3.209 \(\int (c+d x)^4 \tan (a+b x) \, dx\)
Optimal. Leaf size=158 \[ -\frac{3 d^2 (c+d x)^2 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{b^3}-\frac{3 i d^3 (c+d x) \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{b^4}+\frac{2 i d (c+d x)^3 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac{3 d^4 \text{PolyLog}\left (5,-e^{2 i (a+b x)}\right )}{2 b^5}-\frac{(c+d x)^4 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i (c+d x)^5}{5 d} \]
[Out]
((I/5)*(c + d*x)^5)/d - ((c + d*x)^4*Log[1 + E^((2*I)*(a + b*x))])/b + ((2*I)*d*(c + d*x)^3*PolyLog[2, -E^((2*
I)*(a + b*x))])/b^2 - (3*d^2*(c + d*x)^2*PolyLog[3, -E^((2*I)*(a + b*x))])/b^3 - ((3*I)*d^3*(c + d*x)*PolyLog[
4, -E^((2*I)*(a + b*x))])/b^4 + (3*d^4*PolyLog[5, -E^((2*I)*(a + b*x))])/(2*b^5)
________________________________________________________________________________________
Rubi [A] time = 0.209578, antiderivative size = 158, normalized size of antiderivative = 1.,
number of steps used = 7, 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} \[ -\frac{3 d^2 (c+d x)^2 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{b^3}-\frac{3 i d^3 (c+d x) \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{b^4}+\frac{2 i d (c+d x)^3 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac{3 d^4 \text{PolyLog}\left (5,-e^{2 i (a+b x)}\right )}{2 b^5}-\frac{(c+d x)^4 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i (c+d x)^5}{5 d} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)^4*Tan[a + b*x],x]
[Out]
((I/5)*(c + d*x)^5)/d - ((c + d*x)^4*Log[1 + E^((2*I)*(a + b*x))])/b + ((2*I)*d*(c + d*x)^3*PolyLog[2, -E^((2*
I)*(a + b*x))])/b^2 - (3*d^2*(c + d*x)^2*PolyLog[3, -E^((2*I)*(a + b*x))])/b^3 - ((3*I)*d^3*(c + d*x)*PolyLog[
4, -E^((2*I)*(a + b*x))])/b^4 + (3*d^4*PolyLog[5, -E^((2*I)*(a + b*x))])/(2*b^5)
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)^4 \tan (a+b x) \, dx &=\frac{i (c+d x)^5}{5 d}-2 i \int \frac{e^{2 i (a+b x)} (c+d x)^4}{1+e^{2 i (a+b x)}} \, dx\\ &=\frac{i (c+d x)^5}{5 d}-\frac{(c+d x)^4 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{(4 d) \int (c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac{i (c+d x)^5}{5 d}-\frac{(c+d x)^4 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{2 i d (c+d x)^3 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{\left (6 i d^2\right ) \int (c+d x)^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=\frac{i (c+d x)^5}{5 d}-\frac{(c+d x)^4 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{2 i d (c+d x)^3 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{3 d^2 (c+d x)^2 \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{b^3}+\frac{\left (6 d^3\right ) \int (c+d x) \text{Li}_3\left (-e^{2 i (a+b x)}\right ) \, dx}{b^3}\\ &=\frac{i (c+d x)^5}{5 d}-\frac{(c+d x)^4 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{2 i d (c+d x)^3 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{3 d^2 (c+d x)^2 \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{b^3}-\frac{3 i d^3 (c+d x) \text{Li}_4\left (-e^{2 i (a+b x)}\right )}{b^4}+\frac{\left (3 i d^4\right ) \int \text{Li}_4\left (-e^{2 i (a+b x)}\right ) \, dx}{b^4}\\ &=\frac{i (c+d x)^5}{5 d}-\frac{(c+d x)^4 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{2 i d (c+d x)^3 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{3 d^2 (c+d x)^2 \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{b^3}-\frac{3 i d^3 (c+d x) \text{Li}_4\left (-e^{2 i (a+b x)}\right )}{b^4}+\frac{\left (3 d^4\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_4(-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^5}\\ &=\frac{i (c+d x)^5}{5 d}-\frac{(c+d x)^4 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{2 i d (c+d x)^3 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{3 d^2 (c+d x)^2 \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{b^3}-\frac{3 i d^3 (c+d x) \text{Li}_4\left (-e^{2 i (a+b x)}\right )}{b^4}+\frac{3 d^4 \text{Li}_5\left (-e^{2 i (a+b x)}\right )}{2 b^5}\\ \end{align*}
Mathematica [A] time = 0.0687408, size = 157, normalized size = 0.99 \[ -\frac{3 d^2 \left (2 b^2 (c+d x)^2 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )+d \left (2 i b (c+d x) \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )-d \text{PolyLog}\left (5,-e^{2 i (a+b x)}\right )\right )\right )}{2 b^5}+\frac{2 i d (c+d x)^3 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac{(c+d x)^4 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i (c+d x)^5}{5 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x)^4*Tan[a + b*x],x]
[Out]
((I/5)*(c + d*x)^5)/d - ((c + d*x)^4*Log[1 + E^((2*I)*(a + b*x))])/b + ((2*I)*d*(c + d*x)^3*PolyLog[2, -E^((2*
I)*(a + b*x))])/b^2 - (3*d^2*(2*b^2*(c + d*x)^2*PolyLog[3, -E^((2*I)*(a + b*x))] + d*((2*I)*b*(c + d*x)*PolyLo
g[4, -E^((2*I)*(a + b*x))] - d*PolyLog[5, -E^((2*I)*(a + b*x))])))/(2*b^5)
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Maple [B] time = 0.315, size = 616, normalized size = 3.9 \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)^4*sec(b*x+a)*sin(b*x+a),x)
[Out]
3/2*d^4*polylog(5,-exp(2*I*(b*x+a)))/b^5+1/5*I*d^4*x^5+I*c*d^3*x^4-3*I/b^4*c*d^3*polylog(4,-exp(2*I*(b*x+a)))-
3/b^3*c^2*d^2*polylog(3,-exp(2*I*(b*x+a)))-3/b^3*d^4*polylog(3,-exp(2*I*(b*x+a)))*x^2-4/b*c*d^3*ln(exp(2*I*(b*
x+a))+1)*x^3-1/b*c^4*ln(exp(2*I*(b*x+a))+1)-I*c^4*x-6/b*c^2*d^2*ln(exp(2*I*(b*x+a))+1)*x^2-12*I/b^2*a^2*c^2*d^
2*x+8*I/b^3*c*d^3*a^3*x+6*I/b^2*c*d^3*polylog(2,-exp(2*I*(b*x+a)))*x^2+6*I/b^2*c^2*d^2*polylog(2,-exp(2*I*(b*x
+a)))*x-6/b^3*c*d^3*polylog(3,-exp(2*I*(b*x+a)))*x-1/b*d^4*ln(exp(2*I*(b*x+a))+1)*x^4-8/5*I/b^5*d^4*a^5+8*I/b*
a*c^3*d*x-4/b*c^3*d*ln(exp(2*I*(b*x+a))+1)*x-2*I/b^4*d^4*a^4*x-8*I/b^3*a^3*c^2*d^2+4*I/b^2*a^2*c^3*d+6*I/b^4*c
*d^3*a^4+2*I*c^2*d^2*x^3+2*I*c^3*d*x^2+2/b^5*d^4*a^4*ln(exp(I*(b*x+a)))-8/b^2*c^3*d*a*ln(exp(I*(b*x+a)))+12/b^
3*c^2*d^2*a^2*ln(exp(I*(b*x+a)))-8/b^4*c*d^3*a^3*ln(exp(I*(b*x+a)))+2/b*c^4*ln(exp(I*(b*x+a)))+2*I/b^2*c^3*d*p
olylog(2,-exp(2*I*(b*x+a)))-3*I/b^4*d^4*polylog(4,-exp(2*I*(b*x+a)))*x+2*I/b^2*d^4*polylog(2,-exp(2*I*(b*x+a))
)*x^3
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Maxima [B] time = 2.05808, size = 1069, normalized size = 6.77 \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)^4*sec(b*x+a)*sin(b*x+a),x, algorithm="maxima")
[Out]
-1/30*(15*c^4*log(-sin(b*x + a)^2 + 1) - 60*a*c^3*d*log(-sin(b*x + a)^2 + 1)/b + 90*a^2*c^2*d^2*log(-sin(b*x +
a)^2 + 1)/b^2 - 60*a^3*c*d^3*log(-sin(b*x + a)^2 + 1)/b^3 + 15*a^4*d^4*log(-sin(b*x + a)^2 + 1)/b^4 + 2*(-3*I
*(b*x + a)^5*d^4 + (-15*I*b*c*d^3 + 15*I*a*d^4)*(b*x + a)^4 - 45*d^4*polylog(5, -e^(2*I*b*x + 2*I*a)) + (-30*I
*b^2*c^2*d^2 + 60*I*a*b*c*d^3 - 30*I*a^2*d^4)*(b*x + a)^3 + (-30*I*b^3*c^3*d + 90*I*a*b^2*c^2*d^2 - 90*I*a^2*b
*c*d^3 + 30*I*a^3*d^4)*(b*x + a)^2 + (30*I*(b*x + a)^4*d^4 + (80*I*b*c*d^3 - 80*I*a*d^4)*(b*x + a)^3 + (90*I*b
^2*c^2*d^2 - 180*I*a*b*c*d^3 + 90*I*a^2*d^4)*(b*x + a)^2 + (60*I*b^3*c^3*d - 180*I*a*b^2*c^2*d^2 + 180*I*a^2*b
*c*d^3 - 60*I*a^3*d^4)*(b*x + a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + (-30*I*b^3*c^3*d + 90*I*a*
b^2*c^2*d^2 - 90*I*a^2*b*c*d^3 - 60*I*(b*x + a)^3*d^4 + 30*I*a^3*d^4 + (-120*I*b*c*d^3 + 120*I*a*d^4)*(b*x + a
)^2 + (-90*I*b^2*c^2*d^2 + 180*I*a*b*c*d^3 - 90*I*a^2*d^4)*(b*x + a))*dilog(-e^(2*I*b*x + 2*I*a)) + 5*(3*(b*x
+ a)^4*d^4 + 8*(b*c*d^3 - a*d^4)*(b*x + a)^3 + 9*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*(b*x + a)^2 + 6*(b^3*c^
3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - a^3*d^4)*(b*x + a))*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*co
s(2*b*x + 2*a) + 1) + (60*I*b*c*d^3 + 90*I*(b*x + a)*d^4 - 60*I*a*d^4)*polylog(4, -e^(2*I*b*x + 2*I*a)) + 15*(
3*b^2*c^2*d^2 - 6*a*b*c*d^3 + 6*(b*x + a)^2*d^4 + 3*a^2*d^4 + 8*(b*c*d^3 - a*d^4)*(b*x + a))*polylog(3, -e^(2*
I*b*x + 2*I*a)))/b^4)/b
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Fricas [C] time = 0.741923, size = 3355, normalized size = 21.23 \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)^4*sec(b*x+a)*sin(b*x+a),x, algorithm="fricas")
[Out]
1/2*(24*d^4*polylog(5, I*cos(b*x + a) + sin(b*x + a)) + 24*d^4*polylog(5, I*cos(b*x + a) - sin(b*x + a)) + 24*
d^4*polylog(5, -I*cos(b*x + a) + sin(b*x + a)) + 24*d^4*polylog(5, -I*cos(b*x + a) - sin(b*x + a)) + (-4*I*b^3
*d^4*x^3 - 12*I*b^3*c*d^3*x^2 - 12*I*b^3*c^2*d^2*x - 4*I*b^3*c^3*d)*dilog(I*cos(b*x + a) + sin(b*x + a)) + (4*
I*b^3*d^4*x^3 + 12*I*b^3*c*d^3*x^2 + 12*I*b^3*c^2*d^2*x + 4*I*b^3*c^3*d)*dilog(I*cos(b*x + a) - sin(b*x + a))
+ (4*I*b^3*d^4*x^3 + 12*I*b^3*c*d^3*x^2 + 12*I*b^3*c^2*d^2*x + 4*I*b^3*c^3*d)*dilog(-I*cos(b*x + a) + sin(b*x
+ a)) + (-4*I*b^3*d^4*x^3 - 12*I*b^3*c*d^3*x^2 - 12*I*b^3*c^2*d^2*x - 4*I*b^3*c^3*d)*dilog(-I*cos(b*x + a) - s
in(b*x + a)) - (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*log(cos(b*x + a) + I*si
n(b*x + a) + I) - (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*log(cos(b*x + a) - I
*sin(b*x + a) + I) - (b^4*d^4*x^4 + 4*b^4*c*d^3*x^3 + 6*b^4*c^2*d^2*x^2 + 4*b^4*c^3*d*x + 4*a*b^3*c^3*d - 6*a^
2*b^2*c^2*d^2 + 4*a^3*b*c*d^3 - a^4*d^4)*log(I*cos(b*x + a) + sin(b*x + a) + 1) - (b^4*d^4*x^4 + 4*b^4*c*d^3*x
^3 + 6*b^4*c^2*d^2*x^2 + 4*b^4*c^3*d*x + 4*a*b^3*c^3*d - 6*a^2*b^2*c^2*d^2 + 4*a^3*b*c*d^3 - a^4*d^4)*log(I*co
s(b*x + a) - sin(b*x + a) + 1) - (b^4*d^4*x^4 + 4*b^4*c*d^3*x^3 + 6*b^4*c^2*d^2*x^2 + 4*b^4*c^3*d*x + 4*a*b^3*
c^3*d - 6*a^2*b^2*c^2*d^2 + 4*a^3*b*c*d^3 - a^4*d^4)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) - (b^4*d^4*x^4 +
4*b^4*c*d^3*x^3 + 6*b^4*c^2*d^2*x^2 + 4*b^4*c^3*d*x + 4*a*b^3*c^3*d - 6*a^2*b^2*c^2*d^2 + 4*a^3*b*c*d^3 - a^4*
d^4)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) - (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 +
a^4*d^4)*log(-cos(b*x + a) + I*sin(b*x + a) + I) - (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^
3 + a^4*d^4)*log(-cos(b*x + a) - I*sin(b*x + a) + I) + (24*I*b*d^4*x + 24*I*b*c*d^3)*polylog(4, I*cos(b*x + a)
+ sin(b*x + a)) + (-24*I*b*d^4*x - 24*I*b*c*d^3)*polylog(4, I*cos(b*x + a) - sin(b*x + a)) + (-24*I*b*d^4*x -
24*I*b*c*d^3)*polylog(4, -I*cos(b*x + a) + sin(b*x + a)) + (24*I*b*d^4*x + 24*I*b*c*d^3)*polylog(4, -I*cos(b*
x + a) - sin(b*x + a)) - 12*(b^2*d^4*x^2 + 2*b^2*c*d^3*x + b^2*c^2*d^2)*polylog(3, I*cos(b*x + a) + sin(b*x +
a)) - 12*(b^2*d^4*x^2 + 2*b^2*c*d^3*x + b^2*c^2*d^2)*polylog(3, I*cos(b*x + a) - sin(b*x + a)) - 12*(b^2*d^4*x
^2 + 2*b^2*c*d^3*x + b^2*c^2*d^2)*polylog(3, -I*cos(b*x + a) + sin(b*x + a)) - 12*(b^2*d^4*x^2 + 2*b^2*c*d^3*x
+ b^2*c^2*d^2)*polylog(3, -I*cos(b*x + a) - sin(b*x + a)))/b^5
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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)**4*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 )}^{4} \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)^4*sec(b*x+a)*sin(b*x+a),x, algorithm="giac")
[Out]
integrate((d*x + c)^4*sec(b*x + a)*sin(b*x + a), x)