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

Optimal. Leaf size=131 \[ \frac {a^2 (c+d x)^2}{2 d}-\frac {4 i a b (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {2 i a b d \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a b d \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}+\frac {b^2 (c+d x) \tan (e+f x)}{f}+\frac {b^2 d \log (\cos (e+f x))}{f^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 131, 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} \[ \frac {2 i a b d \text {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a b d \text {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}+\frac {a^2 (c+d x)^2}{2 d}-\frac {4 i a b (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {b^2 (c+d x) \tan (e+f x)}{f}+\frac {b^2 d \log (\cos (e+f x))}{f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(c + d*x)^2)/(2*d) - ((4*I)*a*b*(c + d*x)*ArcTan[E^(I*(e + f*x))])/f + (b^2*d*Log[Cos[e + f*x]])/f^2 + ((
2*I)*a*b*d*PolyLog[2, (-I)*E^(I*(e + f*x))])/f^2 - ((2*I)*a*b*d*PolyLog[2, I*E^(I*(e + f*x))])/f^2 + (b^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+b \sec (e+f x))^2 \, dx &=\int \left (a^2 (c+d x)+2 a b (c+d x) \sec (e+f x)+b^2 (c+d x) \sec ^2(e+f x)\right ) \, dx\\ &=\frac {a^2 (c+d x)^2}{2 d}+(2 a b) \int (c+d x) \sec (e+f x) \, dx+b^2 \int (c+d x) \sec ^2(e+f x) \, dx\\ &=\frac {a^2 (c+d x)^2}{2 d}-\frac {4 i a b (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {b^2 (c+d x) \tan (e+f x)}{f}-\frac {(2 a b d) \int \log \left (1-i e^{i (e+f x)}\right ) \, dx}{f}+\frac {(2 a b d) \int \log \left (1+i e^{i (e+f x)}\right ) \, dx}{f}-\frac {\left (b^2 d\right ) \int \tan (e+f x) \, dx}{f}\\ &=\frac {a^2 (c+d x)^2}{2 d}-\frac {4 i a b (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {b^2 d \log (\cos (e+f x))}{f^2}+\frac {b^2 (c+d x) \tan (e+f x)}{f}+\frac {(2 i a b d) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^2}-\frac {(2 i a b d) \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 b (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {b^2 d \log (\cos (e+f x))}{f^2}+\frac {2 i a b d \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a b d \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}+\frac {b^2 (c+d x) \tan (e+f x)}{f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.53, size = 151, normalized size = 1.15 \[ \frac {2 a^2 c f^2 x+a^2 d f^2 x^2+4 a b c f \tanh ^{-1}(\sin (e+f x))+4 i a b d \text {Li}_2\left (-i e^{i (e+f x)}\right )-4 i a b d \text {Li}_2\left (i e^{i (e+f x)}\right )-8 i a b d f x \tan ^{-1}\left (e^{i (e+f x)}\right )+2 b^2 c f \tan (e+f x)+2 b^2 d f x \tan (e+f x)+2 b^2 d \log (\cos (e+f x))}{2 f^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.65, size = 505, normalized size = 3.85 \[ \frac {-2 i \, a b 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 b 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 b 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 b 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 b d e - 2 \, a b c f - b^{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 b d e - 2 \, a b c f + b^{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 b d f x + a b 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 b d f x + a b 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 b d f x + a b 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 b d f x + a b 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 b d e - 2 \, a b c f - b^{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 b d e - 2 \, a b c f + b^{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 (b^{2} d f x + b^{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+b*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.52, size = 271, normalized size = 2.07 \[ \frac {a^{2} d \,x^{2}}{2}-\frac {a^{2} d \,e^{2}}{2 f^{2}}+a^{2} c x +\frac {a^{2} c e}{f}-\frac {2 a b d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) x}{f}+\frac {2 a b d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}-\frac {2 a b d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) e}{f^{2}}+\frac {2 a b d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}-\frac {2 i a b d \dilog \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+\frac {2 i a b d \dilog \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right )}{f^{2}}+\frac {2 a b c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}-\frac {2 a b d e \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f^{2}}+\frac {b^{2} d \tan \left (f x +e \right ) x}{f}+\frac {b^{2} d \ln \left (\cos \left (f x +e \right )\right )}{f^{2}}+\frac {c \,b^{2} \tan \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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+b*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________