3.54 $$\int \frac{(c+d x)^3}{a+b \tan (e+f x)} \, dx$$

Optimal. Leaf size=243 $\frac{3 b d^2 (c+d x) \text{PolyLog}\left (3,-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f^3 \left (a^2+b^2\right )}-\frac{3 i b d (c+d x)^2 \text{PolyLog}\left (2,-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f^2 \left (a^2+b^2\right )}+\frac{3 i b d^3 \text{PolyLog}\left (4,-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{4 f^4 \left (a^2+b^2\right )}+\frac{b (c+d x)^3 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{f \left (a^2+b^2\right )}+\frac{(c+d x)^4}{4 d (a+i b)}$

[Out]

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

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Rubi [A]  time = 0.333558, antiderivative size = 243, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.3, Rules used = {3732, 2190, 2531, 6609, 2282, 6589} $\frac{3 b d^2 (c+d x) \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f^3 \left (a^2+b^2\right )}-\frac{3 i b d (c+d x)^2 \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f^2 \left (a^2+b^2\right )}+\frac{b (c+d x)^3 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{f \left (a^2+b^2\right )}+\frac{3 i b d^3 \text{Li}_4\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{4 f^4 \left (a^2+b^2\right )}+\frac{(c+d x)^4}{4 d (a+i b)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(c + d*x)^3/(a + b*Tan[e + f*x]),x]

[Out]

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

Rule 3732

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(d*
(m + 1)*(a + I*b)), x] + Dist[2*I*b, Int[((c + d*x)^m*E^Simp[2*I*(e + f*x), x])/((a + I*b)^2 + (a^2 + b^2)*E^S
imp[2*I*(e + f*x), x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2, 0] && 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 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 \frac{(c+d x)^3}{a+b \tan (e+f x)} \, dx &=\frac{(c+d x)^4}{4 (a+i b) d}+(2 i b) \int \frac{e^{2 i (e+f x)} (c+d x)^3}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (e+f x)}} \, dx\\ &=\frac{(c+d x)^4}{4 (a+i b) d}+\frac{b (c+d x)^3 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}-\frac{(3 b d) \int (c+d x)^2 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right ) \, dx}{\left (a^2+b^2\right ) f}\\ &=\frac{(c+d x)^4}{4 (a+i b) d}+\frac{b (c+d x)^3 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}-\frac{3 i b d (c+d x)^2 \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) f^2}+\frac{\left (3 i b d^2\right ) \int (c+d x) \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right ) \, dx}{\left (a^2+b^2\right ) f^2}\\ &=\frac{(c+d x)^4}{4 (a+i b) d}+\frac{b (c+d x)^3 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}-\frac{3 i b d (c+d x)^2 \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) f^2}+\frac{3 b d^2 (c+d x) \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) f^3}-\frac{\left (3 b d^3\right ) \int \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right ) \, dx}{2 \left (a^2+b^2\right ) f^3}\\ &=\frac{(c+d x)^4}{4 (a+i b) d}+\frac{b (c+d x)^3 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}-\frac{3 i b d (c+d x)^2 \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) f^2}+\frac{3 b d^2 (c+d x) \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) f^3}+\frac{\left (3 i b d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (-\frac{\left (a^2+b^2\right ) x}{(a+i b)^2}\right )}{x} \, dx,x,e^{2 i (e+f x)}\right )}{4 \left (a^2+b^2\right ) f^4}\\ &=\frac{(c+d x)^4}{4 (a+i b) d}+\frac{b (c+d x)^3 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}-\frac{3 i b d (c+d x)^2 \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) f^2}+\frac{3 b d^2 (c+d x) \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) f^3}+\frac{3 i b d^3 \text{Li}_4\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{4 \left (a^2+b^2\right ) f^4}\\ \end{align*}

Mathematica [A]  time = 1.94944, size = 297, normalized size = 1.22 $\frac{x \cos (e) \left (6 c^2 d x+4 c^3+4 c d^2 x^2+d^3 x^3\right )}{4 (a \cos (e)+b \sin (e))}+\frac{1}{4} b \left (\frac{3 d \left (2 i f^2 (c+d x)^2 \text{PolyLog}\left (2,\frac{(-a-i b) e^{-2 i (e+f x)}}{a-i b}\right )+d \left (2 f (c+d x) \text{PolyLog}\left (3,\frac{(-a-i b) e^{-2 i (e+f x)}}{a-i b}\right )-i d \text{PolyLog}\left (4,\frac{(-a-i b) e^{-2 i (e+f x)}}{a-i b}\right )\right )\right )}{f^4 \left (a^2+b^2\right )}+\frac{4 (c+d x)^3 \log \left (1+\frac{(a+i b) e^{-2 i (e+f x)}}{a-i b}\right )}{f \left (a^2+b^2\right )}-\frac{2 (c+d x)^4}{d (b+i a) \left (a \left (1+e^{2 i e}\right )-i b \left (-1+e^{2 i e}\right )\right )}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(c + d*x)^3/(a + b*Tan[e + f*x]),x]

[Out]

(b*((-2*(c + d*x)^4)/((I*a + b)*d*((-I)*b*(-1 + E^((2*I)*e)) + a*(1 + E^((2*I)*e)))) + (4*(c + d*x)^3*Log[1 +
(a + I*b)/((a - I*b)*E^((2*I)*(e + f*x)))])/((a^2 + b^2)*f) + (3*d*((2*I)*f^2*(c + d*x)^2*PolyLog[2, (-a - I*b
)/((a - I*b)*E^((2*I)*(e + f*x)))] + d*(2*f*(c + d*x)*PolyLog[3, (-a - I*b)/((a - I*b)*E^((2*I)*(e + f*x)))] -
I*d*PolyLog[4, (-a - I*b)/((a - I*b)*E^((2*I)*(e + f*x)))])))/((a^2 + b^2)*f^4)))/4 + (x*(4*c^3 + 6*c^2*d*x +
4*c*d^2*x^2 + d^3*x^3)*Cos[e])/(4*(a*Cos[e] + b*Sin[e]))

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Maple [B]  time = 0.284, size = 1450, normalized size = 6. \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/(a+b*tan(f*x+e)),x)

[Out]

-1/2*b/(I*a+b)*d^3/(-I*b-a)*x^4-3*I*b/(I*a+b)/f*c^2*d/(-I*b-a)*ln(1-(a-I*b)*exp(2*I*(f*x+e))/(-I*b-a))*x-3*I*b
/(I*a+b)/f^2*c^2*d/(-I*b-a)*ln(1-(a-I*b)*exp(2*I*(f*x+e))/(-I*b-a))*e+3*I*b/(I*a+b)/f^3*c*d^2*e^2/(a+I*b)*ln(I
*exp(2*I*(f*x+e))*b-a*exp(2*I*(f*x+e))-I*b-a)-3*I*b/(I*a+b)/f^2*c^2*d*e/(a+I*b)*ln(I*exp(2*I*(f*x+e))*b-a*exp(
2*I*(f*x+e))-I*b-a)+6*I*b/(I*a+b)/f^2*c^2*d*e/(a+I*b)*ln(exp(I*(f*x+e)))-6*I*b/(I*a+b)/f^3*c*d^2*e^2/(a+I*b)*l
n(exp(I*(f*x+e)))+3*I*b/(I*a+b)/f^3*c*d^2*e^2/(-I*b-a)*ln(1-(a-I*b)*exp(2*I*(f*x+e))/(-I*b-a))-3*I*b/(I*a+b)/f
*c*d^2/(-I*b-a)*ln(1-(a-I*b)*exp(2*I*(f*x+e))/(-I*b-a))*x^2-3*b/(I*a+b)/f^2*c*d^2/(-I*b-a)*polylog(2,(a-I*b)*e
xp(2*I*(f*x+e))/(-I*b-a))*x+6*b/(I*a+b)/f^2*c*d^2*e^2/(-I*b-a)*x-I*b/(I*a+b)/f^4*d^3*e^3/(-I*b-a)*ln(1-(a-I*b)
*exp(2*I*(f*x+e))/(-I*b-a))-6*b/(I*a+b)/f*c^2*d/(-I*b-a)*e*x-3/2*I*b/(I*a+b)/f^3*d^3/(-I*b-a)*polylog(3,(a-I*b
)*exp(2*I*(f*x+e))/(-I*b-a))*x+2*I*b/(I*a+b)/f^4*d^3*e^3/(a+I*b)*ln(exp(I*(f*x+e)))-I*b/(I*a+b)/f^4*d^3*e^3/(a
+I*b)*ln(I*exp(2*I*(f*x+e))*b-a*exp(2*I*(f*x+e))-I*b-a)-3/2*I*b/(I*a+b)/f^3*c*d^2/(-I*b-a)*polylog(3,(a-I*b)*e
xp(2*I*(f*x+e))/(-I*b-a))-I*b/(I*a+b)/f*d^3/(-I*b-a)*ln(1-(a-I*b)*exp(2*I*(f*x+e))/(-I*b-a))*x^3-2*b/(I*a+b)*c
*d^2/(-I*b-a)*x^3-3*b/(I*a+b)*c^2*d/(-I*b-a)*x^2-3/2*b/(I*a+b)/f^4*d^3/(-I*b-a)*e^4+3/4*b/(I*a+b)/f^4*d^3/(-I*
b-a)*polylog(4,(a-I*b)*exp(2*I*(f*x+e))/(-I*b-a))-1/4/(I*b-a)*d^3*x^4-1/(I*b-a)*c^3*x-1/(I*b-a)*c*d^2*x^3-3/2/
(I*b-a)*c^2*d*x^2-3*b/(I*a+b)/f^2*c^2*d/(-I*b-a)*e^2+4*b/(I*a+b)/f^3*c*d^2*e^3/(-I*b-a)-2*b/(I*a+b)/f^3*d^3/(-
I*b-a)*e^3*x-3/2*b/(I*a+b)/f^2*c^2*d/(-I*b-a)*polylog(2,(a-I*b)*exp(2*I*(f*x+e))/(-I*b-a))-3/2*b/(I*a+b)/f^2*d
^3/(-I*b-a)*polylog(2,(a-I*b)*exp(2*I*(f*x+e))/(-I*b-a))*x^2+I*b/(I*a+b)/f*c^3/(a+I*b)*ln(I*exp(2*I*(f*x+e))*b
-a*exp(2*I*(f*x+e))-I*b-a)-2*I*b/(I*a+b)/f*c^3/(a+I*b)*ln(exp(I*(f*x+e)))

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Maxima [B]  time = 2.83507, size = 1319, normalized size = 5.43 \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/(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 1.96225, size = 2782, normalized size = 11.45 \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/(a+b*tan(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d x\right )^{3}}{a + b \tan{\left (e + f x \right )}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3/(a+b*tan(f*x+e)),x)

[Out]

Integral((c + d*x)**3/(a + b*tan(e + f*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{3}}{b \tan \left (f x + e\right ) + a}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3/(a+b*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*x + c)^3/(b*tan(f*x + e) + a), x)