3.7 \(\int (c+d x)^2 (a+a \sec (e+f x))^2 \, dx\)

Optimal. Leaf size=262 \[ \frac{4 i a^2 d (c+d x) \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac{4 i a^2 d (c+d x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac{i a^2 d^2 \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^3}-\frac{4 a^2 d^2 \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac{4 a^2 d^2 \text{PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac{2 a^2 d (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f^2}-\frac{4 i a^2 (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{a^2 (c+d x)^2 \tan (e+f x)}{f}-\frac{i a^2 (c+d x)^2}{f}+\frac{a^2 (c+d x)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.31306, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4190, 4181, 2531, 2282, 6589, 4184, 3719, 2190, 2279, 2391} \[ \frac{4 i a^2 d (c+d x) \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac{4 i a^2 d (c+d x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac{i a^2 d^2 \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^3}-\frac{4 a^2 d^2 \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac{4 a^2 d^2 \text{PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac{2 a^2 d (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f^2}-\frac{4 i a^2 (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{a^2 (c+d x)^2 \tan (e+f x)}{f}-\frac{i a^2 (c+d x)^2}{f}+\frac{a^2 (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4190

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

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

Rule 4184

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

Rule 3719

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

Rule 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]

Rubi steps

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

Mathematica [A]  time = 5.63997, size = 505, normalized size = 1.93 \[ \frac{1}{12} a^2 (\cos (e+f x)+1)^2 \sec ^4\left (\frac{1}{2} (e+f x)\right ) \left (-\frac{3 i \left (4 d f (c+d x) \text{PolyLog}(2,-\sin (e+f x)+i \cos (e+f x))-4 d f (c+d x) \text{PolyLog}(2,\sin (e+f x)-i \cos (e+f x))+d^2 \text{PolyLog}(2,-\cos (2 (e+f x))-i \sin (2 (e+f x)))+4 i d^2 \text{PolyLog}(3,-\sin (e+f x)+i \cos (e+f x))-4 i d^2 \text{PolyLog}(3,\sin (e+f x)-i \cos (e+f x))+4 c^2 f^2 \tan ^{-1}(\cos (e+f x)+i \sin (e+f x))+2 i c d f^2 x \tan (e)+8 c d f^2 x \tan ^{-1}(\cos (e+f x)+i \sin (e+f x))+2 i c d f \log (i \sin (2 (e+f x))+\cos (2 (e+f x))+1)+2 c d f^2 x+i d^2 f^2 x^2 \tan (e)+4 d^2 f^2 x^2 \tan ^{-1}(\cos (e+f x)+i \sin (e+f x))+2 i d^2 f x \log (i \sin (2 (e+f x))+\cos (2 (e+f x))+1)+d^2 f^2 x^2\right )}{f^3}+x \left (3 c^2+3 c d x+d^2 x^2\right )+\frac{3 (c+d x)^2 \sin \left (\frac{f x}{2}\right )}{f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{3 (c+d x)^2 \sin \left (\frac{f x}{2}\right )}{f \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*(1 + Cos[e + f*x])^2*Sec[(e + f*x)/2]^4*(x*(3*c^2 + 3*c*d*x + d^2*x^2) + (3*(c + d*x)^2*Sin[(f*x)/2])/(f*
(Cos[e/2] - Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])) + (3*(c + d*x)^2*Sin[(f*x)/2])/(f*(Cos[e/2] + Sin
[e/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])) - ((3*I)*(2*c*d*f^2*x + d^2*f^2*x^2 + 4*c^2*f^2*ArcTan[Cos[e + f
*x] + I*Sin[e + f*x]] + 8*c*d*f^2*x*ArcTan[Cos[e + f*x] + I*Sin[e + f*x]] + 4*d^2*f^2*x^2*ArcTan[Cos[e + f*x]
+ I*Sin[e + f*x]] + (2*I)*c*d*f*Log[1 + Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]] + (2*I)*d^2*f*x*Log[1 + Cos[2*(
e + f*x)] + I*Sin[2*(e + f*x)]] + 4*d*f*(c + d*x)*PolyLog[2, I*Cos[e + f*x] - Sin[e + f*x]] - 4*d*f*(c + d*x)*
PolyLog[2, (-I)*Cos[e + f*x] + Sin[e + f*x]] + d^2*PolyLog[2, -Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)]] + (4*I)*
d^2*PolyLog[3, I*Cos[e + f*x] - Sin[e + f*x]] - (4*I)*d^2*PolyLog[3, (-I)*Cos[e + f*x] + Sin[e + f*x]] + (2*I)
*c*d*f^2*x*Tan[e] + I*d^2*f^2*x^2*Tan[e]))/f^3))/12

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Maple [B]  time = 0.204, size = 668, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.56784, size = 2304, normalized size = 8.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(f*x + e)*a^2*c^2 + (f*x + e)^3*a^2*d^2/f^2 - 3*(f*x + e)^2*a^2*d^2*e/f^2 + 3*(f*x + e)*a^2*d^2*e^2/f^2
+ 3*(f*x + e)^2*a^2*c*d/f - 6*(f*x + e)*a^2*c*d*e/f + 6*a^2*c^2*log(sec(f*x + e) + tan(f*x + e)) + 6*a^2*d^2*
e^2*log(sec(f*x + e) + tan(f*x + e))/f^2 - 12*a^2*c*d*e*log(sec(f*x + e) + tan(f*x + e))/f + 3*(2*a^2*d^2*e^2
- 4*a^2*c*d*e*f + 2*a^2*c^2*f^2 - (2*(f*x + e)^2*a^2*d^2 - 4*(a^2*d^2*e - a^2*c*d*f)*(f*x + e) + 2*((f*x + e)^
2*a^2*d^2 - 2*(a^2*d^2*e - a^2*c*d*f)*(f*x + e))*cos(2*f*x + 2*e) - (-2*I*(f*x + e)^2*a^2*d^2 + (4*I*a^2*d^2*e
- 4*I*a^2*c*d*f)*(f*x + e))*sin(2*f*x + 2*e))*arctan2(cos(f*x + e), sin(f*x + e) + 1) - (2*(f*x + e)^2*a^2*d^
2 - 4*(a^2*d^2*e - a^2*c*d*f)*(f*x + e) + 2*((f*x + e)^2*a^2*d^2 - 2*(a^2*d^2*e - a^2*c*d*f)*(f*x + e))*cos(2*
f*x + 2*e) - (-2*I*(f*x + e)^2*a^2*d^2 + (4*I*a^2*d^2*e - 4*I*a^2*c*d*f)*(f*x + e))*sin(2*f*x + 2*e))*arctan2(
cos(f*x + e), -sin(f*x + e) + 1) + (2*(f*x + e)*a^2*d^2 - 2*a^2*d^2*e + 2*a^2*c*d*f + 2*((f*x + e)*a^2*d^2 - a
^2*d^2*e + a^2*c*d*f)*cos(2*f*x + 2*e) + (2*I*(f*x + e)*a^2*d^2 - 2*I*a^2*d^2*e + 2*I*a^2*c*d*f)*sin(2*f*x + 2
*e))*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 2*((f*x + e)^2*a^2*d^2 - 2*(a^2*d^2*e - a^2*c*d*f)*(f*x
+ e))*cos(2*f*x + 2*e) - (a^2*d^2*cos(2*f*x + 2*e) + I*a^2*d^2*sin(2*f*x + 2*e) + a^2*d^2)*dilog(-e^(2*I*f*x
+ 2*I*e)) - (4*(f*x + e)*a^2*d^2 - 4*a^2*d^2*e + 4*a^2*c*d*f + 4*((f*x + e)*a^2*d^2 - a^2*d^2*e + a^2*c*d*f)*c
os(2*f*x + 2*e) - (-4*I*(f*x + e)*a^2*d^2 + 4*I*a^2*d^2*e - 4*I*a^2*c*d*f)*sin(2*f*x + 2*e))*dilog(I*e^(I*f*x
+ I*e)) + (4*(f*x + e)*a^2*d^2 - 4*a^2*d^2*e + 4*a^2*c*d*f + 4*((f*x + e)*a^2*d^2 - a^2*d^2*e + a^2*c*d*f)*cos
(2*f*x + 2*e) + (4*I*(f*x + e)*a^2*d^2 - 4*I*a^2*d^2*e + 4*I*a^2*c*d*f)*sin(2*f*x + 2*e))*dilog(-I*e^(I*f*x +
I*e)) + (-I*(f*x + e)*a^2*d^2 + I*a^2*d^2*e - I*a^2*c*d*f + (-I*(f*x + e)*a^2*d^2 + I*a^2*d^2*e - I*a^2*c*d*f)
*cos(2*f*x + 2*e) + ((f*x + e)*a^2*d^2 - a^2*d^2*e + a^2*c*d*f)*sin(2*f*x + 2*e))*log(cos(2*f*x + 2*e)^2 + sin
(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1) + (-I*(f*x + e)^2*a^2*d^2 + (2*I*a^2*d^2*e - 2*I*a^2*c*d*f)*(f*x + e
) + (-I*(f*x + e)^2*a^2*d^2 + (2*I*a^2*d^2*e - 2*I*a^2*c*d*f)*(f*x + e))*cos(2*f*x + 2*e) + ((f*x + e)^2*a^2*d
^2 - 2*(a^2*d^2*e - a^2*c*d*f)*(f*x + e))*sin(2*f*x + 2*e))*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x +
e) + 1) + (I*(f*x + e)^2*a^2*d^2 + (-2*I*a^2*d^2*e + 2*I*a^2*c*d*f)*(f*x + e) + (I*(f*x + e)^2*a^2*d^2 + (-2*I
*a^2*d^2*e + 2*I*a^2*c*d*f)*(f*x + e))*cos(2*f*x + 2*e) - ((f*x + e)^2*a^2*d^2 - 2*(a^2*d^2*e - a^2*c*d*f)*(f*
x + e))*sin(2*f*x + 2*e))*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) + (-4*I*a^2*d^2*cos(2*f*x
+ 2*e) + 4*a^2*d^2*sin(2*f*x + 2*e) - 4*I*a^2*d^2)*polylog(3, I*e^(I*f*x + I*e)) + (4*I*a^2*d^2*cos(2*f*x + 2*
e) - 4*a^2*d^2*sin(2*f*x + 2*e) + 4*I*a^2*d^2)*polylog(3, -I*e^(I*f*x + I*e)) + (-2*I*(f*x + e)^2*a^2*d^2 + (4
*I*a^2*d^2*e - 4*I*a^2*c*d*f)*(f*x + e))*sin(2*f*x + 2*e))/(-I*f^2*cos(2*f*x + 2*e) + f^2*sin(2*f*x + 2*e) - I
*f^2))/f

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Fricas [C]  time = 2.53807, size = 2604, normalized size = 9.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*a^2*d^2*cos(f*x + e)*polylog(3, I*cos(f*x + e) + sin(f*x + e)) - 6*a^2*d^2*cos(f*x + e)*polylog(3, I*c
os(f*x + e) - sin(f*x + e)) + 6*a^2*d^2*cos(f*x + e)*polylog(3, -I*cos(f*x + e) + sin(f*x + e)) - 6*a^2*d^2*co
s(f*x + e)*polylog(3, -I*cos(f*x + e) - sin(f*x + e)) - (-6*I*a^2*d^2*f*x - 6*I*a^2*c*d*f + 3*I*a^2*d^2)*cos(f
*x + e)*dilog(I*cos(f*x + e) + sin(f*x + e)) - (-6*I*a^2*d^2*f*x - 6*I*a^2*c*d*f - 3*I*a^2*d^2)*cos(f*x + e)*d
ilog(I*cos(f*x + e) - sin(f*x + e)) - (6*I*a^2*d^2*f*x + 6*I*a^2*c*d*f - 3*I*a^2*d^2)*cos(f*x + e)*dilog(-I*co
s(f*x + e) + sin(f*x + e)) - (6*I*a^2*d^2*f*x + 6*I*a^2*c*d*f + 3*I*a^2*d^2)*cos(f*x + e)*dilog(-I*cos(f*x + e
) - sin(f*x + e)) - 3*(a^2*d^2*e^2 + a^2*c^2*f^2 - a^2*d^2*e - (2*a^2*c*d*e - a^2*c*d)*f)*cos(f*x + e)*log(cos
(f*x + e) + I*sin(f*x + e) + I) + 3*(a^2*d^2*e^2 + a^2*c^2*f^2 + a^2*d^2*e - (2*a^2*c*d*e + a^2*c*d)*f)*cos(f*
x + e)*log(cos(f*x + e) - I*sin(f*x + e) + I) - 3*(a^2*d^2*f^2*x^2 - a^2*d^2*e^2 + 2*a^2*c*d*e*f + a^2*d^2*e +
(2*a^2*c*d*f^2 + a^2*d^2*f)*x)*cos(f*x + e)*log(I*cos(f*x + e) + sin(f*x + e) + 1) + 3*(a^2*d^2*f^2*x^2 - a^2
*d^2*e^2 + 2*a^2*c*d*e*f - a^2*d^2*e + (2*a^2*c*d*f^2 - a^2*d^2*f)*x)*cos(f*x + e)*log(I*cos(f*x + e) - sin(f*
x + e) + 1) - 3*(a^2*d^2*f^2*x^2 - a^2*d^2*e^2 + 2*a^2*c*d*e*f + a^2*d^2*e + (2*a^2*c*d*f^2 + a^2*d^2*f)*x)*co
s(f*x + e)*log(-I*cos(f*x + e) + sin(f*x + e) + 1) + 3*(a^2*d^2*f^2*x^2 - a^2*d^2*e^2 + 2*a^2*c*d*e*f - a^2*d^
2*e + (2*a^2*c*d*f^2 - a^2*d^2*f)*x)*cos(f*x + e)*log(-I*cos(f*x + e) - sin(f*x + e) + 1) - 3*(a^2*d^2*e^2 + a
^2*c^2*f^2 - a^2*d^2*e - (2*a^2*c*d*e - a^2*c*d)*f)*cos(f*x + e)*log(-cos(f*x + e) + I*sin(f*x + e) + I) + 3*(
a^2*d^2*e^2 + a^2*c^2*f^2 + a^2*d^2*e - (2*a^2*c*d*e + a^2*c*d)*f)*cos(f*x + e)*log(-cos(f*x + e) - I*sin(f*x
+ e) + I) - (a^2*d^2*f^3*x^3 + 3*a^2*c*d*f^3*x^2 + 3*a^2*c^2*f^3*x)*cos(f*x + e) - 3*(a^2*d^2*f^2*x^2 + 2*a^2*
c*d*f^2*x + a^2*c^2*f^2)*sin(f*x + e))/(f^3*cos(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int c^{2}\, dx + \int 2 c^{2} \sec{\left (e + f x \right )}\, dx + \int c^{2} \sec ^{2}{\left (e + f x \right )}\, dx + \int d^{2} x^{2}\, dx + \int 2 c d x\, dx + \int 2 d^{2} x^{2} \sec{\left (e + f x \right )}\, dx + \int d^{2} x^{2} \sec ^{2}{\left (e + f x \right )}\, dx + \int 4 c d x \sec{\left (e + f x \right )}\, dx + \int 2 c d x \sec ^{2}{\left (e + f x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(c**2, x) + Integral(2*c**2*sec(e + f*x), x) + Integral(c**2*sec(e + f*x)**2, x) + Integral(d**2
*x**2, x) + Integral(2*c*d*x, x) + Integral(2*d**2*x**2*sec(e + f*x), x) + Integral(d**2*x**2*sec(e + f*x)**2,
x) + Integral(4*c*d*x*sec(e + f*x), x) + Integral(2*c*d*x*sec(e + f*x)**2, x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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