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

Optimal. Leaf size=157 \[ \frac{2 i a d (c+d x) \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac{2 i a d (c+d x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac{2 a d^2 \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac{2 a d^2 \text{PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}-\frac{2 i a (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{a (c+d x)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.14041, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {4190, 4181, 2531, 2282, 6589} \[ \frac{2 i a d (c+d x) \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac{2 i a d (c+d x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac{2 a d^2 \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac{2 a d^2 \text{PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}-\frac{2 i a (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{a (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

Mathematica [A]  time = 0.137477, size = 151, normalized size = 0.96 \[ a \left (\frac{2 i d \left (f (c+d x) \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )+i d \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )\right )}{f^3}+\frac{2 d \left (d \text{PolyLog}\left (3,i e^{i (e+f x)}\right )-i f (c+d x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )\right )}{f^3}-\frac{2 i (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{(c+d x)^3}{3 d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.157, size = 431, normalized size = 2.8 \begin{align*}{\frac{a{d}^{2}{x}^{3}}{3}}+acd{x}^{2}+a{c}^{2}x-2\,{\frac{acd\ln \left ( 1+i{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{f}}-{\frac{a{d}^{2}\ln \left ( 1+i{{\rm e}^{i \left ( fx+e \right ) }} \right ){x}^{2}}{f}}-2\,{\frac{a{d}^{2}{\it polylog} \left ( 3,-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}+2\,{\frac{a{d}^{2}{\it polylog} \left ( 3,i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}-{\frac{a{d}^{2}{e}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}-2\,{\frac{acd\ln \left ( 1+i{{\rm e}^{i \left ( fx+e \right ) }} \right ) e}{{f}^{2}}}+{\frac{4\,iacde\arctan \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-{\frac{2\,ia{c}^{2}\arctan \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{f}}-{\frac{2\,iacd{\it polylog} \left ( 2,i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-{\frac{2\,ia{d}^{2}{e}^{2}\arctan \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}+{\frac{a{d}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ){x}^{2}}{f}}+2\,{\frac{acd\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{f}}-{\frac{2\,ia{d}^{2}{\it polylog} \left ( 2,i{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{{f}^{2}}}+{\frac{a{d}^{2}{e}^{2}\ln \left ( 1+i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}+2\,{\frac{acd\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) e}{{f}^{2}}}+{\frac{2\,iacd{\it polylog} \left ( 2,-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}+{\frac{2\,ia{d}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{{f}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.11467, size = 689, normalized size = 4.39 \begin{align*} \frac{6 \,{\left (f x + e\right )} a c^{2} + \frac{2 \,{\left (f x + e\right )}^{3} a d^{2}}{f^{2}} - \frac{6 \,{\left (f x + e\right )}^{2} a d^{2} e}{f^{2}} + \frac{6 \,{\left (f x + e\right )} a d^{2} e^{2}}{f^{2}} + \frac{6 \,{\left (f x + e\right )}^{2} a c d}{f} - \frac{12 \,{\left (f x + e\right )} a c d e}{f} + 6 \, a c^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + \frac{6 \, a d^{2} e^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{f^{2}} - \frac{12 \, a c d e \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{f} + \frac{3 \,{\left (4 \, a d^{2}{\rm Li}_{3}(i \, e^{\left (i \, f x + i \, e\right )}) - 4 \, a d^{2}{\rm Li}_{3}(-i \, e^{\left (i \, f x + i \, e\right )}) +{\left (-2 i \,{\left (f x + e\right )}^{2} a d^{2} +{\left (4 i \, a d^{2} e - 4 i \, a c d f\right )}{\left (f x + e\right )}\right )} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) +{\left (-2 i \,{\left (f x + e\right )}^{2} a d^{2} +{\left (4 i \, a d^{2} e - 4 i \, a c d f\right )}{\left (f x + e\right )}\right )} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) +{\left (-4 i \,{\left (f x + e\right )} a d^{2} + 4 i \, a d^{2} e - 4 i \, a c d f\right )}{\rm Li}_2\left (i \, e^{\left (i \, f x + i \, e\right )}\right ) +{\left (4 i \,{\left (f x + e\right )} a d^{2} - 4 i \, a d^{2} e + 4 i \, a c d f\right )}{\rm Li}_2\left (-i \, e^{\left (i \, f x + i \, e\right )}\right ) +{\left ({\left (f x + e\right )}^{2} a d^{2} - 2 \,{\left (a d^{2} e - a c d f\right )}{\left (f x + e\right )}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) -{\left ({\left (f x + e\right )}^{2} a d^{2} - 2 \,{\left (a d^{2} e - a c d f\right )}{\left (f x + e\right )}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right )\right )}}{f^{2}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 2.16225, size = 1732, normalized size = 11.03 \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)),x, algorithm="fricas")

[Out]

1/6*(2*a*d^2*f^3*x^3 + 6*a*c*d*f^3*x^2 + 6*a*c^2*f^3*x - 6*a*d^2*polylog(3, I*cos(f*x + e) + sin(f*x + e)) + 6
*a*d^2*polylog(3, I*cos(f*x + e) - sin(f*x + e)) - 6*a*d^2*polylog(3, -I*cos(f*x + e) + sin(f*x + e)) + 6*a*d^
2*polylog(3, -I*cos(f*x + e) - sin(f*x + e)) + (-6*I*a*d^2*f*x - 6*I*a*c*d*f)*dilog(I*cos(f*x + e) + sin(f*x +
e)) + (-6*I*a*d^2*f*x - 6*I*a*c*d*f)*dilog(I*cos(f*x + e) - sin(f*x + e)) + (6*I*a*d^2*f*x + 6*I*a*c*d*f)*dil
og(-I*cos(f*x + e) + sin(f*x + e)) + (6*I*a*d^2*f*x + 6*I*a*c*d*f)*dilog(-I*cos(f*x + e) - sin(f*x + e)) + 3*(
a*d^2*e^2 - 2*a*c*d*e*f + a*c^2*f^2)*log(cos(f*x + e) + I*sin(f*x + e) + I) - 3*(a*d^2*e^2 - 2*a*c*d*e*f + a*c
^2*f^2)*log(cos(f*x + e) - I*sin(f*x + e) + I) + 3*(a*d^2*f^2*x^2 + 2*a*c*d*f^2*x - a*d^2*e^2 + 2*a*c*d*e*f)*l
og(I*cos(f*x + e) + sin(f*x + e) + 1) - 3*(a*d^2*f^2*x^2 + 2*a*c*d*f^2*x - a*d^2*e^2 + 2*a*c*d*e*f)*log(I*cos(
f*x + e) - sin(f*x + e) + 1) + 3*(a*d^2*f^2*x^2 + 2*a*c*d*f^2*x - a*d^2*e^2 + 2*a*c*d*e*f)*log(-I*cos(f*x + e)
+ sin(f*x + e) + 1) - 3*(a*d^2*f^2*x^2 + 2*a*c*d*f^2*x - a*d^2*e^2 + 2*a*c*d*e*f)*log(-I*cos(f*x + e) - sin(f
*x + e) + 1) + 3*(a*d^2*e^2 - 2*a*c*d*e*f + a*c^2*f^2)*log(-cos(f*x + e) + I*sin(f*x + e) + I) - 3*(a*d^2*e^2
- 2*a*c*d*e*f + a*c^2*f^2)*log(-cos(f*x + e) - I*sin(f*x + e) + I))/f^3

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

[Out]

a*(Integral(c**2, x) + Integral(c**2*sec(e + f*x), x) + Integral(d**2*x**2, x) + Integral(2*c*d*x, x) + Integr
al(d**2*x**2*sec(e + f*x), x) + Integral(2*c*d*x*sec(e + f*x), 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 )}\,{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)),x, algorithm="giac")

[Out]

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