### 3.41 $$\int (c+d x) (a+b \tan (e+f x)) \, dx$$

Optimal. Leaf size=84 $\frac{i b d \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{2 f^2}+\frac{a (c+d x)^2}{2 d}-\frac{b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac{i b (c+d x)^2}{2 d}$

[Out]

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

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Rubi [A]  time = 0.119414, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.312, Rules used = {3722, 3719, 2190, 2279, 2391} $\frac{a (c+d x)^2}{2 d}-\frac{b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac{i b (c+d x)^2}{2 d}+\frac{i b d \text{Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(c + d*x)*(a + b*Tan[e + f*x]),x]

[Out]

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

Rule 3722

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

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

Mathematica [A]  time = 0.0142465, size = 87, normalized size = 1.04 $\frac{i b d \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{2 f^2}+a c x+\frac{1}{2} a d x^2-\frac{b c \log (\cos (e+f x))}{f}-\frac{b d x \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac{1}{2} i b d x^2$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.061, size = 143, normalized size = 1.7 \begin{align*}{\frac{i}{2}}bd{x}^{2}-ibcx+{\frac{ad{x}^{2}}{2}}+acx-{\frac{bc\ln \left ({{\rm e}^{2\,i \left ( fx+e \right ) }}+1 \right ) }{f}}+2\,{\frac{bc\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{f}}+{\frac{2\,ibdex}{f}}+{\frac{ibd{e}^{2}}{{f}^{2}}}-{\frac{bd\ln \left ({{\rm e}^{2\,i \left ( fx+e \right ) }}+1 \right ) x}{f}}+{\frac{{\frac{i}{2}}bd{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-2\,{\frac{bde\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.63981, size = 176, normalized size = 2.1 \begin{align*} \frac{{\left (a + i \, b\right )} d f^{2} x^{2} + 2 \,{\left (a + i \, b\right )} c f^{2} x + i \, b d{\rm Li}_2\left (-e^{\left (2 i \, f x + 2 i \, e\right )}\right ) -{\left (2 i \, b d f x + 2 i \, b c f\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) -{\left (b d f x + b c f\right )} \log \left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}{2 \, f^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.62954, size = 420, normalized size = 5. \begin{align*} \frac{2 \, a d f^{2} x^{2} + 4 \, a c f^{2} x - i \, b d{\rm Li}_2\left (\frac{2 \,{\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) + i \, b d{\rm Li}_2\left (\frac{2 \,{\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 2 \,{\left (b d f x + b c f\right )} \log \left (-\frac{2 \,{\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \,{\left (b d f x + b c f\right )} \log \left (-\frac{2 \,{\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \, f^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*d*f^2*x^2 + 4*a*c*f^2*x - I*b*d*dilog(2*(I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1) + 1) + I*b*d*dilog(
2*(-I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1) + 1) - 2*(b*d*f*x + b*c*f)*log(-2*(I*tan(f*x + e) - 1)/(tan(f*x +
e)^2 + 1)) - 2*(b*d*f*x + b*c*f)*log(-2*(-I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)))/f^2

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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