Integrand size = 21, antiderivative size = 196 \[ \int (e x)^m \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(i (1+m)-b d n) (e x)^{1+m}}{b d e (1+m) n}+\frac {i (e x)^{1+m} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d e n \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {2 i (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,-\frac {i (1+m)}{2 b d n},1-\frac {i (1+m)}{2 b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d e n} \]
(I*(1+m)-b*d*n)*(e*x)^(1+m)/b/d/e/(1+m)/n+I*(e*x)^(1+m)*(1-exp(2*I*a*d)*(c *x^n)^(2*I*b*d))/b/d/e/n/(1+exp(2*I*a*d)*(c*x^n)^(2*I*b*d))-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))/b/d/e/n
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(550\) vs. \(2(196)=392\).
Time = 16.13 (sec) , antiderivative size = 550, normalized size of antiderivative = 2.81 \[ \int (e x)^m \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-\frac {x (e x)^m}{1+m}+\frac {x (e x)^m \sec \left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \sec \left (b d n \log (x)+d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \sin (b d n \log (x))}{b d n}-\frac {(1+m) x^{-m} (e x)^m \sec \left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \left (\frac {x^{1+m} \sec \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \sin (b d n \log (x))}{1+m}-\frac {i e^{-\frac {(1+2 m) \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{b n}} \cos \left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \left (-e^{\frac {a+2 a m+b (1+m) n \log (x)+b (1+2 m) \left (-n \log (x)+\log \left (c x^n\right )\right )}{b n}} (1+m+2 i b d n) \operatorname {Hypergeometric2F1}\left (1,-\frac {i (1+m)}{2 b d n},1-\frac {i (1+m)}{2 b d n},-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )+e^{\frac {a (1+2 m+2 i b d n)}{b n}+(1+m+2 i b d n) \log (x)+\frac {(1+2 m+2 i b d n) \left (-n \log (x)+\log \left (c x^n\right )\right )}{n}} (1+m) \operatorname {Hypergeometric2F1}\left (1,-\frac {i (1+m+2 i b d n)}{2 b d n},-\frac {i (1+m+4 i b d n)}{2 b d n},-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )-i e^{\frac {a+2 a m+b (1+m) n \log (x)+b (1+2 m) \left (-n \log (x)+\log \left (c x^n\right )\right )}{b n}} (1+m+2 i b d n) \tan \left (d \left (a+b \log \left (c x^n\right )\right )\right )\right )}{(1+m) (1+m+2 i b d n)}\right )}{b d n} \]
-((x*(e*x)^m)/(1 + m)) + (x*(e*x)^m*Sec[d*(a + b*(-(n*Log[x]) + Log[c*x^n] ))]*Sec[b*d*n*Log[x] + d*(a + b*(-(n*Log[x]) + Log[c*x^n]))]*Sin[b*d*n*Log [x]])/(b*d*n) - ((1 + m)*(e*x)^m*Sec[d*(a + b*(-(n*Log[x]) + Log[c*x^n]))] *((x^(1 + m)*Sec[d*(a + b*Log[c*x^n])]*Sin[b*d*n*Log[x]])/(1 + m) - (I*Cos [d*(a + b*(-(n*Log[x]) + Log[c*x^n]))]*(-(E^((a + 2*a*m + b*(1 + m)*n*Log[ x] + b*(1 + 2*m)*(-(n*Log[x]) + Log[c*x^n]))/(b*n))*(1 + m + (2*I)*b*d*n)* 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^((a*(1 + 2*m + (2*I)*b*d*n))/(b* n) + (1 + m + (2*I)*b*d*n)*Log[x] + ((1 + 2*m + (2*I)*b*d*n)*(-(n*Log[x]) + Log[c*x^n]))/n)*(1 + m)*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)*d*(a + b*Log[c*x^n]))] - I*E^((a + 2*a*m + b*(1 + m)*n*Log[x] + b*(1 + 2*m)*(-(n* Log[x]) + Log[c*x^n]))/(b*n))*(1 + m + (2*I)*b*d*n)*Tan[d*(a + b*Log[c*x^n ])]))/(E^(((1 + 2*m)*(a + b*(-(n*Log[x]) + Log[c*x^n])))/(b*n))*(1 + m)*(1 + m + (2*I)*b*d*n))))/(b*d*n*x^m)
Time = 0.50 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.24, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5008, 5006, 1004, 27, 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.
| \(\displaystyle \int (e x)^m \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\) |
| \(\Big \downarrow \) 5008 |
| \(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (c x^n\right )^{\frac {m+1}{n}-1} \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )d\left (c x^n\right )}{e n}\) |
| \(\Big \downarrow \) 5006 |
| \(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \frac {\left (c x^n\right )^{\frac {m+1}{n}-1} \left (i-i e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^2}{\left (e^{2 i a d} \left (c x^n\right )^{2 i b d}+1\right )^2}d\left (c x^n\right )}{e n}\) |
| \(\Big \downarrow \) 1004 |
| \(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (\frac {i e^{-2 i a d} \int -\frac {2 \left (c x^n\right )^{\frac {m+1}{n}-1} \left (\frac {e^{2 i a d} (m-i b d n+1)}{n}-\frac {e^{4 i a d} (m+i b d n+1) \left (c x^n\right )^{2 i b d}}{n}\right )}{e^{2 i a d} \left (c x^n\right )^{2 i b d}+1}d\left (c x^n\right )}{2 b d}+\frac {i \left (c x^n\right )^{\frac {m+1}{n}} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}\right )}{e n}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (\frac {i \left (c x^n\right )^{\frac {m+1}{n}} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {i e^{-2 i a d} \int \frac {\left (c x^n\right )^{\frac {m+1}{n}-1} \left (\frac {e^{2 i a d} (m-i b d n+1)}{n}-\frac {e^{4 i a d} (m+i b d n+1) \left (c x^n\right )^{2 i b d}}{n}\right )}{e^{2 i a d} \left (c x^n\right )^{2 i b d}+1}d\left (c x^n\right )}{b d}\right )}{e n}\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (\frac {i \left (c x^n\right )^{\frac {m+1}{n}} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {i e^{-2 i a d} \left (\frac {2 (m+1) e^{2 i a d} \int \frac {\left (c x^n\right )^{\frac {m+1}{n}-1}}{e^{2 i a d} \left (c x^n\right )^{2 i b d}+1}d\left (c x^n\right )}{n}-\frac {e^{2 i a d} (i b d n+m+1) \left (c x^n\right )^{\frac {m+1}{n}}}{m+1}\right )}{b d}\right )}{e n}\) |
| \(\Big \downarrow \) 888 |
| \(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (\frac {i \left (c x^n\right )^{\frac {m+1}{n}} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {i e^{-2 i a d} \left (2 e^{2 i a d} \left (c x^n\right )^{\frac {m+1}{n}} \operatorname {Hypergeometric2F1}\left (1,-\frac {i (m+1)}{2 b d n},1-\frac {i (m+1)}{2 b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )-\frac {e^{2 i a d} (i b d n+m+1) \left (c x^n\right )^{\frac {m+1}{n}}}{m+1}\right )}{b d}\right )}{e n}\) |
((e*x)^(1 + m)*((I*(c*x^n)^((1 + m)/n)*(1 - E^((2*I)*a*d)*(c*x^n)^((2*I)*b *d)))/(b*d*(1 + E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))) - (I*(-((E^((2*I)*a*d) *(1 + m + I*b*d*n)*(c*x^n)^((1 + m)/n))/(1 + m)) + 2*E^((2*I)*a*d)*(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))]))/(b*d*E^((2*I)*a*d ))))/(e*n*(c*x^n)^((1 + m)/n))
3.2.76.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[(-(c*b - a*d))*(e*x)^(m + 1)*(a + b*x^n)^(p + 1) *((c + d*x^n)^(q - 1)/(a*b*e*n*(p + 1))), x] + Simp[1/(a*b*n*(p + 1)) Int [(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c *b - a*d)*(m + 1)) + d*(c*b*n*(p + 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^ n, x], x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && Lt Q[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
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])
\[\int \left (e x \right )^{m} {\tan \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{2}d x\]
\[ \int (e x)^m \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]
\[ \int (e x)^m \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \left (e x\right )^{m} \tan ^{2}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \]
\[ \int (e x)^m \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]
-((b*d*cos(2*b*d*log(c))^2 + b*d*sin(2*b*d*log(c))^2)*e^m*n*x*x^m*cos(2*b* d*log(x^n) + 2*a*d)^2 + (b*d*cos(2*b*d*log(c))^2 + b*d*sin(2*b*d*log(c))^2 )*e^m*n*x*x^m*sin(2*b*d*log(x^n) + 2*a*d)^2 + b*d*e^m*n*x*x^m + 2*(b*d*e^m *n*cos(2*b*d*log(c)) - e^m*m*sin(2*b*d*log(c)) - e^m*sin(2*b*d*log(c)))*x* x^m*cos(2*b*d*log(x^n) + 2*a*d) - 2*(b*d*e^m*n*sin(2*b*d*log(c)) + e^m*m*c os(2*b*d*log(c)) + e^m*cos(2*b*d*log(c)))*x*x^m*sin(2*b*d*log(x^n) + 2*a*d ) + 2*(((b^2*d^2*cos(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c))^2)*e^m*m^ 2 + 2*(b^2*d^2*cos(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c))^2)*e^m*m + (b^2*d^2*cos(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c))^2)*e^m)*n^2*cos(2 *b*d*log(x^n) + 2*a*d)^2 + ((b^2*d^2*cos(2*b*d*log(c))^2 + b^2*d^2*sin(2*b *d*log(c))^2)*e^m*m^2 + 2*(b^2*d^2*cos(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d *log(c))^2)*e^m*m + (b^2*d^2*cos(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c ))^2)*e^m)*n^2*sin(2*b*d*log(x^n) + 2*a*d)^2 + 2*(b^2*d^2*e^m*m^2*cos(2*b* d*log(c)) + 2*b^2*d^2*e^m*m*cos(2*b*d*log(c)) + b^2*d^2*e^m*cos(2*b*d*log( c)))*n^2*cos(2*b*d*log(x^n) + 2*a*d) - 2*(b^2*d^2*e^m*m^2*sin(2*b*d*log(c) ) + 2*b^2*d^2*e^m*m*sin(2*b*d*log(c)) + b^2*d^2*e^m*sin(2*b*d*log(c)))*n^2 *sin(2*b*d*log(x^n) + 2*a*d) + (b^2*d^2*e^m*m^2 + 2*b^2*d^2*e^m*m + b^2*d^ 2*e^m)*n^2)*integrate((x^m*cos(2*b*d*log(x^n) + 2*a*d)*sin(2*b*d*log(c)) + x^m*cos(2*b*d*log(c))*sin(2*b*d*log(x^n) + 2*a*d))/(2*b^2*d^2*n^2*cos(2*b *d*log(c))*cos(2*b*d*log(x^n) + 2*a*d) - 2*b^2*d^2*n^2*sin(2*b*d*log(c)...
Timed out. \[ \int (e x)^m \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int (e x)^m \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int {\mathrm {tan}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^2\,{\left (e\,x\right )}^m \,d x \]