Integrand size = 19, antiderivative size = 159 \[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(4 i-b d n) x^4}{4 b d n}+\frac {i x^4 \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {2 i x^4 \operatorname {Hypergeometric2F1}\left (1,-\frac {2 i}{b d n},1-\frac {2 i}{b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n} \]
1/4*(4*I-b*d*n)*x^4/b/d/n+I*x^4*(1-exp(2*I*a*d)*(c*x^n)^(2*I*b*d))/b/d/n/( 1+exp(2*I*a*d)*(c*x^n)^(2*I*b*d))-2*I*x^4*hypergeom([1, -2*I/b/d/n],[1-2*I /b/d/n],-exp(2*I*a*d)*(c*x^n)^(2*I*b*d))/b/d/n
Time = 5.60 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.13 \[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-\frac {x^4 \left (-8 e^{2 i d \left (a+b \log \left (c x^n\right )\right )} \operatorname {Hypergeometric2F1}\left (1,1-\frac {2 i}{b d n},2-\frac {2 i}{b d n},-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )+(-2 i+b d n) \left (b d n+4 i \operatorname {Hypergeometric2F1}\left (1,-\frac {2 i}{b d n},1-\frac {2 i}{b d n},-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )-4 \tan \left (d \left (a+b \log \left (c x^n\right )\right )\right )\right )\right )}{4 b d n (-2 i+b d n)} \]
-1/4*(x^4*(-8*E^((2*I)*d*(a + b*Log[c*x^n]))*Hypergeometric2F1[1, 1 - (2*I )/(b*d*n), 2 - (2*I)/(b*d*n), -E^((2*I)*d*(a + b*Log[c*x^n]))] + (-2*I + b *d*n)*(b*d*n + (4*I)*Hypergeometric2F1[1, (-2*I)/(b*d*n), 1 - (2*I)/(b*d*n ), -E^((2*I)*d*(a + b*Log[c*x^n]))] - 4*Tan[d*(a + b*Log[c*x^n])])))/(b*d* n*(-2*I + b*d*n))
Time = 0.46 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.34, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, 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.
| \(\displaystyle \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\) |
| \(\Big \downarrow \) 5008 |
| \(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \int \left (c x^n\right )^{\frac {4}{n}-1} \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )d\left (c x^n\right )}{n}\) |
| \(\Big \downarrow \) 5006 |
| \(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \int \frac {\left (c x^n\right )^{\frac {4}{n}-1} \left (i-i e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^2}{\left (e^{2 i a d} \left (c x^n\right )^{2 i b d}+1\right )^2}d\left (c x^n\right )}{n}\) |
| \(\Big \downarrow \) 1004 |
| \(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \left (\frac {i e^{-2 i a d} \int -\frac {2 \left (c x^n\right )^{\frac {4}{n}-1} \left (\frac {e^{2 i a d} (4-i b d n)}{n}-\frac {e^{4 i a d} (i b d n+4) \left (c x^n\right )^{2 i b d}}{n}\right )}{e^{2 i a d} \left (c x^n\right )^{2 i b d}+1}d\left (c x^n\right )}{2 b d}+\frac {i \left (c x^n\right )^{4/n} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}\right )}{n}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \left (\frac {i \left (c x^n\right )^{4/n} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {i e^{-2 i a d} \int \frac {\left (c x^n\right )^{\frac {4}{n}-1} \left (\frac {e^{2 i a d} (4-i b d n)}{n}-\frac {e^{4 i a d} (i b d n+4) \left (c x^n\right )^{2 i b d}}{n}\right )}{e^{2 i a d} \left (c x^n\right )^{2 i b d}+1}d\left (c x^n\right )}{b d}\right )}{n}\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \left (\frac {i \left (c x^n\right )^{4/n} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {i e^{-2 i a d} \left (\frac {8 e^{2 i a d} \int \frac {\left (c x^n\right )^{\frac {4}{n}-1}}{e^{2 i a d} \left (c x^n\right )^{2 i b d}+1}d\left (c x^n\right )}{n}-\frac {1}{4} e^{2 i a d} (4+i b d n) \left (c x^n\right )^{4/n}\right )}{b d}\right )}{n}\) |
| \(\Big \downarrow \) 888 |
| \(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \left (\frac {i \left (c x^n\right )^{4/n} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {i e^{-2 i a d} \left (2 e^{2 i a d} \left (c x^n\right )^{4/n} \operatorname {Hypergeometric2F1}\left (1,-\frac {2 i}{b d n},1-\frac {2 i}{b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )-\frac {1}{4} e^{2 i a d} (4+i b d n) \left (c x^n\right )^{4/n}\right )}{b d}\right )}{n}\) |
(x^4*((I*(c*x^n)^(4/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*(-1/4*(E^((2*I)*a*d)*(4 + I*b*d*n )*(c*x^n)^(4/n)) + 2*E^((2*I)*a*d)*(c*x^n)^(4/n)*Hypergeometric2F1[1, (-2* I)/(b*d*n), 1 - (2*I)/(b*d*n), -(E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))]))/(b* d*E^((2*I)*a*d))))/(n*(c*x^n)^(4/n))
3.2.65.3.1 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])
\[\int x^{3} {\tan \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{2}d x\]
\[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x^{3} \tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]
Timed out. \[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \]
\[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x^{3} \tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]
-1/4*((b*d*cos(2*b*d*log(c))^2 + b*d*sin(2*b*d*log(c))^2)*n*x^4*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)* n*x^4*sin(2*b*d*log(x^n) + 2*a*d)^2 + b*d*n*x^4 + 2*(b*d*n*cos(2*b*d*log(c )) - 4*sin(2*b*d*log(c)))*x^4*cos(2*b*d*log(x^n) + 2*a*d) - 2*(b*d*n*sin(2 *b*d*log(c)) + 4*cos(2*b*d*log(c)))*x^4*sin(2*b*d*log(x^n) + 2*a*d) + 32*( 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))*sin(2*b*d*log(x^n) + 2*a*d) + b^2*d^2*n^2 + (b^2*d^2*c os(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c))^2)*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)*n^ 2*sin(2*b*d*log(x^n) + 2*a*d)^2)*integrate((x^3*cos(2*b*d*log(x^n) + 2*a*d )*sin(2*b*d*log(c)) + x^3*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))*sin(2*b*d*log(x^n) + 2*a*d) + b^2*d^2*n^2 + (b^2*d^2*c os(2*b*d*log(c))^2 + b^2*d^2*sin(2*b*d*log(c))^2)*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)*n^ 2*sin(2*b*d*log(x^n) + 2*a*d)^2), x))/(2*b*d*n*cos(2*b*d*log(c))*cos(2*b*d *log(x^n) + 2*a*d) - 2*b*d*n*sin(2*b*d*log(c))*sin(2*b*d*log(x^n) + 2*a*d) + (b*d*cos(2*b*d*log(c))^2 + b*d*sin(2*b*d*log(c))^2)*n*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)*n*sin(2 *b*d*log(x^n) + 2*a*d)^2 + b*d*n)
Timed out. \[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int x^3\,{\mathrm {tan}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^2 \,d x \]