3.31 \(\int (c+d x) \sec (a+b x) \, dx\)

Optimal. Leaf size=75 \[ \frac{i d \text{Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac{i d \text{Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac{2 i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0415003, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4181, 2279, 2391} \[ \frac{i d \text{Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac{i d \text{Li}_2\left (i e^{i (a+b x)}\right )}{b^2}-\frac{2 i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sec[a + b*x],x]

[Out]

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

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

Mathematica [A]  time = 0.0061443, size = 87, normalized size = 1.16 \[ \frac{i d \text{Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac{i d \text{Li}_2\left (i e^{i (a+b x)}\right )}{b^2}+\frac{c \tanh ^{-1}(\sin (a+b x))}{b}-\frac{2 i d x \tan ^{-1}\left (e^{i a+i b x}\right )}{b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)*Sec[a + b*x],x]

[Out]

((-2*I)*d*x*ArcTan[E^(I*a + I*b*x)])/b + (c*ArcTanh[Sin[a + b*x]])/b + (I*d*PolyLog[2, (-I)*E^(I*(a + b*x))])/
b^2 - (I*d*PolyLog[2, I*E^(I*(a + b*x))])/b^2

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Maple [B]  time = 0.368, size = 167, normalized size = 2.2 \begin{align*}{\frac{-2\,ic\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}-{\frac{d\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}-{\frac{d\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{2}}}+{\frac{d\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}+{\frac{d\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{2}}}+{\frac{id{\it dilog} \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{id{\it dilog} \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{2\,ida\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*sec(b*x+a),x)

[Out]

-2*I/b*c*arctan(exp(I*(b*x+a)))-1/b*d*ln(1+I*exp(I*(b*x+a)))*x-1/b^2*d*ln(1+I*exp(I*(b*x+a)))*a+1/b*d*ln(1-I*e
xp(I*(b*x+a)))*x+1/b^2*d*ln(1-I*exp(I*(b*x+a)))*a+I/b^2*d*dilog(1+I*exp(I*(b*x+a)))-I/b^2*d*dilog(1-I*exp(I*(b
*x+a)))+2*I/b^2*d*a*arctan(exp(I*(b*x+a)))

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sec(b*x+a),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.56131, size = 837, normalized size = 11.16 \begin{align*} \frac{-i \, d{\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d{\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + i \, d{\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d{\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) +{\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) -{\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) +{\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) -{\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) +{\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) -{\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) +{\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) -{\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right ) \sec{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*x)*sec(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)*sec(b*x + a), x)