3.34 \(\int \frac {(c+d x)^3}{a+b \sec (e+f x)} \, dx\)

Optimal. Leaf size=526 \[ \frac {6 i b d^2 (c+d x) \text {Li}_3\left (-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a f^3 \sqrt {b^2-a^2}}-\frac {6 i b d^2 (c+d x) \text {Li}_3\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right )}{a f^3 \sqrt {b^2-a^2}}+\frac {3 b d (c+d x)^2 \text {Li}_2\left (-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a f^2 \sqrt {b^2-a^2}}-\frac {3 b d (c+d x)^2 \text {Li}_2\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right )}{a f^2 \sqrt {b^2-a^2}}+\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a f \sqrt {b^2-a^2}}-\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}+b}\right )}{a f \sqrt {b^2-a^2}}-\frac {6 b d^3 \text {Li}_4\left (-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a f^4 \sqrt {b^2-a^2}}+\frac {6 b d^3 \text {Li}_4\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right )}{a f^4 \sqrt {b^2-a^2}}+\frac {(c+d x)^4}{4 a d} \]

[Out]

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

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Rubi [A]  time = 1.04, antiderivative size = 526, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4191, 3321, 2264, 2190, 2531, 6609, 2282, 6589} \[ \frac {6 i b d^2 (c+d x) \text {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a f^3 \sqrt {b^2-a^2}}-\frac {6 i b d^2 (c+d x) \text {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}+b}\right )}{a f^3 \sqrt {b^2-a^2}}+\frac {3 b d (c+d x)^2 \text {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a f^2 \sqrt {b^2-a^2}}-\frac {3 b d (c+d x)^2 \text {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}+b}\right )}{a f^2 \sqrt {b^2-a^2}}-\frac {6 b d^3 \text {PolyLog}\left (4,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a f^4 \sqrt {b^2-a^2}}+\frac {6 b d^3 \text {PolyLog}\left (4,-\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}+b}\right )}{a f^4 \sqrt {b^2-a^2}}+\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a f \sqrt {b^2-a^2}}-\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}+b}\right )}{a f \sqrt {b^2-a^2}}+\frac {(c+d x)^4}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)^4/(4*a*d) + (I*b*(c + d*x)^3*Log[1 + (a*E^(I*(e + f*x)))/(b - Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2
]*f) - (I*b*(c + d*x)^3*Log[1 + (a*E^(I*(e + f*x)))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*f) + (3*b*d*(
c + d*x)^2*PolyLog[2, -((a*E^(I*(e + f*x)))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*f^2) - (3*b*d*(c + d
*x)^2*PolyLog[2, -((a*E^(I*(e + f*x)))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*f^2) + ((6*I)*b*d^2*(c +
d*x)*PolyLog[3, -((a*E^(I*(e + f*x)))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*f^3) - ((6*I)*b*d^2*(c + d
*x)*PolyLog[3, -((a*E^(I*(e + f*x)))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*f^3) - (6*b*d^3*PolyLog[4,
-((a*E^(I*(e + f*x)))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*f^4) + (6*b*d^3*PolyLog[4, -((a*E^(I*(e +
f*x)))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*f^4)

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 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && 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 3321

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c
+ d*x)^m*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(
2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 4191

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& 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 \frac {(c+d x)^3}{a+b \sec (e+f x)} \, dx &=\int \left (\frac {(c+d x)^3}{a}-\frac {b (c+d x)^3}{a (b+a \cos (e+f x))}\right ) \, dx\\ &=\frac {(c+d x)^4}{4 a d}-\frac {b \int \frac {(c+d x)^3}{b+a \cos (e+f x)} \, dx}{a}\\ &=\frac {(c+d x)^4}{4 a d}-\frac {(2 b) \int \frac {e^{i (e+f x)} (c+d x)^3}{a+2 b e^{i (e+f x)}+a e^{2 i (e+f x)}} \, dx}{a}\\ &=\frac {(c+d x)^4}{4 a d}-\frac {(2 b) \int \frac {e^{i (e+f x)} (c+d x)^3}{2 b-2 \sqrt {-a^2+b^2}+2 a e^{i (e+f x)}} \, dx}{\sqrt {-a^2+b^2}}+\frac {(2 b) \int \frac {e^{i (e+f x)} (c+d x)^3}{2 b+2 \sqrt {-a^2+b^2}+2 a e^{i (e+f x)}} \, dx}{\sqrt {-a^2+b^2}}\\ &=\frac {(c+d x)^4}{4 a d}+\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}-\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}-\frac {(3 i b d) \int (c+d x)^2 \log \left (1+\frac {2 a e^{i (e+f x)}}{2 b-2 \sqrt {-a^2+b^2}}\right ) \, dx}{a \sqrt {-a^2+b^2} f}+\frac {(3 i b d) \int (c+d x)^2 \log \left (1+\frac {2 a e^{i (e+f x)}}{2 b+2 \sqrt {-a^2+b^2}}\right ) \, dx}{a \sqrt {-a^2+b^2} f}\\ &=\frac {(c+d x)^4}{4 a d}+\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}-\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}+\frac {3 b d (c+d x)^2 \text {Li}_2\left (-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}-\frac {3 b d (c+d x)^2 \text {Li}_2\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}-\frac {\left (6 b d^2\right ) \int (c+d x) \text {Li}_2\left (-\frac {2 a e^{i (e+f x)}}{2 b-2 \sqrt {-a^2+b^2}}\right ) \, dx}{a \sqrt {-a^2+b^2} f^2}+\frac {\left (6 b d^2\right ) \int (c+d x) \text {Li}_2\left (-\frac {2 a e^{i (e+f x)}}{2 b+2 \sqrt {-a^2+b^2}}\right ) \, dx}{a \sqrt {-a^2+b^2} f^2}\\ &=\frac {(c+d x)^4}{4 a d}+\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}-\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}+\frac {3 b d (c+d x)^2 \text {Li}_2\left (-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}-\frac {3 b d (c+d x)^2 \text {Li}_2\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}+\frac {6 i b d^2 (c+d x) \text {Li}_3\left (-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^3}-\frac {6 i b d^2 (c+d x) \text {Li}_3\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^3}-\frac {\left (6 i b d^3\right ) \int \text {Li}_3\left (-\frac {2 a e^{i (e+f x)}}{2 b-2 \sqrt {-a^2+b^2}}\right ) \, dx}{a \sqrt {-a^2+b^2} f^3}+\frac {\left (6 i b d^3\right ) \int \text {Li}_3\left (-\frac {2 a e^{i (e+f x)}}{2 b+2 \sqrt {-a^2+b^2}}\right ) \, dx}{a \sqrt {-a^2+b^2} f^3}\\ &=\frac {(c+d x)^4}{4 a d}+\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}-\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}+\frac {3 b d (c+d x)^2 \text {Li}_2\left (-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}-\frac {3 b d (c+d x)^2 \text {Li}_2\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}+\frac {6 i b d^2 (c+d x) \text {Li}_3\left (-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^3}-\frac {6 i b d^2 (c+d x) \text {Li}_3\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^3}-\frac {\left (6 b d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {a x}{-b+\sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{a \sqrt {-a^2+b^2} f^4}+\frac {\left (6 b d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {a x}{b+\sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{a \sqrt {-a^2+b^2} f^4}\\ &=\frac {(c+d x)^4}{4 a d}+\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}-\frac {i b (c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}+\frac {3 b d (c+d x)^2 \text {Li}_2\left (-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}-\frac {3 b d (c+d x)^2 \text {Li}_2\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}+\frac {6 i b d^2 (c+d x) \text {Li}_3\left (-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^3}-\frac {6 i b d^2 (c+d x) \text {Li}_3\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^3}-\frac {6 b d^3 \text {Li}_4\left (-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^4}+\frac {6 b d^3 \text {Li}_4\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^4}\\ \end {align*}

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Mathematica [A]  time = 1.13, size = 449, normalized size = 0.85 \[ \frac {\sec (e+f x) (a \cos (e+f x)+b) \left (x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )+\frac {4 i b \left (\frac {3 i d \left (f^2 (c+d x)^2 \text {Li}_2\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right )+2 i d f (c+d x) \text {Li}_3\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right )-2 d^2 \text {Li}_4\left (-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right )\right )}{f^3}+\frac {3 d \left (2 d \left (f (c+d x) \text {Li}_3\left (\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}-b}\right )+i d \text {Li}_4\left (\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}-b}\right )\right )-i f^2 (c+d x)^2 \text {Li}_2\left (\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}-b}\right )\right )}{f^3}+(c+d x)^3 \log \left (1-\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}-b}\right )-(c+d x)^3 \log \left (1+\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}+b}\right )\right )}{f \sqrt {b^2-a^2}}\right )}{4 a (a+b \sec (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cos[e + f*x])*(x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3) + ((4*I)*b*((c + d*x)^3*Log[1 - (a*E^(I*(
e + f*x)))/(-b + Sqrt[-a^2 + b^2])] - (c + d*x)^3*Log[1 + (a*E^(I*(e + f*x)))/(b + Sqrt[-a^2 + b^2])] + (3*d*(
(-I)*f^2*(c + d*x)^2*PolyLog[2, (a*E^(I*(e + f*x)))/(-b + Sqrt[-a^2 + b^2])] + 2*d*(f*(c + d*x)*PolyLog[3, (a*
E^(I*(e + f*x)))/(-b + Sqrt[-a^2 + b^2])] + I*d*PolyLog[4, (a*E^(I*(e + f*x)))/(-b + Sqrt[-a^2 + b^2])])))/f^3
+ ((3*I)*d*(f^2*(c + d*x)^2*PolyLog[2, -((a*E^(I*(e + f*x)))/(b + Sqrt[-a^2 + b^2]))] + (2*I)*d*f*(c + d*x)*P
olyLog[3, -((a*E^(I*(e + f*x)))/(b + Sqrt[-a^2 + b^2]))] - 2*d^2*PolyLog[4, -((a*E^(I*(e + f*x)))/(b + Sqrt[-a
^2 + b^2]))]))/f^3))/(Sqrt[-a^2 + b^2]*f))*Sec[e + f*x])/(4*a*(a + b*Sec[e + f*x]))

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fricas [C]  time = 0.91, size = 2353, normalized size = 4.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((a^2 - b^2)*d^3*f^4*x^4 + 4*(a^2 - b^2)*c*d^2*f^4*x^3 + 6*(a^2 - b^2)*c^2*d*f^4*x^2 + 4*(a^2 - b^2)*c^3*f
^4*x + 12*a*b*d^3*sqrt(-(a^2 - b^2)/a^2)*polylog(4, -1/2*(2*b*cos(f*x + e) + 2*I*b*sin(f*x + e) + 2*(a*cos(f*x
+ e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2))/a) - 12*a*b*d^3*sqrt(-(a^2 - b^2)/a^2)*polylog(4, -1/2*(2*b*
cos(f*x + e) + 2*I*b*sin(f*x + e) - 2*(a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2))/a) + 12*a*b*
d^3*sqrt(-(a^2 - b^2)/a^2)*polylog(4, -1/2*(2*b*cos(f*x + e) - 2*I*b*sin(f*x + e) + 2*(a*cos(f*x + e) - I*a*si
n(f*x + e))*sqrt(-(a^2 - b^2)/a^2))/a) - 12*a*b*d^3*sqrt(-(a^2 - b^2)/a^2)*polylog(4, -1/2*(2*b*cos(f*x + e) -
2*I*b*sin(f*x + e) - 2*(a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2))/a) - 6*(a*b*d^3*f^2*x^2 +
2*a*b*c*d^2*f^2*x + a*b*c^2*d*f^2)*sqrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(f*x + e) + 2*I*b*sin(f*x + e) +
2*(a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a + 1) + 6*(a*b*d^3*f^2*x^2 + 2*a*b*c*d^2*
f^2*x + a*b*c^2*d*f^2)*sqrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(f*x + e) + 2*I*b*sin(f*x + e) - 2*(a*cos(f*x
+ e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a + 1) - 6*(a*b*d^3*f^2*x^2 + 2*a*b*c*d^2*f^2*x + a*b*
c^2*d*f^2)*sqrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(f*x + e) - 2*I*b*sin(f*x + e) + 2*(a*cos(f*x + e) - I*a*
sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a + 1) + 6*(a*b*d^3*f^2*x^2 + 2*a*b*c*d^2*f^2*x + a*b*c^2*d*f^2)*s
qrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(f*x + e) - 2*I*b*sin(f*x + e) - 2*(a*cos(f*x + e) - I*a*sin(f*x + e)
)*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a + 1) - 2*(I*a*b*d^3*e^3 - 3*I*a*b*c*d^2*e^2*f + 3*I*a*b*c^2*d*e*f^2 - I*a*b*
c^3*f^3)*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cos(f*x + e) + 2*I*a*sin(f*x + e) + 2*a*sqrt(-(a^2 - b^2)/a^2) + 2*b)
- 2*(-I*a*b*d^3*e^3 + 3*I*a*b*c*d^2*e^2*f - 3*I*a*b*c^2*d*e*f^2 + I*a*b*c^3*f^3)*sqrt(-(a^2 - b^2)/a^2)*log(2*
a*cos(f*x + e) - 2*I*a*sin(f*x + e) + 2*a*sqrt(-(a^2 - b^2)/a^2) + 2*b) - 2*(I*a*b*d^3*e^3 - 3*I*a*b*c*d^2*e^2
*f + 3*I*a*b*c^2*d*e*f^2 - I*a*b*c^3*f^3)*sqrt(-(a^2 - b^2)/a^2)*log(-2*a*cos(f*x + e) + 2*I*a*sin(f*x + e) +
2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) - 2*(-I*a*b*d^3*e^3 + 3*I*a*b*c*d^2*e^2*f - 3*I*a*b*c^2*d*e*f^2 + I*a*b*c^3*
f^3)*sqrt(-(a^2 - b^2)/a^2)*log(-2*a*cos(f*x + e) - 2*I*a*sin(f*x + e) + 2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) - 2
*(I*a*b*d^3*f^3*x^3 + 3*I*a*b*c*d^2*f^3*x^2 + 3*I*a*b*c^2*d*f^3*x + I*a*b*d^3*e^3 - 3*I*a*b*c*d^2*e^2*f + 3*I*
a*b*c^2*d*e*f^2)*sqrt(-(a^2 - b^2)/a^2)*log(1/2*(2*b*cos(f*x + e) + 2*I*b*sin(f*x + e) + 2*(a*cos(f*x + e) + I
*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a) - 2*(-I*a*b*d^3*f^3*x^3 - 3*I*a*b*c*d^2*f^3*x^2 - 3*I*a*b*c^
2*d*f^3*x - I*a*b*d^3*e^3 + 3*I*a*b*c*d^2*e^2*f - 3*I*a*b*c^2*d*e*f^2)*sqrt(-(a^2 - b^2)/a^2)*log(1/2*(2*b*cos
(f*x + e) + 2*I*b*sin(f*x + e) - 2*(a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a) - 2*(-
I*a*b*d^3*f^3*x^3 - 3*I*a*b*c*d^2*f^3*x^2 - 3*I*a*b*c^2*d*f^3*x - I*a*b*d^3*e^3 + 3*I*a*b*c*d^2*e^2*f - 3*I*a*
b*c^2*d*e*f^2)*sqrt(-(a^2 - b^2)/a^2)*log(1/2*(2*b*cos(f*x + e) - 2*I*b*sin(f*x + e) + 2*(a*cos(f*x + e) - I*a
*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a) - 2*(I*a*b*d^3*f^3*x^3 + 3*I*a*b*c*d^2*f^3*x^2 + 3*I*a*b*c^2*d
*f^3*x + I*a*b*d^3*e^3 - 3*I*a*b*c*d^2*e^2*f + 3*I*a*b*c^2*d*e*f^2)*sqrt(-(a^2 - b^2)/a^2)*log(1/2*(2*b*cos(f*
x + e) - 2*I*b*sin(f*x + e) - 2*(a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a) - 2*(6*I*
a*b*d^3*f*x + 6*I*a*b*c*d^2*f)*sqrt(-(a^2 - b^2)/a^2)*polylog(3, -1/2*(2*b*cos(f*x + e) + 2*I*b*sin(f*x + e) +
2*(a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2))/a) - 2*(-6*I*a*b*d^3*f*x - 6*I*a*b*c*d^2*f)*sqr
t(-(a^2 - b^2)/a^2)*polylog(3, -1/2*(2*b*cos(f*x + e) + 2*I*b*sin(f*x + e) - 2*(a*cos(f*x + e) + I*a*sin(f*x +
e))*sqrt(-(a^2 - b^2)/a^2))/a) - 2*(-6*I*a*b*d^3*f*x - 6*I*a*b*c*d^2*f)*sqrt(-(a^2 - b^2)/a^2)*polylog(3, -1/
2*(2*b*cos(f*x + e) - 2*I*b*sin(f*x + e) + 2*(a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2))/a) -
2*(6*I*a*b*d^3*f*x + 6*I*a*b*c*d^2*f)*sqrt(-(a^2 - b^2)/a^2)*polylog(3, -1/2*(2*b*cos(f*x + e) - 2*I*b*sin(f*x
+ e) - 2*(a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2))/a))/((a^3 - a*b^2)*f^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 2.57, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{3}}{a +b \sec \left (f x +e \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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