\(\int x^3 \tan ^2(d (a+b \log (c x^n))) \, dx\) [165]

Optimal result

Integrand size = 19, antiderivative size = 159 \[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(4 i-b d n) x^4}{4 b d n}+\frac {i x^4 \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {2 i x^4 \operatorname {Hypergeometric2F1}\left (1,-\frac {2 i}{b d n},1-\frac {2 i}{b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {4593, 4591, 516, 470, 371} \[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-\frac {2 i x^4 \operatorname {Hypergeometric2F1}\left (1,-\frac {2 i}{b d n},1-\frac {2 i}{b d n},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n}+\frac {i x^4 \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}+\frac {x^4 (-b d n+4 i)}{4 b d n} \]

[In]

Int[x^3*Tan[d*(a + b*Log[c*x^n])]^2,x]

[Out]

((4*I - b*d*n)*x^4)/(4*b*d*n) + (I*x^4*(1 - E^((2*I)*a*d)*(c*x^n)^((2*I)*b*d)))/(b*d*n*(1 + E^((2*I)*a*d)*(c*x
^n)^((2*I)*b*d))) - ((2*I)*x^4*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*n)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 470

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] - Dist[(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 516

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] + Dist[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] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 4591

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 4593

Int[((e_.)*(x_))^(m_.)*Tan[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(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])

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^4 \left (c x^n\right )^{-4/n}\right ) \text {Subst}\left (\int x^{-1+\frac {4}{n}} \tan ^2(d (a+b \log (x))) \, dx,x,c x^n\right )}{n} \\ & = \frac {\left (x^4 \left (c x^n\right )^{-4/n}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {4}{n}} \left (i-i e^{2 i a d} x^{2 i b d}\right )^2}{\left (1+e^{2 i a d} x^{2 i b d}\right )^2} \, dx,x,c x^n\right )}{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 {\left (i e^{-2 i a d} x^4 \left (c x^n\right )^{-4/n}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {4}{n}} \left (-\frac {2 e^{2 i a d} (4-i b d n)}{n}+\frac {2 e^{4 i a d} (4+i b d n) x^{2 i b d}}{n}\right )}{1+e^{2 i a d} x^{2 i b d}} \, dx,x,c x^n\right )}{2 b d n} \\ & = -\frac {1}{4} \left (1-\frac {4 i}{b d n}\right ) x^4+\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 {\left (8 i x^4 \left (c x^n\right )^{-4/n}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {4}{n}}}{1+e^{2 i a d} x^{2 i b d}} \, dx,x,c x^n\right )}{b d n^2} \\ & = -\frac {1}{4} \left (1-\frac {4 i}{b d n}\right ) x^4+\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.60 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.13 \[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-\frac {x^4 \left (-8 e^{2 i d \left (a+b \log \left (c x^n\right )\right )} \operatorname {Hypergeometric2F1}\left (1,1-\frac {2 i}{b d n},2-\frac {2 i}{b d n},-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )+(-2 i+b d n) \left (b d n+4 i \operatorname {Hypergeometric2F1}\left (1,-\frac {2 i}{b d n},1-\frac {2 i}{b d n},-e^{2 i d \left (a+b \log \left (c x^n\right )\right )}\right )-4 \tan \left (d \left (a+b \log \left (c x^n\right )\right )\right )\right )\right )}{4 b d n (-2 i+b d n)} \]

[In]

Integrate[x^3*Tan[d*(a + b*Log[c*x^n])]^2,x]

[Out]

-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))

Maple [F]

\[\int x^{3} {\tan \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{2}d x\]

[In]

int(x^3*tan(d*(a+b*ln(c*x^n)))^2,x)

[Out]

int(x^3*tan(d*(a+b*ln(c*x^n)))^2,x)

Fricas [F]

\[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x^{3} \tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]

[In]

integrate(x^3*tan(d*(a+b*log(c*x^n)))^2,x, algorithm="fricas")

[Out]

integral(x^3*tan(b*d*log(c*x^n) + a*d)^2, x)

Sympy [F(-1)]

Timed out. \[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \]

[In]

integrate(x**3*tan(d*(a+b*ln(c*x**n)))**2,x)

[Out]

Timed out

Maxima [F]

\[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x^{3} \tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]

[In]

integrate(x^3*tan(d*(a+b*log(c*x^n)))^2,x, algorithm="maxima")

[Out]

-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*cos(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*cos(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)

Giac [F(-1)]

Timed out. \[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \]

[In]

integrate(x^3*tan(d*(a+b*log(c*x^n)))^2,x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int x^3\,{\mathrm {tan}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^2 \,d x \]

[In]

int(x^3*tan(d*(a + b*log(c*x^n)))^2,x)

[Out]

int(x^3*tan(d*(a + b*log(c*x^n)))^2, x)