3.210 \(\int (c+d x)^3 \tan (a+b x) \, dx\)

Optimal. Leaf size=132 \[ -\frac{3 d^2 (c+d x) \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{3 i d (c+d x)^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 i d^3 \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i (c+d x)^4}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.182795, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3719, 2190, 2531, 6609, 2282, 6589} \[ -\frac{3 d^2 (c+d x) \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{3 i d (c+d x)^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 i d^3 \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i (c+d x)^4}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Tan[a + b*x],x]

[Out]

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

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

Mathematica [A]  time = 0.0874103, size = 126, normalized size = 0.95 \[ \frac{1}{4} i \left (\frac{3 d \left (2 b^2 (c+d x)^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )+d \left (2 i b (c+d x) \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )-d \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )\right )\right )}{b^4}+\frac{4 i (c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{(c+d x)^4}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^3*Tan[a + b*x],x]

[Out]

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

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Maple [B]  time = 0.289, size = 423, normalized size = 3.2 \begin{align*} -{\frac{{c}^{3}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{b}}-{\frac{3\,{d}^{2}c{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{3}}}-{\frac{3\,{d}^{3}{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) x}{2\,{b}^{3}}}+{\frac{{\frac{3\,i}{2}}{d}^{3}{a}^{4}}{{b}^{4}}}+{\frac{i}{4}}{d}^{3}{x}^{4}+ic{d}^{2}{x}^{3}-i{c}^{3}x+{\frac{3\,i}{2}}{c}^{2}d{x}^{2}+{\frac{{\frac{3\,i}{2}}{c}^{2}d{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-3\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{b}}-3\,{\frac{{d}^{2}c\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ){x}^{2}}{b}}-{\frac{{d}^{3}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ){x}^{3}}{b}}+{\frac{{\frac{3\,i}{2}}{d}^{3}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ){x}^{2}}{{b}^{2}}}+6\,{\frac{{d}^{2}c{a}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-6\,{\frac{{c}^{2}da\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{4\,ic{d}^{2}{a}^{3}}{{b}^{3}}}+{\frac{2\,i{d}^{3}{a}^{3}x}{{b}^{3}}}+{\frac{3\,i{a}^{2}{c}^{2}d}{{b}^{2}}}-{\frac{{\frac{3\,i}{4}}{d}^{3}{\it polylog} \left ( 4,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}+{\frac{3\,ic{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}-2\,{\frac{{a}^{3}{d}^{3}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}+{\frac{6\,ia{c}^{2}dx}{b}}-{\frac{6\,ic{d}^{2}{a}^{2}x}{{b}^{2}}}+2\,{\frac{{c}^{3}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*sec(b*x+a)*sin(b*x+a),x)

[Out]

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

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Maxima [B]  time = 1.857, size = 662, normalized size = 5.02 \begin{align*} -\frac{6 \, c^{3} \log \left (-\sin \left (b x + a\right )^{2} + 1\right ) - \frac{18 \, a c^{2} d \log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{b} + \frac{18 \, a^{2} c d^{2} \log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{b^{2}} - \frac{6 \, a^{3} d^{3} \log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{b^{3}} + \frac{-3 i \,{\left (b x + a\right )}^{4} d^{3} +{\left (-12 i \, b c d^{2} + 12 i \, a d^{3}\right )}{\left (b x + a\right )}^{3} + 12 i \, d^{3}{\rm Li}_{4}(-e^{\left (2 i \, b x + 2 i \, a\right )}) +{\left (-18 i \, b^{2} c^{2} d + 36 i \, a b c d^{2} - 18 i \, a^{2} d^{3}\right )}{\left (b x + a\right )}^{2} +{\left (16 i \,{\left (b x + a\right )}^{3} d^{3} +{\left (36 i \, b c d^{2} - 36 i \, a d^{3}\right )}{\left (b x + a\right )}^{2} +{\left (36 i \, b^{2} c^{2} d - 72 i \, a b c d^{2} + 36 i \, a^{2} d^{3}\right )}{\left (b x + a\right )}\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) +{\left (-18 i \, b^{2} c^{2} d + 36 i \, a b c d^{2} - 24 i \,{\left (b x + a\right )}^{2} d^{3} - 18 i \, a^{2} d^{3} +{\left (-36 i \, b c d^{2} + 36 i \, a d^{3}\right )}{\left (b x + a\right )}\right )}{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 2 \,{\left (4 \,{\left (b x + a\right )}^{3} d^{3} + 9 \,{\left (b c d^{2} - a d^{3}\right )}{\left (b x + a\right )}^{2} + 9 \,{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )}{\left (b x + a\right )}\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 6 \,{\left (3 \, b c d^{2} + 4 \,{\left (b x + a\right )} d^{3} - 3 \, a d^{3}\right )}{\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )})}{b^{3}}}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*sec(b*x+a)*sin(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 0.664148, size = 2367, normalized size = 17.93 \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*sec(b*x+a)*sin(b*x+a),x, algorithm="fricas")

[Out]

1/2*(6*I*d^3*polylog(4, I*cos(b*x + a) + sin(b*x + a)) - 6*I*d^3*polylog(4, I*cos(b*x + a) - sin(b*x + a)) - 6
*I*d^3*polylog(4, -I*cos(b*x + a) + sin(b*x + a)) + 6*I*d^3*polylog(4, -I*cos(b*x + a) - sin(b*x + a)) + (-3*I
*b^2*d^3*x^2 - 6*I*b^2*c*d^2*x - 3*I*b^2*c^2*d)*dilog(I*cos(b*x + a) + sin(b*x + a)) + (3*I*b^2*d^3*x^2 + 6*I*
b^2*c*d^2*x + 3*I*b^2*c^2*d)*dilog(I*cos(b*x + a) - sin(b*x + a)) + (3*I*b^2*d^3*x^2 + 6*I*b^2*c*d^2*x + 3*I*b
^2*c^2*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) + (-3*I*b^2*d^3*x^2 - 6*I*b^2*c*d^2*x - 3*I*b^2*c^2*d)*dilog(-
I*cos(b*x + a) - sin(b*x + a)) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(cos(b*x + a) + I*sin(
b*x + a) + I) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(cos(b*x + a) - I*sin(b*x + a) + I) - (
b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(I*cos(b*x + a) +
sin(b*x + a) + 1) - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*
log(I*cos(b*x + a) - sin(b*x + a) + 1) - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^
2*b*c*d^2 + a^3*d^3)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x
+ 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) - (b^3*c^3 - 3*a*b^2*c^2*d
+ 3*a^2*b*c*d^2 - a^3*d^3)*log(-cos(b*x + a) + I*sin(b*x + a) + I) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2
- a^3*d^3)*log(-cos(b*x + a) - I*sin(b*x + a) + I) - 6*(b*d^3*x + b*c*d^2)*polylog(3, I*cos(b*x + a) + sin(b*x
+ a)) - 6*(b*d^3*x + b*c*d^2)*polylog(3, I*cos(b*x + a) - sin(b*x + a)) - 6*(b*d^3*x + b*c*d^2)*polylog(3, -I
*cos(b*x + a) + sin(b*x + a)) - 6*(b*d^3*x + b*c*d^2)*polylog(3, -I*cos(b*x + a) - sin(b*x + a)))/b^4

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*sec(b*x+a)*sin(b*x+a),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*sec(b*x+a)*sin(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^3*sec(b*x + a)*sin(b*x + a), x)