\(\int (c+d x)^3 (a+b \sec (e+f x)) \, dx\) [24]

Optimal result

Integrand size = 18, antiderivative size = 227 \[ \int (c+d x)^3 (a+b \sec (e+f x)) \, dx=\frac {a (c+d x)^4}{4 d}-\frac {2 i b (c+d x)^3 \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {6 b d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac {6 b d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}-\frac {6 i b d^3 \operatorname {PolyLog}\left (4,-i e^{i (e+f x)}\right )}{f^4}+\frac {6 i b d^3 \operatorname {PolyLog}\left (4,i e^{i (e+f x)}\right )}{f^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4275, 4266, 2611, 6744, 2320, 6724} \[ \int (c+d x)^3 (a+b \sec (e+f x)) \, dx=\frac {a (c+d x)^4}{4 d}-\frac {2 i b (c+d x)^3 \arctan \left (e^{i (e+f x)}\right )}{f}-\frac {6 b d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac {6 b d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {6 i b d^3 \operatorname {PolyLog}\left (4,-i e^{i (e+f x)}\right )}{f^4}+\frac {6 i b d^3 \operatorname {PolyLog}\left (4,i e^{i (e+f x)}\right )}{f^4} \]

[In]

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

[Out]

(a*(c + d*x)^4)/(4*d) - ((2*I)*b*(c + d*x)^3*ArcTan[E^(I*(e + f*x))])/f + ((3*I)*b*d*(c + d*x)^2*PolyLog[2, (-
I)*E^(I*(e + f*x))])/f^2 - ((3*I)*b*d*(c + d*x)^2*PolyLog[2, I*E^(I*(e + f*x))])/f^2 - (6*b*d^2*(c + d*x)*Poly
Log[3, (-I)*E^(I*(e + f*x))])/f^3 + (6*b*d^2*(c + d*x)*PolyLog[3, I*E^(I*(e + f*x))])/f^3 - ((6*I)*b*d^3*PolyL
og[4, (-I)*E^(I*(e + f*x))])/f^4 + ((6*I)*b*d^3*PolyLog[4, I*E^(I*(e + f*x))])/f^4

Rule 2320

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 2611

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 4266

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 4275

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 6724

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 6744

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*} \text {integral}& = \int \left (a (c+d x)^3+b (c+d x)^3 \sec (e+f x)\right ) \, dx \\ & = \frac {a (c+d x)^4}{4 d}+b \int (c+d x)^3 \sec (e+f x) \, dx \\ & = \frac {a (c+d x)^4}{4 d}-\frac {2 i b (c+d x)^3 \arctan \left (e^{i (e+f x)}\right )}{f}-\frac {(3 b d) \int (c+d x)^2 \log \left (1-i e^{i (e+f x)}\right ) \, dx}{f}+\frac {(3 b d) \int (c+d x)^2 \log \left (1+i e^{i (e+f x)}\right ) \, dx}{f} \\ & = \frac {a (c+d x)^4}{4 d}-\frac {2 i b (c+d x)^3 \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {\left (6 i b d^2\right ) \int (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right ) \, dx}{f^2}+\frac {\left (6 i b d^2\right ) \int (c+d x) \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right ) \, dx}{f^2} \\ & = \frac {a (c+d x)^4}{4 d}-\frac {2 i b (c+d x)^3 \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {6 b d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac {6 b d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac {\left (6 b d^3\right ) \int \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right ) \, dx}{f^3}-\frac {\left (6 b d^3\right ) \int \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right ) \, dx}{f^3} \\ & = \frac {a (c+d x)^4}{4 d}-\frac {2 i b (c+d x)^3 \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {6 b d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac {6 b d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}-\frac {\left (6 i b d^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^4}+\frac {\left (6 i b d^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^4} \\ & = \frac {a (c+d x)^4}{4 d}-\frac {2 i b (c+d x)^3 \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {6 b d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac {6 b d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}-\frac {6 i b d^3 \operatorname {PolyLog}\left (4,-i e^{i (e+f x)}\right )}{f^4}+\frac {6 i b d^3 \operatorname {PolyLog}\left (4,i e^{i (e+f x)}\right )}{f^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.61 \[ \int (c+d x)^3 (a+b \sec (e+f x)) \, dx=\frac {4 a c^3 f^4 x+6 a c^2 d f^4 x^2+4 a c d^2 f^4 x^3+a d^3 f^4 x^4-24 i b c^2 d f^3 x \arctan \left (e^{i (e+f x)}\right )-24 i b c d^2 f^3 x^2 \arctan \left (e^{i (e+f x)}\right )-8 i b d^3 f^3 x^3 \arctan \left (e^{i (e+f x)}\right )+4 b c^3 f^3 \text {arctanh}(\sin (e+f x))+12 i b d f^2 (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )-12 i b d f^2 (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )-24 b c d^2 f \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )-24 b d^3 f x \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )+24 b c d^2 f \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )+24 b d^3 f x \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )-24 i b d^3 \operatorname {PolyLog}\left (4,-i e^{i (e+f x)}\right )+24 i b d^3 \operatorname {PolyLog}\left (4,i e^{i (e+f x)}\right )}{4 f^4} \]

[In]

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

[Out]

(4*a*c^3*f^4*x + 6*a*c^2*d*f^4*x^2 + 4*a*c*d^2*f^4*x^3 + a*d^3*f^4*x^4 - (24*I)*b*c^2*d*f^3*x*ArcTan[E^(I*(e +
f*x))] - (24*I)*b*c*d^2*f^3*x^2*ArcTan[E^(I*(e + f*x))] - (8*I)*b*d^3*f^3*x^3*ArcTan[E^(I*(e + f*x))] + 4*b*c
^3*f^3*ArcTanh[Sin[e + f*x]] + (12*I)*b*d*f^2*(c + d*x)^2*PolyLog[2, (-I)*E^(I*(e + f*x))] - (12*I)*b*d*f^2*(c
+ d*x)^2*PolyLog[2, I*E^(I*(e + f*x))] - 24*b*c*d^2*f*PolyLog[3, (-I)*E^(I*(e + f*x))] - 24*b*d^3*f*x*PolyLog
[3, (-I)*E^(I*(e + f*x))] + 24*b*c*d^2*f*PolyLog[3, I*E^(I*(e + f*x))] + 24*b*d^3*f*x*PolyLog[3, I*E^(I*(e + f
*x))] - (24*I)*b*d^3*PolyLog[4, (-I)*E^(I*(e + f*x))] + (24*I)*b*d^3*PolyLog[4, I*E^(I*(e + f*x))])/(4*f^4)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 755 vs. \(2 (200 ) = 400\).

Time = 1.01 (sec) , antiderivative size = 756, normalized size of antiderivative = 3.33

method	result	size
risch	\(-\frac {6 i b c \,d^{2} e^{2} \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}+\frac {6 i b \,c^{2} d e \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {6 i b \,d^{2} c \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{2}}+\frac {6 i b \,d^{2} c \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{2}}+\frac {3 b \,d^{2} c \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{2}}{f}+\frac {3 b \,e^{2} c \,d^{2} \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {3 b \,d^{2} c \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{2}}{f}-\frac {3 b \,e^{2} c \,d^{2} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}+\frac {3 b \,c^{2} d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}+\frac {3 b \,c^{2} d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}-\frac {3 b \,c^{2} d \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}-\frac {3 b \,c^{2} d \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}-\frac {3 i b \,d^{3} \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{2}}{f^{2}}+\frac {3 i b \,d^{3} \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{2}}{f^{2}}-\frac {3 i b \,c^{2} d \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+\frac {3 i b \,c^{2} d \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+\frac {2 i b \,d^{3} e^{3} \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{4}}+a \,d^{2} c \,x^{3}+\frac {3 a d \,c^{2} x^{2}}{2}+a \,c^{3} x +\frac {a \,d^{3} x^{4}}{4}+\frac {a \,c^{4}}{4 d}+\frac {6 i b \,d^{3} \operatorname {polylog}\left (4, i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{4}}+\frac {6 b \,d^{2} c \operatorname {polylog}\left (3, i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {b \,e^{3} d^{3} \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{4}}-\frac {6 b \,d^{3} \operatorname {polylog}\left (3, -i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{3}}+\frac {b \,d^{3} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{3}}{f}-\frac {b \,d^{3} \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{3}}{f}-\frac {6 b \,d^{2} c \operatorname {polylog}\left (3, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}+\frac {6 b \,d^{3} \operatorname {polylog}\left (3, i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{3}}+\frac {b \,e^{3} d^{3} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{4}}-\frac {6 i b \,d^{3} \operatorname {polylog}\left (4, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{4}}-\frac {2 i b \,c^{3} \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}\)	\(756\)

[In]

int((d*x+c)^3*(a+b*sec(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

3/f*b*d^2*c*ln(1-I*exp(I*(f*x+e)))*x^2+3/f^3*b*e^2*c*d^2*ln(1+I*exp(I*(f*x+e)))-3/f*b*d^2*c*ln(1+I*exp(I*(f*x+
e)))*x^2-3/f^3*b*e^2*c*d^2*ln(1-I*exp(I*(f*x+e)))+6*I*b*d^3*polylog(4,I*exp(I*(f*x+e)))/f^4+6/f^3*b*d^2*c*poly
log(3,I*exp(I*(f*x+e)))-1/f^4*b*e^3*d^3*ln(1+I*exp(I*(f*x+e)))-6/f^3*b*d^3*polylog(3,-I*exp(I*(f*x+e)))*x+1/f*
b*d^3*ln(1-I*exp(I*(f*x+e)))*x^3-1/f*b*d^3*ln(1+I*exp(I*(f*x+e)))*x^3-6/f^3*b*d^2*c*polylog(3,-I*exp(I*(f*x+e)
))+6/f^3*b*d^3*polylog(3,I*exp(I*(f*x+e)))*x+1/f^4*b*e^3*d^3*ln(1-I*exp(I*(f*x+e)))+a*d^2*c*x^3+3/2*a*d*c^2*x^
2+a*c^3*x-6*I/f^3*b*c*d^2*e^2*arctan(exp(I*(f*x+e)))+6*I/f^2*b*c^2*d*e*arctan(exp(I*(f*x+e)))-6*I/f^2*b*d^2*c*
polylog(2,I*exp(I*(f*x+e)))*x+1/4*a*d^3*x^4+1/4*a/d*c^4+3/f*b*c^2*d*ln(1-I*exp(I*(f*x+e)))*x+3/f^2*b*c^2*d*ln(
1-I*exp(I*(f*x+e)))*e-3/f*b*c^2*d*ln(1+I*exp(I*(f*x+e)))*x-3/f^2*b*c^2*d*ln(1+I*exp(I*(f*x+e)))*e-3*I/f^2*b*d^
3*polylog(2,I*exp(I*(f*x+e)))*x^2+6*I/f^2*b*d^2*c*polylog(2,-I*exp(I*(f*x+e)))*x+3*I/f^2*b*d^3*polylog(2,-I*ex
p(I*(f*x+e)))*x^2-3*I/f^2*b*c^2*d*polylog(2,I*exp(I*(f*x+e)))+3*I/f^2*b*c^2*d*polylog(2,-I*exp(I*(f*x+e)))+2*I
/f^4*b*d^3*e^3*arctan(exp(I*(f*x+e)))-6*I*b*d^3*polylog(4,-I*exp(I*(f*x+e)))/f^4-2*I/f*b*c^3*arctan(exp(I*(f*x
+e)))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1085 vs. \(2 (187) = 374\).

Time = 0.34 (sec) , antiderivative size = 1085, normalized size of antiderivative = 4.78 \[ \int (c+d x)^3 (a+b \sec (e+f x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/4*(a*d^3*f^4*x^4 + 4*a*c*d^2*f^4*x^3 + 6*a*c^2*d*f^4*x^2 + 4*a*c^3*f^4*x + 12*I*b*d^3*polylog(4, I*cos(f*x +
e) + sin(f*x + e)) + 12*I*b*d^3*polylog(4, I*cos(f*x + e) - sin(f*x + e)) - 12*I*b*d^3*polylog(4, -I*cos(f*x
+ e) + sin(f*x + e)) - 12*I*b*d^3*polylog(4, -I*cos(f*x + e) - sin(f*x + e)) - 6*(I*b*d^3*f^2*x^2 + 2*I*b*c*d^
2*f^2*x + I*b*c^2*d*f^2)*dilog(I*cos(f*x + e) + sin(f*x + e)) - 6*(I*b*d^3*f^2*x^2 + 2*I*b*c*d^2*f^2*x + I*b*c
^2*d*f^2)*dilog(I*cos(f*x + e) - sin(f*x + e)) - 6*(-I*b*d^3*f^2*x^2 - 2*I*b*c*d^2*f^2*x - I*b*c^2*d*f^2)*dilo
g(-I*cos(f*x + e) + sin(f*x + e)) - 6*(-I*b*d^3*f^2*x^2 - 2*I*b*c*d^2*f^2*x - I*b*c^2*d*f^2)*dilog(-I*cos(f*x
+ e) - sin(f*x + e)) - 2*(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(cos(f*x + e) + I*sin(
f*x + e) + I) + 2*(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(cos(f*x + e) - I*sin(f*x + e
) + I) + 2*(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(I*cos(f*x + e) + sin(f*x + e) + 1) - 2*(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(I*cos(f*x + e) - sin(f*x + e) + 1) + 2*(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(-I*cos(f*x + e) + sin(f*x + e) +
1) - 2*(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)*l
og(-I*cos(f*x + e) - sin(f*x + e) + 1) - 2*(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(-co
s(f*x + e) + I*sin(f*x + e) + I) + 2*(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(-cos(f*x
+ e) - I*sin(f*x + e) + I) - 12*(b*d^3*f*x + b*c*d^2*f)*polylog(3, I*cos(f*x + e) + sin(f*x + e)) + 12*(b*d^3*
f*x + b*c*d^2*f)*polylog(3, I*cos(f*x + e) - sin(f*x + e)) - 12*(b*d^3*f*x + b*c*d^2*f)*polylog(3, -I*cos(f*x
+ e) + sin(f*x + e)) + 12*(b*d^3*f*x + b*c*d^2*f)*polylog(3, -I*cos(f*x + e) - sin(f*x + e)))/f^4

Sympy [F]

\[ \int (c+d x)^3 (a+b \sec (e+f x)) \, dx=\int \left (a + b \sec {\left (e + f x \right )}\right ) \left (c + d x\right )^{3}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 936 vs. \(2 (187) = 374\).

Time = 0.42 (sec) , antiderivative size = 936, normalized size of antiderivative = 4.12 \[ \int (c+d x)^3 (a+b \sec (e+f x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Giac [F]

\[ \int (c+d x)^3 (a+b \sec (e+f x)) \, dx=\int { {\left (d x + c\right )}^{3} {\left (b \sec \left (f x + e\right ) + a\right )} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 (a+b \sec (e+f x)) \, dx=\int \left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )\,{\left (c+d\,x\right )}^3 \,d x \]

[In]

int((a + b/cos(e + f*x))*(c + d*x)^3,x)

[Out]

int((a + b/cos(e + f*x))*(c + d*x)^3, x)