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

Optimal. Leaf size=343 \[ \frac {i f^2 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{3 a d^3}-\frac {i f^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{3 a d^3}-\frac {2 i f^2 \text {Li}_2\left (-e^{2 i (c+d x)}\right )}{3 a d^3}+\frac {f^2 \tan (c+d x)}{3 a d^3}-\frac {f^2 \sec (c+d x)}{3 a d^3}+\frac {4 f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{3 a d^2}-\frac {2 i f (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{3 a d^2}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}+\frac {f (e+f x) \tan (c+d x) \sec (c+d x)}{3 a d^2}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {(e+f x)^2 \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {2 i (e+f x)^2}{3 a d} \]

[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

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Rubi [A]  time = 0.38, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4531, 4186, 3767, 8, 4184, 3719, 2190, 2279, 2391, 4409, 4185, 4181} \[ \frac {i f^2 \text {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{3 a d^3}-\frac {i f^2 \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{3 a d^3}-\frac {2 i f^2 \text {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{3 a d^3}+\frac {4 f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{3 a d^2}-\frac {2 i f (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{3 a d^2}-\frac {f (e+f x) \sec ^2(c+d x)}{3 a d^2}+\frac {f (e+f x) \tan (c+d x) \sec (c+d x)}{3 a d^2}+\frac {f^2 \tan (c+d x)}{3 a d^3}-\frac {f^2 \sec (c+d x)}{3 a d^3}+\frac {2 (e+f x)^2 \tan (c+d x)}{3 a d}-\frac {(e+f x)^2 \sec ^3(c+d x)}{3 a d}+\frac {(e+f x)^2 \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {2 i (e+f x)^2}{3 a d} \]

Antiderivative was successfully verified.

[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 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 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 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 3767

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 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 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 4185

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] && G
tQ[n, 1] && NeQ[n, 2]

Rule 4186

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]) /; FreeQ[
{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 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 4531

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*} \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=\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 \operatorname {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) \tan ^{-1}\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) \tan ^{-1}\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 ) \operatorname {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 ) \operatorname {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) \tan ^{-1}\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 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{3 a d^3}-\frac {i f^2 \text {Li}_2\left (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 ) \operatorname {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) \tan ^{-1}\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 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{3 a d^3}-\frac {i f^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{3 a d^3}-\frac {2 i f^2 \text {Li}_2\left (-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*}

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Mathematica [A]  time = 7.04, size = 637, normalized size = 1.86 \[ \frac {\frac {d^2 e^2 \sin (2 (c+d x))+2 d^2 e^2 \cos (c+d x)-4 d^2 e^2 \cos (c+2 d x)+2 d^2 e f x \sin (2 (c+d x))+4 d^2 e f x \cos (c+d x)-8 d^2 e f x \cos (c+2 d x)+d^2 f^2 x^2 \sin (2 (c+d x))+2 d^2 f^2 x^2 \cos (c+d x)-4 d^2 f^2 x^2 \cos (c+2 d x)-2 d e f \cos (2 c+d x)+2 f^2 \sin (2 (c+d x))-2 f^2 \sin (2 c+d x)-2 d f^2 x \cos (2 c+d x)+4 f^2 \cos (c+d x)-2 f^2 \cos (c+2 d x)-2 f^2 \cos (c)+8 d^2 e^2 \sin (d x)+16 d^2 e f x \sin (d x)+8 d^2 f^2 x^2 \sin (d x)-2 d f \cos (d x) (e+f x)+2 f^2 \sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {12 d^2 f (\cos (c)+i \sin (c)) \left (\frac {f (\cos (c)-i (\sin (c)-1)) \text {Li}_2(i \cos (c+d x)+\sin (c+d x))}{d^2}+\frac {(-\sin (c)-i \cos (c)+1) (e+f x) \log (-\sin (c+d x)-i \cos (c+d x)+1)}{d}+\frac {(\cos (c)-i \sin (c)) (e+f x)^2}{2 f}\right )}{\cos (c)+i (\sin (c)-1)}-\frac {20 d^2 f (\cos (c)+i \sin (c)) \left (\frac {f (\cos (c)-i (\sin (c)+1)) \text {Li}_2(-i \cos (c+d x)-\sin (c+d x))}{d^2}-\frac {(\sin (c)+i \cos (c)+1) (e+f x) \log (\sin (c+d x)+i \cos (c+d x)+1)}{d}+\frac {(\cos (c)-i \sin (c)) (e+f x)^2}{2 f}\right )}{\cos (c)+i (\sin (c)+1)}}{12 a d^3} \]

Warning: Unable to verify antiderivative.

[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)

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fricas [B]  time = 0.56, size = 855, normalized size = 2.49 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*sec(d*x+c)^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) + 3*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) - 5*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) - 3*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) + 5*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))*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*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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*sec(d*x+c)^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)

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maple [A]  time = 0.38, size = 573, normalized size = 1.67 \[ -\frac {2 \left (4 d^{2} e^{2} {\mathrm e}^{i \left (d x +c \right )}+2 i d^{2} e^{2}+f^{2} {\mathrm e}^{3 i \left (d x +c \right )}+i f^{2} {\mathrm e}^{2 i \left (d x +c \right )}+i f^{2}+4 d^{2} f^{2} x^{2} {\mathrm e}^{i \left (d x +c \right )}+2 i d^{2} f^{2} x^{2}+f^{2} {\mathrm e}^{i \left (d x +c \right )}+8 d^{2} e f x \,{\mathrm e}^{i \left (d x +c \right )}+i d \,f^{2} x \,{\mathrm e}^{i \left (d x +c \right )}+i d e f \,{\mathrm e}^{i \left (d x +c \right )}+i d \,f^{2} x \,{\mathrm e}^{3 i \left (d x +c \right )}+i d e f \,{\mathrm e}^{3 i \left (d x +c \right )}+4 i d^{2} e f x \right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{3} d^{3} a}+\frac {e f \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a \,d^{2}}+\frac {5 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e}{3 a \,d^{2}}-\frac {8 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right ) e}{3 a \,d^{2}}-\frac {f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a \,d^{3}}-\frac {5 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{3 a \,d^{3}}+\frac {8 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{3 a \,d^{3}}-\frac {5 i f^{2} \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{3 a \,d^{3}}-\frac {i f^{2} \polylog \left (2, -i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}-\frac {4 i f^{2} c^{2}}{3 a \,d^{3}}+\frac {f^{2} \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}+\frac {f^{2} \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{a \,d^{3}}-\frac {8 i f^{2} c x}{3 a \,d^{2}}+\frac {5 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{3 a \,d^{2}}+\frac {5 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{3 a \,d^{3}}-\frac {4 i f^{2} x^{2}}{3 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.87, size = 1332, normalized size = 3.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*sec(d*x+c)^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) + 20*I*d*
e*f*cos(3*d*x + 3*c) + 20*I*d*e*f*cos(d*x + c) + 10*I*d*e*f*sin(4*d*x + 4*c) - 20*d*e*f*sin(3*d*x + 3*c) - 20*
d*e*f*sin(d*x + c) - 10*d*e*f)*arctan2(sin(d*x + c) + 1, cos(d*x + c)) - (6*d*e*f*cos(4*d*x + 4*c) + 12*I*d*e*
f*cos(3*d*x + 3*c) + 12*I*d*e*f*cos(d*x + c) + 6*I*d*e*f*sin(4*d*x + 4*c) - 12*d*e*f*sin(3*d*x + 3*c) - 12*d*e
*f*sin(d*x + c) - 6*d*e*f)*arctan2(sin(d*x + c) - 1, cos(d*x + c)) + (10*d*f^2*x*cos(4*d*x + 4*c) + 20*I*d*f^2
*x*cos(3*d*x + 3*c) + 20*I*d*f^2*x*cos(d*x + c) + 10*I*d*f^2*x*sin(4*d*x + 4*c) - 20*d*f^2*x*sin(3*d*x + 3*c)
- 20*d*f^2*x*sin(d*x + c) - 10*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)
+ 12*I*d*f^2*x*cos(3*d*x + 3*c) + 12*I*d*f^2*x*cos(d*x + c) + 6*I*d*f^2*x*sin(4*d*x + 4*c) - 12*d*f^2*x*sin(3*
d*x + 3*c) - 12*d*f^2*x*sin(d*x + c) - 6*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) - (-16*I*d^2*f^2*x^2 - 4*d*e*f + 4*I*f^2 - 4*(8*I*d^2*e*f + d*f^2)*x)*cos(3*d*x
+ 3*c) - (16*I*d^2*e^2 - 4*d*f^2*x - 4*d*e*f + 4*I*f^2)*cos(d*x + c) + (10*f^2*cos(4*d*x + 4*c) + 20*I*f^2*cos
(3*d*x + 3*c) + 20*I*f^2*cos(d*x + c) + 10*I*f^2*sin(4*d*x + 4*c) - 20*f^2*sin(3*d*x + 3*c) - 20*f^2*sin(d*x +
c) - 10*f^2)*dilog(I*e^(I*d*x + I*c)) + (6*f^2*cos(4*d*x + 4*c) + 12*I*f^2*cos(3*d*x + 3*c) + 12*I*f^2*cos(d*
x + c) + 6*I*f^2*sin(4*d*x + 4*c) - 12*f^2*sin(3*d*x + 3*c) - 12*f^2*sin(d*x + c) - 6*f^2)*dilog(-I*e^(I*d*x +
I*c)) - (5*I*d*f^2*x + 5*I*d*e*f + (-5*I*d*f^2*x - 5*I*d*e*f)*cos(4*d*x + 4*c) + 10*(d*f^2*x + d*e*f)*cos(3*d
*x + 3*c) + 10*(d*f^2*x + d*e*f)*cos(d*x + c) + 5*(d*f^2*x + d*e*f)*sin(4*d*x + 4*c) + (10*I*d*f^2*x + 10*I*d*
e*f)*sin(3*d*x + 3*c) + (10*I*d*f^2*x + 10*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 + 3*I*d*e*f + (-3*I*d*f^2*x - 3*I*d*e*f)*cos(4*d*x + 4*c) + 6*(d*f^2*x + d*e*f)*c
os(3*d*x + 3*c) + 6*(d*f^2*x + d*e*f)*cos(d*x + c) + 3*(d*f^2*x + d*e*f)*sin(4*d*x + 4*c) + (6*I*d*f^2*x + 6*I
*d*e*f)*sin(3*d*x + 3*c) + (6*I*d*f^2*x + 6*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) - (-8*I*d^2*f^2*x^2 - 16*I*d^2*e*f*x)*sin(4*d*x + 4*c) - (16*d^2*f^2*x^2 - 4*I*d*e*f - 4*f^2 +
(32*d^2*e*f - 4*I*d*f^2)*x)*sin(3*d*x + 3*c) + (16*d^2*e^2 + 4*I*d*f^2*x + 4*I*d*e*f + 4*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)

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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)^2*(a + a*sin(c + d*x))),x)

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {e^{2} \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{2} x^{2} \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {2 e f x \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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