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

Optimal. Leaf size=278 \[ -\frac {f^2 \text {Li}_3\left (-i e^{i (c+d x)}\right )}{a d^3}+\frac {f^2 \text {Li}_3\left (i e^{i (c+d x)}\right )}{a d^3}+\frac {f^2 \tanh ^{-1}(\sin (c+d x))}{a d^3}+\frac {f^2 \log (\cos (c+d x))}{a d^3}+\frac {i f (e+f x) \text {Li}_2\left (-i e^{i (c+d x)}\right )}{a d^2}-\frac {i f (e+f x) \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^2}+\frac {f (e+f x) \tan (c+d x)}{a d^2}-\frac {f (e+f x) \sec (c+d x)}{a d^2}-\frac {i (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {(e+f x)^2 \sec ^2(c+d x)}{2 a d}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 a d} \]

[Out]

-I*(f*x+e)^2*arctan(exp(I*(d*x+c)))/a/d+f^2*arctanh(sin(d*x+c))/a/d^3+f^2*ln(cos(d*x+c))/a/d^3+I*f*(f*x+e)*pol
ylog(2,-I*exp(I*(d*x+c)))/a/d^2-I*f*(f*x+e)*polylog(2,I*exp(I*(d*x+c)))/a/d^2-f^2*polylog(3,-I*exp(I*(d*x+c)))
/a/d^3+f^2*polylog(3,I*exp(I*(d*x+c)))/a/d^3-f*(f*x+e)*sec(d*x+c)/a/d^2-1/2*(f*x+e)^2*sec(d*x+c)^2/a/d+f*(f*x+
e)*tan(d*x+c)/a/d^2+1/2*(f*x+e)^2*sec(d*x+c)*tan(d*x+c)/a/d

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Rubi [A]  time = 0.27, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {4531, 4186, 3770, 4181, 2531, 2282, 6589, 4409, 4184, 3475} \[ \frac {i f (e+f x) \text {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{a d^2}-\frac {i f (e+f x) \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^2}-\frac {f^2 \text {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{a d^3}+\frac {f^2 \text {PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^3}+\frac {f (e+f x) \tan (c+d x)}{a d^2}-\frac {f (e+f x) \sec (c+d x)}{a d^2}+\frac {f^2 \tanh ^{-1}(\sin (c+d x))}{a d^3}+\frac {f^2 \log (\cos (c+d x))}{a d^3}-\frac {i (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {(e+f x)^2 \sec ^2(c+d x)}{2 a d}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(e + f*x)^2*ArcTan[E^(I*(c + d*x))])/(a*d) + (f^2*ArcTanh[Sin[c + d*x]])/(a*d^3) + (f^2*Log[Cos[c + d*x]
])/(a*d^3) + (I*f*(e + f*x)*PolyLog[2, (-I)*E^(I*(c + d*x))])/(a*d^2) - (I*f*(e + f*x)*PolyLog[2, I*E^(I*(c +
d*x))])/(a*d^2) - (f^2*PolyLog[3, (-I)*E^(I*(c + d*x))])/(a*d^3) + (f^2*PolyLog[3, I*E^(I*(c + d*x))])/(a*d^3)
- (f*(e + f*x)*Sec[c + d*x])/(a*d^2) - ((e + f*x)^2*Sec[c + d*x]^2)/(2*a*d) + (f*(e + f*x)*Tan[c + d*x])/(a*d
^2) + ((e + f*x)^2*Sec[c + d*x]*Tan[c + d*x])/(2*a*d)

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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 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]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

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

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Mathematica [B]  time = 8.21, size = 670, normalized size = 2.41 \[ -\frac {\frac {6 f \left (i d (e+f x) \text {Li}_2\left (i e^{-i (c+d x)}\right )+f \text {Li}_3\left (i e^{-i (c+d x)}\right )\right )}{d^3}+\frac {3 (e+f x)^2 \log \left (1-i e^{-i (c+d x)}\right )}{d}+\frac {(e+f x)^3}{\left (e^{i c}-i\right ) f}}{6 a}-\frac {(\cos (c)+i \sin (c)) \left (x (\cos (c)-i \sin (c)) \left (d^2 e^2+4 f^2\right )+\frac {(\sin (c)+i \cos (c)) (\cos (c)+i (\sin (c)+1)) \left (d^2 e^2+4 f^2\right ) (d x+i \log (\cos (c+d x)+i (\sin (c+d x)+1)))}{d}-i d^2 e f x^2 \sin (c)+d^2 e f x^2 \cos (c)+\frac {1}{3} d^2 f^2 x^3 (\cos (c)-i \sin (c))+2 e f (\cos (c)-i (\sin (c)+1)) \text {Li}_2(-i \cos (c+d x)-\sin (c+d x))-2 d e f x (\cos (c)-i \sin (c)) (\cos (c)+i (\sin (c)+1)) \log (\sin (c+d x)+i \cos (c+d x)+1)+\frac {2 f^2 (\cos (c)-i \sin (c)) (\sin (c)-i \cos (c)+1) (d x \text {Li}_2(-i \cos (c+d x)-\sin (c+d x))-i \text {Li}_3(-i \cos (c+d x)-\sin (c+d x)))}{d}-d f^2 x^2 (\cos (c)-i \sin (c)) (\cos (c)+i (\sin (c)+1)) \log (\sin (c+d x)+i \cos (c+d x)+1)\right )}{2 a d^2 (\cos (c)+i (\sin (c)+1))}+\frac {2 \left (e f \sin \left (\frac {d x}{2}\right )+f^2 x \sin \left (\frac {d x}{2}\right )\right )}{a d^2 \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}-\frac {(e+f x)^2}{2 a d \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {x \left (3 e^2+3 e f x+f^2 x^2\right )}{6 a \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/6*((e + f*x)^3/((-I + E^(I*c))*f) + (3*(e + f*x)^2*Log[1 - I/E^(I*(c + d*x))])/d + (6*f*(I*d*(e + f*x)*Poly
Log[2, I/E^(I*(c + d*x))] + f*PolyLog[3, I/E^(I*(c + d*x))]))/d^3)/a + (x*(3*e^2 + 3*e*f*x + f^2*x^2))/(6*a*(C
os[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])) - ((Cos[c] + I*Sin[c])*(d^2*e*f*x^2*Cos[c] + (d^2*e^2 + 4*f^2)*x*(C
os[c] - I*Sin[c]) + (d^2*f^2*x^3*(Cos[c] - I*Sin[c]))/3 - I*d^2*e*f*x^2*Sin[c] + (2*f^2*(d*x*PolyLog[2, (-I)*C
os[c + d*x] - Sin[c + d*x]] - I*PolyLog[3, (-I)*Cos[c + d*x] - Sin[c + d*x]])*(Cos[c] - I*Sin[c])*(1 - I*Cos[c
] + Sin[c]))/d + 2*e*f*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*(1 + Sin[c])) - 2*d*e*f*x*Log[
1 + I*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] - I*Sin[c])*(Cos[c] + I*(1 + Sin[c])) - d*f^2*x^2*Log[1 + I*Cos[c +
d*x] + Sin[c + d*x]]*(Cos[c] - I*Sin[c])*(Cos[c] + I*(1 + Sin[c])) + ((d^2*e^2 + 4*f^2)*(d*x + I*Log[Cos[c +
d*x] + I*(1 + Sin[c + d*x])])*(I*Cos[c] + Sin[c])*(Cos[c] + I*(1 + Sin[c])))/d))/(2*a*d^2*(Cos[c] + I*(1 + Sin
[c]))) - (e + f*x)^2/(2*a*d*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (2*(e*f*Sin[(d*x)/2] + f^2*x*Sin[(d
*x)/2]))/(a*d^2*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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fricas [C]  time = 0.58, size = 1064, normalized size = 3.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*d^2*f^2*x^2 + 4*d^2*e*f*x + 2*d^2*e^2 + 4*(d*f^2*x + d*e*f)*cos(d*x + c) - (-2*I*d*f^2*x - 2*I*d*e*f +
(-2*I*d*f^2*x - 2*I*d*e*f)*sin(d*x + c))*dilog(I*cos(d*x + c) + sin(d*x + c)) - (-2*I*d*f^2*x - 2*I*d*e*f + (
-2*I*d*f^2*x - 2*I*d*e*f)*sin(d*x + c))*dilog(I*cos(d*x + c) - sin(d*x + c)) - (2*I*d*f^2*x + 2*I*d*e*f + (2*I
*d*f^2*x + 2*I*d*e*f)*sin(d*x + c))*dilog(-I*cos(d*x + c) + sin(d*x + c)) - (2*I*d*f^2*x + 2*I*d*e*f + (2*I*d*
f^2*x + 2*I*d*e*f)*sin(d*x + c))*dilog(-I*cos(d*x + c) - sin(d*x + c)) - (d^2*e^2 - 2*c*d*e*f + (c^2 + 4)*f^2
+ (d^2*e^2 - 2*c*d*e*f + (c^2 + 4)*f^2)*sin(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) + (d^2*e^2 - 2*c*
d*e*f + c^2*f^2 + (d^2*e^2 - 2*c*d*e*f + c^2*f^2)*sin(d*x + c))*log(cos(d*x + c) - I*sin(d*x + c) + I) - (d^2*
f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2 + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2)*sin(d*x + c))*
log(I*cos(d*x + c) + sin(d*x + c) + 1) + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2 + (d^2*f^2*x^2 + 2*d
^2*e*f*x + 2*c*d*e*f - c^2*f^2)*sin(d*x + c))*log(I*cos(d*x + c) - sin(d*x + c) + 1) - (d^2*f^2*x^2 + 2*d^2*e*
f*x + 2*c*d*e*f - c^2*f^2 + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2)*sin(d*x + c))*log(-I*cos(d*x + c
) + sin(d*x + c) + 1) + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2 + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*
e*f - c^2*f^2)*sin(d*x + c))*log(-I*cos(d*x + c) - sin(d*x + c) + 1) - (d^2*e^2 - 2*c*d*e*f + (c^2 + 4)*f^2 +
(d^2*e^2 - 2*c*d*e*f + (c^2 + 4)*f^2)*sin(d*x + c))*log(-cos(d*x + c) + I*sin(d*x + c) + I) + (d^2*e^2 - 2*c*d
*e*f + c^2*f^2 + (d^2*e^2 - 2*c*d*e*f + c^2*f^2)*sin(d*x + c))*log(-cos(d*x + c) - I*sin(d*x + c) + I) + 2*(f^
2*sin(d*x + c) + f^2)*polylog(3, I*cos(d*x + c) + sin(d*x + c)) - 2*(f^2*sin(d*x + c) + f^2)*polylog(3, I*cos(
d*x + c) - sin(d*x + c)) + 2*(f^2*sin(d*x + c) + f^2)*polylog(3, -I*cos(d*x + c) + sin(d*x + c)) - 2*(f^2*sin(
d*x + c) + f^2)*polylog(3, -I*cos(d*x + c) - sin(d*x + c)))/(a*d^3*sin(d*x + c) + a*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.31, size = 677, normalized size = 2.44 \[ -\frac {i \left (d \,f^{2} x^{2} {\mathrm e}^{i \left (d x +c \right )}+2 d e f x \,{\mathrm e}^{i \left (d x +c \right )}+d \,e^{2} {\mathrm e}^{i \left (d x +c \right )}+2 f^{2} x -2 i f^{2} x \,{\mathrm e}^{i \left (d x +c \right )}+2 e f -2 i e f \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} a}+\frac {2 f^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d^{3} a}-\frac {2 f^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{d^{3} a}-\frac {e^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d a}+\frac {f e \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{d a}+\frac {f e \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{d^{2} a}-\frac {f e c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d^{2} a}+\frac {f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x^{2}}{2 d a}-\frac {f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c^{2}}{2 d^{3} a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e^{2}}{2 d a}+\frac {\ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) c^{2} f^{2}}{2 d^{3} a}-\frac {f^{2} c^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d^{3} a}-\frac {\ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) f^{2} x^{2}}{2 d a}+\frac {i \polylog \left (2, -i {\mathrm e}^{i \left (d x +c \right )}\right ) f^{2} x}{d^{2} a}+\frac {i e f \polylog \left (2, -i {\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} a}-\frac {\ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) e f x}{d a}-\frac {\ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) c e f}{d^{2} a}+\frac {e f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d^{2} a}-\frac {i f^{2} \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{d^{2} a}-\frac {i f e \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} a}+\frac {f^{2} c^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d^{3} a}-\frac {f^{2} \polylog \left (3, -i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {f^{2} \polylog \left (3, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.84, size = 1923, normalized size = 6.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*c*e*f*(2/(a*d*sin(d*x + c) + a*d) - log(sin(d*x + c) + 1)/(a*d) + log(sin(d*x + c) - 1)/(a*d)) + e^2*(l
og(sin(d*x + c) + 1)/a - log(sin(d*x + c) - 1)/a - 2/(a*sin(d*x + c) + a)) - 4*(8*(d*x + c)*f^2*cos(2*d*x + 2*
c) + 8*I*(d*x + c)*f^2*sin(2*d*x + 2*c) + 8*d*e*f - 8*c*f^2 - (2*(c^2 + 4)*f^2*cos(2*d*x + 2*c) + (4*I*c^2 + 1
6*I)*f^2*cos(d*x + c) + (2*I*c^2 + 8*I)*f^2*sin(2*d*x + 2*c) - 4*(c^2 + 4)*f^2*sin(d*x + c) - 2*(c^2 + 4)*f^2)
*arctan2(sin(d*x + c) + 1, cos(d*x + c)) + (2*c^2*f^2*cos(2*d*x + 2*c) + 4*I*c^2*f^2*cos(d*x + c) + 2*I*c^2*f^
2*sin(2*d*x + 2*c) - 4*c^2*f^2*sin(d*x + c) - 2*c^2*f^2)*arctan2(sin(d*x + c) - 1, cos(d*x + c)) - (2*(d*x + c
)^2*f^2 + 4*(d*e*f - c*f^2)*(d*x + c) - 2*((d*x + c)^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*cos(2*d*x + 2*c) + (
-4*I*(d*x + c)^2*f^2 + (-8*I*d*e*f + 8*I*c*f^2)*(d*x + c))*cos(d*x + c) + (-2*I*(d*x + c)^2*f^2 + (-4*I*d*e*f
+ 4*I*c*f^2)*(d*x + c))*sin(2*d*x + 2*c) + 4*((d*x + c)^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*sin(d*x + c))*arc
tan2(cos(d*x + c), sin(d*x + c) + 1) - (2*(d*x + c)^2*f^2 + 4*(d*e*f - c*f^2)*(d*x + c) - 2*((d*x + c)^2*f^2 +
2*(d*e*f - c*f^2)*(d*x + c))*cos(2*d*x + 2*c) + (-4*I*(d*x + c)^2*f^2 + (-8*I*d*e*f + 8*I*c*f^2)*(d*x + c))*c
os(d*x + c) + (-2*I*(d*x + c)^2*f^2 + (-4*I*d*e*f + 4*I*c*f^2)*(d*x + c))*sin(2*d*x + 2*c) + 4*((d*x + c)^2*f^
2 + 2*(d*e*f - c*f^2)*(d*x + c))*sin(d*x + c))*arctan2(cos(d*x + c), -sin(d*x + c) + 1) + (4*(d*x + c)^2*f^2 -
8*I*d*e*f + 4*(c^2 + 2*I*c)*f^2 + (8*d*e*f - (8*c - 8*I)*f^2)*(d*x + c))*cos(d*x + c) - (4*d*e*f + 4*(d*x + c
)*f^2 - 4*c*f^2 - 4*(d*e*f + (d*x + c)*f^2 - c*f^2)*cos(2*d*x + 2*c) + (-8*I*d*e*f - 8*I*(d*x + c)*f^2 + 8*I*c
*f^2)*cos(d*x + c) + (-4*I*d*e*f - 4*I*(d*x + c)*f^2 + 4*I*c*f^2)*sin(2*d*x + 2*c) + 8*(d*e*f + (d*x + c)*f^2
- c*f^2)*sin(d*x + c))*dilog(I*e^(I*d*x + I*c)) + (4*d*e*f + 4*(d*x + c)*f^2 - 4*c*f^2 - 4*(d*e*f + (d*x + c)*
f^2 - c*f^2)*cos(2*d*x + 2*c) - (8*I*d*e*f + 8*I*(d*x + c)*f^2 - 8*I*c*f^2)*cos(d*x + c) - (4*I*d*e*f + 4*I*(d
*x + c)*f^2 - 4*I*c*f^2)*sin(2*d*x + 2*c) + 8*(d*e*f + (d*x + c)*f^2 - c*f^2)*sin(d*x + c))*dilog(-I*e^(I*d*x
+ I*c)) - (I*(d*x + c)^2*f^2 + (I*c^2 + 4*I)*f^2 + (2*I*d*e*f - 2*I*c*f^2)*(d*x + c) + (-I*(d*x + c)^2*f^2 + (
-I*c^2 - 4*I)*f^2 + (-2*I*d*e*f + 2*I*c*f^2)*(d*x + c))*cos(2*d*x + 2*c) + 2*((d*x + c)^2*f^2 + (c^2 + 4)*f^2
+ 2*(d*e*f - c*f^2)*(d*x + c))*cos(d*x + c) + ((d*x + c)^2*f^2 + (c^2 + 4)*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*
sin(2*d*x + 2*c) + (2*I*(d*x + c)^2*f^2 + (2*I*c^2 + 8*I)*f^2 + (4*I*d*e*f - 4*I*c*f^2)*(d*x + c))*sin(d*x + c
))*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - (-I*(d*x + c)^2*f^2 - I*c^2*f^2 + (-2*I*d*e*f +
2*I*c*f^2)*(d*x + c) + (I*(d*x + c)^2*f^2 + I*c^2*f^2 + (2*I*d*e*f - 2*I*c*f^2)*(d*x + c))*cos(2*d*x + 2*c) -
2*((d*x + c)^2*f^2 + c^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*cos(d*x + c) - ((d*x + c)^2*f^2 + c^2*f^2 + 2*(d*
e*f - c*f^2)*(d*x + c))*sin(2*d*x + 2*c) + (-2*I*(d*x + c)^2*f^2 - 2*I*c^2*f^2 + (-4*I*d*e*f + 4*I*c*f^2)*(d*x
+ c))*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) - (-4*I*f^2*cos(2*d*x + 2*c) +
8*f^2*cos(d*x + c) + 4*f^2*sin(2*d*x + 2*c) + 8*I*f^2*sin(d*x + c) + 4*I*f^2)*polylog(3, I*e^(I*d*x + I*c)) -
(4*I*f^2*cos(2*d*x + 2*c) - 8*f^2*cos(d*x + c) - 4*f^2*sin(2*d*x + 2*c) - 8*I*f^2*sin(d*x + c) - 4*I*f^2)*poly
log(3, -I*e^(I*d*x + I*c)) - (-4*I*(d*x + c)^2*f^2 - 8*d*e*f + (-4*I*c^2 + 8*c)*f^2 + (-8*I*d*e*f - 8*(-I*c -
1)*f^2)*(d*x + c))*sin(d*x + c))/(-4*I*a*d^2*cos(2*d*x + 2*c) + 8*a*d^2*cos(d*x + c) + 4*a*d^2*sin(2*d*x + 2*c
) + 8*I*a*d^2*sin(d*x + c) + 4*I*a*d^2))/d

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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 + a*sin(c + d*x))),x)

[Out]

\text{Hanged}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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