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

Optimal. Leaf size=659 \[ \frac{2 b^2 f (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}-\frac{2 b^2 f (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}+\frac{2 i b^2 f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^3 \left (a^2-b^2\right )^{3/2}}-\frac{2 i b^2 f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d^3 \left (a^2-b^2\right )^{3/2}}+\frac{2 i b f^2 \text{PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}-\frac{2 i b f^2 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}-\frac{i a f^2 \text{PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}+\frac{2 a f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}-\frac{4 i b f (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}+\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac{a (e+f x)^2 \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac{b (e+f x)^2 \sec (c+d x)}{d \left (a^2-b^2\right )}-\frac{i a (e+f x)^2}{d \left (a^2-b^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.43355, antiderivative size = 659, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 14, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4533, 3323, 2264, 2190, 2531, 2282, 6589, 6742, 4184, 3719, 2279, 2391, 4409, 4181} \[ \frac{2 b^2 f (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}-\frac{2 b^2 f (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}+\frac{2 i b^2 f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^3 \left (a^2-b^2\right )^{3/2}}-\frac{2 i b^2 f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d^3 \left (a^2-b^2\right )^{3/2}}+\frac{2 i b f^2 \text{PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}-\frac{2 i b f^2 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}-\frac{i a f^2 \text{PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}+\frac{2 a f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}-\frac{4 i b f (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}+\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac{a (e+f x)^2 \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac{b (e+f x)^2 \sec (c+d x)}{d \left (a^2-b^2\right )}-\frac{i a (e+f x)^2}{d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4533

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sec[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbo
l] :> -Dist[b^2/(a^2 - b^2), Int[((e + f*x)^m*Sec[c + d*x]^(n - 2))/(a + b*Sin[c + d*x]), x], x] + Dist[1/(a^2
- b^2), Int[(e + f*x)^m*Sec[c + d*x]^n*(a - b*Sin[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m
, 0] && NeQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 3323

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4409

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
((c + d*x)^m*Sec[a + b*x]^n)/(b*n), x] - Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && 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]

Rubi steps

\begin{align*} \int \frac{(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{\int (e+f x)^2 \sec ^2(c+d x) (a-b \sin (c+d x)) \, dx}{a^2-b^2}-\frac{b^2 \int \frac{(e+f x)^2}{a+b \sin (c+d x)} \, dx}{a^2-b^2}\\ &=\frac{\int \left (a (e+f x)^2 \sec ^2(c+d x)-b (e+f x)^2 \sec (c+d x) \tan (c+d x)\right ) \, dx}{a^2-b^2}-\frac{\left (2 b^2\right ) \int \frac{e^{i (c+d x)} (e+f x)^2}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{a^2-b^2}\\ &=\frac{\left (2 i b^3\right ) \int \frac{e^{i (c+d x)} (e+f x)^2}{2 a-2 \sqrt{a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}-\frac{\left (2 i b^3\right ) \int \frac{e^{i (c+d x)} (e+f x)^2}{2 a+2 \sqrt{a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac{a \int (e+f x)^2 \sec ^2(c+d x) \, dx}{a^2-b^2}-\frac{b \int (e+f x)^2 \sec (c+d x) \tan (c+d x) \, dx}{a^2-b^2}\\ &=\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac{b (e+f x)^2 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a (e+f x)^2 \tan (c+d x)}{\left (a^2-b^2\right ) d}-\frac{\left (2 i b^2 f\right ) \int (e+f x) \log \left (1-\frac{2 i b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d}+\frac{\left (2 i b^2 f\right ) \int (e+f x) \log \left (1-\frac{2 i b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d}-\frac{(2 a f) \int (e+f x) \tan (c+d x) \, dx}{\left (a^2-b^2\right ) d}+\frac{(2 b f) \int (e+f x) \sec (c+d x) \, dx}{\left (a^2-b^2\right ) d}\\ &=-\frac{i a (e+f x)^2}{\left (a^2-b^2\right ) d}-\frac{4 i b f (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac{2 b^2 f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac{2 b^2 f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac{b (e+f x)^2 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a (e+f x)^2 \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac{(4 i a f) \int \frac{e^{2 i (c+d x)} (e+f x)}{1+e^{2 i (c+d x)}} \, dx}{\left (a^2-b^2\right ) d}-\frac{\left (2 b^2 f^2\right ) \int \text{Li}_2\left (\frac{2 i b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d^2}+\frac{\left (2 b^2 f^2\right ) \int \text{Li}_2\left (\frac{2 i b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d^2}-\frac{\left (2 b f^2\right ) \int \log \left (1-i e^{i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d^2}+\frac{\left (2 b f^2\right ) \int \log \left (1+i e^{i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d^2}\\ &=-\frac{i a (e+f x)^2}{\left (a^2-b^2\right ) d}-\frac{4 i b f (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac{2 a f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac{2 b^2 f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac{2 b^2 f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac{b (e+f x)^2 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a (e+f x)^2 \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac{\left (2 i b^2 f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{i b x}{a-\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac{\left (2 i b^2 f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{i b x}{a+\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right )^{3/2} d^3}+\frac{\left (2 i b f^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac{\left (2 i b f^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac{\left (2 a f^2\right ) \int \log \left (1+e^{2 i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d^2}\\ &=-\frac{i a (e+f x)^2}{\left (a^2-b^2\right ) d}-\frac{4 i b f (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac{2 a f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac{2 i b f^2 \text{Li}_2\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac{2 i b f^2 \text{Li}_2\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac{2 b^2 f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac{2 b^2 f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}+\frac{2 i b^2 f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac{2 i b^2 f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac{b (e+f x)^2 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a (e+f x)^2 \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac{\left (i a f^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}\\ &=-\frac{i a (e+f x)^2}{\left (a^2-b^2\right ) d}-\frac{4 i b f (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac{i b^2 (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac{2 a f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac{2 i b f^2 \text{Li}_2\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac{2 i b f^2 \text{Li}_2\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac{2 b^2 f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac{2 b^2 f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac{i a f^2 \text{Li}_2\left (-e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac{2 i b^2 f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac{2 i b^2 f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac{b (e+f x)^2 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a (e+f x)^2 \tan (c+d x)}{\left (a^2-b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 7.9013, size = 1122, normalized size = 1.7 \[ \frac{i \left (-2 \sqrt{a^2-b^2} d f (e+f x) \text{PolyLog}\left (2,\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )+2 \sqrt{a^2-b^2} d f (e+f x) \text{PolyLog}\left (2,-\frac{b e^{i (c+d x)}}{i a+\sqrt{b^2-a^2}}\right )-i \left (\left (2 \sqrt{b^2-a^2} \tan ^{-1}\left (\frac{i a+b e^{i (c+d x)}}{\sqrt{a^2-b^2}}\right ) e^2+\sqrt{a^2-b^2} f x (2 e+f x) \left (\log \left (1-\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )-\log \left (\frac{e^{i (c+d x)} b}{i a+\sqrt{b^2-a^2}}+1\right )\right )\right ) d^2+2 \sqrt{a^2-b^2} f^2 \text{PolyLog}\left (3,\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )-2 \sqrt{a^2-b^2} f^2 \text{PolyLog}\left (3,-\frac{b e^{i (c+d x)}}{i a+\sqrt{b^2-a^2}}\right )\right )\right ) b^2}{\sqrt{-\left (a^2-b^2\right )^2} \left (b^2-a^2\right ) d^3}+\frac{(e+f x)^2 \sec (c) b}{\left (b^2-a^2\right ) d}+\frac{2 f^2 \left (\frac{2 \tan ^{-1}(\cot (c)) \tanh ^{-1}\left (\frac{\sin (c)+\cos (c) \tan \left (\frac{d x}{2}\right )}{\sqrt{\cos ^2(c)+\sin ^2(c)}}\right )}{\sqrt{\cos ^2(c)+\sin ^2(c)}}-\frac{\csc (c) \left (\left (d x-\tan ^{-1}(\cot (c))\right ) \left (\log \left (1-e^{i \left (d x-\tan ^{-1}(\cot (c))\right )}\right )-\log \left (1+e^{i \left (d x-\tan ^{-1}(\cot (c))\right )}\right )\right )+i \left (\text{PolyLog}\left (2,-e^{i \left (d x-\tan ^{-1}(\cot (c))\right )}\right )-\text{PolyLog}\left (2,e^{i \left (d x-\tan ^{-1}(\cot (c))\right )}\right )\right )\right )}{\sqrt{\cot ^2(c)+1}}\right ) b}{\left (a^2-b^2\right ) d^3}+\frac{4 i e f \tan ^{-1}\left (\frac{-i \sin (c)-i \cos (c) \tan \left (\frac{d x}{2}\right )}{\sqrt{\cos ^2(c)+\sin ^2(c)}}\right ) b}{\left (a^2-b^2\right ) d^2 \sqrt{\cos ^2(c)+\sin ^2(c)}}+\frac{a f^2 \csc (c) \left (d^2 e^{-i \tan ^{-1}(\cot (c))} x^2-\frac{\cot (c) \left (i d x \left (-2 \tan ^{-1}(\cot (c))-\pi \right )-\pi \log \left (1+e^{-2 i d x}\right )-2 \left (d x-\tan ^{-1}(\cot (c))\right ) \log \left (1-e^{2 i \left (d x-\tan ^{-1}(\cot (c))\right )}\right )+\pi \log (\cos (d x))-2 \tan ^{-1}(\cot (c)) \log \left (\sin \left (d x-\tan ^{-1}(\cot (c))\right )\right )+i \text{PolyLog}\left (2,e^{2 i \left (d x-\tan ^{-1}(\cot (c))\right )}\right )\right )}{\sqrt{\cot ^2(c)+1}}\right ) \sec (c)}{\left (a^2-b^2\right ) d^3 \sqrt{\csc ^2(c) \left (\cos ^2(c)+\sin ^2(c)\right )}}+\frac{2 a e f \sec (c) (\cos (c) \log (\cos (c) \cos (d x)-\sin (c) \sin (d x))+d x \sin (c))}{\left (a^2-b^2\right ) d^2 \left (\cos ^2(c)+\sin ^2(c)\right )}+\frac{\sin \left (\frac{d x}{2}\right ) e^2+2 f x \sin \left (\frac{d x}{2}\right ) e+f^2 x^2 \sin \left (\frac{d x}{2}\right )}{(a+b) d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}+\frac{\sin \left (\frac{d x}{2}\right ) e^2+2 f x \sin \left (\frac{d x}{2}\right ) e+f^2 x^2 \sin \left (\frac{d x}{2}\right )}{(a-b) d \left (\cos \left (\frac{c}{2}\right )+\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(I*b^2*(-2*Sqrt[a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + 2*Sqrt[
a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))] - I*(d^2*(2*Sqrt[-a^2 + b
^2]*e^2*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]] + Sqrt[a^2 - b^2]*f*x*(2*e + f*x)*(Log[1 - (b*E^(I*(
c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])])) + 2*Sqrt[a^2
- b^2]*f^2*PolyLog[3, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - 2*Sqrt[a^2 - b^2]*f^2*PolyLog[3, -((
b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))])))/(Sqrt[-(a^2 - b^2)^2]*(-a^2 + b^2)*d^3) + (b*(e + f*x)^2*Sec[
c])/((-a^2 + b^2)*d) + (2*a*e*f*Sec[c]*(Cos[c]*Log[Cos[c]*Cos[d*x] - Sin[c]*Sin[d*x]] + d*x*Sin[c]))/((a^2 - b
^2)*d^2*(Cos[c]^2 + Sin[c]^2)) + ((4*I)*b*e*f*ArcTan[((-I)*Sin[c] - I*Cos[c]*Tan[(d*x)/2])/Sqrt[Cos[c]^2 + Sin
[c]^2]])/((a^2 - b^2)*d^2*Sqrt[Cos[c]^2 + Sin[c]^2]) + (a*f^2*Csc[c]*((d^2*x^2)/E^(I*ArcTan[Cot[c]]) - (Cot[c]
*(I*d*x*(-Pi - 2*ArcTan[Cot[c]]) - Pi*Log[1 + E^((-2*I)*d*x)] - 2*(d*x - ArcTan[Cot[c]])*Log[1 - E^((2*I)*(d*x
- ArcTan[Cot[c]]))] + Pi*Log[Cos[d*x]] - 2*ArcTan[Cot[c]]*Log[Sin[d*x - ArcTan[Cot[c]]]] + I*PolyLog[2, E^((2
*I)*(d*x - ArcTan[Cot[c]]))]))/Sqrt[1 + Cot[c]^2])*Sec[c])/((a^2 - b^2)*d^3*Sqrt[Csc[c]^2*(Cos[c]^2 + Sin[c]^2
)]) + (2*b*f^2*(-((Csc[c]*((d*x - ArcTan[Cot[c]])*(Log[1 - E^(I*(d*x - ArcTan[Cot[c]]))] - Log[1 + E^(I*(d*x -
ArcTan[Cot[c]]))]) + I*(PolyLog[2, -E^(I*(d*x - ArcTan[Cot[c]]))] - PolyLog[2, E^(I*(d*x - ArcTan[Cot[c]]))])
))/Sqrt[1 + Cot[c]^2]) + (2*ArcTan[Cot[c]]*ArcTanh[(Sin[c] + Cos[c]*Tan[(d*x)/2])/Sqrt[Cos[c]^2 + Sin[c]^2]])/
Sqrt[Cos[c]^2 + Sin[c]^2]))/((a^2 - b^2)*d^3) + (e^2*Sin[(d*x)/2] + 2*e*f*x*Sin[(d*x)/2] + f^2*x^2*Sin[(d*x)/2
])/((a + b)*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (e^2*Sin[(d*x)/2] + 2*e*f*x*S
in[(d*x)/2] + f^2*x^2*Sin[(d*x)/2])/((a - b)*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])
)

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Maple [F]  time = 3.475, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{a+b\sin \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 5.22192, size = 6276, normalized size = 9.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*b^3*f^2*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*polylog(3, 1/2*(2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) + 2*(
b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 4*b^3*f^2*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*
polylog(3, 1/2*(2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2
)/b^2))/b) - 4*b^3*f^2*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) + (b
*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 4*b^3*f^2*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*p
olylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b)
- 4*(a^2*b - b^3)*d^2*f^2*x^2 - 8*(a^2*b - b^3)*d^2*e*f*x - 4*(a^2*b - b^3)*d^2*e^2 + 4*I*(a^3 - a^2*b - a*b^
2 + b^3)*f^2*cos(d*x + c)*dilog(I*cos(d*x + c) + sin(d*x + c)) - 4*I*(a^3 + a^2*b - a*b^2 - b^3)*f^2*cos(d*x +
c)*dilog(I*cos(d*x + c) - sin(d*x + c)) - 4*I*(a^3 - a^2*b - a*b^2 + b^3)*f^2*cos(d*x + c)*dilog(-I*cos(d*x +
c) + sin(d*x + c)) + 4*I*(a^3 + a^2*b - a*b^2 - b^3)*f^2*cos(d*x + c)*dilog(-I*cos(d*x + c) - sin(d*x + c)) +
2*(-2*I*b^3*d*f^2*x - 2*I*b^3*d*e*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a
*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*(2*I*b^3*d*f^2*
x + 2*I*b^3*d*e*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(
b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*(2*I*b^3*d*f^2*x + 2*I*b^3*d*e*f)*
sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I
*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*(-2*I*b^3*d*f^2*x - 2*I*b^3*d*e*f)*sqrt(-(a^2 - b^2)
/b^2)*cos(d*x + c)*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*
sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - 2*(b^3*d^2*e^2 - 2*b^3*c*d*e*f + b^3*c^2*f^2)*sqrt(-(a^2 - b^2)/b^2)*co
s(d*x + c)*log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) - 2*(b^3*d^2*e^2 -
2*b^3*c*d*e*f + b^3*c^2*f^2)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2
*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + 2*(b^3*d^2*e^2 - 2*b^3*c*d*e*f + b^3*c^2*f^2)*sqrt(-(a^2 - b^2)/b^2)*cos(
d*x + c)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + 2*(b^3*d^2*e^2 - 2
*b^3*c*d*e*f + b^3*c^2*f^2)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(-2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2
*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) - 2*(b^3*d^2*f^2*x^2 + 2*b^3*d^2*e*f*x + 2*b^3*c*d*e*f - b^3*c^2*f^2)*sqrt(
-(a^2 - b^2)/b^2)*cos(d*x + c)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*
x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*(b^3*d^2*f^2*x^2 + 2*b^3*d^2*e*f*x + 2*b^3*c*d*e*f - b^3*c^2*f^2)
*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*
sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - 2*(b^3*d^2*f^2*x^2 + 2*b^3*d^2*e*f*x + 2*b^3*c*d*e*f - b^3*c^
2*f^2)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c)
+ I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*(b^3*d^2*f^2*x^2 + 2*b^3*d^2*e*f*x + 2*b^3*c*d*e*f -
b^3*c^2*f^2)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d
*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 4*((a^3 + a^2*b - a*b^2 - b^3)*d*e*f - (a^3 + a
^2*b - a*b^2 - b^3)*c*f^2)*cos(d*x + c)*log(cos(d*x + c) + I*sin(d*x + c) + I) + 4*((a^3 - a^2*b - a*b^2 + b^3
)*d*e*f - (a^3 - a^2*b - a*b^2 + b^3)*c*f^2)*cos(d*x + c)*log(cos(d*x + c) - I*sin(d*x + c) + I) + 4*((a^3 + a
^2*b - a*b^2 - b^3)*d*f^2*x + (a^3 + a^2*b - a*b^2 - b^3)*c*f^2)*cos(d*x + c)*log(I*cos(d*x + c) + sin(d*x + c
) + 1) + 4*((a^3 - a^2*b - a*b^2 + b^3)*d*f^2*x + (a^3 - a^2*b - a*b^2 + b^3)*c*f^2)*cos(d*x + c)*log(I*cos(d*
x + c) - sin(d*x + c) + 1) + 4*((a^3 + a^2*b - a*b^2 - b^3)*d*f^2*x + (a^3 + a^2*b - a*b^2 - b^3)*c*f^2)*cos(d
*x + c)*log(-I*cos(d*x + c) + sin(d*x + c) + 1) + 4*((a^3 - a^2*b - a*b^2 + b^3)*d*f^2*x + (a^3 - a^2*b - a*b^
2 + b^3)*c*f^2)*cos(d*x + c)*log(-I*cos(d*x + c) - sin(d*x + c) + 1) + 4*((a^3 + a^2*b - a*b^2 - b^3)*d*e*f -
(a^3 + a^2*b - a*b^2 - b^3)*c*f^2)*cos(d*x + c)*log(-cos(d*x + c) + I*sin(d*x + c) + I) + 4*((a^3 - a^2*b - a*
b^2 + b^3)*d*e*f - (a^3 - a^2*b - a*b^2 + b^3)*c*f^2)*cos(d*x + c)*log(-cos(d*x + c) - I*sin(d*x + c) + I) + 4
*((a^3 - a*b^2)*d^2*f^2*x^2 + 2*(a^3 - a*b^2)*d^2*e*f*x + (a^3 - a*b^2)*d^2*e^2)*sin(d*x + c))/((a^4 - 2*a^2*b
^2 + b^4)*d^3*cos(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out