3.212 \(\int (c+d x) \tan (a+b x) \, dx\)

Optimal. Leaf size=66 \[ \frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i (c+d x)^2}{2 d} \]

[Out]

((I/2)*(c + d*x)^2)/d - ((c + d*x)*Log[1 + E^((2*I)*(a + b*x))])/b + ((I/2)*d*PolyLog[2, -E^((2*I)*(a + b*x))]
)/b^2

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Rubi [A]  time = 0.0925894, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3719, 2190, 2279, 2391} \[ \frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i (c+d x)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Tan[a + b*x],x]

[Out]

((I/2)*(c + d*x)^2)/d - ((c + d*x)*Log[1 + E^((2*I)*(a + b*x))])/b + ((I/2)*d*PolyLog[2, -E^((2*I)*(a + b*x))]
)/b^2

Rule 3719

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 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) \tan (a+b x) \, dx &=\frac{i (c+d x)^2}{2 d}-2 i \int \frac{e^{2 i (a+b x)} (c+d x)}{1+e^{2 i (a+b x)}} \, dx\\ &=\frac{i (c+d x)^2}{2 d}-\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{d \int \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac{i (c+d x)^2}{2 d}-\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{(i d) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^2}\\ &=\frac{i (c+d x)^2}{2 d}-\frac{(c+d x) \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i d \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0147743, size = 70, normalized size = 1.06 \[ \frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{c \log (\cos (a+b x))}{b}-\frac{d x \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{1}{2} i d x^2 \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Tan[a + b*x],x]

[Out]

(I/2)*d*x^2 - (d*x*Log[1 + E^((2*I)*(a + b*x))])/b - (c*Log[Cos[a + b*x]])/b + ((I/2)*d*PolyLog[2, -E^((2*I)*(
a + b*x))])/b^2

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Maple [B]  time = 0.252, size = 123, normalized size = 1.9 \begin{align*}{\frac{i}{2}}d{x}^{2}-icx+2\,{\frac{c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}-{\frac{c\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{b}}+{\frac{2\,idax}{b}}+{\frac{id{a}^{2}}{{b}^{2}}}-{\frac{d\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{b}}+{\frac{{\frac{i}{2}}d{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-2\,{\frac{ad\ln \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)*sin(b*x+a),x)

[Out]

1/2*I*d*x^2-I*c*x+2/b*c*ln(exp(I*(b*x+a)))-1/b*c*ln(exp(2*I*(b*x+a))+1)+2*I/b*d*a*x+I/b^2*d*a^2-1/b*d*ln(exp(2
*I*(b*x+a))+1)*x+1/2*I*d*polylog(2,-exp(2*I*(b*x+a)))/b^2-2/b^2*d*a*ln(exp(I*(b*x+a)))

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Maxima [B]  time = 1.72612, size = 154, normalized size = 2.33 \begin{align*} -\frac{-i \, b^{2} d x^{2} - 2 i \, b^{2} c x +{\left (2 i \, b d x + 2 i \, b c\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - i \, d{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) +{\left (b d x + b c\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\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)*sin(b*x+a),x, algorithm="maxima")

[Out]

-1/2*(-I*b^2*d*x^2 - 2*I*b^2*c*x + (2*I*b*d*x + 2*I*b*c)*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - I*d
*dilog(-e^(2*I*b*x + 2*I*a)) + (b*d*x + b*c)*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a)
+ 1))/b^2

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Fricas [B]  time = 0.571796, size = 837, normalized size = 12.68 \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)*sin(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 ) \sin{\left (a + b 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)*sin(b*x+a),x)

[Out]

Integral((c + d*x)*sin(a + b*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 ) \sin \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)*sin(b*x+a),x, algorithm="giac")

[Out]

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