\(\int \frac {\tan ^3(x)}{a+b \cos (x)} \, dx\) [11]

Optimal result

Integrand size = 13, antiderivative size = 57 \[ \int \frac {\tan ^3(x)}{a+b \cos (x)} \, dx=\frac {\left (a^2-b^2\right ) \log (\cos (x))}{a^3}-\frac {\left (a^2-b^2\right ) \log (a+b \cos (x))}{a^3}-\frac {b \sec (x)}{a^2}+\frac {\sec ^2(x)}{2 a} \]

[Out]

(a^2-b^2)*ln(cos(x))/a^3-(a^2-b^2)*ln(a+b*cos(x))/a^3-b*sec(x)/a^2+1/2*sec(x)^2/a

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2800, 908} \[ \int \frac {\tan ^3(x)}{a+b \cos (x)} \, dx=-\frac {b \sec (x)}{a^2}+\frac {\left (a^2-b^2\right ) \log (\cos (x))}{a^3}-\frac {\left (a^2-b^2\right ) \log (a+b \cos (x))}{a^3}+\frac {\sec ^2(x)}{2 a} \]

[In]

Int[Tan[x]^3/(a + b*Cos[x]),x]

[Out]

((a^2 - b^2)*Log[Cos[x]])/a^3 - ((a^2 - b^2)*Log[a + b*Cos[x]])/a^3 - (b*Sec[x])/a^2 + Sec[x]^2/(2*a)

Rule 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2800

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {b^2-x^2}{x^3 (a+x)} \, dx,x,b \cos (x)\right ) \\ & = -\text {Subst}\left (\int \left (\frac {b^2}{a x^3}-\frac {b^2}{a^2 x^2}+\frac {-a^2+b^2}{a^3 x}+\frac {a^2-b^2}{a^3 (a+x)}\right ) \, dx,x,b \cos (x)\right ) \\ & = \frac {\left (a^2-b^2\right ) \log (\cos (x))}{a^3}-\frac {\left (a^2-b^2\right ) \log (a+b \cos (x))}{a^3}-\frac {b \sec (x)}{a^2}+\frac {\sec ^2(x)}{2 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.81 \[ \int \frac {\tan ^3(x)}{a+b \cos (x)} \, dx=\frac {2 \left (a^2-b^2\right ) (\log (\cos (x))-\log (a+b \cos (x)))-2 a b \sec (x)+a^2 \sec ^2(x)}{2 a^3} \]

[In]

Integrate[Tan[x]^3/(a + b*Cos[x]),x]

[Out]

(2*(a^2 - b^2)*(Log[Cos[x]] - Log[a + b*Cos[x]]) - 2*a*b*Sec[x] + a^2*Sec[x]^2)/(2*a^3)

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.02

method	result	size
default	\(-\frac {b}{a^{2} \cos \left (x \right )}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (\cos \left (x \right )\right )}{a^{3}}+\frac {1}{2 a \cos \left (x \right )^{2}}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (a +\cos \left (x \right ) b \right )}{a^{3}}\)	\(58\)
risch	\(-\frac {2 \left ({\mathrm e}^{3 i x} b -a \,{\mathrm e}^{2 i x}+{\mathrm e}^{i x} b \right )}{\left ({\mathrm e}^{2 i x}+1\right )^{2} a^{2}}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 a \,{\mathrm e}^{i x}}{b}+1\right )}{a}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 a \,{\mathrm e}^{i x}}{b}+1\right ) b^{2}}{a^{3}}+\frac {\ln \left ({\mathrm e}^{2 i x}+1\right )}{a}-\frac {\ln \left ({\mathrm e}^{2 i x}+1\right ) b^{2}}{a^{3}}\)	\(117\)

[In]

int(tan(x)^3/(a+cos(x)*b),x,method=_RETURNVERBOSE)

[Out]

-1/a^2*b/cos(x)+(a^2-b^2)*ln(cos(x))/a^3+1/2/a/cos(x)^2-(a^2-b^2)*ln(a+cos(x)*b)/a^3

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.16 \[ \int \frac {\tan ^3(x)}{a+b \cos (x)} \, dx=-\frac {2 \, {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} \log \left (-b \cos \left (x\right ) - a\right ) - 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} \log \left (-\cos \left (x\right )\right ) + 2 \, a b \cos \left (x\right ) - a^{2}}{2 \, a^{3} \cos \left (x\right )^{2}} \]

[In]

integrate(tan(x)^3/(a+b*cos(x)),x, algorithm="fricas")

[Out]

-1/2*(2*(a^2 - b^2)*cos(x)^2*log(-b*cos(x) - a) - 2*(a^2 - b^2)*cos(x)^2*log(-cos(x)) + 2*a*b*cos(x) - a^2)/(a
^3*cos(x)^2)

Sympy [F]

\[ \int \frac {\tan ^3(x)}{a+b \cos (x)} \, dx=\int \frac {\tan ^{3}{\left (x \right )}}{a + b \cos {\left (x \right )}}\, dx \]

[In]

integrate(tan(x)**3/(a+b*cos(x)),x)

[Out]

Integral(tan(x)**3/(a + b*cos(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98 \[ \int \frac {\tan ^3(x)}{a+b \cos (x)} \, dx=-\frac {{\left (a^{2} - b^{2}\right )} \log \left (b \cos \left (x\right ) + a\right )}{a^{3}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (\cos \left (x\right )\right )}{a^{3}} - \frac {2 \, b \cos \left (x\right ) - a}{2 \, a^{2} \cos \left (x\right )^{2}} \]

[In]

integrate(tan(x)^3/(a+b*cos(x)),x, algorithm="maxima")

[Out]

-(a^2 - b^2)*log(b*cos(x) + a)/a^3 + (a^2 - b^2)*log(cos(x))/a^3 - 1/2*(2*b*cos(x) - a)/(a^2*cos(x)^2)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.16 \[ \int \frac {\tan ^3(x)}{a+b \cos (x)} \, dx=\frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | \cos \left (x\right ) \right |}\right )}{a^{3}} - \frac {{\left (a^{2} b - b^{3}\right )} \log \left ({\left | b \cos \left (x\right ) + a \right |}\right )}{a^{3} b} - \frac {2 \, a b \cos \left (x\right ) - a^{2}}{2 \, a^{3} \cos \left (x\right )^{2}} \]

[In]

integrate(tan(x)^3/(a+b*cos(x)),x, algorithm="giac")

[Out]

(a^2 - b^2)*log(abs(cos(x)))/a^3 - (a^2*b - b^3)*log(abs(b*cos(x) + a))/(a^3*b) - 1/2*(2*a*b*cos(x) - a^2)/(a^
3*cos(x)^2)

Mupad [B] (verification not implemented)

Time = 14.54 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.02 \[ \int \frac {\tan ^3(x)}{a+b \cos (x)} \, dx=-\frac {2\,a^2\,\mathrm {atanh}\left (\frac {a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{-b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+a+b}\right )-2\,b^2\,\mathrm {atanh}\left (\frac {a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{-b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+a+b}\right )}{a^3}-\frac {2\,a\,b-{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,a^2+2\,b\,a\right )}{a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+a^3} \]

[In]

int(tan(x)^3/(a + b*cos(x)),x)

[Out]

- (2*a^2*atanh((a*tan(x/2)^2)/(a + b - b*tan(x/2)^2)) - 2*b^2*atanh((a*tan(x/2)^2)/(a + b - b*tan(x/2)^2)))/a^
3 - (2*a*b - tan(x/2)^2*(2*a*b + 2*a^2))/(a^3 - 2*a^3*tan(x/2)^2 + a^3*tan(x/2)^4)