Integrand size = 10, antiderivative size = 77 \[ \int x^2 \tan (a+b x) \, dx=\frac {i x^3}{3}-\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac {\operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3} \] Output:
1/3*I*x^3-x^2*ln(1+exp(2*I*(b*x+a)))/b+I*x*polylog(2,-exp(2*I*(b*x+a)))/b^ 2-1/2*polylog(3,-exp(2*I*(b*x+a)))/b^3
Time = 0.01 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int x^2 \tan (a+b x) \, dx=\frac {i x^3}{3}-\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac {\operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3} \] Input:
Integrate[x^2*Tan[a + b*x],x]
Output:
(I/3)*x^3 - (x^2*Log[1 + E^((2*I)*(a + b*x))])/b + (I*x*PolyLog[2, -E^((2* I)*(a + b*x))])/b^2 - PolyLog[3, -E^((2*I)*(a + b*x))]/(2*b^3)
Time = 0.47 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.25, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4202, 2620, 3011, 2720, 7143}
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.
\(\displaystyle \int x^2 \tan (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int x^2 \tan (a+b x)dx\) |
\(\Big \downarrow \) 4202 |
\(\displaystyle \frac {i x^3}{3}-2 i \int \frac {e^{2 i (a+b x)} x^2}{1+e^{2 i (a+b x)}}dx\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {i x^3}{3}-2 i \left (\frac {i \int x \log \left (1+e^{2 i (a+b x)}\right )dx}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {i x^3}{3}-2 i \left (\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \int \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{2 b}\right )}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {i x^3}{3}-2 i \left (\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {\int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {i x^3}{3}-2 i \left (\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {\operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{4 b^2}\right )}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\) |
Input:
Int[x^2*Tan[a + b*x],x]
Output:
(I/3)*x^3 - (2*I)*(((-1/2*I)*x^2*Log[1 + E^((2*I)*(a + b*x))])/b + (I*(((I /2)*x*PolyLog[2, -E^((2*I)*(a + b*x))])/b - PolyLog[3, -E^((2*I)*(a + b*x) )]/(4*b^2)))/b)
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I *((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I Int[(c + d*x)^m*(E^(2*I*( e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt Q[m, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.34
method | result | size |
risch | \(\frac {i x^{3}}{3}-\frac {2 i a^{2} x}{b^{2}}-\frac {4 i a^{3}}{3 b^{3}}-\frac {x^{2} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b}+\frac {i x \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{2}}-\frac {\operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}\) | \(103\) |
Input:
int(x^2*tan(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/3*I*x^3-2*I/b^2*a^2*x-4/3*I/b^3*a^3-x^2*ln(1+exp(2*I*(b*x+a)))/b+I*x*pol ylog(2,-exp(2*I*(b*x+a)))/b^2-1/2*polylog(3,-exp(2*I*(b*x+a)))/b^3+2/b^3*a ^2*ln(exp(I*(b*x+a)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (62) = 124\).
Time = 0.09 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.60 \[ \int x^2 \tan (a+b x) \, dx=-\frac {2 \, b^{2} x^{2} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, b^{2} x^{2} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 i \, b x {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 2 i \, b x {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right )}{4 \, b^{3}} \] Input:
integrate(x^2*tan(b*x+a),x, algorithm="fricas")
Output:
-1/4*(2*b^2*x^2*log(-2*(I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) + 2*b^2* x^2*log(-2*(-I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) + 2*I*b*x*dilog(2*( I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1) + 1) - 2*I*b*x*dilog(2*(-I*tan(b* x + a) - 1)/(tan(b*x + a)^2 + 1) + 1) + polylog(3, (tan(b*x + a)^2 + 2*I*t an(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) + polylog(3, (tan(b*x + a)^2 - 2*I* tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)))/b^3
\[ \int x^2 \tan (a+b x) \, dx=\int x^{2} \tan {\left (a + b x \right )}\, dx \] Input:
integrate(x**2*tan(b*x+a),x)
Output:
Integral(x**2*tan(a + b*x), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (62) = 124\).
Time = 0.16 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.12 \[ \int x^2 \tan (a+b x) \, dx=-\frac {-2 i \, {\left (b x + a\right )}^{3} + 6 i \, {\left (b x + a\right )}^{2} a - 6 i \, b x {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 6 \, a^{2} \log \left (\sec \left (b x + a\right )\right ) - 6 \, {\left (-i \, {\left (b x + a\right )}^{2} + 2 i \, {\left (b x + a\right )} a\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )})}{6 \, b^{3}} \] Input:
integrate(x^2*tan(b*x+a),x, algorithm="maxima")
Output:
-1/6*(-2*I*(b*x + a)^3 + 6*I*(b*x + a)^2*a - 6*I*b*x*dilog(-e^(2*I*b*x + 2 *I*a)) - 6*a^2*log(sec(b*x + a)) - 6*(-I*(b*x + a)^2 + 2*I*(b*x + a)*a)*ar ctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + 3*((b*x + a)^2 - 2*(b*x + a)*a)*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1 ) + 3*polylog(3, -e^(2*I*b*x + 2*I*a)))/b^3
\[ \int x^2 \tan (a+b x) \, dx=\int { x^{2} \tan \left (b x + a\right ) \,d x } \] Input:
integrate(x^2*tan(b*x+a),x, algorithm="giac")
Output:
integrate(x^2*tan(b*x + a), x)
Timed out. \[ \int x^2 \tan (a+b x) \, dx=\int x^2\,\mathrm {tan}\left (a+b\,x\right ) \,d x \] Input:
int(x^2*tan(a + b*x),x)
Output:
int(x^2*tan(a + b*x), x)
\[ \int x^2 \tan (a+b x) \, dx=\int \tan \left (b x +a \right ) x^{2}d x \] Input:
int(x^2*tan(b*x+a),x)
Output:
int(tan(a + b*x)*x**2,x)