Integrand size = 19, antiderivative size = 101 \[ \int (e x)^m \tan \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-\frac {i (e x)^{1+m}}{e (1+m)}+\frac {2 i (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,-\frac {i (1+m)}{2 b d n},1-\frac {i (1+m)}{2 b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{e (1+m)} \]
-I*(e*x)^(1+m)/e/(1+m)+2*I*(e*x)^(1+m)*hypergeom([1, -1/2*I*(1+m)/b/d/n],[ 1-1/2*I*(1+m)/b/d/n],-exp(2*I*a*d)*(c*x^n)^(2*I*b*d))/e/(1+m)
Time = 12.54 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.84 \[ \int (e x)^m \tan \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {i x (e x)^m \left (\operatorname {Hypergeometric2F1}\left (1,-\frac {i (1+m)}{2 b d n},1-\frac {i (1+m)}{2 b d n},-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )-\frac {e^{2 i a d} (1+m) \left (c x^n\right )^{2 i b d} \operatorname {Hypergeometric2F1}\left (1,-\frac {i (1+m+2 i b d n)}{2 b d n},-\frac {i (1+m+4 i b d n)}{2 b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+m+2 i b d n}\right )}{1+m} \]
(I*x*(e*x)^m*(Hypergeometric2F1[1, ((-1/2*I)*(1 + m))/(b*d*n), 1 - ((I/2)* (1 + m))/(b*d*n), -E^((2*I)*d*(a + b*Log[c*x^n]))] - (E^((2*I)*a*d)*(1 + m )*(c*x^n)^((2*I)*b*d)*Hypergeometric2F1[1, ((-1/2*I)*(1 + m + (2*I)*b*d*n) )/(b*d*n), ((-1/2*I)*(1 + m + (4*I)*b*d*n))/(b*d*n), -(E^((2*I)*a*d)*(c*x^ n)^((2*I)*b*d))])/(1 + m + (2*I)*b*d*n)))/(1 + m)
Time = 0.33 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.36, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {5008, 5006, 959, 888}
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.
\(\displaystyle \int (e x)^m \tan \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\) |
\(\Big \downarrow \) 5008 |
\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (c x^n\right )^{\frac {m+1}{n}-1} \tan \left (d \left (a+b \log \left (c x^n\right )\right )\right )d\left (c x^n\right )}{e n}\) |
\(\Big \downarrow \) 5006 |
\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \frac {\left (c x^n\right )^{\frac {m+1}{n}-1} \left (i-i e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{e^{2 i a d} \left (c x^n\right )^{2 i b d}+1}d\left (c x^n\right )}{e n}\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (2 i \int \frac {\left (c x^n\right )^{\frac {m+1}{n}-1}}{e^{2 i a d} \left (c x^n\right )^{2 i b d}+1}d\left (c x^n\right )-\frac {i n \left (c x^n\right )^{\frac {m+1}{n}}}{m+1}\right )}{e n}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (\frac {2 i n \left (c x^n\right )^{\frac {m+1}{n}} \operatorname {Hypergeometric2F1}\left (1,-\frac {i (m+1)}{2 b d n},1-\frac {i (m+1)}{2 b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{m+1}-\frac {i n \left (c x^n\right )^{\frac {m+1}{n}}}{m+1}\right )}{e n}\) |
((e*x)^(1 + m)*(((-I)*n*(c*x^n)^((1 + m)/n))/(1 + m) + ((2*I)*n*(c*x^n)^(( 1 + m)/n)*Hypergeometric2F1[1, ((-1/2*I)*(1 + m))/(b*d*n), 1 - ((I/2)*(1 + m))/(b*d*n), -(E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))])/(1 + m)))/(e*n*(c*x^n )^((1 + m)/n))
3.2.75.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
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]
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])
\[\int \left (e x \right )^{m} \tan \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )d x\]
\[ \int (e x)^m \tan \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right ) \,d x } \]
\[ \int (e x)^m \tan \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \left (e x\right )^{m} \tan {\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \]
\[ \int (e x)^m \tan \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right ) \,d x } \]
Timed out. \[ \int (e x)^m \tan \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int (e x)^m \tan \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \mathrm {tan}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )\,{\left (e\,x\right )}^m \,d x \]