3.11 \(\int x^3 \tan ^3(a+b x) \, dx\)

Optimal. Leaf size=205 \[ \frac {3 i \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^4}+\frac {3 i \text {Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}+\frac {3 x \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}-\frac {3 i x^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {x^3 \tan ^2(a+b x)}{2 b}+\frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4} \]

[Out]

3/2*I*x^2/b^2+1/2*x^3/b-1/4*I*x^4-3*x*ln(1+exp(2*I*(b*x+a)))/b^3+x^3*ln(1+exp(2*I*(b*x+a)))/b+3/2*I*polylog(2,
-exp(2*I*(b*x+a)))/b^4-3/2*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^4-3/2*x^2*tan(b*x+a)/b^2+1/2*x^3*tan(b*x+a)^2/b

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Rubi [A]  time = 0.29, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {3720, 3719, 2190, 2279, 2391, 30, 2531, 6609, 2282, 6589} \[ -\frac {3 i x^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {3 x \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 i \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^4}+\frac {3 i \text {Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}-\frac {3 x^2 \tan (a+b x)}{2 b^2}-\frac {3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {x^3 \tan ^2(a+b x)}{2 b}+\frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Tan[a + b*x]^3,x]

[Out]

(((3*I)/2)*x^2)/b^2 + x^3/(2*b) - (I/4)*x^4 - (3*x*Log[1 + E^((2*I)*(a + b*x))])/b^3 + (x^3*Log[1 + E^((2*I)*(
a + b*x))])/b + (((3*I)/2)*PolyLog[2, -E^((2*I)*(a + b*x))])/b^4 - (((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
- (3*x^2*Tan[a + b*x])/(2*b^2) + (x^3*Tan[a + b*x]^2)/(2*b)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(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]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2531

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] + Dist[(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]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[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] &&
IGtQ[m, 0]

Rule 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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]

Rule 6609

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] - Dist[(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]

Rubi steps

\begin {align*} \int x^3 \tan ^3(a+b x) \, dx &=\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {3 \int x^2 \tan ^2(a+b x) \, dx}{2 b}-\int x^3 \tan (a+b x) \, dx\\ &=-\frac {i x^4}{4}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \tan ^2(a+b x)}{2 b}+2 i \int \frac {e^{2 i (a+b x)} x^3}{1+e^{2 i (a+b x)}} \, dx+\frac {3 \int x \tan (a+b x) \, dx}{b^2}+\frac {3 \int x^2 \, dx}{2 b}\\ &=\frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {(6 i) \int \frac {e^{2 i (a+b x)} x}{1+e^{2 i (a+b x)}} \, dx}{b^2}-\frac {3 \int x^2 \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4}-\frac {3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {3 i x^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \tan ^2(a+b x)}{2 b}+\frac {3 \int \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b^3}+\frac {(3 i) \int x \text {Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=\frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4}-\frac {3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {3 i x^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {3 x \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \tan ^2(a+b x)}{2 b}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^4}-\frac {3 \int \text {Li}_3\left (-e^{2 i (a+b x)}\right ) \, dx}{2 b^3}\\ &=\frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4}-\frac {3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^4}-\frac {3 i x^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {3 x \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \tan ^2(a+b x)}{2 b}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{4 b^4}\\ &=\frac {3 i x^2}{2 b^2}+\frac {x^3}{2 b}-\frac {i x^4}{4}-\frac {3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^4}-\frac {3 i x^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {3 x \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 i \text {Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}-\frac {3 x^2 \tan (a+b x)}{2 b^2}+\frac {x^3 \tan ^2(a+b x)}{2 b}\\ \end {align*}

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Mathematica [A]  time = 6.67, size = 359, normalized size = 1.75 \[ -\frac {3 x^2 \sec (a) \sin (b x) \sec (a+b x)}{2 b^2}-\frac {3 \csc (a) \sec (a) \left (b^2 x^2 e^{-i \tan ^{-1}(\cot (a))}-\frac {\cot (a) \left (i \text {Li}_2\left (e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )+i b x \left (-2 \tan ^{-1}(\cot (a))-\pi \right )-2 \left (b x-\tan ^{-1}(\cot (a))\right ) \log \left (1-e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-2 \tan ^{-1}(\cot (a)) \log \left (\sin \left (b x-\tan ^{-1}(\cot (a))\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt {\cot ^2(a)+1}}\right )}{2 b^4 \sqrt {\csc ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}+\frac {1}{8} i e^{i a} \sec (a) \left (\frac {3 e^{-2 i a} \left (1+e^{2 i a}\right ) \left (2 b^2 x^2 \text {Li}_2\left (-e^{-2 i (a+b x)}\right )-2 i b x \text {Li}_3\left (-e^{-2 i (a+b x)}\right )-\text {Li}_4\left (-e^{-2 i (a+b x)}\right )\right )}{b^4}-\frac {4 i \left (1+e^{-2 i a}\right ) x^3 \log \left (1+e^{-2 i (a+b x)}\right )}{b}+2 e^{-2 i a} x^4\right )+\frac {x^3 \sec ^2(a+b x)}{2 b}-\frac {1}{4} x^4 \tan (a) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3*Tan[a + b*x]^3,x]

[Out]

(I/8)*E^(I*a)*((2*x^4)/E^((2*I)*a) - ((4*I)*(1 + E^((-2*I)*a))*x^3*Log[1 + E^((-2*I)*(a + b*x))])/b + (3*(1 +
E^((2*I)*a))*(2*b^2*x^2*PolyLog[2, -E^((-2*I)*(a + b*x))] - (2*I)*b*x*PolyLog[3, -E^((-2*I)*(a + b*x))] - Poly
Log[4, -E^((-2*I)*(a + b*x))]))/(b^4*E^((2*I)*a)))*Sec[a] + (x^3*Sec[a + b*x]^2)/(2*b) - (3*Csc[a]*((b^2*x^2)/
E^(I*ArcTan[Cot[a]]) - (Cot[a]*(I*b*x*(-Pi - 2*ArcTan[Cot[a]]) - Pi*Log[1 + E^((-2*I)*b*x)] - 2*(b*x - ArcTan[
Cot[a]])*Log[1 - E^((2*I)*(b*x - ArcTan[Cot[a]]))] + Pi*Log[Cos[b*x]] - 2*ArcTan[Cot[a]]*Log[Sin[b*x - ArcTan[
Cot[a]]]] + I*PolyLog[2, E^((2*I)*(b*x - ArcTan[Cot[a]]))]))/Sqrt[1 + Cot[a]^2])*Sec[a])/(2*b^4*Sqrt[Csc[a]^2*
(Cos[a]^2 + Sin[a]^2)]) - (3*x^2*Sec[a]*Sec[a + b*x]*Sin[b*x])/(2*b^2) - (x^4*Tan[a])/4

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fricas [C]  time = 0.57, size = 342, normalized size = 1.67 \[ \frac {4 \, b^{3} x^{3} \tan \left (b x + a\right )^{2} + 4 \, b^{3} x^{3} - 12 \, b^{2} x^{2} \tan \left (b x + a\right ) + 6 \, b x {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 6 \, b x {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + {\left (6 i \, b^{2} x^{2} - 6 i\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + {\left (-6 i \, b^{2} x^{2} + 6 i\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 4 \, {\left (b^{3} x^{3} - 3 \, b x\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 4 \, {\left (b^{3} x^{3} - 3 \, b x\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 3 i \, {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 3 i \, {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right )}{8 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*tan(b*x+a)^3,x, algorithm="fricas")

[Out]

1/8*(4*b^3*x^3*tan(b*x + a)^2 + 4*b^3*x^3 - 12*b^2*x^2*tan(b*x + a) + 6*b*x*polylog(3, (tan(b*x + a)^2 + 2*I*t
an(b*x + a) - 1)/(tan(b*x + a)^2 + 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*I*b^2*x^2 - 6*I)*dilog(2*(I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1) + 1) + (-6*I*b^2*x^2 + 6*I)*
dilog(2*(-I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1) + 1) + 4*(b^3*x^3 - 3*b*x)*log(-2*(I*tan(b*x + a) - 1)/(tan
(b*x + a)^2 + 1)) + 4*(b^3*x^3 - 3*b*x)*log(-2*(-I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) - 3*I*polylog(4, (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)))/b^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \tan \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*tan(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x^3*tan(b*x + a)^3, x)

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maple [A]  time = 0.32, size = 251, normalized size = 1.22 \[ -\frac {2 i a^{3} x}{b^{3}}+\frac {x^{2} \left (2 b x \,{\mathrm e}^{2 i \left (b x +a \right )}-3 i {\mathrm e}^{2 i \left (b x +a \right )}-3 i\right )}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}+\frac {3 i \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{4}}+\frac {3 i \polylog \left (4, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{4 b^{4}}+\frac {3 i x^{2}}{b^{2}}-\frac {3 i a^{4}}{2 b^{4}}-\frac {3 i x^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{2}}-\frac {i x^{4}}{4}+\frac {3 i a^{2}}{b^{4}}+\frac {6 i a x}{b^{3}}+\frac {x^{3} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}+\frac {3 x \polylog \left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}-\frac {3 x \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b^{3}}-\frac {6 a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {2 a^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*tan(b*x+a)^3,x)

[Out]

-2*I/b^3*a^3*x+x^2*(2*b*x*exp(2*I*(b*x+a))-3*I*exp(2*I*(b*x+a))-3*I)/b^2/(exp(2*I*(b*x+a))+1)^2+3/2*I*polylog(
2,-exp(2*I*(b*x+a)))/b^4+3/4*I*polylog(4,-exp(2*I*(b*x+a)))/b^4+3*I/b^2*x^2-3/2*I/b^4*a^4-3/2*I*x^2*polylog(2,
-exp(2*I*(b*x+a)))/b^2-1/4*I*x^4+3*I/b^4*a^2+6*I/b^3*a*x+x^3*ln(exp(2*I*(b*x+a))+1)/b+3/2*x*polylog(3,-exp(2*I
*(b*x+a)))/b^3-3*x*ln(exp(2*I*(b*x+a))+1)/b^3-6/b^4*a*ln(exp(I*(b*x+a)))+2/b^4*a^3*ln(exp(I*(b*x+a)))

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maxima [B]  time = 2.58, size = 1203, normalized size = 5.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*tan(b*x+a)^3,x, algorithm="maxima")

[Out]

1/2*(a^3*(1/(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2 - 1)) - 2*(3*(b*x + a)^4 - 12*(b*x + a)^3*a + 18*(b*x +
a)^2*a^2 + 36*a^2 - (16*(b*x + a)^3 - 36*(b*x + a)^2*a + 36*(a^2 - 1)*(b*x + a) + 4*(4*(b*x + a)^3 - 9*(b*x +
a)^2*a + 9*(a^2 - 1)*(b*x + a) + 9*a)*cos(4*b*x + 4*a) + 8*(4*(b*x + a)^3 - 9*(b*x + a)^2*a + 9*(a^2 - 1)*(b*x
+ a) + 9*a)*cos(2*b*x + 2*a) + (16*I*(b*x + a)^3 - 36*I*(b*x + a)^2*a + (36*I*a^2 - 36*I)*(b*x + a) + 36*I*a)
*sin(4*b*x + 4*a) + (32*I*(b*x + a)^3 - 72*I*(b*x + a)^2*a + (72*I*a^2 - 72*I)*(b*x + a) + 72*I*a)*sin(2*b*x +
2*a) + 36*a)*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + 3*((b*x + a)^4 - 4*(b*x + a)^3*a + 6*(a^2 - 2)
*(b*x + a)^2 + 24*(b*x + a)*a)*cos(4*b*x + 4*a) + (6*(b*x + a)^4 - (b*x + a)^3*(24*a - 24*I) + 36*(a^2 - 2*I*a
- 1)*(b*x + a)^2 - (-72*I*a^2 - 72*a)*(b*x + a) + 36*a^2)*cos(2*b*x + 2*a) + (24*(b*x + a)^2 - 36*(b*x + a)*a
+ 18*a^2 + 6*(4*(b*x + a)^2 - 6*(b*x + a)*a + 3*a^2 - 3)*cos(4*b*x + 4*a) + 12*(4*(b*x + a)^2 - 6*(b*x + a)*a
+ 3*a^2 - 3)*cos(2*b*x + 2*a) - (-24*I*(b*x + a)^2 + 36*I*(b*x + a)*a - 18*I*a^2 + 18*I)*sin(4*b*x + 4*a) - (
-48*I*(b*x + a)^2 + 72*I*(b*x + a)*a - 36*I*a^2 + 36*I)*sin(2*b*x + 2*a) - 18)*dilog(-e^(2*I*b*x + 2*I*a)) - (
-8*I*(b*x + a)^3 + 18*I*(b*x + a)^2*a + (-18*I*a^2 + 18*I)*(b*x + a) + (-8*I*(b*x + a)^3 + 18*I*(b*x + a)^2*a
+ (-18*I*a^2 + 18*I)*(b*x + a) - 18*I*a)*cos(4*b*x + 4*a) + (-16*I*(b*x + a)^3 + 36*I*(b*x + a)^2*a + (-36*I*a
^2 + 36*I)*(b*x + a) - 36*I*a)*cos(2*b*x + 2*a) + 2*(4*(b*x + a)^3 - 9*(b*x + a)^2*a + 9*(a^2 - 1)*(b*x + a) +
9*a)*sin(4*b*x + 4*a) + 4*(4*(b*x + a)^3 - 9*(b*x + a)^2*a + 9*(a^2 - 1)*(b*x + a) + 9*a)*sin(2*b*x + 2*a) -
18*I*a)*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) - (12*cos(4*b*x + 4*a) + 24*cos(
2*b*x + 2*a) + 12*I*sin(4*b*x + 4*a) + 24*I*sin(2*b*x + 2*a) + 12)*polylog(4, -e^(2*I*b*x + 2*I*a)) - (-24*I*b
*x + (-24*I*b*x - 6*I*a)*cos(4*b*x + 4*a) + (-48*I*b*x - 12*I*a)*cos(2*b*x + 2*a) + 6*(4*b*x + a)*sin(4*b*x +
4*a) + 12*(4*b*x + a)*sin(2*b*x + 2*a) - 6*I*a)*polylog(3, -e^(2*I*b*x + 2*I*a)) - (-3*I*(b*x + a)^4 + 12*I*(b
*x + a)^3*a + (-18*I*a^2 + 36*I)*(b*x + a)^2 - 72*I*(b*x + a)*a)*sin(4*b*x + 4*a) - (-6*I*(b*x + a)^4 - 24*(b*
x + a)^3*(-I*a - 1) + (-36*I*a^2 - 72*a + 36*I)*(b*x + a)^2 + 72*(a^2 - I*a)*(b*x + a) - 36*I*a^2)*sin(2*b*x +
2*a))/(-12*I*cos(4*b*x + 4*a) - 24*I*cos(2*b*x + 2*a) + 12*sin(4*b*x + 4*a) + 24*sin(2*b*x + 2*a) - 12*I))/b^
4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,{\mathrm {tan}\left (a+b\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*tan(a + b*x)^3,x)

[Out]

int(x^3*tan(a + b*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \tan ^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*tan(b*x+a)**3,x)

[Out]

Integral(x**3*tan(a + b*x)**3, x)

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