3.1.12 \(\int \frac {(c+d x)^2}{a+a \sec (e+f x)} \, dx\) [12]

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

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {4276, 3399, 4269, 3800, 2221, 2317, 2438} \begin {gather*} -\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}+\frac {4 i d^2 \text {Li}_2\left (-e^{i (e+f x)}\right )}{a f^3} \end {gather*}

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 3399

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) + 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 3800

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 4269

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 4276

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]

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 ) \text {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*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(528\) vs. \(2(119)=238\).
time = 6.37, size = 528, normalized size = 4.44 \begin {gather*} \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+a \sec (e+f x))}-\frac {8 c d \cos ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \sec \left (\frac {e}{2}\right ) \sec (e+f x) \left (\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 )+\frac {1}{2} f x \sin \left (\frac {e}{2}\right )\right )}{f^2 (a+a \sec (e+f x)) \left (\cos ^2\left (\frac {e}{2}\right )+\sin ^2\left (\frac {e}{2}\right )\right )}-\frac {8 d^2 \cos ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \csc \left (\frac {e}{2}\right ) \left (\frac {1}{4} e^{-i \text {ArcTan}\left (\cot \left (\frac {e}{2}\right )\right )} f^2 x^2-\frac {\cot \left (\frac {e}{2}\right ) \left (\frac {1}{2} i f x \left (-\pi -2 \text {ArcTan}\left (\cot \left (\frac {e}{2}\right )\right )\right )-\pi \log \left (1+e^{-i f x}\right )-2 \left (\frac {f x}{2}-\text {ArcTan}\left (\cot \left (\frac {e}{2}\right )\right )\right ) \log \left (1-e^{2 i \left (\frac {f x}{2}-\text {ArcTan}\left (\cot \left (\frac {e}{2}\right )\right )\right )}\right )+\pi \log \left (\cos \left (\frac {f x}{2}\right )\right )-2 \text {ArcTan}\left (\cot \left (\frac {e}{2}\right )\right ) \log \left (\sin \left (\frac {f x}{2}-\text {ArcTan}\left (\cot \left (\frac {e}{2}\right )\right )\right )\right )+i \text {PolyLog}\left (2,e^{2 i \left (\frac {f x}{2}-\text {ArcTan}\left (\cot \left (\frac {e}{2}\right )\right )\right )}\right )\right )}{\sqrt {1+\cot ^2\left (\frac {e}{2}\right )}}\right ) \sec \left (\frac {e}{2}\right ) \sec (e+f x)}{f^3 (a+a \sec (e+f x)) \sqrt {\csc ^2\left (\frac {e}{2}\right ) \left (\cos ^2\left (\frac {e}{2}\right )+\sin ^2\left (\frac {e}{2}\right )\right )}}-\frac {2 \cos \left (\frac {e}{2}+\frac {f x}{2}\right ) \sec \left (\frac {e}{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+a \sec (e+f x))} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (107 ) = 214\).
time = 0.66, size = 236, normalized size = 1.98

method	result	size
risch	\(\frac {d^{2} x^{3}}{3 a}+\frac {d c \,x^{2}}{a}+\frac {c^{2} x}{a}+\frac {c^{3}}{3 a d}-\frac {2 i \left (d^{2} x^{2}+2 c d x +c^{2}\right )}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}+\frac {4 d c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{2}}-\frac {4 d c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{a \,f^{2}}+\frac {2 i d^{2} x^{2}}{a f}+\frac {4 i d^{2} e x}{a \,f^{2}}+\frac {2 i d^{2} e^{2}}{a \,f^{3}}-\frac {4 d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) x}{a \,f^{2}}+\frac {4 i d^{2} \polylog \left (2, -{\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}-\frac {4 d^{2} e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}\)	\(236\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 393 vs. \(2 (107) = 214\).
time = 0.67, size = 393, normalized size = 3.30 \begin {gather*} -\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} + 12 \, {\left (d^{2} f x + c d f + {\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right ) - {\left (-i \, d^{2} f x - 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 ) - 12 \, {\left (d^{2} \cos \left (f x + e\right ) + i \, d^{2} \sin \left (f x + e\right ) + d^{2}\right )} {\rm Li}_2\left (-e^{\left (i \, f x + i \, e\right )}\right ) - 6 \, {\left (i \, d^{2} f x + i \, c d f + {\left (i \, d^{2} f x + i \, c d f\right )} \cos \left (f x + e\right ) - {\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} + 3 \, {\left (c d f^{3} + 2 i \, d^{2} f^{2}\right )} x^{2} + 3 \, {\left (c^{2} f^{3} + 4 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 {gather*}

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 + c*d*f + (d^2*f*x + c*d*f)*cos(f*
x + e) - (-I*d^2*f*x - 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) + I*d^2*sin(f*x
+ e) + d^2)*dilog(-e^(I*f*x + I*e)) - 6*(I*d^2*f*x + I*c*d*f + (I*d^2*f*x + I*c*d*f)*cos(f*x + e) - (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 +
2*I*d^2*f^2)*x^2 + 3*(c^2*f^3 + 4*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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 307 vs. \(2 (107) = 214\).
time = 3.30, size = 307, normalized size = 2.58 \begin {gather*} \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 ) - 6 \, {\left (i \, d^{2} \cos \left (f x + e\right ) + i \, d^{2}\right )} {\rm Li}_2\left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - 6 \, {\left (-i \, d^{2} \cos \left (f x + e\right ) - 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 {gather*}

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) + I*d^2)*dilog(-cos(f*x + e) + I*sin(f*x + e)) - 6*(-I*d^2*cos(f*x + e) - I*d^2)*dilog(-co
s(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*sin(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^2}{a+\frac {a}{\cos \left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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