3.209 \(\int (c+d x)^4 \tan (a+b x) \, dx\)

Optimal. Leaf size=158 \[ \frac {3 d^4 \text {Li}_5\left (-e^{2 i (a+b x)}\right )}{2 b^5}-\frac {3 i d^3 (c+d x) \text {Li}_4\left (-e^{2 i (a+b x)}\right )}{b^4}-\frac {3 d^2 (c+d x)^2 \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{b^3}+\frac {2 i d (c+d x)^3 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {(c+d x)^4 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i (c+d x)^5}{5 d} \]

[Out]

1/5*I*(d*x+c)^5/d-(d*x+c)^4*ln(1+exp(2*I*(b*x+a)))/b+2*I*d*(d*x+c)^3*polylog(2,-exp(2*I*(b*x+a)))/b^2-3*d^2*(d
*x+c)^2*polylog(3,-exp(2*I*(b*x+a)))/b^3-3*I*d^3*(d*x+c)*polylog(4,-exp(2*I*(b*x+a)))/b^4+3/2*d^4*polylog(5,-e
xp(2*I*(b*x+a)))/b^5

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Rubi [A]  time = 0.21, antiderivative size = 158, normalized size of antiderivative = 1.00, 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 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 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 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 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 (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*}

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Mathematica [A]  time = 0.07, size = 157, normalized size = 0.99 \[ \frac {2 i d (c+d x)^3 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {3 d^2 \left (2 b^2 (c+d x)^2 \text {Li}_3\left (-e^{2 i (a+b x)}\right )+d \left (2 i b (c+d x) \text {Li}_4\left (-e^{2 i (a+b x)}\right )-d \text {Li}_5\left (-e^{2 i (a+b x)}\right )\right )\right )}{2 b^5}-\frac {(c+d x)^4 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i (c+d x)^5}{5 d} \]

Antiderivative was successfully verified.

[In]

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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fricas [C]  time = 0.57, size = 1402, normalized size = 8.87 \[ \text {result too large to display} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{4} \sec \left (b x + a\right ) \sin \left (b x + a\right )\,{d x} \]

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)

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maple [B]  time = 0.11, size = 616, normalized size = 3.90 \[ -\frac {3 c^{2} d^{2} \polylog \left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{3}}-\frac {3 d^{4} \polylog \left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{2}}{b^{3}}-\frac {d^{4} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{4}}{b}-i c^{4} x +\frac {2 c^{4} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {i d^{4} x^{5}}{5}+\frac {6 i c \,d^{3} \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{2}}{b^{2}}+\frac {6 i c^{2} d^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 d^{4} a^{4} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{5}}-\frac {8 i d^{4} a^{5}}{5 b^{5}}+i c \,d^{3} x^{4}+2 i c^{2} d^{2} x^{3}+2 i c^{3} d \,x^{2}-\frac {c^{4} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b}+\frac {3 d^{4} \polylog \left (5, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{5}}-\frac {4 c^{3} d \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b}-\frac {6 c^{2} d^{2} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {4 c \,d^{3} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{3}}{b}-\frac {6 c \,d^{3} \polylog \left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{3}}-\frac {3 i d^{4} \polylog \left (4, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{4}}+\frac {2 i d^{4} \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{3}}{b^{2}}-\frac {3 i c \,d^{3} \polylog \left (4, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{4}}+\frac {2 i c^{3} d \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 i d^{4} a^{4} x}{b^{4}}+\frac {4 i c^{3} d \,a^{2}}{b^{2}}-\frac {8 i c^{2} d^{2} a^{3}}{b^{3}}+\frac {6 i c \,d^{3} a^{4}}{b^{4}}-\frac {8 c^{3} d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {8 c \,d^{3} a^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {12 c^{2} d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {8 i c \,d^{3} a^{3} x}{b^{3}}-\frac {12 i c^{2} d^{2} a^{2} x}{b^{2}}+\frac {8 i c^{3} d a x}{b} \]

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+2/b^5*d^4*a^4*ln(exp(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+1/5*I*d^4*x^5+I*c*d^3*x^4+8*I/b*a*c^3*d*x-12*I/b^2*a^2*c^2*d
^2*x+8*I/b^3*c*d^3*a^3*x+6*I/b^4*c*d^3*a^4-8*I/b^3*a^3*c^2*d^2-2*I/b^4*d^4*a^4*x+4*I/b^2*a^2*c^3*d-4/b*c*d^3*l
n(1+exp(2*I*(b*x+a)))*x^3+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-I*c^4*x-1/b*c^4*ln(1+exp(2*I*(b*x+a)))+2/b*c^4*ln(exp(I*(b*x+a)))-1/b*d^4*ln(1+exp(2*I*(b*x+a)))*x^4
-8/b^2*c^3*d*a*ln(exp(I*(b*x+a)))-8/b^4*c*d^3*a^3*ln(exp(I*(b*x+a)))+12/b^3*c^2*d^2*a^2*ln(exp(I*(b*x+a)))+2*I
*c^2*d^2*x^3+2*I*c^3*d*x^2-4/b*c^3*d*ln(1+exp(2*I*(b*x+a)))*x-6/b^3*c*d^3*polylog(3,-exp(2*I*(b*x+a)))*x-6/b*c
^2*d^2*ln(1+exp(2*I*(b*x+a)))*x^2-8/5*I/b^5*d^4*a^5-3*I/b^4*d^4*polylog(4,-exp(2*I*(b*x+a)))*x+2*I/b^2*d^4*pol
ylog(2,-exp(2*I*(b*x+a)))*x^3-3*I/b^4*c*d^3*polylog(4,-exp(2*I*(b*x+a)))+2*I/b^2*c^3*d*polylog(2,-exp(2*I*(b*x
+a)))

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maxima [B]  time = 0.60, size = 792, normalized size = 5.01 \[ -\frac {15 \, c^{4} \log \left (-\sin \left (b x + a\right )^{2} + 1\right ) - \frac {60 \, a c^{3} d \log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{b} + \frac {90 \, a^{2} c^{2} d^{2} \log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{b^{2}} - \frac {60 \, a^{3} c d^{3} \log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{b^{3}} + \frac {15 \, a^{4} d^{4} \log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{b^{4}} + \frac {2 \, {\left (-3 i \, {\left (b x + a\right )}^{5} d^{4} + {\left (-15 i \, b c d^{3} + 15 i \, a d^{4}\right )} {\left (b x + a\right )}^{4} - 45 \, d^{4} {\rm Li}_{5}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + {\left (-30 i \, b^{2} c^{2} d^{2} + 60 i \, a b c d^{3} - 30 i \, a^{2} d^{4}\right )} {\left (b x + a\right )}^{3} + {\left (-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}\right )} {\left (b x + a\right )}^{2} + {\left (30 i \, {\left (b x + a\right )}^{4} d^{4} + {\left (80 i \, b c d^{3} - 80 i \, a d^{4}\right )} {\left (b x + a\right )}^{3} + {\left (90 i \, b^{2} c^{2} d^{2} - 180 i \, a b c d^{3} + 90 i \, a^{2} d^{4}\right )} {\left (b x + a\right )}^{2} + {\left (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}\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 (-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 \, {\left (b x + a\right )}^{3} d^{4} + 30 i \, a^{3} d^{4} + {\left (-120 i \, b c d^{3} + 120 i \, a d^{4}\right )} {\left (b x + a\right )}^{2} + {\left (-90 i \, b^{2} c^{2} d^{2} + 180 i \, a b c d^{3} - 90 i \, a^{2} d^{4}\right )} {\left (b x + a\right )}\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 5 \, {\left (3 \, {\left (b x + a\right )}^{4} d^{4} + 8 \, {\left (b c d^{3} - a d^{4}\right )} {\left (b x + a\right )}^{3} + 9 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} {\left (b x + a\right )}^{2} + 6 \, {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - a^{3} d^{4}\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 ) + {\left (60 i \, b c d^{3} + 90 i \, {\left (b x + a\right )} d^{4} - 60 i \, a d^{4}\right )} {\rm Li}_{4}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + 15 \, {\left (3 \, b^{2} c^{2} d^{2} - 6 \, a b c d^{3} + 6 \, {\left (b x + a\right )}^{2} d^{4} + 3 \, a^{2} d^{4} + 8 \, {\left (b c d^{3} - a d^{4}\right )} {\left (b x + a\right )}\right )} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )})\right )}}{b^{4}}}{30 \, b} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (a+b\,x\right )\,{\left (c+d\,x\right )}^4}{\cos \left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(a + b*x)*(c + d*x)^4)/cos(a + b*x),x)

[Out]

int((sin(a + b*x)*(c + d*x)^4)/cos(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{4} \sin {\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \]

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]

Integral((c + d*x)**4*sin(a + b*x)*sec(a + b*x), x)

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