\(\int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [276]

Optimal result

Integrand size = 28, antiderivative size = 343 \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 i (e+f x)^2}{3 a d}-\frac {2 i f (e+f x) \arctan \left (e^{i (c+d x)}\right )}{3 a d^2}+\frac {4 f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{3 a d^2}+\frac {i f^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{3 a d^3}-\frac {i f^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{3 a d^3}-\frac {2 i f^2 \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{3 a d^3}-\frac {f^2 \sec (c+d x)}{3 a d^3}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {f^2 \tan (c+d x)}{3 a d^3}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}+\frac {f (e+f x) \sec (c+d x) \tan (c+d x)}{3 a d^2}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d} \]

[Out]

-2/3*I*(f*x+e)^2/a/d-2/3*I*f*(f*x+e)*arctan(exp(I*(d*x+c)))/a/d^2+4/3*f*(f*x+e)*ln(1+exp(2*I*(d*x+c)))/a/d^2+1
/3*I*f^2*polylog(2,-I*exp(I*(d*x+c)))/a/d^3-1/3*I*f^2*polylog(2,I*exp(I*(d*x+c)))/a/d^3-2/3*I*f^2*polylog(2,-e
xp(2*I*(d*x+c)))/a/d^3-1/3*f^2*sec(d*x+c)/a/d^3-1/3*f*(f*x+e)*sec(d*x+c)^2/a/d^2-1/3*(f*x+e)^2*sec(d*x+c)^3/a/
d+1/3*f^2*tan(d*x+c)/a/d^3+2/3*(f*x+e)^2*tan(d*x+c)/a/d+1/3*f*(f*x+e)*sec(d*x+c)*tan(d*x+c)/a/d^2+1/3*(f*x+e)^
2*sec(d*x+c)^2*tan(d*x+c)/a/d

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4627, 4271, 3852, 8, 4269, 3800, 2221, 2317, 2438, 4494, 4270, 4266} \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 i f (e+f x) \arctan \left (e^{i (c+d x)}\right )}{3 a d^2}+\frac {i f^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{3 a d^3}-\frac {i f^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{3 a d^3}-\frac {2 i f^2 \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{3 a d^3}+\frac {f^2 \tan (c+d x)}{3 a d^3}-\frac {f^2 \sec (c+d x)}{3 a d^3}+\frac {4 f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{3 a d^2}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}+\frac {f (e+f x) \tan (c+d x) \sec (c+d x)}{3 a d^2}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {(e+f x)^2 \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {2 i (e+f x)^2}{3 a d} \]

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2317

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 2438

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 3800

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 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 4266

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 4269

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 4270

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

Rule 4271

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-b^2)*(c + d*x)^m*Cot[e
+ f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), x] + (Dist[b^2*d^2*m*((m - 1)/(f^2*(n - 1)*(n - 2))), Int[(c +
d*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[b^2*((n - 2)/(n - 1)), Int[(c + d*x)^m*(b*Csc[e + f*x])^
(n - 2), x], x] - Simp[b^2*d*m*(c + d*x)^(m - 1)*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x]) /; Free
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rule 4494

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 4627

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int (e+f x)^2 \sec ^4(c+d x) \, dx}{a}-\frac {\int (e+f x)^2 \sec ^3(c+d x) \tan (c+d x) \, dx}{a} \\ & = -\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {2 \int (e+f x)^2 \sec ^2(c+d x) \, dx}{3 a}+\frac {(2 f) \int (e+f x) \sec ^3(c+d x) \, dx}{3 a d}+\frac {f^2 \int \sec ^2(c+d x) \, dx}{3 a d^2} \\ & = -\frac {f^2 \sec (c+d x)}{3 a d^3}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}+\frac {f (e+f x) \sec (c+d x) \tan (c+d x)}{3 a d^2}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {f \int (e+f x) \sec (c+d x) \, dx}{3 a d}-\frac {(4 f) \int (e+f x) \tan (c+d x) \, dx}{3 a d}-\frac {f^2 \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a d^3} \\ & = -\frac {2 i (e+f x)^2}{3 a d}-\frac {2 i f (e+f x) \arctan \left (e^{i (c+d x)}\right )}{3 a d^2}-\frac {f^2 \sec (c+d x)}{3 a d^3}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {f^2 \tan (c+d x)}{3 a d^3}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}+\frac {f (e+f x) \sec (c+d x) \tan (c+d x)}{3 a d^2}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {(8 i f) \int \frac {e^{2 i (c+d x)} (e+f x)}{1+e^{2 i (c+d x)}} \, dx}{3 a d}-\frac {f^2 \int \log \left (1-i e^{i (c+d x)}\right ) \, dx}{3 a d^2}+\frac {f^2 \int \log \left (1+i e^{i (c+d x)}\right ) \, dx}{3 a d^2} \\ & = -\frac {2 i (e+f x)^2}{3 a d}-\frac {2 i f (e+f x) \arctan \left (e^{i (c+d x)}\right )}{3 a d^2}+\frac {4 f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{3 a d^2}-\frac {f^2 \sec (c+d x)}{3 a d^3}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {f^2 \tan (c+d x)}{3 a d^3}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}+\frac {f (e+f x) \sec (c+d x) \tan (c+d x)}{3 a d^2}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\left (i f^2\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{3 a d^3}-\frac {\left (i f^2\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{3 a d^3}-\frac {\left (4 f^2\right ) \int \log \left (1+e^{2 i (c+d x)}\right ) \, dx}{3 a d^2} \\ & = -\frac {2 i (e+f x)^2}{3 a d}-\frac {2 i f (e+f x) \arctan \left (e^{i (c+d x)}\right )}{3 a d^2}+\frac {4 f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{3 a d^2}+\frac {i f^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{3 a d^3}-\frac {i f^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{3 a d^3}-\frac {f^2 \sec (c+d x)}{3 a d^3}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {f^2 \tan (c+d x)}{3 a d^3}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}+\frac {f (e+f x) \sec (c+d x) \tan (c+d x)}{3 a d^2}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac {\left (2 i f^2\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i (c+d x)}\right )}{3 a d^3} \\ & = -\frac {2 i (e+f x)^2}{3 a d}-\frac {2 i f (e+f x) \arctan \left (e^{i (c+d x)}\right )}{3 a d^2}+\frac {4 f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{3 a d^2}+\frac {i f^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{3 a d^3}-\frac {i f^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{3 a d^3}-\frac {2 i f^2 \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{3 a d^3}-\frac {f^2 \sec (c+d x)}{3 a d^3}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {f^2 \tan (c+d x)}{3 a d^3}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}+\frac {f (e+f x) \sec (c+d x) \tan (c+d x)}{3 a d^2}+\frac {(e+f x)^2 \sec ^2(c+d x) \tan (c+d x)}{3 a d} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 5.07 (sec) , antiderivative size = 637, normalized size of antiderivative = 1.86 \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {12 d^2 f \left (\frac {f \operatorname {PolyLog}(2,i \cos (c+d x)+\sin (c+d x)) (\cos (c)-i (-1+\sin (c)))}{d^2}+\frac {(e+f x) \log (1-i \cos (c+d x)-\sin (c+d x)) (1-i \cos (c)-\sin (c))}{d}+\frac {(e+f x)^2 (\cos (c)-i \sin (c))}{2 f}\right ) (\cos (c)+i \sin (c))}{\cos (c)+i (-1+\sin (c))}-\frac {20 d^2 f (\cos (c)+i \sin (c)) \left (\frac {(e+f x)^2 (\cos (c)-i \sin (c))}{2 f}-\frac {(e+f x) \log (1+i \cos (c+d x)+\sin (c+d x)) (1+i \cos (c)+\sin (c))}{d}+\frac {f \operatorname {PolyLog}(2,-i \cos (c+d x)-\sin (c+d x)) (\cos (c)-i (1+\sin (c)))}{d^2}\right )}{\cos (c)+i (1+\sin (c))}+\frac {-2 f^2 \cos (c)-2 d f (e+f x) \cos (d x)+2 d^2 e^2 \cos (c+d x)+4 f^2 \cos (c+d x)+4 d^2 e f x \cos (c+d x)+2 d^2 f^2 x^2 \cos (c+d x)-2 d e f \cos (2 c+d x)-2 d f^2 x \cos (2 c+d x)-4 d^2 e^2 \cos (c+2 d x)-2 f^2 \cos (c+2 d x)-8 d^2 e f x \cos (c+2 d x)-4 d^2 f^2 x^2 \cos (c+2 d x)+8 d^2 e^2 \sin (d x)+2 f^2 \sin (d x)+16 d^2 e f x \sin (d x)+8 d^2 f^2 x^2 \sin (d x)+d^2 e^2 \sin (2 (c+d x))+2 f^2 \sin (2 (c+d x))+2 d^2 e f x \sin (2 (c+d x))+d^2 f^2 x^2 \sin (2 (c+d x))-2 f^2 \sin (2 c+d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}}{12 a d^3} \]

[In]

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

[Out]

((12*d^2*f*((f*PolyLog[2, I*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] - I*(-1 + Sin[c])))/d^2 + ((e + f*x)*Log[1 -
I*Cos[c + d*x] - Sin[c + d*x]]*(1 - I*Cos[c] - Sin[c]))/d + ((e + f*x)^2*(Cos[c] - I*Sin[c]))/(2*f))*(Cos[c] +
I*Sin[c]))/(Cos[c] + I*(-1 + Sin[c])) - (20*d^2*f*(Cos[c] + I*Sin[c])*(((e + f*x)^2*(Cos[c] - I*Sin[c]))/(2*f
) - ((e + f*x)*Log[1 + I*Cos[c + d*x] + Sin[c + d*x]]*(1 + I*Cos[c] + Sin[c]))/d + (f*PolyLog[2, (-I)*Cos[c +
d*x] - Sin[c + d*x]]*(Cos[c] - I*(1 + Sin[c])))/d^2))/(Cos[c] + I*(1 + Sin[c])) + (-2*f^2*Cos[c] - 2*d*f*(e +
f*x)*Cos[d*x] + 2*d^2*e^2*Cos[c + d*x] + 4*f^2*Cos[c + d*x] + 4*d^2*e*f*x*Cos[c + d*x] + 2*d^2*f^2*x^2*Cos[c +
d*x] - 2*d*e*f*Cos[2*c + d*x] - 2*d*f^2*x*Cos[2*c + d*x] - 4*d^2*e^2*Cos[c + 2*d*x] - 2*f^2*Cos[c + 2*d*x] -
8*d^2*e*f*x*Cos[c + 2*d*x] - 4*d^2*f^2*x^2*Cos[c + 2*d*x] + 8*d^2*e^2*Sin[d*x] + 2*f^2*Sin[d*x] + 16*d^2*e*f*x
*Sin[d*x] + 8*d^2*f^2*x^2*Sin[d*x] + d^2*e^2*Sin[2*(c + d*x)] + 2*f^2*Sin[2*(c + d*x)] + 2*d^2*e*f*x*Sin[2*(c
+ d*x)] + d^2*f^2*x^2*Sin[2*(c + d*x)] - 2*f^2*Sin[2*c + d*x])/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(C
os[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3))/(12*a*d^3)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 568, normalized size of antiderivative = 1.66

method	result	size
risch	\(-\frac {2 \left (4 \,{\mathrm e}^{i \left (d x +c \right )} d^{2} f^{2} x^{2}+2 i d^{2} x^{2} f^{2}+f^{2} {\mathrm e}^{3 i \left (d x +c \right )}+i f^{2}+i f^{2} {\mathrm e}^{2 i \left (d x +c \right )}+8 \,{\mathrm e}^{i \left (d x +c \right )} d^{2} e f x +i d e f \,{\mathrm e}^{i \left (d x +c \right )}+i d \,f^{2} x \,{\mathrm e}^{3 i \left (d x +c \right )}+i d e f \,{\mathrm e}^{3 i \left (d x +c \right )}+4 i d^{2} e f x +i d \,f^{2} x \,{\mathrm e}^{i \left (d x +c \right )}+f^{2} {\mathrm e}^{i \left (d x +c \right )}+4 \,{\mathrm e}^{i \left (d x +c \right )} d^{2} e^{2}+2 i d^{2} e^{2}\right )}{3 \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{3} d^{3} a}+\frac {4 f e \ln \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 a \,d^{2}}-\frac {4 f^{2} c \ln \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 a \,d^{3}}-\frac {5 i f^{2} \operatorname {Li}_{2}\left (i {\mathrm e}^{i \left (d x +c \right )}\right )}{3 a \,d^{3}}+\frac {5 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{3 a \,d^{3}}+\frac {8 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{3 a \,d^{3}}+\frac {2 i f^{2} c \arctan \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{3 a \,d^{3}}-\frac {4 i f^{2} x^{2}}{3 a d}-\frac {4 i f^{2} c^{2}}{3 a \,d^{3}}+\frac {f^{2} \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{d^{3} a}-\frac {8 f e \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{3 a \,d^{2}}-\frac {i f^{2} \operatorname {Li}_{2}\left (-i {\mathrm e}^{i \left (d x +c \right )}\right )}{d^{3} a}+\frac {f^{2} \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{d^{2} a}+\frac {5 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{3 a \,d^{2}}-\frac {8 i c \,f^{2} x}{3 d^{2} a}-\frac {2 i e f \arctan \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{3 a \,d^{2}}\)	\(568\)

[In]

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

[Out]

-2/3*(4*exp(I*(d*x+c))*d^2*f^2*x^2+2*I*d^2*x^2*f^2+f^2*exp(3*I*(d*x+c))+I*f^2+I*f^2*exp(2*I*(d*x+c))+8*exp(I*(
d*x+c))*d^2*e*f*x+I*d*e*f*exp(I*(d*x+c))+I*d*f^2*x*exp(3*I*(d*x+c))+I*d*e*f*exp(3*I*(d*x+c))+4*I*d^2*e*f*x+I*d
*f^2*x*exp(I*(d*x+c))+f^2*exp(I*(d*x+c))+4*exp(I*(d*x+c))*d^2*e^2+2*I*d^2*e^2)/(-I+exp(I*(d*x+c)))/(exp(I*(d*x
+c))+I)^3/d^3/a+4/3/d^2/a*e*f*ln(1+exp(2*I*(d*x+c)))-4/3/d^3/a*f^2*c*ln(1+exp(2*I*(d*x+c)))-5/3*I/d^3/a*f^2*po
lylog(2,I*exp(I*(d*x+c)))+5/3/d^3/a*f^2*ln(1-I*exp(I*(d*x+c)))*c+8/3/d^3/a*f^2*c*ln(exp(I*(d*x+c)))+2/3*I/d^3/
a*f^2*c*arctan(exp(I*(d*x+c)))-4/3*I/d/a*f^2*x^2-4/3*I/d^3/a*f^2*c^2+1/d^3/a*f^2*ln(1+I*exp(I*(d*x+c)))*c-8/3/
d^2/a*e*f*ln(exp(I*(d*x+c)))-I/d^3/a*f^2*polylog(2,-I*exp(I*(d*x+c)))+1/d^2/a*f^2*ln(1+I*exp(I*(d*x+c)))*x+5/3
/d^2/a*f^2*ln(1-I*exp(I*(d*x+c)))*x-8/3*I/d^2/a*c*f^2*x-2/3*I/d^2/a*e*f*arctan(exp(I*(d*x+c)))

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. 859 vs. \(2 (290) = 580\).

Time = 0.31 (sec) , antiderivative size = 859, normalized size of antiderivative = 2.50 \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/6*(2*d^2*f^2*x^2 + 4*d^2*e*f*x + 2*d^2*e^2 - 2*(2*d^2*f^2*x^2 + 4*d^2*e*f*x + 2*d^2*e^2 + f^2)*cos(d*x + c)^
2 - 2*(d*f^2*x + d*e*f)*cos(d*x + c) - 3*(-I*f^2*cos(d*x + c)*sin(d*x + c) - I*f^2*cos(d*x + c))*dilog(I*cos(d
*x + c) + sin(d*x + c)) - 5*(I*f^2*cos(d*x + c)*sin(d*x + c) + I*f^2*cos(d*x + c))*dilog(I*cos(d*x + c) - sin(
d*x + c)) - 3*(I*f^2*cos(d*x + c)*sin(d*x + c) + I*f^2*cos(d*x + c))*dilog(-I*cos(d*x + c) + sin(d*x + c)) - 5
*(-I*f^2*cos(d*x + c)*sin(d*x + c) - I*f^2*cos(d*x + c))*dilog(-I*cos(d*x + c) - sin(d*x + c)) + 5*((d*e*f - c
*f^2)*cos(d*x + c)*sin(d*x + c) + (d*e*f - c*f^2)*cos(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) + 3*((d
*e*f - c*f^2)*cos(d*x + c)*sin(d*x + c) + (d*e*f - c*f^2)*cos(d*x + c))*log(cos(d*x + c) - I*sin(d*x + c) + I)
+ 5*((d*f^2*x + c*f^2)*cos(d*x + c)*sin(d*x + c) + (d*f^2*x + c*f^2)*cos(d*x + c))*log(I*cos(d*x + c) + sin(d
*x + c) + 1) + 3*((d*f^2*x + c*f^2)*cos(d*x + c)*sin(d*x + c) + (d*f^2*x + c*f^2)*cos(d*x + c))*log(I*cos(d*x
+ c) - sin(d*x + c) + 1) + 5*((d*f^2*x + c*f^2)*cos(d*x + c)*sin(d*x + c) + (d*f^2*x + c*f^2)*cos(d*x + c))*lo
g(-I*cos(d*x + c) + sin(d*x + c) + 1) + 3*((d*f^2*x + c*f^2)*cos(d*x + c)*sin(d*x + c) + (d*f^2*x + c*f^2)*cos
(d*x + c))*log(-I*cos(d*x + c) - sin(d*x + c) + 1) + 5*((d*e*f - c*f^2)*cos(d*x + c)*sin(d*x + c) + (d*e*f - c
*f^2)*cos(d*x + c))*log(-cos(d*x + c) + I*sin(d*x + c) + I) + 3*((d*e*f - c*f^2)*cos(d*x + c)*sin(d*x + c) + (
d*e*f - c*f^2)*cos(d*x + c))*log(-cos(d*x + c) - I*sin(d*x + c) + I) + 4*(d^2*f^2*x^2 + 2*d^2*e*f*x + d^2*e^2)
*sin(d*x + c))/(a*d^3*cos(d*x + c)*sin(d*x + c) + a*d^3*cos(d*x + c))

Sympy [F]

\[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {e^{2} \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{2} x^{2} \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {2 e f x \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

(Integral(e**2*sec(c + d*x)**2/(sin(c + d*x) + 1), x) + Integral(f**2*x**2*sec(c + d*x)**2/(sin(c + d*x) + 1),
x) + Integral(2*e*f*x*sec(c + d*x)**2/(sin(c + d*x) + 1), x))/a

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1328 vs. \(2 (290) = 580\).

Time = 0.57 (sec) , antiderivative size = 1328, normalized size of antiderivative = 3.87 \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-(8*d^2*e^2 + 4*f^2*cos(2*d*x + 2*c) + 4*I*f^2*sin(2*d*x + 2*c) + 4*f^2 - 10*(d*e*f*cos(4*d*x + 4*c) + 2*I*d*e
*f*cos(3*d*x + 3*c) + 2*I*d*e*f*cos(d*x + c) + I*d*e*f*sin(4*d*x + 4*c) - 2*d*e*f*sin(3*d*x + 3*c) - 2*d*e*f*s
in(d*x + c) - d*e*f)*arctan2(sin(d*x + c) + 1, cos(d*x + c)) - 6*(d*e*f*cos(4*d*x + 4*c) + 2*I*d*e*f*cos(3*d*x
+ 3*c) + 2*I*d*e*f*cos(d*x + c) + I*d*e*f*sin(4*d*x + 4*c) - 2*d*e*f*sin(3*d*x + 3*c) - 2*d*e*f*sin(d*x + c)
- d*e*f)*arctan2(sin(d*x + c) - 1, cos(d*x + c)) + 10*(d*f^2*x*cos(4*d*x + 4*c) + 2*I*d*f^2*x*cos(3*d*x + 3*c)
+ 2*I*d*f^2*x*cos(d*x + c) + I*d*f^2*x*sin(4*d*x + 4*c) - 2*d*f^2*x*sin(3*d*x + 3*c) - 2*d*f^2*x*sin(d*x + c)
- d*f^2*x)*arctan2(cos(d*x + c), sin(d*x + c) + 1) - 6*(d*f^2*x*cos(4*d*x + 4*c) + 2*I*d*f^2*x*cos(3*d*x + 3*
c) + 2*I*d*f^2*x*cos(d*x + c) + I*d*f^2*x*sin(4*d*x + 4*c) - 2*d*f^2*x*sin(3*d*x + 3*c) - 2*d*f^2*x*sin(d*x +
c) - d*f^2*x)*arctan2(cos(d*x + c), -sin(d*x + c) + 1) + 8*(d^2*f^2*x^2 + 2*d^2*e*f*x)*cos(4*d*x + 4*c) + 4*(4
*I*d^2*f^2*x^2 + d*e*f - I*f^2 + (8*I*d^2*e*f + d*f^2)*x)*cos(3*d*x + 3*c) + 4*(-4*I*d^2*e^2 + d*f^2*x + d*e*f
- I*f^2)*cos(d*x + c) + 10*(f^2*cos(4*d*x + 4*c) + 2*I*f^2*cos(3*d*x + 3*c) + 2*I*f^2*cos(d*x + c) + I*f^2*si
n(4*d*x + 4*c) - 2*f^2*sin(3*d*x + 3*c) - 2*f^2*sin(d*x + c) - f^2)*dilog(I*e^(I*d*x + I*c)) + 6*(f^2*cos(4*d*
x + 4*c) + 2*I*f^2*cos(3*d*x + 3*c) + 2*I*f^2*cos(d*x + c) + I*f^2*sin(4*d*x + 4*c) - 2*f^2*sin(3*d*x + 3*c) -
2*f^2*sin(d*x + c) - f^2)*dilog(-I*e^(I*d*x + I*c)) + 5*(-I*d*f^2*x - I*d*e*f + (I*d*f^2*x + I*d*e*f)*cos(4*d
*x + 4*c) - 2*(d*f^2*x + d*e*f)*cos(3*d*x + 3*c) - 2*(d*f^2*x + d*e*f)*cos(d*x + c) - (d*f^2*x + d*e*f)*sin(4*
d*x + 4*c) + 2*(-I*d*f^2*x - I*d*e*f)*sin(3*d*x + 3*c) + 2*(-I*d*f^2*x - I*d*e*f)*sin(d*x + c))*log(cos(d*x +
c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) + 3*(-I*d*f^2*x - I*d*e*f + (I*d*f^2*x + I*d*e*f)*cos(4*d*x + 4*c)
- 2*(d*f^2*x + d*e*f)*cos(3*d*x + 3*c) - 2*(d*f^2*x + d*e*f)*cos(d*x + c) - (d*f^2*x + d*e*f)*sin(4*d*x + 4*c
) + 2*(-I*d*f^2*x - I*d*e*f)*sin(3*d*x + 3*c) + 2*(-I*d*f^2*x - I*d*e*f)*sin(d*x + c))*log(cos(d*x + c)^2 + si
n(d*x + c)^2 - 2*sin(d*x + c) + 1) + 8*(I*d^2*f^2*x^2 + 2*I*d^2*e*f*x)*sin(4*d*x + 4*c) - 4*(4*d^2*f^2*x^2 - I
*d*e*f - f^2 + (8*d^2*e*f - I*d*f^2)*x)*sin(3*d*x + 3*c) + 4*(4*d^2*e^2 + I*d*f^2*x + I*d*e*f + f^2)*sin(d*x +
c))/(-6*I*a*d^3*cos(4*d*x + 4*c) + 12*a*d^3*cos(3*d*x + 3*c) + 12*a*d^3*cos(d*x + c) + 6*a*d^3*sin(4*d*x + 4*
c) + 12*I*a*d^3*sin(3*d*x + 3*c) + 12*I*a*d^3*sin(d*x + c) + 6*I*a*d^3)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Hanged} \]

[In]

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

[Out]

\text{Hanged}