3.29 \(\int (c+d x)^3 \sec (a+b x) \, dx\)

Optimal. Leaf size=205 \[ -\frac {6 i d^3 \text {Li}_4\left (-i e^{i (a+b x)}\right )}{b^4}+\frac {6 i d^3 \text {Li}_4\left (i e^{i (a+b x)}\right )}{b^4}-\frac {6 d^2 (c+d x) \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^3}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

-2*I*(d*x+c)^3*arctan(exp(I*(b*x+a)))/b+3*I*d*(d*x+c)^2*polylog(2,-I*exp(I*(b*x+a)))/b^2-3*I*d*(d*x+c)^2*polyl
og(2,I*exp(I*(b*x+a)))/b^2-6*d^2*(d*x+c)*polylog(3,-I*exp(I*(b*x+a)))/b^3+6*d^2*(d*x+c)*polylog(3,I*exp(I*(b*x
+a)))/b^3-6*I*d^3*polylog(4,-I*exp(I*(b*x+a)))/b^4+6*I*d^3*polylog(4,I*exp(I*(b*x+a)))/b^4

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4181, 2531, 6609, 2282, 6589} \[ -\frac {6 d^2 (c+d x) \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^3}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {6 i d^3 \text {Li}_4\left (-i e^{i (a+b x)}\right )}{b^4}+\frac {6 i d^3 \text {Li}_4\left (i e^{i (a+b x)}\right )}{b^4}-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Sec[a + b*x],x]

[Out]

((-2*I)*(c + d*x)^3*ArcTan[E^(I*(a + b*x))])/b + ((3*I)*d*(c + d*x)^2*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^2 -
((3*I)*d*(c + d*x)^2*PolyLog[2, I*E^(I*(a + b*x))])/b^2 - (6*d^2*(c + d*x)*PolyLog[3, (-I)*E^(I*(a + b*x))])/b
^3 + (6*d^2*(c + d*x)*PolyLog[3, I*E^(I*(a + b*x))])/b^3 - ((6*I)*d^3*PolyLog[4, (-I)*E^(I*(a + b*x))])/b^4 +
((6*I)*d^3*PolyLog[4, I*E^(I*(a + b*x))])/b^4

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 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && 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)^3 \sec (a+b x) \, dx &=-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {(3 d) \int (c+d x)^2 \log \left (1-i e^{i (a+b x)}\right ) \, dx}{b}+\frac {(3 d) \int (c+d x)^2 \log \left (1+i e^{i (a+b x)}\right ) \, dx}{b}\\ &=-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {\left (6 i d^2\right ) \int (c+d x) \text {Li}_2\left (-i e^{i (a+b x)}\right ) \, dx}{b^2}+\frac {\left (6 i d^2\right ) \int (c+d x) \text {Li}_2\left (i e^{i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {6 d^2 (c+d x) \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^3}+\frac {\left (6 d^3\right ) \int \text {Li}_3\left (-i e^{i (a+b x)}\right ) \, dx}{b^3}-\frac {\left (6 d^3\right ) \int \text {Li}_3\left (i e^{i (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {6 d^2 (c+d x) \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^3}-\frac {\left (6 i d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}+\frac {\left (6 i d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}\\ &=-\frac {2 i (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac {6 d^2 (c+d x) \text {Li}_3\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \text {Li}_3\left (i e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^3 \text {Li}_4\left (-i e^{i (a+b x)}\right )}{b^4}+\frac {6 i d^3 \text {Li}_4\left (i e^{i (a+b x)}\right )}{b^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.19, size = 196, normalized size = 0.96 \[ -\frac {i \left (2 b^3 (c+d x)^3 \tan ^{-1}\left (e^{i (a+b x)}\right )-3 d \left (b^2 (c+d x)^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )+2 i b d (c+d x) \text {Li}_3\left (-i e^{i (a+b x)}\right )-2 d^2 \text {Li}_4\left (-i e^{i (a+b x)}\right )\right )+3 d \left (b^2 (c+d x)^2 \text {Li}_2\left (i e^{i (a+b x)}\right )+2 i b d (c+d x) \text {Li}_3\left (i e^{i (a+b x)}\right )-2 d^2 \text {Li}_4\left (i e^{i (a+b x)}\right )\right )\right )}{b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^3*Sec[a + b*x],x]

[Out]

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

________________________________________________________________________________________

fricas [C]  time = 1.03, size = 966, normalized size = 4.71 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*sec(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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{3} \sec \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.22, size = 685, normalized size = 3.34 \[ -\frac {6 i c \,d^{2} a^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {6 i c^{2} d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {6 i d^{2} c \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {6 i d^{2} c \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {a^{3} d^{3} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {a^{3} d^{3} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {d^{3} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{3}}{b}+\frac {d^{3} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{3}}{b}-\frac {2 i c^{3} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {6 i d^{3} \polylog \left (4, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {6 i d^{3} \polylog \left (4, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {6 d^{2} c \polylog \left (3, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {6 d^{3} \polylog \left (3, i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}-\frac {6 d^{2} c \polylog \left (3, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 d^{3} \polylog \left (3, -i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {3 i d^{3} \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {3 i d^{3} \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {3 i c^{2} d \polylog \left (2, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {3 i c^{2} d \polylog \left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {3 d^{2} c \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {3 d^{2} c \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}+\frac {3 c^{2} d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {3 c^{2} d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {3 a^{2} c \,d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {3 a^{2} c \,d^{2} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {3 c^{2} d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {3 c^{2} d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {2 i d^{3} a^{3} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 1.14, size = 712, normalized size = 3.47 \[ \frac {2 \, c^{3} \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right ) - \frac {6 \, a c^{2} d \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right )}{b} + \frac {6 \, a^{2} c d^{2} \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right )}{b^{2}} - \frac {2 \, a^{3} d^{3} \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right )}{b^{3}} + \frac {12 i \, d^{3} {\rm Li}_{4}(i \, e^{\left (i \, b x + i \, a\right )}) - 12 i \, d^{3} {\rm Li}_{4}(-i \, e^{\left (i \, b x + i \, a\right )}) + {\left (-2 i \, {\left (b x + a\right )}^{3} d^{3} + {\left (-6 i \, b c d^{2} + 6 i \, a d^{3}\right )} {\left (b x + a\right )}^{2} + {\left (-6 i \, b^{2} c^{2} d + 12 i \, a b c d^{2} - 6 i \, a^{2} d^{3}\right )} {\left (b x + a\right )}\right )} \arctan \left (\cos \left (b x + a\right ), \sin \left (b x + a\right ) + 1\right ) + {\left (-2 i \, {\left (b x + a\right )}^{3} d^{3} + {\left (-6 i \, b c d^{2} + 6 i \, a d^{3}\right )} {\left (b x + a\right )}^{2} + {\left (-6 i \, b^{2} c^{2} d + 12 i \, a b c d^{2} - 6 i \, a^{2} d^{3}\right )} {\left (b x + a\right )}\right )} \arctan \left (\cos \left (b x + a\right ), -\sin \left (b x + a\right ) + 1\right ) + {\left (-6 i \, b^{2} c^{2} d + 12 i \, a b c d^{2} - 6 i \, {\left (b x + a\right )}^{2} d^{3} - 6 i \, a^{2} d^{3} + {\left (-12 i \, b c d^{2} + 12 i \, a d^{3}\right )} {\left (b x + a\right )}\right )} {\rm Li}_2\left (i \, e^{\left (i \, b x + i \, a\right )}\right ) + {\left (6 i \, b^{2} c^{2} d - 12 i \, a b c d^{2} + 6 i \, {\left (b x + a\right )}^{2} d^{3} + 6 i \, a^{2} d^{3} + {\left (12 i \, b c d^{2} - 12 i \, a d^{3}\right )} {\left (b x + a\right )}\right )} {\rm Li}_2\left (-i \, e^{\left (i \, b x + i \, a\right )}\right ) + {\left ({\left (b x + a\right )}^{3} d^{3} + 3 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}^{2} + 3 \, {\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 (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \sin \left (b x + a\right ) + 1\right ) - {\left ({\left (b x + a\right )}^{3} d^{3} + 3 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}^{2} + 3 \, {\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 (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \sin \left (b x + a\right ) + 1\right ) + 12 \, {\left (b c d^{2} + {\left (b x + a\right )} d^{3} - a d^{3}\right )} {\rm Li}_{3}(i \, e^{\left (i \, b x + i \, a\right )}) - 12 \, {\left (b c d^{2} + {\left (b x + a\right )} d^{3} - a d^{3}\right )} {\rm Li}_{3}(-i \, e^{\left (i \, b x + i \, a\right )})}{b^{3}}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*c^3*log(sec(b*x + a) + tan(b*x + a)) - 6*a*c^2*d*log(sec(b*x + a) + tan(b*x + a))/b + 6*a^2*c*d^2*log(s
ec(b*x + a) + tan(b*x + a))/b^2 - 2*a^3*d^3*log(sec(b*x + a) + tan(b*x + a))/b^3 + (12*I*d^3*polylog(4, I*e^(I
*b*x + I*a)) - 12*I*d^3*polylog(4, -I*e^(I*b*x + I*a)) + (-2*I*(b*x + a)^3*d^3 + (-6*I*b*c*d^2 + 6*I*a*d^3)*(b
*x + a)^2 + (-6*I*b^2*c^2*d + 12*I*a*b*c*d^2 - 6*I*a^2*d^3)*(b*x + a))*arctan2(cos(b*x + a), sin(b*x + a) + 1)
+ (-2*I*(b*x + a)^3*d^3 + (-6*I*b*c*d^2 + 6*I*a*d^3)*(b*x + a)^2 + (-6*I*b^2*c^2*d + 12*I*a*b*c*d^2 - 6*I*a^2
*d^3)*(b*x + a))*arctan2(cos(b*x + a), -sin(b*x + a) + 1) + (-6*I*b^2*c^2*d + 12*I*a*b*c*d^2 - 6*I*(b*x + a)^2
*d^3 - 6*I*a^2*d^3 + (-12*I*b*c*d^2 + 12*I*a*d^3)*(b*x + a))*dilog(I*e^(I*b*x + I*a)) + (6*I*b^2*c^2*d - 12*I*
a*b*c*d^2 + 6*I*(b*x + a)^2*d^3 + 6*I*a^2*d^3 + (12*I*b*c*d^2 - 12*I*a*d^3)*(b*x + a))*dilog(-I*e^(I*b*x + I*a
)) + ((b*x + a)^3*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(b*x + a))*log
(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*sin(b*x + a) + 1) - ((b*x + a)^3*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 +
3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1) + 1
2*(b*c*d^2 + (b*x + a)*d^3 - a*d^3)*polylog(3, I*e^(I*b*x + I*a)) - 12*(b*c*d^2 + (b*x + a)*d^3 - a*d^3)*polyl
og(3, -I*e^(I*b*x + I*a)))/b^3)/b

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,x\right )}^3}{\cos \left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^3/cos(a + b*x),x)

[Out]

int((c + d*x)^3/cos(a + b*x), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{3} \sec {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________