3.3 \(\int (c+d x) (a+a \sec (e+f x)) \, dx\)

Optimal. Leaf size=93 \[ -\frac {2 i a (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {a (c+d x)^2}{2 d}+\frac {i a d \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {i a d \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4190, 4181, 2279, 2391} \[ \frac {i a d \text {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {i a d \text {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {a (c+d x)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*(a + a*Sec[e + f*x]),x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 87, normalized size = 0.94 \[ \frac {a \left (f \left (f x (2 c+d x)-4 i (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )\right )+2 i d \text {Li}_2\left (-i e^{i (e+f x)}\right )-2 i d \text {Li}_2\left (i e^{i (e+f x)}\right )\right )}{2 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.65, size = 343, normalized size = 3.69 \[ \frac {a d f^{2} x^{2} + 2 \, a c f^{2} x - i \, a d {\rm Li}_2\left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) - i \, a d {\rm Li}_2\left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) + i \, a d {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) + i \, a d {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) - {\left (a d e - a c f\right )} \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) + {\left (a d e - a c f\right )} \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right ) + {\left (a d f x + a d e\right )} \log \left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) - {\left (a d f x + a d e\right )} \log \left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) + {\left (a d f x + a d e\right )} \log \left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) - {\left (a d f x + a d e\right )} \log \left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) - {\left (a d e - a c f\right )} \log \left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) + {\left (a d e - a c f\right )} \log \left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right )}{2 \, f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.27, size = 208, normalized size = 2.24 \[ \frac {a d \,x^{2}}{2}-\frac {a d \,e^{2}}{2 f^{2}}+c a x +\frac {a c e}{f}-\frac {a d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) x}{f}+\frac {a d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}+\frac {i a d \dilog \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right )}{f^{2}}-\frac {i a d \dilog \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {a d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) e}{f^{2}}+\frac {a d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}+\frac {c a \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}-\frac {a d e \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*d*x^2-1/2/f^2*a*d*e^2+c*a*x+1/f*a*c*e-1/f*a*d*ln(I*exp(I*(f*x+e))+1)*x+1/f*a*d*ln(1-I*exp(I*(f*x+e)))*x+
I/f^2*a*d*dilog(I*exp(I*(f*x+e))+1)-I/f^2*a*d*dilog(1-I*exp(I*(f*x+e)))-1/f^2*a*d*ln(I*exp(I*(f*x+e))+1)*e+1/f
^2*a*d*ln(1-I*exp(I*(f*x+e)))*e+1/f*c*a*ln(sec(f*x+e)+tan(f*x+e))-1/f^2*a*d*e*ln(sec(f*x+e)+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+a*sec(f*x+e)),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))*(c + d*x),x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________