3.1.3 \(\int x \tan (a+b x) \, dx\) [3]

Optimal. Leaf size=54 \[ \frac {i x^2}{2}-\frac {x \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i \text {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3800, 2221, 2317, 2438} \begin {gather*} \frac {i \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {x \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Tan[a + b*x],x]

[Out]

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 3800

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]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 54, normalized size = 1.00 \begin {gather*} \frac {i x^2}{2}-\frac {x \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i \text {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Tan[a + b*x],x]

[Out]

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

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Maple [A]
time = 0.08, size = 78, normalized size = 1.44

method	result	size
risch	\(\frac {i x^{2}}{2}+\frac {2 i a x}{b}+\frac {i a^{2}}{b^{2}}-\frac {x \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}+\frac {i \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{2}}-\frac {2 a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*tan(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (41) = 82\).
time = 0.55, size = 92, normalized size = 1.70 \begin {gather*} -\frac {-i \, b^{2} x^{2} + 2 i \, b x \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + b x \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 ) - i \, {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right )}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*tan(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (41) = 82\).
time = 0.36, size = 122, normalized size = 2.26 \begin {gather*} -\frac {2 \, b x \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, b x \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + i \, {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - i \, {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right )}{4 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*tan(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \tan {\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*tan(b*x+a),x)

[Out]

Integral(x*tan(a + b*x), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*tan(b*x+a),x, algorithm="giac")

[Out]

integrate(x*tan(b*x + a), x)

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Mupad [B]
time = 2.46, size = 129, normalized size = 2.39 \begin {gather*} -\frac {\pi \,\ln \left (\cos \left (b\,x\right )\right )+\mathrm {polylog}\left (2,-{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-b\,x\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}-\pi \,\ln \left ({\mathrm {e}}^{-a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-b\,x\,2{}\mathrm {i}}+1\right )+2\,a\,\ln \left ({\mathrm {e}}^{-a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-b\,x\,2{}\mathrm {i}}+1\right )-\pi \,\ln \left ({\mathrm {e}}^{b\,x\,2{}\mathrm {i}}+1\right )+b^2\,x^2\,1{}\mathrm {i}-\ln \left (\cos \left (a+b\,x\right )\right )\,\left (2\,a-\pi \right )+2\,b\,x\,\ln \left ({\mathrm {e}}^{-a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-b\,x\,2{}\mathrm {i}}+1\right )+a\,b\,x\,2{}\mathrm {i}}{2\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*tan(a + b*x),x)

[Out]

-(polylog(2, -exp(-a*2i)*exp(-b*x*2i))*1i - pi*log(exp(b*x*2i) + 1) - pi*log(exp(-a*2i)*exp(-b*x*2i) + 1) + 2*
a*log(exp(-a*2i)*exp(-b*x*2i) + 1) + pi*log(cos(b*x)) + b^2*x^2*1i - log(cos(a + b*x))*(2*a - pi) + 2*b*x*log(
exp(-a*2i)*exp(-b*x*2i) + 1) + a*b*x*2i)/(2*b^2)

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