\(\int (e x)^m \tan ^p(d (a+b \log (c x^n))) \, dx\) [179]

Optimal result

Integrand size = 21, antiderivative size = 210 \[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^{1+m} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{-p} \left (\frac {i \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+e^{2 i a d} \left (c x^n\right )^{2 i b d}}\right )^p \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \operatorname {AppellF1}\left (-\frac {i (1+m)}{2 b d n},-p,p,1-\frac {i (1+m)}{2 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{e (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.98 \[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x (e x)^m \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{-p} \left (-\frac {i \left (-1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+e^{2 i a d} \left (c x^n\right )^{2 i b d}}\right )^p \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \operatorname {AppellF1}\left (-\frac {i (1+m)}{2 b d n},-p,p,1-\frac {i (1+m)}{2 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+m} \] Input:

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

Output:

(x*(e*x)^m*(((-I)*(-1 + E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d)))/(1 + E^((2*I)* 
a*d)*(c*x^n)^((2*I)*b*d)))^p*(1 + E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))^p*App 
ellF1[((-1/2*I)*(1 + m))/(b*d*n), -p, p, 1 - ((I/2)*(1 + m))/(b*d*n), E^(( 
2*I)*a*d)*(c*x^n)^((2*I)*b*d), -(E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))])/((1 
+ m)*(1 - E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))^p)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5008, 5006, 2058, 1013, 1012}

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.

\(\displaystyle \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\)

\(\Big \downarrow \) 5008

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (c x^n\right )^{\frac {m+1}{n}-1} \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right )d\left (c x^n\right )}{e n}\)

\(\Big \downarrow \) 5006

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (c x^n\right )^{\frac {m+1}{n}-1} \left (\frac {i-i e^{2 i a d} \left (c x^n\right )^{2 i b d}}{e^{2 i a d} \left (c x^n\right )^{2 i b d}+1}\right )^pd\left (c x^n\right )}{e n}\)

\(\Big \downarrow \) 2058

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (i-i e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{-p} \left (\frac {i \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+e^{2 i a d} \left (c x^n\right )^{2 i b d}}\right )^p \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \int \left (c x^n\right )^{\frac {m+1}{n}-1} \left (i-i e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \left (e^{2 i a d} \left (c x^n\right )^{2 i b d}+1\right )^{-p}d\left (c x^n\right )}{e n}\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{-p} \left (\frac {i \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+e^{2 i a d} \left (c x^n\right )^{2 i b d}}\right )^p \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \int \left (c x^n\right )^{\frac {m+1}{n}-1} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \left (e^{2 i a d} \left (c x^n\right )^{2 i b d}+1\right )^{-p}d\left (c x^n\right )}{e n}\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {(e x)^{m+1} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{-p} \left (\frac {i \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+e^{2 i a d} \left (c x^n\right )^{2 i b d}}\right )^p \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \operatorname {AppellF1}\left (-\frac {i (m+1)}{2 b d n},-p,p,1-\frac {i (m+1)}{2 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{e (m+1)}\)

Input:

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

Output:

((e*x)^(1 + m)*((I*(1 - E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d)))/(1 + E^((2*I)* 
a*d)*(c*x^n)^((2*I)*b*d)))^p*(1 + E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))^p*App 
ellF1[((-1/2*I)*(1 + m))/(b*d*n), -p, p, 1 - ((I/2)*(1 + m))/(b*d*n), E^(( 
2*I)*a*d)*(c*x^n)^((2*I)*b*d), -(E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))])/(e*( 
1 + m)*(1 - E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))^p)
 

Defintions of rubi rules used

rule 1012

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] /; 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]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1013

Int[((e_.)*(x_))^(m_.)*((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[(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 2058

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]
 

rule 5006

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]
 

rule 5008

Int[((e_.)*(x_))^(m_.)*Tan[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_ 
.), x_Symbol] :> Simp[(e*x)^(m + 1)/(e*n*(c*x^n)^((m + 1)/n))   Subst[Int[x 
^((m + 1)/n - 1)*Tan[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, b, 
 c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])
 

Maple [F]

\[\int \left (e x \right )^{m} {\tan \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{p}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{p} \,d x } \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{p} \,d x } \] Input:

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

Output:

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

Giac [F(-1)]

Timed out. \[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int {\mathrm {tan}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^p\,{\left (e\,x\right )}^m \,d x \] Input:

int(tan(d*(a + b*log(c*x^n)))^p*(e*x)^m,x)
 

Output:

int(tan(d*(a + b*log(c*x^n)))^p*(e*x)^m, x)
 

Reduce [F]

\[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=e^{m} \left (\int x^{m} {\tan \left (\mathrm {log}\left (x^{n} c \right ) b d +a d \right )}^{p}d x \right ) \] Input:

int((e*x)^m*tan(d*(a+b*log(c*x^n)))^p,x)
 

Output:

e**m*int(x**m*tan(log(x**n*c)*b*d + a*d)**p,x)