Integrand size = 12, antiderivative size = 98 \[ \int x^3 \tan ^2(a+b x) \, dx=-\frac {i x^3}{b}-\frac {x^4}{4}+\frac {3 x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}-\frac {3 i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^3}+\frac {3 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^4}+\frac {x^3 \tan (a+b x)}{b} \] Output:
-I*x^3/b-1/4*x^4+3*x^2*ln(1+exp(2*I*(b*x+a)))/b^2-3*I*x*polylog(2,-exp(2*I *(b*x+a)))/b^3+3/2*polylog(3,-exp(2*I*(b*x+a)))/b^4+x^3*tan(b*x+a)/b
Time = 0.56 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.17 \[ \int x^3 \tan ^2(a+b x) \, dx=-\frac {x^4}{4}+\frac {2 b^2 x^2 \left (\frac {2 i b x}{1+e^{2 i a}}+3 \log \left (1+e^{-2 i (a+b x)}\right )\right )+6 i b x \operatorname {PolyLog}\left (2,-e^{-2 i (a+b x)}\right )+3 \operatorname {PolyLog}\left (3,-e^{-2 i (a+b x)}\right )}{2 b^4}+\frac {x^3 \sec (a) \sec (a+b x) \sin (b x)}{b} \] Input:
Integrate[x^3*Tan[a + b*x]^2,x]
Output:
-1/4*x^4 + (2*b^2*x^2*(((2*I)*b*x)/(1 + E^((2*I)*a)) + 3*Log[1 + E^((-2*I) *(a + b*x))]) + (6*I)*b*x*PolyLog[2, -E^((-2*I)*(a + b*x))] + 3*PolyLog[3, -E^((-2*I)*(a + b*x))])/(2*b^4) + (x^3*Sec[a]*Sec[a + b*x]*Sin[b*x])/b
Time = 0.60 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.24, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 4203, 15, 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.
\(\displaystyle \int x^3 \tan ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int x^3 \tan (a+b x)^2dx\) |
\(\Big \downarrow \) 4203 |
\(\displaystyle -\frac {3 \int x^2 \tan (a+b x)dx}{b}-\int x^3dx+\frac {x^3 \tan (a+b x)}{b}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {3 \int x^2 \tan (a+b x)dx}{b}+\frac {x^3 \tan (a+b x)}{b}-\frac {x^4}{4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 \int x^2 \tan (a+b x)dx}{b}+\frac {x^3 \tan (a+b x)}{b}-\frac {x^4}{4}\) |
\(\Big \downarrow \) 4202 |
\(\displaystyle -\frac {3 \left (\frac {i x^3}{3}-2 i \int \frac {e^{2 i (a+b x)} x^2}{1+e^{2 i (a+b x)}}dx\right )}{b}+\frac {x^3 \tan (a+b x)}{b}-\frac {x^4}{4}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -\frac {3 \left (\frac {i x^3}{3}-2 i \left (\frac {i \int x \log \left (1+e^{2 i (a+b x)}\right )dx}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\right )}{b}+\frac {x^3 \tan (a+b x)}{b}-\frac {x^4}{4}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle -\frac {3 \left (\frac {i x^3}{3}-2 i \left (\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i \int \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{2 b}\right )}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\right )}{b}+\frac {x^3 \tan (a+b x)}{b}-\frac {x^4}{4}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle -\frac {3 \left (\frac {i x^3}{3}-2 i \left (\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {\int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\right )}{b}+\frac {x^3 \tan (a+b x)}{b}-\frac {x^4}{4}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle -\frac {3 \left (\frac {i x^3}{3}-2 i \left (\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {\operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{4 b^2}\right )}{b}-\frac {i x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\right )}{b}+\frac {x^3 \tan (a+b x)}{b}-\frac {x^4}{4}\) |
Input:
Int[x^3*Tan[a + b*x]^2,x]
Output:
-1/4*x^4 - (3*((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)))/b + (x^3*Tan[a + b*x])/b
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[((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 = 0.33 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.36
method | result | size |
risch | \(-\frac {x^{4}}{4}+\frac {2 i x^{3}}{b \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}-\frac {2 i x^{3}}{b}+\frac {6 i a^{2} x}{b^{3}}+\frac {4 i a^{3}}{b^{4}}+\frac {3 x^{2} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{2}}-\frac {3 i x \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{3}}+\frac {3 \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{4}}-\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}\) | \(133\) |
Input:
int(x^3*tan(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
-1/4*x^4+2*I*x^3/b/(1+exp(2*I*(b*x+a)))-2*I/b*x^3+6*I/b^3*a^2*x+4*I/b^4*a^ 3+3*x^2*ln(1+exp(2*I*(b*x+a)))/b^2-3*I*x*polylog(2,-exp(2*I*(b*x+a)))/b^3+ 3/2*polylog(3,-exp(2*I*(b*x+a)))/b^4-6/b^4*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. 225 vs. \(2 (83) = 166\).
Time = 0.08 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.30 \[ \int x^3 \tan ^2(a+b x) \, dx=-\frac {b^{4} x^{4} - 4 \, b^{3} x^{3} \tan \left (b x + a\right ) - 6 \, b^{2} x^{2} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 6 \, b^{2} x^{2} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 6 i \, b x {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 6 i \, b x {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 3 \, {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 3 \, {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right )}{4 \, b^{4}} \] Input:
integrate(x^3*tan(b*x+a)^2,x, algorithm="fricas")
Output:
-1/4*(b^4*x^4 - 4*b^3*x^3*tan(b*x + a) - 6*b^2*x^2*log(-2*(I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) - 6*b^2*x^2*log(-2*(-I*tan(b*x + a) - 1)/(tan(b *x + a)^2 + 1)) - 6*I*b*x*dilog(2*(I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1 ) + 1) + 6*I*b*x*dilog(2*(-I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1) + 1) - 3*polylog(3, (tan(b*x + a)^2 + 2*I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1) ) - 3*polylog(3, (tan(b*x + a)^2 - 2*I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)))/b^4
\[ \int x^3 \tan ^2(a+b x) \, dx=\int x^{3} \tan ^{2}{\left (a + b x \right )}\, dx \] Input:
integrate(x**3*tan(b*x+a)**2,x)
Output:
Integral(x**3*tan(a + b*x)**2, x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 639 vs. \(2 (83) = 166\).
Time = 0.24 (sec) , antiderivative size = 639, normalized size of antiderivative = 6.52 \[ \int x^3 \tan ^2(a+b x) \, dx =\text {Too large to display} \] Input:
integrate(x^3*tan(b*x+a)^2,x, algorithm="maxima")
Output:
1/2*(2*(b*x + a - tan(b*x + a))*a^3 - 3*((b*x + a)^2*cos(2*b*x + 2*a)^2 + (b*x + a)^2*sin(2*b*x + 2*a)^2 + 2*(b*x + a)^2*cos(2*b*x + 2*a) + (b*x + a )^2 - (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*l og(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) - 4*( b*x + a)*sin(2*b*x + 2*a))*a^2/(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 2*(I*(b*x + a)^4 - 4*I*(b*x + a)^3*a + 12*((b*x + a)^2 - 2*(b*x + a)*a + ((b*x + a)^2 - 2*(b*x + a)*a)*cos(2*b*x + 2*a) - (-I*(b*x + a)^2 + 2*I*(b*x + a)*a)*sin(2*b*x + 2*a))*arctan2(sin(2*b*x + 2 *a), cos(2*b*x + 2*a) + 1) + (I*(b*x + a)^4 - 4*(b*x + a)^3*(I*a + 2) + 24 *(b*x + a)^2*a)*cos(2*b*x + 2*a) - 12*(b*x*cos(2*b*x + 2*a) + I*b*x*sin(2* b*x + 2*a) + b*x)*dilog(-e^(2*I*b*x + 2*I*a)) - 6*(I*(b*x + a)^2 - 2*I*(b* x + a)*a + (I*(b*x + a)^2 - 2*I*(b*x + a)*a)*cos(2*b*x + 2*a) - ((b*x + a) ^2 - 2*(b*x + a)*a)*sin(2*b*x + 2*a))*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) - 6*(I*cos(2*b*x + 2*a) - sin(2*b*x + 2* a) + I)*polylog(3, -e^(2*I*b*x + 2*I*a)) - ((b*x + a)^4 - 4*(b*x + a)^3*(a - 2*I) - 24*I*(b*x + a)^2*a)*sin(2*b*x + 2*a))/(-4*I*cos(2*b*x + 2*a) + 4 *sin(2*b*x + 2*a) - 4*I))/b^4
\[ \int x^3 \tan ^2(a+b x) \, dx=\int { x^{3} \tan \left (b x + a\right )^{2} \,d x } \] Input:
integrate(x^3*tan(b*x+a)^2,x, algorithm="giac")
Output:
integrate(x^3*tan(b*x + a)^2, x)
Timed out. \[ \int x^3 \tan ^2(a+b x) \, dx=\int x^3\,{\mathrm {tan}\left (a+b\,x\right )}^2 \,d x \] Input:
int(x^3*tan(a + b*x)^2,x)
Output:
int(x^3*tan(a + b*x)^2, x)
\[ \int x^3 \tan ^2(a+b x) \, dx=\frac {-12 \left (\int \tan \left (b x +a \right ) x^{2}d x \right )+4 \tan \left (b x +a \right ) x^{3}-b \,x^{4}}{4 b} \] Input:
int(x^3*tan(b*x+a)^2,x)
Output:
( - 12*int(tan(a + b*x)*x**2,x) + 4*tan(a + b*x)*x**3 - b*x**4)/(4*b)