\(\int \frac {(c+d x)^2}{a+b \sec (e+f x)} \, dx\) [35]

Optimal result

Integrand size = 20, antiderivative size = 394 \[ \int \frac {(c+d x)^2}{a+b \sec (e+f x)} \, dx=\frac {(c+d x)^3}{3 a d}+\frac {i b (c+d x)^2 \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}-\frac {i b (c+d x)^2 \log \left (1+\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}+\frac {2 b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}-\frac {2 b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}+\frac {2 i b d^2 \operatorname {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^3}-\frac {2 i b d^2 \operatorname {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^3} \]

[Out]

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

Rubi [A] (verified)

Time = 1.37 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {4276, 3402, 2296, 2221, 2611, 2320, 6724} \[ \int \frac {(c+d x)^2}{a+b \sec (e+f x)} \, dx=\frac {2 b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a f^2 \sqrt {b^2-a^2}}-\frac {2 b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right )}{a f^2 \sqrt {b^2-a^2}}+\frac {i b (c+d x)^2 \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a f \sqrt {b^2-a^2}}-\frac {i b (c+d x)^2 \log \left (1+\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}+b}\right )}{a f \sqrt {b^2-a^2}}+\frac {2 i b d^2 \operatorname {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a f^3 \sqrt {b^2-a^2}}-\frac {2 i b d^2 \operatorname {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right )}{a f^3 \sqrt {b^2-a^2}}+\frac {(c+d x)^3}{3 a d} \]

[In]

Int[(c + d*x)^2/(a + b*Sec[e + f*x]),x]

[Out]

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 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 3402

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 4276

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 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]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(c+d x)^2}{a}-\frac {b (c+d x)^2}{a (b+a \cos (e+f x))}\right ) \, dx \\ & = \frac {(c+d x)^3}{3 a d}-\frac {b \int \frac {(c+d x)^2}{b+a \cos (e+f x)} \, dx}{a} \\ & = \frac {(c+d x)^3}{3 a d}-\frac {(2 b) \int \frac {e^{i (e+f x)} (c+d x)^2}{a+2 b e^{i (e+f x)}+a e^{2 i (e+f x)}} \, dx}{a} \\ & = \frac {(c+d x)^3}{3 a d}-\frac {(2 b) \int \frac {e^{i (e+f x)} (c+d x)^2}{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)^2}{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)^3}{3 a d}+\frac {i b (c+d x)^2 \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)^2 \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 {(2 i b d) \int (c+d x) \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 {(2 i b d) \int (c+d x) \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)^3}{3 a d}+\frac {i b (c+d x)^2 \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)^2 \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 {2 b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}-\frac {2 b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}-\frac {\left (2 b d^2\right ) \int \operatorname {PolyLog}\left (2,-\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 (2 b d^2\right ) \int \operatorname {PolyLog}\left (2,-\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)^3}{3 a d}+\frac {i b (c+d x)^2 \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)^2 \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 {2 b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}-\frac {2 b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}+\frac {\left (2 i b d^2\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,\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^3}-\frac {\left (2 i b d^2\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,-\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^3} \\ & = \frac {(c+d x)^3}{3 a d}+\frac {i b (c+d x)^2 \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)^2 \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 {2 b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}-\frac {2 b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}+\frac {2 i b d^2 \operatorname {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^3}-\frac {2 i b d^2 \operatorname {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 338, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x)^2}{a+b \sec (e+f x)} \, dx=\frac {(b+a \cos (e+f x)) \left (x \left (3 c^2+3 c d x+d^2 x^2\right )+\frac {3 i b \left ((c+d x)^2 \log \left (1-\frac {a e^{i (e+f x)}}{-b+\sqrt {-a^2+b^2}}\right )-(c+d x)^2 \log \left (1+\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )+\frac {2 d \left (-i f (c+d x) \operatorname {PolyLog}\left (2,\frac {a e^{i (e+f x)}}{-b+\sqrt {-a^2+b^2}}\right )+d \operatorname {PolyLog}\left (3,\frac {a e^{i (e+f x)}}{-b+\sqrt {-a^2+b^2}}\right )\right )}{f^2}+\frac {2 i d \left (f (c+d x) \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )+i d \operatorname {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )\right )}{f^2}\right )}{\sqrt {-a^2+b^2} f}\right ) \sec (e+f x)}{3 a (a+b \sec (e+f x))} \]

[In]

Integrate[(c + d*x)^2/(a + b*Sec[e + f*x]),x]

[Out]

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

Maple [F]

\[\int \frac {\left (d x +c \right )^{2}}{a +b \sec \left (f x +e \right )}d x\]

[In]

int((d*x+c)^2/(a+b*sec(f*x+e)),x)

[Out]

int((d*x+c)^2/(a+b*sec(f*x+e)),x)

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. 1625 vs. \(2 (346) = 692\).

Time = 0.46 (sec) , antiderivative size = 1625, normalized size of antiderivative = 4.12 \[ \int \frac {(c+d x)^2}{a+b \sec (e+f x)} \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)^2/(a+b*sec(f*x+e)),x, algorithm="fricas")

[Out]

1/6*(2*(a^2 - b^2)*d^2*f^3*x^3 + 6*(a^2 - b^2)*c*d*f^3*x^2 + 6*(a^2 - b^2)*c^2*f^3*x - 6*I*a*b*d^2*sqrt(-(a^2
- b^2)/a^2)*polylog(3, -(b*cos(f*x + e) + I*b*sin(f*x + e) + (a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 -
b^2)/a^2))/a) + 6*I*a*b*d^2*sqrt(-(a^2 - b^2)/a^2)*polylog(3, -(b*cos(f*x + e) + I*b*sin(f*x + e) - (a*cos(f*x
+ e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2))/a) + 6*I*a*b*d^2*sqrt(-(a^2 - b^2)/a^2)*polylog(3, -(b*cos(f
*x + e) - I*b*sin(f*x + e) + (a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2))/a) - 6*I*a*b*d^2*sqrt
(-(a^2 - b^2)/a^2)*polylog(3, -(b*cos(f*x + e) - I*b*sin(f*x + e) - (a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-
(a^2 - b^2)/a^2))/a) - 6*(a*b*d^2*f*x + a*b*c*d*f)*sqrt(-(a^2 - b^2)/a^2)*dilog(-(b*cos(f*x + e) + I*b*sin(f*x
+ e) + (a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) + 6*(a*b*d^2*f*x + a*b*c*d*f)*s
qrt(-(a^2 - b^2)/a^2)*dilog(-(b*cos(f*x + e) + I*b*sin(f*x + e) - (a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a
^2 - b^2)/a^2) + a)/a + 1) - 6*(a*b*d^2*f*x + a*b*c*d*f)*sqrt(-(a^2 - b^2)/a^2)*dilog(-(b*cos(f*x + e) - I*b*s
in(f*x + e) + (a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) + 6*(a*b*d^2*f*x + a*b*c*
d*f)*sqrt(-(a^2 - b^2)/a^2)*dilog(-(b*cos(f*x + e) - I*b*sin(f*x + e) - (a*cos(f*x + e) - I*a*sin(f*x + e))*sq
rt(-(a^2 - b^2)/a^2) + a)/a + 1) + 3*(I*a*b*d^2*e^2 - 2*I*a*b*c*d*e*f + I*a*b*c^2*f^2)*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) + 3*(-I*a*b*d^2*e^2 + 2*I*a*b*c*
d*e*f - I*a*b*c^2*f^2)*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) + 3*(I*a*b*d^2*e^2 - 2*I*a*b*c*d*e*f + I*a*b*c^2*f^2)*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) + 3*(-I*a*b*d^2*e^2 + 2*I*a*b*c*d*e*f - I*a*b*c^2*
f^2)*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) + 3
*(-I*a*b*d^2*f^2*x^2 - 2*I*a*b*c*d*f^2*x + I*a*b*d^2*e^2 - 2*I*a*b*c*d*e*f)*sqrt(-(a^2 - b^2)/a^2)*log((b*cos(
f*x + e) + I*b*sin(f*x + e) + (a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + a)/a) + 3*(I*a*b*d^
2*f^2*x^2 + 2*I*a*b*c*d*f^2*x - I*a*b*d^2*e^2 + 2*I*a*b*c*d*e*f)*sqrt(-(a^2 - b^2)/a^2)*log((b*cos(f*x + e) +
I*b*sin(f*x + e) - (a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + a)/a) + 3*(I*a*b*d^2*f^2*x^2 +
2*I*a*b*c*d*f^2*x - I*a*b*d^2*e^2 + 2*I*a*b*c*d*e*f)*sqrt(-(a^2 - b^2)/a^2)*log((b*cos(f*x + e) - I*b*sin(f*x
+ e) + (a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + a)/a) + 3*(-I*a*b*d^2*f^2*x^2 - 2*I*a*b*c
*d*f^2*x + I*a*b*d^2*e^2 - 2*I*a*b*c*d*e*f)*sqrt(-(a^2 - b^2)/a^2)*log((b*cos(f*x + e) - I*b*sin(f*x + e) - (a
*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + a)/a))/((a^3 - a*b^2)*f^3)

Sympy [F]

\[ \int \frac {(c+d x)^2}{a+b \sec (e+f x)} \, dx=\int \frac {\left (c + d x\right )^{2}}{a + b \sec {\left (e + f x \right )}}\, dx \]

[In]

integrate((d*x+c)**2/(a+b*sec(f*x+e)),x)

[Out]

Integral((c + d*x)**2/(a + b*sec(e + f*x)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^2}{a+b \sec (e+f x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((d*x+c)^2/(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 de

Giac [F]

\[ \int \frac {(c+d x)^2}{a+b \sec (e+f x)} \, dx=\int { \frac {{\left (d x + c\right )}^{2}}{b \sec \left (f x + e\right ) + a} \,d x } \]

[In]

integrate((d*x+c)^2/(a+b*sec(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*x + c)^2/(b*sec(f*x + e) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2}{a+b \sec (e+f x)} \, dx=\int \frac {{\left (c+d\,x\right )}^2}{a+\frac {b}{\cos \left (e+f\,x\right )}} \,d x \]

[In]

int((c + d*x)^2/(a + b/cos(e + f*x)),x)

[Out]

int((c + d*x)^2/(a + b/cos(e + f*x)), x)