3.2.65 \(\int x^3 \tan ^2(d (a+b \log (c x^n))) \, dx\) [165]
Optimal. Leaf size=159 \[ \frac {(4 i-b d n) x^4}{4 b d n}+\frac {i x^4 \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}-\frac {2 i x^4 \, _2F_1\left (1,-\frac {2 i}{b d n};1-\frac {2 i}{b d n};-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{b d n} \]
[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
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Rubi [A]
time = 0.12, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {4593, 4591,
516, 470, 371} \begin {gather*} -\frac {2 i x^4 \, _2F_1\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} \end {gather*}
Antiderivative was successfully verified.
[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*} \int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx &=\int x^3 \tan ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\\ \end {align*}
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Mathematica [A]
time = 7.08, size = 179, normalized size = 1.13 \begin {gather*} -\frac {x^4 \left (-8 e^{2 i d \left (a+b \log \left (c x^n\right )\right )} \, _2F_1\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 \, _2F_1\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)} \end {gather*}
Antiderivative was successfully verified.
[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))
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int x^{3} \left (\tan ^{2}\left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \tan ^{2}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**3*tan(d*(a+b*ln(c*x**n)))**2,x)
[Out]
Integral(x**3*tan(a*d + b*d*log(c*x**n))**2, x)
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*tan(d*(a+b*log(c*x^n)))^2,x, algorithm="giac")
[Out]
Timed out
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,{\mathrm {tan}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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