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

Optimal. Leaf size=394 \[ \frac{2 b d (c+d x) \text{PolyLog}\left (2,-\frac{a e^{i (e+f x)}}{b-\sqrt{b^2-a^2}}\right )}{a f^2 \sqrt{b^2-a^2}}-\frac{2 b d (c+d x) \text{PolyLog}\left (2,-\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )}{a f^2 \sqrt{b^2-a^2}}+\frac{2 i b d^2 \text{PolyLog}\left (3,-\frac{a e^{i (e+f x)}}{b-\sqrt{b^2-a^2}}\right )}{a f^3 \sqrt{b^2-a^2}}-\frac{2 i b d^2 \text{PolyLog}\left (3,-\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )}{a f^3 \sqrt{b^2-a^2}}+\frac{i b (c+d x)^2 \log \left (1+\frac{a e^{i (e+f x)}}{b-\sqrt{b^2-a^2}}\right )}{a f \sqrt{b^2-a^2}}-\frac{i b (c+d x)^2 \log \left (1+\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )}{a f \sqrt{b^2-a^2}}+\frac{(c+d x)^3}{3 a d} \]

[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)

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Rubi [A]  time = 0.867481, antiderivative size = 394, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {4191, 3321, 2264, 2190, 2531, 2282, 6589} \[ \frac{2 b d (c+d x) \text{PolyLog}\left (2,-\frac{a e^{i (e+f x)}}{b-\sqrt{b^2-a^2}}\right )}{a f^2 \sqrt{b^2-a^2}}-\frac{2 b d (c+d x) \text{PolyLog}\left (2,-\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )}{a f^2 \sqrt{b^2-a^2}}+\frac{2 i b d^2 \text{PolyLog}\left (3,-\frac{a e^{i (e+f x)}}{b-\sqrt{b^2-a^2}}\right )}{a f^3 \sqrt{b^2-a^2}}-\frac{2 i b d^2 \text{PolyLog}\left (3,-\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )}{a f^3 \sqrt{b^2-a^2}}+\frac{i b (c+d x)^2 \log \left (1+\frac{a e^{i (e+f x)}}{b-\sqrt{b^2-a^2}}\right )}{a f \sqrt{b^2-a^2}}-\frac{i b (c+d x)^2 \log \left (1+\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )}{a f \sqrt{b^2-a^2}}+\frac{(c+d x)^3}{3 a d} \]

Antiderivative was successfully verified.

[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 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 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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{(c+d x)^2}{a+b \sec (e+f x)} \, dx &=\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) \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{2 b d (c+d x) \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 (2 b d^2\right ) \int \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 (2 b d^2\right ) \int \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)^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) \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{2 b d (c+d x) \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 (2 i b d^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\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^3}-\frac{\left (2 i b d^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\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^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) \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{2 b d (c+d x) \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{2 i b d^2 \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{2 i b d^2 \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}\\ \end{align*}

Mathematica [A]  time = 0.769527, size = 338, normalized size = 0.86 \[ \frac{\sec (e+f x) (a \cos (e+f x)+b) \left (x \left (3 c^2+3 c d x+d^2 x^2\right )+\frac{3 i b \left (\frac{2 d \left (d \text{PolyLog}\left (3,\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}-b}\right )-i f (c+d x) \text{PolyLog}\left (2,\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}-b}\right )\right )}{f^2}+\frac{2 i d \left (f (c+d x) \text{PolyLog}\left (2,-\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )+i d \text{PolyLog}\left (3,-\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )\right )}{f^2}+(c+d x)^2 \log \left (1-\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}-b}\right )-(c+d x)^2 \log \left (1+\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )\right )}{f \sqrt{b^2-a^2}}\right )}{3 a (a+b \sec (e+f x))} \]

Antiderivative was successfully verified.

[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]))

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Maple [F]  time = 0.441, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{2}}{a+b\sec \left ( fx+e \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.76575, size = 4018, normalized size = 10.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(a^2 - b^2)*d^2*f^3*x^3 + 12*(a^2 - b^2)*c*d*f^3*x^2 + 12*(a^2 - b^2)*c^2*f^3*x - 12*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) + 12*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*co
s(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2))/a) + 12*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) - 12*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) - 12*(a*b*d^2*f*x + a*b*c*d*f)*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) + 12*(a*b*d
^2*f*x + a*b*c*d*f)*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) - 12*(a*b*d^2*f*x + a*b*c*d*f)*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) + 12*(a*b*d^2*f*x + a*b*c*d*f)*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) - 2*(-3*I*a*b*d^2*
e^2 + 6*I*a*b*c*d*e*f - 3*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) - 2*(3*I*a*b*d^2*e^2 - 6*I*a*b*c*d*e*f + 3*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) - 2*(-3*I*a*b*d^2*e^2 + 6*I*a
*b*c*d*e*f - 3*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) - 2*(3*I*a*b*d^2*e^2 - 6*I*a*b*c*d*e*f + 3*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) - 2*(3*I*a*b*d^2*f^2*x^2 + 6*I*a*b*c*d
*f^2*x - 3*I*a*b*d^2*e^2 + 6*I*a*b*c*d*e*f)*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*(-3*I*a*b*d^2*f^2*x^2 - 6*I*a
*b*c*d*f^2*x + 3*I*a*b*d^2*e^2 - 6*I*a*b*c*d*e*f)*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*(-3*I*a*b*d^2*f^2*x^2 -
6*I*a*b*c*d*f^2*x + 3*I*a*b*d^2*e^2 - 6*I*a*b*c*d*e*f)*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*(3*I*a*b*d^2*f^2*
x^2 + 6*I*a*b*c*d*f^2*x - 3*I*a*b*d^2*e^2 + 6*I*a*b*c*d*e*f)*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))/((a^3 - a*b^2)*
f^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d x\right )^{2}}{a + b \sec{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{2}}{b \sec \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)