3.304 \(\int (c+d x)^3 \tan ^3(a+b x) \, dx\)
Optimal. Leaf size=259 \[ \frac{3 d^2 (c+d x) \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i d (c+d x)^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 i d^3 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^4}+\frac{3 i d^3 \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{3 d^2 (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^3}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{(c+d x)^3 \tan ^2(a+b x)}{2 b}+\frac{3 i d (c+d x)^2}{2 b^2}+\frac{(c+d x)^3}{2 b}-\frac{i (c+d x)^4}{4 d} \]
[Out]
(((3*I)/2)*d*(c + d*x)^2)/b^2 + (c + d*x)^3/(2*b) - ((I/4)*(c + d*x)^4)/d - (3*d^2*(c + d*x)*Log[1 + E^((2*I)*
(a + b*x))])/b^3 + ((c + d*x)^3*Log[1 + E^((2*I)*(a + b*x))])/b + (((3*I)/2)*d^3*PolyLog[2, -E^((2*I)*(a + b*x
))])/b^4 - (((3*I)/2)*d*(c + d*x)^2*PolyLog[2, -E^((2*I)*(a + b*x))])/b^2 + (3*d^2*(c + d*x)*PolyLog[3, -E^((2
*I)*(a + b*x))])/(2*b^3) + (((3*I)/4)*d^3*PolyLog[4, -E^((2*I)*(a + b*x))])/b^4 - (3*d*(c + d*x)^2*Tan[a + b*x
])/(2*b^2) + ((c + d*x)^3*Tan[a + b*x]^2)/(2*b)
________________________________________________________________________________________
Rubi [A] time = 0.355707, antiderivative size = 259, normalized size of antiderivative
= 1., number of steps used = 13, number of rules used = 10, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\)
= 0.625, Rules used = {3720, 3719, 2190, 2279, 2391, 32, 2531, 6609, 2282, 6589}
\[ \frac{3 d^2 (c+d x) \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i d (c+d x)^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 i d^3 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^4}+\frac{3 i d^3 \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{3 d^2 (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^3}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{(c+d x)^3 \tan ^2(a+b x)}{2 b}+\frac{3 i d (c+d x)^2}{2 b^2}+\frac{(c+d x)^3}{2 b}-\frac{i (c+d x)^4}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)^3*Tan[a + b*x]^3,x]
[Out]
(((3*I)/2)*d*(c + d*x)^2)/b^2 + (c + d*x)^3/(2*b) - ((I/4)*(c + d*x)^4)/d - (3*d^2*(c + d*x)*Log[1 + E^((2*I)*
(a + b*x))])/b^3 + ((c + d*x)^3*Log[1 + E^((2*I)*(a + b*x))])/b + (((3*I)/2)*d^3*PolyLog[2, -E^((2*I)*(a + b*x
))])/b^4 - (((3*I)/2)*d*(c + d*x)^2*PolyLog[2, -E^((2*I)*(a + b*x))])/b^2 + (3*d^2*(c + d*x)*PolyLog[3, -E^((2
*I)*(a + b*x))])/(2*b^3) + (((3*I)/4)*d^3*PolyLog[4, -E^((2*I)*(a + b*x))])/b^4 - (3*d*(c + d*x)^2*Tan[a + b*x
])/(2*b^2) + ((c + d*x)^3*Tan[a + b*x]^2)/(2*b)
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 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 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 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 32
Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -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 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]
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 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]
Rubi steps
\begin{align*} \int (c+d x)^3 \tan ^3(a+b x) \, dx &=\frac{(c+d x)^3 \tan ^2(a+b x)}{2 b}-\frac{(3 d) \int (c+d x)^2 \tan ^2(a+b x) \, dx}{2 b}-\int (c+d x)^3 \tan (a+b x) \, dx\\ &=-\frac{i (c+d x)^4}{4 d}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}+\frac{(c+d x)^3 \tan ^2(a+b x)}{2 b}+2 i \int \frac{e^{2 i (a+b x)} (c+d x)^3}{1+e^{2 i (a+b x)}} \, dx+\frac{(3 d) \int (c+d x)^2 \, dx}{2 b}+\frac{\left (3 d^2\right ) \int (c+d x) \tan (a+b x) \, dx}{b^2}\\ &=\frac{3 i d (c+d x)^2}{2 b^2}+\frac{(c+d x)^3}{2 b}-\frac{i (c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}+\frac{(c+d x)^3 \tan ^2(a+b x)}{2 b}-\frac{(3 d) \int (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}-\frac{\left (6 i d^2\right ) \int \frac{e^{2 i (a+b x)} (c+d x)}{1+e^{2 i (a+b x)}} \, dx}{b^2}\\ &=\frac{3 i d (c+d x)^2}{2 b^2}+\frac{(c+d x)^3}{2 b}-\frac{i (c+d x)^4}{4 d}-\frac{3 d^2 (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}+\frac{(c+d x)^3 \tan ^2(a+b x)}{2 b}+\frac{\left (3 i d^2\right ) \int (c+d x) \text{Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}+\frac{\left (3 d^3\right ) \int \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b^3}\\ &=\frac{3 i d (c+d x)^2}{2 b^2}+\frac{(c+d x)^3}{2 b}-\frac{i (c+d x)^4}{4 d}-\frac{3 d^2 (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 d^2 (c+d x) \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}+\frac{(c+d x)^3 \tan ^2(a+b x)}{2 b}-\frac{\left (3 i d^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^4}-\frac{\left (3 d^3\right ) \int \text{Li}_3\left (-e^{2 i (a+b x)}\right ) \, dx}{2 b^3}\\ &=\frac{3 i d (c+d x)^2}{2 b^2}+\frac{(c+d x)^3}{2 b}-\frac{i (c+d x)^4}{4 d}-\frac{3 d^2 (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 i d^3 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^4}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 d^2 (c+d x) \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}+\frac{(c+d x)^3 \tan ^2(a+b x)}{2 b}+\frac{\left (3 i d^3\right ) \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 d (c+d x)^2}{2 b^2}+\frac{(c+d x)^3}{2 b}-\frac{i (c+d x)^4}{4 d}-\frac{3 d^2 (c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 i d^3 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^4}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 d^2 (c+d x) \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{3 i d^3 \text{Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{3 d (c+d x)^2 \tan (a+b x)}{2 b^2}+\frac{(c+d x)^3 \tan ^2(a+b x)}{2 b}\\ \end{align*}
Mathematica [B] time = 6.9231, size = 803, normalized size = 3.1 \[ \frac{\sec (a) (\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x))+b x \sin (a)) c^3}{b \left (\cos ^2(a)+\sin ^2(a)\right )}+\frac{3 d \csc (a) \left (b^2 e^{-i \tan ^{-1}(\cot (a))} x^2-\frac{\cot (a) \left (i b x \left (-2 \tan ^{-1}(\cot (a))-\pi \right )-\pi \log \left (1+e^{-2 i b x}\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 )+\pi \log (\cos (b x))-2 \tan ^{-1}(\cot (a)) \log \left (\sin \left (b x-\tan ^{-1}(\cot (a))\right )\right )+i \text{PolyLog}\left (2,e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )\right )}{\sqrt{\cot ^2(a)+1}}\right ) \sec (a) c^2}{2 b^2 \sqrt{\csc ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}}+\frac{i d^2 e^{-i a} \left (2 b^2 \left (2 b x-3 i \left (1+e^{2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right )\right ) x^2+6 b \left (1+e^{2 i a}\right ) \text{PolyLog}\left (2,-e^{-2 i (a+b x)}\right ) x-3 i \left (1+e^{2 i a}\right ) \text{PolyLog}\left (3,-e^{-2 i (a+b x)}\right )\right ) \sec (a) c}{4 b^3}-\frac{3 d^2 \sec (a) (\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x))+b x \sin (a)) c}{b^3 \left (\cos ^2(a)+\sin ^2(a)\right )}+\frac{(c+d x)^3 \sec ^2(a+b x)}{2 b}+\frac{1}{8} i d^3 e^{i a} \left (2 e^{-2 i a} x^4-\frac{4 i \left (1+e^{-2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right ) x^3}{b}+\frac{3 e^{-2 i a} \left (1+e^{2 i a}\right ) \left (2 b^2 \text{PolyLog}\left (2,-e^{-2 i (a+b x)}\right ) x^2-2 i b \text{PolyLog}\left (3,-e^{-2 i (a+b x)}\right ) x-\text{PolyLog}\left (4,-e^{-2 i (a+b x)}\right )\right )}{b^4}\right ) \sec (a)-\frac{3 \sec (a) \sec (a+b x) \left (x^2 \sin (b x) d^3+2 c x \sin (b x) d^2+c^2 \sin (b x) d\right )}{2 b^2}-\frac{1}{4} x \left (4 c^3+6 d x c^2+4 d^2 x^2 c+d^3 x^3\right ) \tan (a)-\frac{3 d^3 \csc (a) \left (b^2 e^{-i \tan ^{-1}(\cot (a))} x^2-\frac{\cot (a) \left (i b x \left (-2 \tan ^{-1}(\cot (a))-\pi \right )-\pi \log \left (1+e^{-2 i b x}\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 )+\pi \log (\cos (b x))-2 \tan ^{-1}(\cot (a)) \log \left (\sin \left (b x-\tan ^{-1}(\cot (a))\right )\right )+i \text{PolyLog}\left (2,e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )\right )}{\sqrt{\cot ^2(a)+1}}\right ) \sec (a)}{2 b^4 \sqrt{\csc ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(c + d*x)^3*Tan[a + b*x]^3,x]
[Out]
((I/4)*c*d^2*(2*b^2*x^2*(2*b*x - (3*I)*(1 + E^((2*I)*a))*Log[1 + E^((-2*I)*(a + b*x))]) + 6*b*(1 + E^((2*I)*a)
)*x*PolyLog[2, -E^((-2*I)*(a + b*x))] - (3*I)*(1 + E^((2*I)*a))*PolyLog[3, -E^((-2*I)*(a + b*x))])*Sec[a])/(b^
3*E^(I*a)) + (I/8)*d^3*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))] - PolyLog[4, -E^((-2*I)*(a + b*x))]))/(b^4*E^((2*I)*a)))*Sec[a] + ((c + d*x)^3*Sec[a + b*x]^2)/(2*b
) + (c^3*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] - Sin[a]*Sin[b*x]] + b*x*Sin[a]))/(b*(Cos[a]^2 + Sin[a]^2)) - (3*c
*d^2*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] - Sin[a]*Sin[b*x]] + b*x*Sin[a]))/(b^3*(Cos[a]^2 + Sin[a]^2)) + (3*c^2
*d*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^2*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)]) - (3*d^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 - ArcTa
n[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*Sec[a]*Se
c[a + b*x]*(c^2*d*Sin[b*x] + 2*c*d^2*x*Sin[b*x] + d^3*x^2*Sin[b*x]))/(2*b^2) - (x*(4*c^3 + 6*c^2*d*x + 4*c*d^2
*x^2 + d^3*x^3)*Tan[a])/4
________________________________________________________________________________________
Maple [B] time = 0.368, size = 720, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x+c)^3*tan(b*x+a)^3,x)
[Out]
3/4*I*d^3*polylog(4,-exp(2*I*(b*x+a)))/b^4+1/b*c^3*ln(exp(2*I*(b*x+a))+1)+3/2/b^3*c*d^2*polylog(3,-exp(2*I*(b*
x+a)))+3/2/b^3*d^3*polylog(3,-exp(2*I*(b*x+a)))*x+(2*b*d^3*x^3*exp(2*I*(b*x+a))-3*I*d^3*x^2*exp(2*I*(b*x+a))+6
*b*c*d^2*x^2*exp(2*I*(b*x+a))-6*I*c*d^2*x*exp(2*I*(b*x+a))+6*b*c^2*d*x*exp(2*I*(b*x+a))-3*I*c^2*d*exp(2*I*(b*x
+a))-3*I*d^3*x^2+2*b*c^3*exp(2*I*(b*x+a))-6*I*c*d^2*x-3*I*c^2*d)/b^2/(exp(2*I*(b*x+a))+1)^2-3*d^2/b^3*c*ln(exp
(2*I*(b*x+a))+1)-3*d^3/b^3*ln(exp(2*I*(b*x+a))+1)*x+3*I*d^3/b^2*x^2+3*I*d^3/b^4*a^2-3/2*I/b^4*d^3*a^4+3/2*I*d^
3*polylog(2,-exp(2*I*(b*x+a)))/b^4-3/2*I/b^2*c^2*d*polylog(2,-exp(2*I*(b*x+a)))-3/2*I/b^2*d^3*polylog(2,-exp(2
*I*(b*x+a)))*x^2+4*I/b^3*a^3*c*d^2-3*I/b^2*a^2*c^2*d+3/b*c^2*d*ln(exp(2*I*(b*x+a))+1)*x+3/b*c*d^2*ln(exp(2*I*(
b*x+a))+1)*x^2+1/b*d^3*ln(exp(2*I*(b*x+a))+1)*x^3-2*I/b^3*d^3*a^3*x-6/b^4*d^3*a*ln(exp(I*(b*x+a)))+6/b^3*d^2*c
*ln(exp(I*(b*x+a)))+6*I*d^3/b^3*a*x-6/b^3*c*d^2*a^2*ln(exp(I*(b*x+a)))+6/b^2*c^2*d*a*ln(exp(I*(b*x+a)))+I*c^3*
x-I*c*d^2*x^3-3/2*I*c^2*d*x^2+2/b^4*d^3*a^3*ln(exp(I*(b*x+a)))-3*I/b^2*polylog(2,-exp(2*I*(b*x+a)))*c*d^2*x-6*
I/b*a*c^2*d*x+6*I/b^2*a^2*c*d^2*x-2/b*c^3*ln(exp(I*(b*x+a)))-1/4*I*d^3*x^4
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Maxima [B] time = 4.48058, size = 3247, normalized size = 12.54 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^3*tan(b*x+a)^3,x, algorithm="maxima")
[Out]
-1/2*(c^3*(1/(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2 - 1)) - 3*a*c^2*d*(1/(sin(b*x + a)^2 - 1) - log(sin(b*x
+ a)^2 - 1))/b + 3*a^2*c*d^2*(1/(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2 - 1))/b^2 - a^3*d^3*(1/(sin(b*x + a
)^2 - 1) - log(sin(b*x + a)^2 - 1))/b^3 + 2*(3*(b*x + a)^4*d^3 + 36*b^2*c^2*d - 72*a*b*c*d^2 + 36*a^2*d^3 + 12
*(b*c*d^2 - a*d^3)*(b*x + a)^3 + 18*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(b*x + a)^2 - (16*(b*x + a)^3*d^3 - 36
*b*c*d^2 + 36*a*d^3 + 36*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 36*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 1)*d^3)*(b*x + a
) + 4*(4*(b*x + a)^3*d^3 - 9*b*c*d^2 + 9*a*d^3 + 9*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 9*(b^2*c^2*d - 2*a*b*c*d^2
+ (a^2 - 1)*d^3)*(b*x + a))*cos(4*b*x + 4*a) + 8*(4*(b*x + a)^3*d^3 - 9*b*c*d^2 + 9*a*d^3 + 9*(b*c*d^2 - a*d^3
)*(b*x + a)^2 + 9*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 1)*d^3)*(b*x + a))*cos(2*b*x + 2*a) + (16*I*(b*x + a)^3*d^
3 - 36*I*b*c*d^2 + 36*I*a*d^3 + (36*I*b*c*d^2 - 36*I*a*d^3)*(b*x + a)^2 + (36*I*b^2*c^2*d - 72*I*a*b*c*d^2 + (
36*I*a^2 - 36*I)*d^3)*(b*x + a))*sin(4*b*x + 4*a) + (32*I*(b*x + a)^3*d^3 - 72*I*b*c*d^2 + 72*I*a*d^3 + (72*I*
b*c*d^2 - 72*I*a*d^3)*(b*x + a)^2 + (72*I*b^2*c^2*d - 144*I*a*b*c*d^2 + (72*I*a^2 - 72*I)*d^3)*(b*x + a))*sin(
2*b*x + 2*a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + 3*((b*x + a)^4*d^3 + 4*(b*c*d^2 - a*d^3)*(b*x
+ a)^3 + 6*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 2)*d^3)*(b*x + a)^2 - 24*(b*c*d^2 - a*d^3)*(b*x + a))*cos(4*b*x +
4*a) + (6*(b*x + a)^4*d^3 + 36*b^2*c^2*d - 72*a*b*c*d^2 + 36*a^2*d^3 + (24*b*c*d^2 - (24*a - 24*I)*d^3)*(b*x
+ a)^3 + (36*b^2*c^2*d - (72*a - 72*I)*b*c*d^2 + 36*(a^2 - 2*I*a - 1)*d^3)*(b*x + a)^2 - (-72*I*b^2*c^2*d - 72
*(-2*I*a - 1)*b*c*d^2 + (-72*I*a^2 - 72*a)*d^3)*(b*x + a))*cos(2*b*x + 2*a) + (18*b^2*c^2*d - 36*a*b*c*d^2 + 2
4*(b*x + a)^2*d^3 + 18*(a^2 - 1)*d^3 + 36*(b*c*d^2 - a*d^3)*(b*x + a) + 6*(3*b^2*c^2*d - 6*a*b*c*d^2 + 4*(b*x
+ a)^2*d^3 + 3*(a^2 - 1)*d^3 + 6*(b*c*d^2 - a*d^3)*(b*x + a))*cos(4*b*x + 4*a) + 12*(3*b^2*c^2*d - 6*a*b*c*d^2
+ 4*(b*x + a)^2*d^3 + 3*(a^2 - 1)*d^3 + 6*(b*c*d^2 - a*d^3)*(b*x + a))*cos(2*b*x + 2*a) - (-18*I*b^2*c^2*d +
36*I*a*b*c*d^2 - 24*I*(b*x + a)^2*d^3 + (-18*I*a^2 + 18*I)*d^3 + (-36*I*b*c*d^2 + 36*I*a*d^3)*(b*x + a))*sin(4
*b*x + 4*a) - (-36*I*b^2*c^2*d + 72*I*a*b*c*d^2 - 48*I*(b*x + a)^2*d^3 + (-36*I*a^2 + 36*I)*d^3 + (-72*I*b*c*d
^2 + 72*I*a*d^3)*(b*x + a))*sin(2*b*x + 2*a))*dilog(-e^(2*I*b*x + 2*I*a)) - (-8*I*(b*x + a)^3*d^3 + 18*I*b*c*d
^2 - 18*I*a*d^3 + (-18*I*b*c*d^2 + 18*I*a*d^3)*(b*x + a)^2 + (-18*I*b^2*c^2*d + 36*I*a*b*c*d^2 + (-18*I*a^2 +
18*I)*d^3)*(b*x + a) + (-8*I*(b*x + a)^3*d^3 + 18*I*b*c*d^2 - 18*I*a*d^3 + (-18*I*b*c*d^2 + 18*I*a*d^3)*(b*x +
a)^2 + (-18*I*b^2*c^2*d + 36*I*a*b*c*d^2 + (-18*I*a^2 + 18*I)*d^3)*(b*x + a))*cos(4*b*x + 4*a) + (-16*I*(b*x
+ a)^3*d^3 + 36*I*b*c*d^2 - 36*I*a*d^3 + (-36*I*b*c*d^2 + 36*I*a*d^3)*(b*x + a)^2 + (-36*I*b^2*c^2*d + 72*I*a*
b*c*d^2 + (-36*I*a^2 + 36*I)*d^3)*(b*x + a))*cos(2*b*x + 2*a) + 2*(4*(b*x + a)^3*d^3 - 9*b*c*d^2 + 9*a*d^3 + 9
*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 9*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 1)*d^3)*(b*x + a))*sin(4*b*x + 4*a) + 4*(
4*(b*x + a)^3*d^3 - 9*b*c*d^2 + 9*a*d^3 + 9*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 9*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2
- 1)*d^3)*(b*x + 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) -
(12*d^3*cos(4*b*x + 4*a) + 24*d^3*cos(2*b*x + 2*a) + 12*I*d^3*sin(4*b*x + 4*a) + 24*I*d^3*sin(2*b*x + 2*a) +
12*d^3)*polylog(4, -e^(2*I*b*x + 2*I*a)) - (-18*I*b*c*d^2 - 24*I*(b*x + a)*d^3 + 18*I*a*d^3 + (-18*I*b*c*d^2 -
24*I*(b*x + a)*d^3 + 18*I*a*d^3)*cos(4*b*x + 4*a) + (-36*I*b*c*d^2 - 48*I*(b*x + a)*d^3 + 36*I*a*d^3)*cos(2*b
*x + 2*a) + 6*(3*b*c*d^2 + 4*(b*x + a)*d^3 - 3*a*d^3)*sin(4*b*x + 4*a) + 12*(3*b*c*d^2 + 4*(b*x + a)*d^3 - 3*a
*d^3)*sin(2*b*x + 2*a))*polylog(3, -e^(2*I*b*x + 2*I*a)) - (-3*I*(b*x + a)^4*d^3 + (-12*I*b*c*d^2 + 12*I*a*d^3
)*(b*x + a)^3 + (-18*I*b^2*c^2*d + 36*I*a*b*c*d^2 + (-18*I*a^2 + 36*I)*d^3)*(b*x + a)^2 + (72*I*b*c*d^2 - 72*I
*a*d^3)*(b*x + a))*sin(4*b*x + 4*a) - (-6*I*(b*x + a)^4*d^3 - 36*I*b^2*c^2*d + 72*I*a*b*c*d^2 - 36*I*a^2*d^3 +
(-24*I*b*c*d^2 - 24*(-I*a - 1)*d^3)*(b*x + a)^3 + (-36*I*b^2*c^2*d - 72*(-I*a - 1)*b*c*d^2 + (-36*I*a^2 - 72*
a + 36*I)*d^3)*(b*x + a)^2 + (72*b^2*c^2*d - (144*a - 72*I)*b*c*d^2 + 72*(a^2 - I*a)*d^3)*(b*x + a))*sin(2*b*x
+ 2*a))/(-12*I*b^3*cos(4*b*x + 4*a) - 24*I*b^3*cos(2*b*x + 2*a) + 12*b^3*sin(4*b*x + 4*a) + 24*b^3*sin(2*b*x
+ 2*a) - 12*I*b^3))/b
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Fricas [C] time = 0.537343, size = 1443, normalized size = 5.57 \begin{align*} \frac{4 \, b^{3} d^{3} x^{3} + 12 \, b^{3} c d^{2} x^{2} + 12 \, b^{3} c^{2} d x - 3 i \, d^{3}{\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 \, d^{3}{\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 ) + 4 \,{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \tan \left (b x + a\right )^{2} +{\left (6 i \, b^{2} d^{3} x^{2} + 12 i \, b^{2} c d^{2} x + 6 i \, b^{2} c^{2} d - 6 i \, d^{3}\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} d^{3} x^{2} - 12 i \, b^{2} c d^{2} x - 6 i \, b^{2} c^{2} d + 6 i \, d^{3}\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} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + b^{3} c^{3} - 3 \, b c d^{2} + 3 \,{\left (b^{3} c^{2} d - b d^{3}\right )} 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} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + b^{3} c^{3} - 3 \, b c d^{2} + 3 \,{\left (b^{3} c^{2} d - b d^{3}\right )} 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 ) + 6 \,{\left (b d^{3} x + b c d^{2}\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 ) + 6 \,{\left (b d^{3} x + b c d^{2}\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 ) - 12 \,{\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \tan \left (b x + a\right )}{8 \, b^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^3*tan(b*x+a)^3,x, algorithm="fricas")
[Out]
1/8*(4*b^3*d^3*x^3 + 12*b^3*c*d^2*x^2 + 12*b^3*c^2*d*x - 3*I*d^3*polylog(4, (tan(b*x + a)^2 + 2*I*tan(b*x + a)
- 1)/(tan(b*x + a)^2 + 1)) + 3*I*d^3*polylog(4, (tan(b*x + a)^2 - 2*I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1))
+ 4*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*tan(b*x + a)^2 + (6*I*b^2*d^3*x^2 + 12*I*b^2*c*
d^2*x + 6*I*b^2*c^2*d - 6*I*d^3)*dilog(2*(I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1) + 1) + (-6*I*b^2*d^3*x^2 -
12*I*b^2*c*d^2*x - 6*I*b^2*c^2*d + 6*I*d^3)*dilog(2*(-I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1) + 1) + 4*(b^3*d
^3*x^3 + 3*b^3*c*d^2*x^2 + b^3*c^3 - 3*b*c*d^2 + 3*(b^3*c^2*d - b*d^3)*x)*log(-2*(I*tan(b*x + a) - 1)/(tan(b*x
+ a)^2 + 1)) + 4*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + b^3*c^3 - 3*b*c*d^2 + 3*(b^3*c^2*d - b*d^3)*x)*log(-2*(-I*t
an(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) + 6*(b*d^3*x + b*c*d^2)*polylog(3, (tan(b*x + a)^2 + 2*I*tan(b*x + a) -
1)/(tan(b*x + a)^2 + 1)) + 6*(b*d^3*x + b*c*d^2)*polylog(3, (tan(b*x + a)^2 - 2*I*tan(b*x + a) - 1)/(tan(b*x
+ a)^2 + 1)) - 12*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)*tan(b*x + a))/b^4
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{3} \tan ^{3}{\left (a + b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)**3*tan(b*x+a)**3,x)
[Out]
Integral((c + d*x)**3*tan(a + b*x)**3, x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{3} \tan \left (b x + a\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^3*tan(b*x+a)^3,x, algorithm="giac")
[Out]
integrate((d*x + c)^3*tan(b*x + a)^3, x)