Integrand size = 10, antiderivative size = 106 \[ \int x^3 \tan (a+b x) \, dx=\frac {i x^4}{4}-\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 i \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4} \]
1/4*I*x^4-x^3*ln(1+exp(2*I*(b*x+a)))/b+3/2*I*x^2*polylog(2,-exp(2*I*(b*x+a )))/b^2-3/2*x*polylog(3,-exp(2*I*(b*x+a)))/b^3-3/4*I*polylog(4,-exp(2*I*(b *x+a)))/b^4
Time = 0.01 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00 \[ \int x^3 \tan (a+b x) \, dx=\frac {i x^4}{4}-\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 i \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4} \]
(I/4)*x^4 - (x^3*Log[1 + E^((2*I)*(a + b*x))])/b + (((3*I)/2)*x^2*PolyLog[ 2, -E^((2*I)*(a + b*x))])/b^2 - (3*x*PolyLog[3, -E^((2*I)*(a + b*x))])/(2* b^3) - (((3*I)/4)*PolyLog[4, -E^((2*I)*(a + b*x))])/b^4
Time = 0.61 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.25, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {3042, 4202, 2620, 3011, 7163, 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.
\(\displaystyle \int x^3 \tan (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int x^3 \tan (a+b x)dx\) |
\(\Big \downarrow \) 4202 |
\(\displaystyle \frac {i x^4}{4}-2 i \int \frac {e^{2 i (a+b x)} x^3}{1+e^{2 i (a+b x)}}dx\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {i x^4}{4}-2 i \left (\frac {3 i \int x^2 \log \left (1+e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {i x^4}{4}-2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {i x^4}{4}-2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \left (\frac {i \int \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {i x^4}{4}-2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \left (\frac {\int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {i x^4}{4}-2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \left (\frac {\operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\) |
(I/4)*x^4 - (2*I)*(((-1/2*I)*x^3*Log[1 + E^((2*I)*(a + b*x))])/b + (((3*I) /2)*(((I/2)*x^2*PolyLog[2, -E^((2*I)*(a + b*x))])/b - (I*(((-1/2*I)*x*Poly Log[3, -E^((2*I)*(a + b*x))])/b + PolyLog[4, -E^((2*I)*(a + b*x))]/(4*b^2) ))/b))/b)
3.1.1.3.1 Defintions of rubi rules used
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]
Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. )*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F])) Int[(e + f*x) ^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c , d, e, f, n, p}, x] && GtQ[m, 0]
Time = 0.49 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.18
method | result | size |
risch | \(\frac {i x^{4}}{4}-\frac {3 x \,\operatorname {Li}_{3}\left (-{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}+\frac {2 i a^{3} x}{b^{3}}+\frac {3 i x^{2} \operatorname {Li}_{2}\left (-{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{2}}+\frac {3 i a^{4}}{2 b^{4}}-\frac {3 i \operatorname {Li}_{4}\left (-{\mathrm e}^{2 i \left (b x +a \right )}\right )}{4 b^{4}}-\frac {2 a^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {x^{3} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}\) | \(125\) |
1/4*I*x^4-3/2*x*polylog(3,-exp(2*I*(b*x+a)))/b^3+2*I/b^3*a^3*x+3/2*I*x^2*p olylog(2,-exp(2*I*(b*x+a)))/b^2+3/2*I/b^4*a^4-3/4*I*polylog(4,-exp(2*I*(b* x+a)))/b^4-2/b^4*a^3*ln(exp(I*(b*x+a)))-x^3*ln(exp(2*I*(b*x+a))+1)/b
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (83) = 166\).
Time = 0.28 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.70 \[ \int x^3 \tan (a+b x) \, dx=-\frac {4 \, b^{3} x^{3} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 4 \, b^{3} x^{3} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 6 \, b x {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 6 \, b x {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 3 i \, {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 3 i \, {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right )}{8 \, b^{4}} \]
-1/8*(4*b^3*x^3*log(-2*(I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) + 4*b^3* x^3*log(-2*(-I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) + 6*I*b^2*x^2*dilog (2*(I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1) + 1) - 6*I*b^2*x^2*dilog(2*(- I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1) + 1) + 6*b*x*polylog(3, (tan(b*x + a)^2 + 2*I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) + 6*b*x*polylog(3, (t an(b*x + a)^2 - 2*I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) - 3*I*polylog( 4, (tan(b*x + a)^2 + 2*I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) + 3*I*pol ylog(4, (tan(b*x + a)^2 - 2*I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)))/b^4
\[ \int x^3 \tan (a+b x) \, dx=\int x^{3} \tan {\left (a + b x \right )}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (83) = 166\).
Time = 0.60 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.29 \[ \int x^3 \tan (a+b x) \, dx=-\frac {-3 i \, {\left (b x + a\right )}^{4} + 12 i \, {\left (b x + a\right )}^{3} a - 18 i \, {\left (b x + a\right )}^{2} a^{2} + 12 \, a^{3} \log \left (\sec \left (b x + a\right )\right ) - 4 \, {\left (-4 i \, {\left (b x + a\right )}^{3} + 9 i \, {\left (b x + a\right )}^{2} a - 9 i \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 6 \, {\left (4 i \, {\left (b x + a\right )}^{2} - 6 i \, {\left (b x + a\right )} a + 3 i \, a^{2}\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 2 \, {\left (4 \, {\left (b x + a\right )}^{3} - 9 \, {\left (b x + a\right )}^{2} a + 9 \, {\left (b x + a\right )} a^{2}\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 6 \, {\left (4 \, b x + a\right )} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + 12 i \, {\rm Li}_{4}(-e^{\left (2 i \, b x + 2 i \, a\right )})}{12 \, b^{4}} \]
-1/12*(-3*I*(b*x + a)^4 + 12*I*(b*x + a)^3*a - 18*I*(b*x + a)^2*a^2 + 12*a ^3*log(sec(b*x + a)) - 4*(-4*I*(b*x + a)^3 + 9*I*(b*x + a)^2*a - 9*I*(b*x + a)*a^2)*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 6*(4*I*(b*x + a)^2 - 6*I*(b*x + a)*a + 3*I*a^2)*dilog(-e^(2*I*b*x + 2*I*a)) + 2*(4*(b*x + a)^3 - 9*(b*x + a)^2*a + 9*(b*x + a)*a^2)*log(cos(2*b*x + 2*a)^2 + sin(2 *b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 6*(4*b*x + a)*polylog(3, -e^(2*I *b*x + 2*I*a)) + 12*I*polylog(4, -e^(2*I*b*x + 2*I*a)))/b^4
\[ \int x^3 \tan (a+b x) \, dx=\int { x^{3} \tan \left (b x + a\right ) \,d x } \]
Timed out. \[ \int x^3 \tan (a+b x) \, dx=\int x^3\,\mathrm {tan}\left (a+b\,x\right ) \,d x \]