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

Optimal. Leaf size=134 \[ -\frac {4 i a^2 (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {a^2 (c+d x) \tan (e+f x)}{f}+\frac {a^2 (c+d x)^2}{2 d}+\frac {2 i a^2 d \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a^2 d \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}+\frac {a^2 d \log (\cos (e+f x))}{f^2} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4190, 4181, 2279, 2391, 4184, 3475} \[ \frac {2 i a^2 d \text {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a^2 d \text {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {4 i a^2 (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {a^2 (c+d x) \tan (e+f x)}{f}+\frac {a^2 (c+d x)^2}{2 d}+\frac {a^2 d \log (\cos (e+f x))}{f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 3475

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

Rubi steps

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

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Mathematica [B]  time = 5.61, size = 330, normalized size = 2.46 \[ \frac {a^2 (\cos (e+f x)+1)^2 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \left (f x (2 c f+2 d \tan (e)+d f x)+\frac {2 f (c+d x) \sin \left (\frac {f x}{2}\right )}{\left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}+\frac {2 f (c+d x) \sin \left (\frac {f x}{2}\right )}{\left (\sin \left (\frac {e}{2}\right )+\cos \left (\frac {e}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}+8 c f \tanh ^{-1}\left (\cos (e) \tan \left (\frac {f x}{2}\right )+\sin (e)\right )-\frac {4 d \csc (e) \left (i \text {Li}_2\left (-e^{i \left (f x-\tan ^{-1}(\cot (e))\right )}\right )-i \text {Li}_2\left (e^{i \left (f x-\tan ^{-1}(\cot (e))\right )}\right )+\left (f x-\tan ^{-1}(\cot (e))\right ) \left (\log \left (1-e^{i \left (f x-\tan ^{-1}(\cot (e))\right )}\right )-\log \left (1+e^{i \left (f x-\tan ^{-1}(\cot (e))\right )}\right )\right )\right )}{\sqrt {\csc ^2(e)}}-2 d f x \tan (e)+2 d (f x \tan (e)+\log (\cos (e+f x)))+8 d \tan ^{-1}(\cot (e)) \tanh ^{-1}\left (\cos (e) \tan \left (\frac {f x}{2}\right )+\sin (e)\right )\right )}{8 f^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [B]  time = 0.85, size = 525, normalized size = 3.92 \[ \frac {-2 i \, a^{2} d \cos \left (f x + e\right ) {\rm Li}_2\left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) - 2 i \, a^{2} d \cos \left (f x + e\right ) {\rm Li}_2\left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) + 2 i \, a^{2} d \cos \left (f x + e\right ) {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) + 2 i \, a^{2} d \cos \left (f x + e\right ) {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) - {\left (2 \, a^{2} d e - 2 \, a^{2} c f - a^{2} d\right )} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) + {\left (2 \, a^{2} d e - 2 \, a^{2} c f + a^{2} d\right )} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right ) + 2 \, {\left (a^{2} d f x + a^{2} d e\right )} \cos \left (f x + e\right ) \log \left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (a^{2} d f x + a^{2} d e\right )} \cos \left (f x + e\right ) \log \left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (a^{2} d f x + a^{2} d e\right )} \cos \left (f x + e\right ) \log \left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (a^{2} d f x + a^{2} d e\right )} \cos \left (f x + e\right ) \log \left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) - {\left (2 \, a^{2} d e - 2 \, a^{2} c f - a^{2} d\right )} \cos \left (f x + e\right ) \log \left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) + {\left (2 \, a^{2} d e - 2 \, a^{2} c f + a^{2} d\right )} \cos \left (f x + e\right ) \log \left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right ) + {\left (a^{2} d f^{2} x^{2} + 2 \, a^{2} c f^{2} x\right )} \cos \left (f x + e\right ) + 2 \, {\left (a^{2} d f x + a^{2} c f\right )} \sin \left (f x + e\right )}{2 \, f^{2} \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-2*I*a^2*d*cos(f*x + e)*dilog(I*cos(f*x + e) + sin(f*x + e)) - 2*I*a^2*d*cos(f*x + e)*dilog(I*cos(f*x + e
) - sin(f*x + e)) + 2*I*a^2*d*cos(f*x + e)*dilog(-I*cos(f*x + e) + sin(f*x + e)) + 2*I*a^2*d*cos(f*x + e)*dilo
g(-I*cos(f*x + e) - sin(f*x + e)) - (2*a^2*d*e - 2*a^2*c*f - a^2*d)*cos(f*x + e)*log(cos(f*x + e) + I*sin(f*x
+ e) + I) + (2*a^2*d*e - 2*a^2*c*f + a^2*d)*cos(f*x + e)*log(cos(f*x + e) - I*sin(f*x + e) + I) + 2*(a^2*d*f*x
+ a^2*d*e)*cos(f*x + e)*log(I*cos(f*x + e) + sin(f*x + e) + 1) - 2*(a^2*d*f*x + a^2*d*e)*cos(f*x + e)*log(I*c
os(f*x + e) - sin(f*x + e) + 1) + 2*(a^2*d*f*x + a^2*d*e)*cos(f*x + e)*log(-I*cos(f*x + e) + sin(f*x + e) + 1)
- 2*(a^2*d*f*x + a^2*d*e)*cos(f*x + e)*log(-I*cos(f*x + e) - sin(f*x + e) + 1) - (2*a^2*d*e - 2*a^2*c*f - a^2
*d)*cos(f*x + e)*log(-cos(f*x + e) + I*sin(f*x + e) + I) + (2*a^2*d*e - 2*a^2*c*f + a^2*d)*cos(f*x + e)*log(-c
os(f*x + e) - I*sin(f*x + e) + I) + (a^2*d*f^2*x^2 + 2*a^2*c*f^2*x)*cos(f*x + e) + 2*(a^2*d*f*x + a^2*c*f)*sin
(f*x + e))/(f^2*cos(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.53, size = 279, normalized size = 2.08 \[ \frac {a^{2} d \,x^{2}}{2}-\frac {a^{2} d \,e^{2}}{2 f^{2}}+a^{2} c x +\frac {a^{2} c e}{f}-\frac {2 a^{2} d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) x}{f}+\frac {2 a^{2} d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}-\frac {2 i a^{2} d \dilog \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+\frac {2 i a^{2} d \dilog \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right )}{f^{2}}-\frac {2 a^{2} d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) e}{f^{2}}+\frac {2 a^{2} d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}+\frac {2 a^{2} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}-\frac {2 a^{2} d e \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f^{2}}+\frac {a^{2} d \tan \left (f x +e \right ) x}{f}+\frac {a^{2} d \ln \left (\cos \left (f x +e \right )\right )}{f^{2}}+\frac {a^{2} c \tan \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*d*x^2-1/2/f^2*a^2*d*e^2+a^2*c*x+1/f*a^2*c*e-2/f*a^2*d*ln(I*exp(I*(f*x+e))+1)*x+2/f*a^2*d*ln(1-I*exp(I*
(f*x+e)))*x-2*I/f^2*a^2*d*dilog(1-I*exp(I*(f*x+e)))+2*I/f^2*a^2*d*dilog(I*exp(I*(f*x+e))+1)-2/f^2*a^2*d*ln(I*e
xp(I*(f*x+e))+1)*e+2/f^2*a^2*d*ln(1-I*exp(I*(f*x+e)))*e+2/f*a^2*c*ln(sec(f*x+e)+tan(f*x+e))-2/f^2*a^2*d*e*ln(s
ec(f*x+e)+tan(f*x+e))+1/f*a^2*d*tan(f*x+e)*x+a^2*d*ln(cos(f*x+e))/f^2+a^2*c*tan(f*x+e)/f

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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