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

Optimal. Leaf size=667 \[ -\frac {2 a f^2 \text {Li}_3\left (-i e^{i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}+\frac {2 a f^2 \text {Li}_3\left (i e^{i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}-\frac {2 b f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^3 \left (a^2-b^2\right )}-\frac {2 b f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^3 \left (a^2-b^2\right )}+\frac {b f^2 \text {Li}_3\left (-e^{2 i (c+d x)}\right )}{2 d^3 \left (a^2-b^2\right )}+\frac {2 i a f (e+f x) \text {Li}_2\left (-i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}-\frac {2 i a f (e+f x) \text {Li}_2\left (i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}+\frac {2 i b f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )}+\frac {2 i b f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )}-\frac {i b f (e+f x) \text {Li}_2\left (-e^{2 i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}-\frac {b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )}-\frac {b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )}+\frac {b (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{d \left (a^2-b^2\right )}-\frac {2 i a (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{d \left (a^2-b^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.14, antiderivative size = 667, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {4533, 4519, 2190, 2531, 2282, 6589, 6742, 4181, 3719} \[ \frac {2 i a f (e+f x) \text {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}-\frac {2 i a f (e+f x) \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}+\frac {2 i b f (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )}+\frac {2 i b f (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )}-\frac {i b f (e+f x) \text {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}-\frac {2 a f^2 \text {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}+\frac {2 a f^2 \text {PolyLog}\left (3,i e^{i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}-\frac {2 b f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^3 \left (a^2-b^2\right )}-\frac {2 b f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d^3 \left (a^2-b^2\right )}+\frac {b f^2 \text {PolyLog}\left (3,-e^{2 i (c+d x)}\right )}{2 d^3 \left (a^2-b^2\right )}-\frac {b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )}-\frac {b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )}+\frac {b (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{d \left (a^2-b^2\right )}-\frac {2 i a (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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]

Rule 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

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 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]]

Rubi steps

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

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Mathematica [B]  time = 5.86, size = 1561, normalized size = 2.34 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(((2*I)*b*(e + f*x)^3)/f - (3*(a - b)*(1 + E^((2*I)*c))*(e + f*x)^2*Log[1 - I/E^(I*(c + d*x))])/d + (3*(a
+ b)*(1 + E^((2*I)*c))*(e + f*x)^2*Log[1 + I/E^(I*(c + d*x))])/d + (6*(a + b)*(1 + E^((2*I)*c))*f*(I*d*(e + f*
x)*PolyLog[2, (-I)/E^(I*(c + d*x))] + f*PolyLog[3, (-I)/E^(I*(c + d*x))]))/d^3 - ((6*I)*(a - b)*(1 + E^((2*I)*
c))*f*(d*(e + f*x)*PolyLog[2, I/E^(I*(c + d*x))] - I*f*PolyLog[3, I/E^(I*(c + d*x))]))/d^3))/((a^2 - b^2)*(1 +
E^((2*I)*c))) + (b*((-12*I)*d^3*e^2*E^((2*I)*c)*x - (12*I)*d^3*e*E^((2*I)*c)*f*x^2 - (4*I)*d^3*E^((2*I)*c)*f^
2*x^3 - (6*I)*d^2*e^2*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))] + (6*I)*d^2*e^2*E^((2*I)*c)
*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))] - 3*d^2*e^2*Log[4*a^2*E^((2*I)*(c + d*x)) + b^2*
(-1 + E^((2*I)*(c + d*x)))^2] + 3*d^2*e^2*E^((2*I)*c)*Log[4*a^2*E^((2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c +
d*x)))^2] - 12*d^2*e*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 12*d^
2*e*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^2*f^2*
x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^2*E^((2*I)*c)*f^2*x^2*
Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 12*d^2*e*f*x*Log[1 + (b*E^(I*(
2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 12*d^2*e*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c +
d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^2*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(
I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^2*E^((2*I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c)
+ Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - (12*I)*d*(-1 + E^((2*I)*c))*f*(e + f*x)*PolyLog[2, (I*b*E^(I*(2*c + d*x)
))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - (12*I)*d*(-1 + E^((2*I)*c))*f*(e + f*x)*PolyLog[2, -((b*E
^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] - 12*f^2*PolyLog[3, (I*b*E^(I*(2*c + d*x)))
/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 12*E^((2*I)*c)*f^2*PolyLog[3, (I*b*E^(I*(2*c + d*x)))/(a*E^
(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 12*f^2*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a
^2 + b^2)*E^((2*I)*c)]))] + 12*E^((2*I)*c)*f^2*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 +
b^2)*E^((2*I)*c)]))]))/((-a^2 + b^2)*d^3*(-1 + E^((2*I)*c))) - (8*b*x*(3*e^2 + 3*e*f*x + f^2*x^2)*Csc[c]^3)/((
a - b)*(a + b)*(Csc[c/2] - Sec[c/2])*(Csc[c/2] + Sec[c/2])))/6

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fricas [C]  time = 0.67, size = 2041, normalized size = 3.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*f^2*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))*sq
rt(-(a^2 - b^2)/b^2))/b) + 2*b*f^2*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) + 2*b*f^2*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) + 2*b*f^2*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) + 2*(a - b)*f^2*poly
log(3, I*cos(d*x + c) + sin(d*x + c)) - 2*(a + b)*f^2*polylog(3, I*cos(d*x + c) - sin(d*x + c)) + 2*(a - b)*f^
2*polylog(3, -I*cos(d*x + c) + sin(d*x + c)) - 2*(a + b)*f^2*polylog(3, -I*cos(d*x + c) - sin(d*x + c)) + (2*I
*b*d*f^2*x + 2*I*b*d*e*f)*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*I*b*d*f^2*x + 2*I*b*d*e*f)*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*I*b*d*f^2*x
- 2*I*b*d*e*f)*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*I*b*d*f^2*x - 2*I*b*d*e*f)*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*I*(a - b)*d*f^2*x + 2
*I*(a - b)*d*e*f)*dilog(I*cos(d*x + c) + sin(d*x + c)) + (2*I*(a + b)*d*f^2*x + 2*I*(a + b)*d*e*f)*dilog(I*cos
(d*x + c) - sin(d*x + c)) + (-2*I*(a - b)*d*f^2*x - 2*I*(a - b)*d*e*f)*dilog(-I*cos(d*x + c) + sin(d*x + c)) +
(-2*I*(a + b)*d*f^2*x - 2*I*(a + b)*d*e*f)*dilog(-I*cos(d*x + c) - sin(d*x + c)) + (b*d^2*e^2 - 2*b*c*d*e*f +
b*c^2*f^2)*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) + (b*d^2*e^2 - 2*b
*c*d*e*f + b*c^2*f^2)*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) + (b*d^2
*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*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) + (b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*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) + (b*d^2*f^2*x^2 + 2*b*d^2*e*f*x + 2*b*c*d*e*f - b*c^2*f^2)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*si
n(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) + (b*d^2*f^2*x^2 + 2*b*d^2
*e*f*x + 2*b*c*d*e*f - b*c^2*f^2)*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) + (b*d^2*f^2*x^2 + 2*b*d^2*e*f*x + 2*b*c*d*e*f - b*c^2*f^2)*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) + (b*d^2*f^2*x^2 + 2*b*d^2*e*f*x + 2*b*c*d*e*f - b*c^2*f^2)*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) - ((a + b)*d^2*e^2 - 2*(a + b)*c*d
*e*f + (a + b)*c^2*f^2)*log(cos(d*x + c) + I*sin(d*x + c) + I) + ((a - b)*d^2*e^2 - 2*(a - b)*c*d*e*f + (a - b
)*c^2*f^2)*log(cos(d*x + c) - I*sin(d*x + c) + I) - ((a + b)*d^2*f^2*x^2 + 2*(a + b)*d^2*e*f*x + 2*(a + b)*c*d
*e*f - (a + b)*c^2*f^2)*log(I*cos(d*x + c) + sin(d*x + c) + 1) + ((a - b)*d^2*f^2*x^2 + 2*(a - b)*d^2*e*f*x +
2*(a - b)*c*d*e*f - (a - b)*c^2*f^2)*log(I*cos(d*x + c) - sin(d*x + c) + 1) - ((a + b)*d^2*f^2*x^2 + 2*(a + b)
*d^2*e*f*x + 2*(a + b)*c*d*e*f - (a + b)*c^2*f^2)*log(-I*cos(d*x + c) + sin(d*x + c) + 1) + ((a - b)*d^2*f^2*x
^2 + 2*(a - b)*d^2*e*f*x + 2*(a - b)*c*d*e*f - (a - b)*c^2*f^2)*log(-I*cos(d*x + c) - sin(d*x + c) + 1) - ((a
+ b)*d^2*e^2 - 2*(a + b)*c*d*e*f + (a + b)*c^2*f^2)*log(-cos(d*x + c) + I*sin(d*x + c) + I) + ((a - b)*d^2*e^2
- 2*(a - b)*c*d*e*f + (a - b)*c^2*f^2)*log(-cos(d*x + c) - I*sin(d*x + c) + I))/((a^2 - b^2)*d^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{2} \sec \left (d x + c\right )}{b \sin \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \sec \left (d x +c \right )}{a +b \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x+e)^2*sec(d*x+c)/(a+b*sin(d*x+c)),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((f*x+e)^2*sec(d*x+c)/(a+b*sin(d*x+c)),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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e + f x\right )^{2} \sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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