Integrand size = 12, antiderivative size = 128 \[ \int x^2 \tan ^3(a+b x) \, dx=\frac {x^2}{2 b}-\frac {i x^3}{3}+\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {\log (\cos (a+b x))}{b^3}-\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac {\operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {x \tan (a+b x)}{b^2}+\frac {x^2 \tan ^2(a+b x)}{2 b} \]
1/2*x^2/b-1/3*I*x^3+x^2*ln(1+exp(2*I*(b*x+a)))/b-ln(cos(b*x+a))/b^3-I*x*po lylog(2,-exp(2*I*(b*x+a)))/b^2+1/2*polylog(3,-exp(2*I*(b*x+a)))/b^3-x*tan( b*x+a)/b^2+1/2*x^2*tan(b*x+a)^2/b
Time = 2.28 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.34 \[ \int x^2 \tan ^3(a+b x) \, dx=\frac {e^{-i a} \left (2 b^2 x^2 \left (2 i b x+3 \left (1+e^{2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right )\right )+6 i b \left (1+e^{2 i a}\right ) x \operatorname {PolyLog}\left (2,-e^{-2 i (a+b x)}\right )+3 \left (1+e^{2 i a}\right ) \operatorname {PolyLog}\left (3,-e^{-2 i (a+b x)}\right )\right ) \sec (a)+6 b^2 x^2 \sec ^2(a+b x)-12 b x \sec (a) \sec (a+b x) \sin (b x)-4 b^3 x^3 \tan (a)-12 (\log (\cos (a+b x))+b x \tan (a))}{12 b^3} \]
(((2*b^2*x^2*((2*I)*b*x + 3*(1 + E^((2*I)*a))*Log[1 + E^((-2*I)*(a + b*x)) ]) + (6*I)*b*(1 + E^((2*I)*a))*x*PolyLog[2, -E^((-2*I)*(a + b*x))] + 3*(1 + E^((2*I)*a))*PolyLog[3, -E^((-2*I)*(a + b*x))])*Sec[a])/E^(I*a) + 6*b^2* x^2*Sec[a + b*x]^2 - 12*b*x*Sec[a]*Sec[a + b*x]*Sin[b*x] - 4*b^3*x^3*Tan[a ] - 12*(Log[Cos[a + b*x]] + b*x*Tan[a]))/(12*b^3)
Time = 0.85 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.16, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4203, 3042, 4202, 2620, 3011, 2720, 4203, 15, 3042, 3956, 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.
| \(\displaystyle \int x^2 \tan ^3(a+b x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int x^2 \tan (a+b x)^3dx\) |
| \(\Big \downarrow \) 4203 |
| \(\displaystyle -\int x^2 \tan (a+b x)dx-\frac {\int x \tan ^2(a+b x)dx}{b}+\frac {x^2 \tan ^2(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\int x^2 \tan (a+b x)dx-\frac {\int x \tan (a+b x)^2dx}{b}+\frac {x^2 \tan ^2(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 4202 |
| \(\displaystyle 2 i \int \frac {e^{2 i (a+b x)} x^2}{1+e^{2 i (a+b x)}}dx-\frac {\int x \tan (a+b x)^2dx}{b}+\frac {x^2 \tan ^2(a+b x)}{2 b}-\frac {i x^3}{3}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle 2 i \left (\frac {i \int x \log \left (1+e^{2 i (a+b x)}\right )dx}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {\int x \tan (a+b x)^2dx}{b}+\frac {x^2 \tan ^2(a+b x)}{2 b}-\frac {i x^3}{3}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle 2 i \left (\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \int \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{2 b}\right )}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {\int x \tan (a+b x)^2dx}{b}+\frac {x^2 \tan ^2(a+b x)}{2 b}-\frac {i x^3}{3}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle 2 i \left (\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {\int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {\int x \tan (a+b x)^2dx}{b}+\frac {x^2 \tan ^2(a+b x)}{2 b}-\frac {i x^3}{3}\) |
| \(\Big \downarrow \) 4203 |
| \(\displaystyle 2 i \left (\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {\int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {-\frac {\int \tan (a+b x)dx}{b}-\int xdx+\frac {x \tan (a+b x)}{b}}{b}+\frac {x^2 \tan ^2(a+b x)}{2 b}-\frac {i x^3}{3}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle 2 i \left (\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {\int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {-\frac {\int \tan (a+b x)dx}{b}+\frac {x \tan (a+b x)}{b}-\frac {x^2}{2}}{b}+\frac {x^2 \tan ^2(a+b x)}{2 b}-\frac {i x^3}{3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 i \left (\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {\int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {-\frac {\int \tan (a+b x)dx}{b}+\frac {x \tan (a+b x)}{b}-\frac {x^2}{2}}{b}+\frac {x^2 \tan ^2(a+b x)}{2 b}-\frac {i x^3}{3}\) |
| \(\Big \downarrow \) 3956 |
| \(\displaystyle 2 i \left (\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {\int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {\frac {\log (\cos (a+b x))}{b^2}+\frac {x \tan (a+b x)}{b}-\frac {x^2}{2}}{b}+\frac {x^2 \tan ^2(a+b x)}{2 b}-\frac {i x^3}{3}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle 2 i \left (\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {\operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{4 b^2}\right )}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\frac {\frac {\log (\cos (a+b x))}{b^2}+\frac {x \tan (a+b x)}{b}-\frac {x^2}{2}}{b}+\frac {x^2 \tan ^2(a+b x)}{2 b}-\frac {i x^3}{3}\) |
(-1/3*I)*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) + (x^2*Tan[a + b*x]^2)/(2*b) - (-1/2*x^2 + Log[Cos[a + b*x]]/b^2 + (x*Tan[a + b*x])/b)/b
3.1.12.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
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[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symb ol] :> Simp[b*(c + d*x)^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] + (-Si mp[b*d*(m/(f*(n - 1))) Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1), x] , x] - Simp[b^2 Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; Free Q[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[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 = 0.30 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.41
| method | result | size |
| risch | \(-\frac {i x^{3}}{3}+\frac {2 x \left (b x \,{\mathrm e}^{2 i \left (b x +a \right )}-i {\mathrm e}^{2 i \left (b x +a \right )}-i\right )}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {4 i a^{3}}{3 b^{3}}+\frac {2 i a^{2} x}{b^{2}}+\frac {x^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}-\frac {i x \,\operatorname {Li}_{2}\left (-{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{2}}+\frac {\operatorname {Li}_{3}\left (-{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}-\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b^{3}}+\frac {2 \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}\) | \(180\) |
-1/3*I*x^3+2*x*(b*x*exp(2*I*(b*x+a))-I*exp(2*I*(b*x+a))-I)/b^2/(exp(2*I*(b *x+a))+1)^2-2/b^3*a^2*ln(exp(I*(b*x+a)))+4/3*I/b^3*a^3+2*I/b^2*a^2*x+x^2*l n(exp(2*I*(b*x+a))+1)/b-I*x*polylog(2,-exp(2*I*(b*x+a)))/b^2+1/2*polylog(3 ,-exp(2*I*(b*x+a)))/b^3-1/b^3*ln(exp(2*I*(b*x+a))+1)+2/b^3*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. 240 vs. \(2 (109) = 218\).
Time = 0.26 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.88 \[ \int x^2 \tan ^3(a+b x) \, dx=\frac {2 \, b^{2} x^{2} \tan \left (b x + a\right )^{2} + 2 \, b^{2} x^{2} + 2 i \, b x {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 2 i \, b x {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 4 \, b x \tan \left (b x + a\right ) + 2 \, {\left (b^{2} x^{2} - 1\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, {\left (b^{2} x^{2} - 1\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right )}{4 \, b^{3}} \]
1/4*(2*b^2*x^2*tan(b*x + a)^2 + 2*b^2*x^2 + 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) - 4*b*x*tan(b*x + a) + 2*(b^2*x^2 - 1)*log(-2*(I* tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) + 2*(b^2*x^2 - 1)*log(-2*(-I*tan(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)) + polylog(3, (tan(b*x + a)^2 - 2*I*tan (b*x + a) - 1)/(tan(b*x + a)^2 + 1)))/b^3
\[ \int x^2 \tan ^3(a+b x) \, dx=\int x^{2} \tan ^{3}{\left (a + b x \right )}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 736 vs. \(2 (109) = 218\).
Time = 0.66 (sec) , antiderivative size = 736, normalized size of antiderivative = 5.75 \[ \int x^2 \tan ^3(a+b x) \, dx=\text {Too large to display} \]
-1/2*(a^2*(1/(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2 - 1)) + 2*(2*(b*x + a)^3 - 6*(b*x + a)^2*a - 6*((b*x + a)^2 - 2*(b*x + a)*a + ((b*x + a)^2 - 2*(b*x + a)*a - 1)*cos(4*b*x + 4*a) + 2*((b*x + a)^2 - 2*(b*x + a)*a - 1)* cos(2*b*x + 2*a) - (-I*(b*x + a)^2 + 2*I*(b*x + a)*a + I)*sin(4*b*x + 4*a) - 2*(-I*(b*x + a)^2 + 2*I*(b*x + a)*a + I)*sin(2*b*x + 2*a) - 1)*arctan2( sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + 2*((b*x + a)^3 - 3*(b*x + a)^2*a - 6*b*x - 6*a)*cos(4*b*x + 4*a) + 4*((b*x + a)^3 - 3*(b*x + a)^2*(a - I) + 3*(b*x + a)*(-2*I*a - 1) - 3*a)*cos(2*b*x + 2*a) + 6*(b*x*cos(4*b*x + 4* a) + 2*b*x*cos(2*b*x + 2*a) + I*b*x*sin(4*b*x + 4*a) + 2*I*b*x*sin(2*b*x + 2*a) + b*x)*dilog(-e^(2*I*b*x + 2*I*a)) + 3*(I*(b*x + a)^2 - 2*I*(b*x + a )*a + (I*(b*x + a)^2 - 2*I*(b*x + a)*a - I)*cos(4*b*x + 4*a) + 2*(I*(b*x + a)^2 - 2*I*(b*x + a)*a - I)*cos(2*b*x + 2*a) - ((b*x + a)^2 - 2*(b*x + a) *a - 1)*sin(4*b*x + 4*a) - 2*((b*x + a)^2 - 2*(b*x + a)*a - 1)*sin(2*b*x + 2*a) - I)*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a ) + 1) + 3*(I*cos(4*b*x + 4*a) + 2*I*cos(2*b*x + 2*a) - sin(4*b*x + 4*a) - 2*sin(2*b*x + 2*a) + I)*polylog(3, -e^(2*I*b*x + 2*I*a)) + 2*(I*(b*x + a) ^3 - 3*I*(b*x + a)^2*a - 6*I*b*x - 6*I*a)*sin(4*b*x + 4*a) + 4*(I*(b*x + a )^3 + 3*(b*x + a)^2*(-I*a - 1) + 3*(b*x + a)*(2*a - I) - 3*I*a)*sin(2*b*x + 2*a) - 12*a)/(-6*I*cos(4*b*x + 4*a) - 12*I*cos(2*b*x + 2*a) + 6*sin(4*b* x + 4*a) + 12*sin(2*b*x + 2*a) - 6*I))/b^3
\[ \int x^2 \tan ^3(a+b x) \, dx=\int { x^{2} \tan \left (b x + a\right )^{3} \,d x } \]
Timed out. \[ \int x^2 \tan ^3(a+b x) \, dx=\int x^2\,{\mathrm {tan}\left (a+b\,x\right )}^3 \,d x \]