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

Optimal. Leaf size=119 \[ \frac{4 i d^2 \text{PolyLog}\left (2,-e^{i (e+f x)}\right )}{a f^3}-\frac{4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{a f^2}-\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}+\frac{i (c+d x)^2}{a f}+\frac{(c+d x)^3}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.245432, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {4191, 3318, 4184, 3719, 2190, 2279, 2391} \[ \frac{4 i d^2 \text{PolyLog}\left (2,-e^{i (e+f x)}\right )}{a f^3}-\frac{4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{a f^2}-\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}+\frac{i (c+d x)^2}{a f}+\frac{(c+d x)^3}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4191

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& IGtQ[m, 0]

Rule 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || 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 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 \frac{(c+d x)^2}{a+a \sec (e+f x)} \, dx &=\int \left (\frac{(c+d x)^2}{a}-\frac{(c+d x)^2}{a+a \cos (e+f x)}\right ) \, dx\\ &=\frac{(c+d x)^3}{3 a d}-\int \frac{(c+d x)^2}{a+a \cos (e+f x)} \, dx\\ &=\frac{(c+d x)^3}{3 a d}-\frac{\int (c+d x)^2 \csc ^2\left (\frac{e+\pi }{2}+\frac{f x}{2}\right ) \, dx}{2 a}\\ &=\frac{(c+d x)^3}{3 a d}-\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}+\frac{(2 d) \int (c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right ) \, dx}{a f}\\ &=\frac{i (c+d x)^2}{a f}+\frac{(c+d x)^3}{3 a d}-\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}-\frac{(4 i d) \int \frac{e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )} (c+d x)}{1+e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}} \, dx}{a f}\\ &=\frac{i (c+d x)^2}{a f}+\frac{(c+d x)^3}{3 a d}-\frac{4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{a f^2}-\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}+\frac{\left (4 d^2\right ) \int \log \left (1+e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}\right ) \, dx}{a f^2}\\ &=\frac{i (c+d x)^2}{a f}+\frac{(c+d x)^3}{3 a d}-\frac{4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{a f^2}-\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}-\frac{\left (4 i d^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}\right )}{a f^3}\\ &=\frac{i (c+d x)^2}{a f}+\frac{(c+d x)^3}{3 a d}-\frac{4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{a f^2}+\frac{4 i d^2 \text{Li}_2\left (-e^{i (e+f x)}\right )}{a f^3}-\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}\\ \end{align*}

Mathematica [B]  time = 6.47128, size = 528, normalized size = 4.44 \[ -\frac{8 d^2 \csc \left (\frac{e}{2}\right ) \sec \left (\frac{e}{2}\right ) \cos ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \sec (e+f x) \left (\frac{1}{4} f^2 x^2 e^{-i \tan ^{-1}\left (\cot \left (\frac{e}{2}\right )\right )}-\frac{\cot \left (\frac{e}{2}\right ) \left (i \text{PolyLog}\left (2,e^{2 i \left (\frac{f x}{2}-\tan ^{-1}\left (\cot \left (\frac{e}{2}\right )\right )\right )}\right )+\frac{1}{2} i f x \left (-2 \tan ^{-1}\left (\cot \left (\frac{e}{2}\right )\right )-\pi \right )-2 \left (\frac{f x}{2}-\tan ^{-1}\left (\cot \left (\frac{e}{2}\right )\right )\right ) \log \left (1-e^{2 i \left (\frac{f x}{2}-\tan ^{-1}\left (\cot \left (\frac{e}{2}\right )\right )\right )}\right )-2 \tan ^{-1}\left (\cot \left (\frac{e}{2}\right )\right ) \log \left (\sin \left (\frac{f x}{2}-\tan ^{-1}\left (\cot \left (\frac{e}{2}\right )\right )\right )\right )-\pi \log \left (1+e^{-i f x}\right )+\pi \log \left (\cos \left (\frac{f x}{2}\right )\right )\right )}{\sqrt{\cot ^2\left (\frac{e}{2}\right )+1}}\right )}{f^3 \sqrt{\csc ^2\left (\frac{e}{2}\right ) \left (\sin ^2\left (\frac{e}{2}\right )+\cos ^2\left (\frac{e}{2}\right )\right )} (a \sec (e+f x)+a)}+\frac{2 x \left (3 c^2+3 c d x+d^2 x^2\right ) \cos ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \sec (e+f x)}{3 (a \sec (e+f x)+a)}-\frac{2 \sec \left (\frac{e}{2}\right ) \cos \left (\frac{e}{2}+\frac{f x}{2}\right ) \sec (e+f x) \left (c^2 \sin \left (\frac{f x}{2}\right )+2 c d x \sin \left (\frac{f x}{2}\right )+d^2 x^2 \sin \left (\frac{f x}{2}\right )\right )}{f (a \sec (e+f x)+a)}-\frac{8 c d \sec \left (\frac{e}{2}\right ) \cos ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \sec (e+f x) \left (\frac{1}{2} f x \sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right ) \log \left (\cos \left (\frac{e}{2}\right ) \cos \left (\frac{f x}{2}\right )-\sin \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right )\right )\right )}{f^2 \left (\sin ^2\left (\frac{e}{2}\right )+\cos ^2\left (\frac{e}{2}\right )\right ) (a \sec (e+f x)+a)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*x*(3*c^2 + 3*c*d*x + d^2*x^2)*Cos[e/2 + (f*x)/2]^2*Sec[e + f*x])/(3*(a + a*Sec[e + f*x])) - (8*c*d*Cos[e/2
+ (f*x)/2]^2*Sec[e/2]*Sec[e + f*x]*(Cos[e/2]*Log[Cos[e/2]*Cos[(f*x)/2] - Sin[e/2]*Sin[(f*x)/2]] + (f*x*Sin[e/2
])/2))/(f^2*(a + a*Sec[e + f*x])*(Cos[e/2]^2 + Sin[e/2]^2)) - (8*d^2*Cos[e/2 + (f*x)/2]^2*Csc[e/2]*((f^2*x^2)/
(4*E^(I*ArcTan[Cot[e/2]])) - (Cot[e/2]*((I/2)*f*x*(-Pi - 2*ArcTan[Cot[e/2]]) - Pi*Log[1 + E^((-I)*f*x)] - 2*((
f*x)/2 - ArcTan[Cot[e/2]])*Log[1 - E^((2*I)*((f*x)/2 - ArcTan[Cot[e/2]]))] + Pi*Log[Cos[(f*x)/2]] - 2*ArcTan[C
ot[e/2]]*Log[Sin[(f*x)/2 - ArcTan[Cot[e/2]]]] + I*PolyLog[2, E^((2*I)*((f*x)/2 - ArcTan[Cot[e/2]]))]))/Sqrt[1
+ Cot[e/2]^2])*Sec[e/2]*Sec[e + f*x])/(f^3*(a + a*Sec[e + f*x])*Sqrt[Csc[e/2]^2*(Cos[e/2]^2 + Sin[e/2]^2)]) -
(2*Cos[e/2 + (f*x)/2]*Sec[e/2]*Sec[e + f*x]*(c^2*Sin[(f*x)/2] + 2*c*d*x*Sin[(f*x)/2] + d^2*x^2*Sin[(f*x)/2]))/
(f*(a + a*Sec[e + f*x]))

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Maple [B]  time = 0.129, size = 225, normalized size = 1.9 \begin{align*}{\frac{{d}^{2}{x}^{3}}{3\,a}}+{\frac{cd{x}^{2}}{a}}+{\frac{{c}^{2}x}{a}}-{\frac{2\,i \left ({d}^{2}{x}^{2}+2\,cdx+{c}^{2} \right ) }{fa \left ({{\rm e}^{i \left ( fx+e \right ) }}+1 \right ) }}-4\,{\frac{cd\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}+1 \right ) }{a{f}^{2}}}+4\,{\frac{cd\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{a{f}^{2}}}+{\frac{2\,i{d}^{2}{x}^{2}}{fa}}+{\frac{4\,i{d}^{2}ex}{a{f}^{2}}}+{\frac{2\,i{d}^{2}{e}^{2}}{{f}^{3}a}}-4\,{\frac{{d}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}+1 \right ) x}{a{f}^{2}}}+{\frac{4\,i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}a}}-4\,{\frac{{d}^{2}e\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}a}} \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+1/a*c*d*x^2+1/a*c^2*x-2*I*(d^2*x^2+2*c*d*x+c^2)/f/a/(exp(I*(f*x+e))+1)-4*d/f^2/a*c*ln(exp(I*(f*x
+e))+1)+4*d/f^2/a*c*ln(exp(I*(f*x+e)))+2*I*d^2/f/a*x^2+4*I*d^2/f^2/a*e*x+2*I*d^2/f^3/a*e^2-4*d^2/f^2/a*ln(exp(
I*(f*x+e))+1)*x+4*I*d^2*polylog(2,-exp(I*(f*x+e)))/a/f^3-4*d^2/f^3/a*e*ln(exp(I*(f*x+e)))

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Maxima [B]  time = 2.21611, size = 513, normalized size = 4.31 \begin{align*} -\frac{i \, d^{2} f^{3} x^{3} + 3 i \, c d f^{3} x^{2} + 3 i \, c^{2} f^{3} x + 6 \, c^{2} f^{2} +{\left (12 \, d^{2} f x + 12 \, c d f + 12 \,{\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right ) +{\left (12 i \, d^{2} f x + 12 i \, c d f\right )} \sin \left (f x + e\right )\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) + 1\right ) +{\left (i \, d^{2} f^{3} x^{3} - 3 \,{\left (-i \, c d f^{3} + 2 \, d^{2} f^{2}\right )} x^{2} - 3 \,{\left (-i \, c^{2} f^{3} + 4 \, c d f^{2}\right )} x\right )} \cos \left (f x + e\right ) -{\left (12 \, d^{2} \cos \left (f x + e\right ) + 12 i \, d^{2} \sin \left (f x + e\right ) + 12 \, d^{2}\right )}{\rm Li}_2\left (-e^{\left (i \, f x + i \, e\right )}\right ) +{\left (-6 i \, d^{2} f x - 6 i \, c d f +{\left (-6 i \, d^{2} f x - 6 i \, c d f\right )} \cos \left (f x + e\right ) + 6 \,{\left (d^{2} f x + c d f\right )} \sin \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1\right ) -{\left (d^{2} f^{3} x^{3} +{\left (3 \, c d f^{3} + 6 i \, d^{2} f^{2}\right )} x^{2} +{\left (3 \, c^{2} f^{3} + 12 i \, c d f^{2}\right )} x\right )} \sin \left (f x + e\right )}{-3 i \, a f^{3} \cos \left (f x + e\right ) + 3 \, a f^{3} \sin \left (f x + e\right ) - 3 i \, a f^{3}} \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]

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

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Fricas [B]  time = 1.7907, size = 711, normalized size = 5.97 \begin{align*} \frac{d^{2} f^{3} x^{3} + 3 \, c d f^{3} x^{2} + 3 \, c^{2} f^{3} x +{\left (d^{2} f^{3} x^{3} + 3 \, c d f^{3} x^{2} + 3 \, c^{2} f^{3} x\right )} \cos \left (f x + e\right ) +{\left (-6 i \, d^{2} \cos \left (f x + e\right ) - 6 i \, d^{2}\right )}{\rm Li}_2\left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) +{\left (6 i \, d^{2} \cos \left (f x + e\right ) + 6 i \, d^{2}\right )}{\rm Li}_2\left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 6 \,{\left (d^{2} f x + c d f +{\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + 1\right ) - 6 \,{\left (d^{2} f x + c d f +{\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2}\right )} \sin \left (f x + e\right )}{3 \,{\left (a f^{3} \cos \left (f x + e\right ) + a f^{3}\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, algorithm="fricas")

[Out]

1/3*(d^2*f^3*x^3 + 3*c*d*f^3*x^2 + 3*c^2*f^3*x + (d^2*f^3*x^3 + 3*c*d*f^3*x^2 + 3*c^2*f^3*x)*cos(f*x + e) + (-
6*I*d^2*cos(f*x + e) - 6*I*d^2)*dilog(-cos(f*x + e) + I*sin(f*x + e)) + (6*I*d^2*cos(f*x + e) + 6*I*d^2)*dilog
(-cos(f*x + e) - I*sin(f*x + e)) - 6*(d^2*f*x + c*d*f + (d^2*f*x + c*d*f)*cos(f*x + e))*log(cos(f*x + e) + I*s
in(f*x + e) + 1) - 6*(d^2*f*x + c*d*f + (d^2*f*x + c*d*f)*cos(f*x + e))*log(cos(f*x + e) - I*sin(f*x + e) + 1)
- 3*(d^2*f^2*x^2 + 2*c*d*f^2*x + c^2*f^2)*sin(f*x + e))/(a*f^3*cos(f*x + e) + a*f^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c^{2}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{d^{2} x^{2}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{2 c d x}{\sec{\left (e + f x \right )} + 1}\, dx}{a} \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]

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

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