3.2.70 \(\int \frac {\tan ^2(d (a+b \log (c x^n)))}{x^2} \, dx\) [170]

3.2.70.1 Optimal result
3.2.70.2 Mathematica [A] (verified)
3.2.70.3 Rubi [A] (verified)
3.2.70.4 Maple [F]
3.2.70.5 Fricas [F]
3.2.70.6 Sympy [F]
3.2.70.7 Maxima [F]
3.2.70.8 Giac [F(-1)]
3.2.70.9 Mupad [F(-1)]

3.2.70.1 Optimal result

Integrand size = 19, antiderivative size = 157 \[ \int \frac {\tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\frac {1+\frac {i}{b d n}}{x}+\frac {i \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n x \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {2 i \operatorname {Hypergeometric2F1}\left (1,\frac {i}{2 b d n},1+\frac {i}{2 b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n x} \]

output
(1+I/b/d/n)/x+I*(1-exp(2*I*a*d)*(c*x^n)^(2*I*b*d))/b/d/n/x/(1+exp(2*I*a*d) 
*(c*x^n)^(2*I*b*d))-2*I*hypergeom([1, 1/2*I/b/d/n],[1+1/2*I/b/d/n],-exp(2* 
I*a*d)*(c*x^n)^(2*I*b*d))/b/d/n/x
 
3.2.70.2 Mathematica [A] (verified)

Time = 3.38 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.17 \[ \int \frac {\tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\frac {-e^{2 i d \left (a+b \log \left (c x^n\right )\right )} \operatorname {Hypergeometric2F1}\left (1,1+\frac {i}{2 b d n},2+\frac {i}{2 b d n},-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )+(i+2 b d n) \left (b d n-i \operatorname {Hypergeometric2F1}\left (1,\frac {i}{2 b d n},1+\frac {i}{2 b d n},-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )+\tan \left (d \left (a+b \log \left (c x^n\right )\right )\right )\right )}{b d n (i+2 b d n) x} \]

input
Integrate[Tan[d*(a + b*Log[c*x^n])]^2/x^2,x]
 
output
(-(E^((2*I)*d*(a + b*Log[c*x^n]))*Hypergeometric2F1[1, 1 + (I/2)/(b*d*n), 
2 + (I/2)/(b*d*n), -E^((2*I)*d*(a + b*Log[c*x^n]))]) + (I + 2*b*d*n)*(b*d* 
n - I*Hypergeometric2F1[1, (I/2)/(b*d*n), 1 + (I/2)/(b*d*n), -E^((2*I)*d*( 
a + b*Log[c*x^n]))] + Tan[d*(a + b*Log[c*x^n])]))/(b*d*n*(I + 2*b*d*n)*x)
 
3.2.70.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.35, 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.

\(\displaystyle \int \frac {\tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx\)

\(\Big \downarrow \) 5008

\(\displaystyle \frac {\left (c x^n\right )^{\frac {1}{n}} \int \left (c x^n\right )^{-1-\frac {1}{n}} \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )d\left (c x^n\right )}{n x}\)

\(\Big \downarrow \) 5006

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

\(\Big \downarrow \) 1004

\(\displaystyle \frac {\left (c x^n\right )^{\frac {1}{n}} \left (\frac {i e^{-2 i a d} \int \frac {2 \left (c x^n\right )^{-1-\frac {1}{n}} \left (\frac {e^{2 i a d} (i b d n+1)}{n}-\frac {e^{4 i a d} (1-i b d n) \left (c x^n\right )^{2 i b d}}{n}\right )}{e^{2 i a d} \left (c x^n\right )^{2 i b d}+1}d\left (c x^n\right )}{2 b d}+\frac {i \left (c x^n\right )^{-1/n} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}\right )}{n x}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (c x^n\right )^{\frac {1}{n}} \left (\frac {i e^{-2 i a d} \int \frac {\left (c x^n\right )^{-1-\frac {1}{n}} \left (\frac {e^{2 i a d} (i b d n+1)}{n}-\frac {e^{4 i a d} (1-i b d n) \left (c x^n\right )^{2 i b d}}{n}\right )}{e^{2 i a d} \left (c x^n\right )^{2 i b d}+1}d\left (c x^n\right )}{b d}+\frac {i \left (c x^n\right )^{-1/n} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}\right )}{n x}\)

\(\Big \downarrow \) 959

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

\(\Big \downarrow \) 888

\(\displaystyle \frac {\left (c x^n\right )^{\frac {1}{n}} \left (\frac {i e^{-2 i a d} \left (e^{2 i a d} (1-i b d n) \left (c x^n\right )^{-1/n}-2 e^{2 i a d} \left (c x^n\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (1,\frac {i}{2 b d n},1+\frac {i}{2 b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )\right )}{b d}+\frac {i \left (c x^n\right )^{-1/n} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}\right )}{n x}\)

input
Int[Tan[d*(a + b*Log[c*x^n])]^2/x^2,x]
 
output
((c*x^n)^n^(-1)*((I*(1 - E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d)))/(b*d*(c*x^n)^ 
n^(-1)*(1 + E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))) + (I*((E^((2*I)*a*d)*(1 - 
I*b*d*n))/(c*x^n)^n^(-1) - (2*E^((2*I)*a*d)*Hypergeometric2F1[1, (I/2)/(b* 
d*n), 1 + (I/2)/(b*d*n), -(E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d))])/(c*x^n)^n^ 
(-1)))/(b*d*E^((2*I)*a*d))))/(n*x)
 

3.2.70.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 888
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])
 

rule 959
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]
 

rule 1004
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]
 

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])
 
3.2.70.4 Maple [F]

\[\int \frac {{\tan \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{2}}{x^{2}}d x\]

input
int(tan(d*(a+b*ln(c*x^n)))^2/x^2,x)
 
output
int(tan(d*(a+b*ln(c*x^n)))^2/x^2,x)
 
3.2.70.5 Fricas [F]

\[ \int \frac {\tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\int { \frac {\tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2}}{x^{2}} \,d x } \]

input
integrate(tan(d*(a+b*log(c*x^n)))^2/x^2,x, algorithm="fricas")
 
output
integral(tan(b*d*log(c*x^n) + a*d)^2/x^2, x)
 
3.2.70.6 Sympy [F]

\[ \int \frac {\tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\int \frac {\tan ^{2}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}}{x^{2}}\, dx \]

input
integrate(tan(d*(a+b*ln(c*x**n)))**2/x**2,x)
 
output
Integral(tan(a*d + b*d*log(c*x**n))**2/x**2, x)
 
3.2.70.7 Maxima [F]

\[ \int \frac {\tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\int { \frac {\tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2}}{x^{2}} \,d x } \]

input
integrate(tan(d*(a+b*log(c*x^n)))^2/x^2,x, algorithm="maxima")
 
output
((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 + 2*(b*d*n*cos(2*b*d*log(c)) + sin(2*b*d*lo 
g(c)))*cos(2*b*d*log(x^n) + 2*a*d) + 2*(2*b^2*d^2*n^2*x*cos(2*b*d*log(c))* 
cos(2*b*d*log(x^n) + 2*a*d) - 2*b^2*d^2*n^2*x*sin(2*b*d*log(c))*sin(2*b*d* 
log(x^n) + 2*a*d) + b^2*d^2*n^2*x + (b^2*d^2*cos(2*b*d*log(c))^2 + b^2*d^2 
*sin(2*b*d*log(c))^2)*n^2*x*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*x*sin(2*b*d*log(x^n) + 2 
*a*d)^2)*integrate((cos(2*b*d*log(x^n) + 2*a*d)*sin(2*b*d*log(c)) + cos(2* 
b*d*log(c))*sin(2*b*d*log(x^n) + 2*a*d))/(2*b^2*d^2*n^2*x^2*cos(2*b*d*log( 
c))*cos(2*b*d*log(x^n) + 2*a*d) - 2*b^2*d^2*n^2*x^2*sin(2*b*d*log(c))*sin( 
2*b*d*log(x^n) + 2*a*d) + b^2*d^2*n^2*x^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*x^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*x^2*sin(2*b*d* 
log(x^n) + 2*a*d)^2), x) - 2*(b*d*n*sin(2*b*d*log(c)) - cos(2*b*d*log(c))) 
*sin(2*b*d*log(x^n) + 2*a*d))/(2*b*d*n*x*cos(2*b*d*log(c))*cos(2*b*d*log(x 
^n) + 2*a*d) - 2*b*d*n*x*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*x*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*sin(2 
*b*d*log(x^n) + 2*a*d)^2 + b*d*n*x)
 
3.2.70.8 Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\text {Timed out} \]

input
integrate(tan(d*(a+b*log(c*x^n)))^2/x^2,x, algorithm="giac")
 
output
Timed out
 
3.2.70.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{x^2} \, dx=\int \frac {{\mathrm {tan}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^2}{x^2} \,d x \]

input
int(tan(d*(a + b*log(c*x^n)))^2/x^2,x)
 
output
int(tan(d*(a + b*log(c*x^n)))^2/x^2, x)