b},e^{2 i a} x^{2 i b},-e^{2 i a} x^{2 i b}\right )-2 b e^{2 i a} p x^{2 i b} \operatorname {AppellF1}\left (1-\frac {i}{2 b},-p,1+p,2-\frac {i}{2 b},e^{2 i a} x^{2 i b},-e^{2 i a} x^{2 i b}\right )+(-i+2 b) \operatorname {AppellF1}\left (-\frac {i}{2 b},-p,p,1-\frac {i}{2 b},e^{2 i a} x^{2 i b},-e^{2 i a} x^{2 i b}\right )} \] Input:
Integrate[Tan[a + b*Log[x]]^p,x]
Output:
((-I + 2*b)*x*(((-I)*(-1 + E^((2*I)*a)*x^((2*I)*b)))/(1 + E^((2*I)*a)*x^(( 2*I)*b)))^p*AppellF1[(-1/2*I)/b, -p, p, 1 - (I/2)/b, E^((2*I)*a)*x^((2*I)* b), -(E^((2*I)*a)*x^((2*I)*b))])/(-2*b*E^((2*I)*a)*p*x^((2*I)*b)*AppellF1[ 1 - (I/2)/b, 1 - p, p, 2 - (I/2)/b, E^((2*I)*a)*x^((2*I)*b), -(E^((2*I)*a) *x^((2*I)*b))] - 2*b*E^((2*I)*a)*p*x^((2*I)*b)*AppellF1[1 - (I/2)/b, -p, 1 + p, 2 - (I/2)/b, E^((2*I)*a)*x^((2*I)*b), -(E^((2*I)*a)*x^((2*I)*b))] + (-I + 2*b)*AppellF1[(-1/2*I)/b, -p, p, 1 - (I/2)/b, E^((2*I)*a)*x^((2*I)*b ), -(E^((2*I)*a)*x^((2*I)*b))])
Time = 0.30 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {5002, 2058, 937, 936}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \tan ^p(a+b \log (x)) \, dx\) |
\(\Big \downarrow \) 5002 |
\(\displaystyle \int \left (\frac {i-i e^{2 i a} x^{2 i b}}{1+e^{2 i a} x^{2 i b}}\right )^pdx\) |
\(\Big \downarrow \) 2058 |
\(\displaystyle \left (i-i e^{2 i a} x^{2 i b}\right )^{-p} \left (\frac {i \left (1-e^{2 i a} x^{2 i b}\right )}{1+e^{2 i a} x^{2 i b}}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^p \int \left (i-i e^{2 i a} x^{2 i b}\right )^p \left (e^{2 i a} x^{2 i b}+1\right )^{-p}dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \left (1-e^{2 i a} x^{2 i b}\right )^{-p} \left (\frac {i \left (1-e^{2 i a} x^{2 i b}\right )}{1+e^{2 i a} x^{2 i b}}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^p \int \left (1-e^{2 i a} x^{2 i b}\right )^p \left (e^{2 i a} x^{2 i b}+1\right )^{-p}dx\) |
\(\Big \downarrow \) 936 |
\(\displaystyle x \left (1-e^{2 i a} x^{2 i b}\right )^{-p} \left (\frac {i \left (1-e^{2 i a} x^{2 i b}\right )}{1+e^{2 i a} x^{2 i b}}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^p \operatorname {AppellF1}\left (-\frac {i}{2 b},-p,p,1-\frac {i}{2 b},e^{2 i a} x^{2 i b},-e^{2 i a} x^{2 i b}\right )\) |
Input:
Int[Tan[a + b*Log[x]]^p,x]
Output:
(x*((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)/b, -p, p, 1 - (I/2)/b, E^(( 2*I)*a)*x^((2*I)*b), -(E^((2*I)*a)*x^((2*I)*b))])/(1 - E^((2*I)*a)*x^((2*I )*b))^p
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) ], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q }, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && !(IntegerQ[p] || GtQ[a, 0])
Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^ (r_.))^(p_), x_Symbol] :> Simp[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]
Int[Tan[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Int[((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 , p}, x]
\[\int \tan \left (a +b \ln \left (x \right )\right )^{p}d x\]
Input:
int(tan(a+b*ln(x))^p,x)
Output:
int(tan(a+b*ln(x))^p,x)
\[ \int \tan ^p(a+b \log (x)) \, dx=\int { \tan \left (b \log \left (x\right ) + a\right )^{p} \,d x } \] Input:
integrate(tan(a+b*log(x))^p,x, algorithm="fricas")
Output:
integral(tan(b*log(x) + a)^p, x)
\[ \int \tan ^p(a+b \log (x)) \, dx=\int \tan ^{p}{\left (a + b \log {\left (x \right )} \right )}\, dx \] Input:
integrate(tan(a+b*ln(x))**p,x)
Output:
Integral(tan(a + b*log(x))**p, x)
\[ \int \tan ^p(a+b \log (x)) \, dx=\int { \tan \left (b \log \left (x\right ) + a\right )^{p} \,d x } \] Input:
integrate(tan(a+b*log(x))^p,x, algorithm="maxima")
Output:
integrate(tan(b*log(x) + a)^p, x)
\[ \int \tan ^p(a+b \log (x)) \, dx=\int { \tan \left (b \log \left (x\right ) + a\right )^{p} \,d x } \] Input:
integrate(tan(a+b*log(x))^p,x, algorithm="giac")
Output:
integrate(tan(b*log(x) + a)^p, x)
Timed out. \[ \int \tan ^p(a+b \log (x)) \, dx=\int {\mathrm {tan}\left (a+b\,\ln \left (x\right )\right )}^p \,d x \] Input:
int(tan(a + b*log(x))^p,x)
Output:
int(tan(a + b*log(x))^p, x)
\[ \int \tan ^p(a+b \log (x)) \, dx=\int \tan \left (\mathrm {log}\left (x \right ) b +a \right )^{p}d x \] Input:
int(tan(a+b*log(x))^p,x)
Output:
int(tan(log(x)*b + a)**p,x)