Integrand size = 13, antiderivative size = 33 \[ \int \frac {\tan ^4(x)}{a+a \cos (x)} \, dx=\frac {\text {arctanh}(\sin (x))}{2 a}-\frac {\sec (x) \tan (x)}{2 a}+\frac {\tan ^3(x)}{3 a} \] Output:
1/2*arctanh(sin(x))/a-1/2*sec(x)*tan(x)/a+1/3*tan(x)^3/a
Leaf count is larger than twice the leaf count of optimal. \(105\) vs. \(2(33)=66\).
Time = 0.35 (sec) , antiderivative size = 105, normalized size of antiderivative = 3.18 \[ \int \frac {\tan ^4(x)}{a+a \cos (x)} \, dx=-\frac {\sec ^3(x) \left (9 \cos (x) \left (\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )\right )+3 \cos (3 x) \left (\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )\right )+2 (-3 \sin (x)+3 \sin (2 x)+\sin (3 x))\right )}{24 a} \] Input:
Integrate[Tan[x]^4/(a + a*Cos[x]),x]
Output:
-1/24*(Sec[x]^3*(9*Cos[x]*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2] + Sin[x /2]]) + 3*Cos[3*x]*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2] + Sin[x/2]]) + 2*(-3*Sin[x] + 3*Sin[2*x] + Sin[3*x])))/a
Time = 0.36 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {3042, 3185, 25, 3042, 3087, 15, 3091, 3042, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\tan ^4(x)}{a \cos (x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan \left (x-\frac {\pi }{2}\right )^4 \left (a-a \sin \left (x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 3185 |
\(\displaystyle \frac {\int \sec ^2(x) \tan ^2(x)dx}{a}+\frac {\int -\sec (x) \tan ^2(x)dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \sec ^2(x) \tan ^2(x)dx}{a}-\frac {\int \sec (x) \tan ^2(x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sec (x)^2 \tan (x)^2dx}{a}-\frac {\int \sec (x) \tan (x)^2dx}{a}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {\int \tan ^2(x)d\tan (x)}{a}-\frac {\int \sec (x) \tan (x)^2dx}{a}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\tan ^3(x)}{3 a}-\frac {\int \sec (x) \tan (x)^2dx}{a}\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle \frac {\tan ^3(x)}{3 a}-\frac {\frac {1}{2} \tan (x) \sec (x)-\frac {\int \sec (x)dx}{2}}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan ^3(x)}{3 a}-\frac {\frac {1}{2} \tan (x) \sec (x)-\frac {1}{2} \int \csc \left (x+\frac {\pi }{2}\right )dx}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\tan ^3(x)}{3 a}-\frac {\frac {1}{2} \tan (x) \sec (x)-\frac {1}{2} \text {arctanh}(\sin (x))}{a}\) |
Input:
Int[Tan[x]^4/(a + a*Cos[x]),x]
Output:
Tan[x]^3/(3*a) - (-1/2*ArcTanh[Sin[x]] + (Sec[x]*Tan[x])/2)/a
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[1/a Int[Sec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x ] - Simp[1/(b*g) Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /; Fre eQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.06
method | result | size |
risch | \(\frac {i \left (3 \,{\mathrm e}^{5 i x}-6 \,{\mathrm e}^{4 i x}-3 \,{\mathrm e}^{i x}-2\right )}{3 \left ({\mathrm e}^{2 i x}+1\right )^{3} a}-\frac {\ln \left ({\mathrm e}^{i x}-i\right )}{2 a}+\frac {\ln \left ({\mathrm e}^{i x}+i\right )}{2 a}\) | \(68\) |
default | \(\frac {-\frac {1}{3 \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {1}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {1}{2 \left (\tan \left (\frac {x}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{2}-\frac {1}{3 \left (\tan \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {1}{\left (\tan \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {1}{2 \left (\tan \left (\frac {x}{2}\right )-1\right )}-\frac {\ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{2}}{a}\) | \(85\) |
Input:
int(tan(x)^4/(a+a*cos(x)),x,method=_RETURNVERBOSE)
Output:
1/3*I*(3*exp(5*I*x)-6*exp(4*I*x)-3*exp(I*x)-2)/(exp(2*I*x)+1)^3/a-1/2/a*ln (exp(I*x)-I)+1/2/a*ln(exp(I*x)+I)
Time = 0.09 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52 \[ \int \frac {\tan ^4(x)}{a+a \cos (x)} \, dx=\frac {3 \, \cos \left (x\right )^{3} \log \left (\sin \left (x\right ) + 1\right ) - 3 \, \cos \left (x\right )^{3} \log \left (-\sin \left (x\right ) + 1\right ) - 2 \, {\left (2 \, \cos \left (x\right )^{2} + 3 \, \cos \left (x\right ) - 2\right )} \sin \left (x\right )}{12 \, a \cos \left (x\right )^{3}} \] Input:
integrate(tan(x)^4/(a+a*cos(x)),x, algorithm="fricas")
Output:
1/12*(3*cos(x)^3*log(sin(x) + 1) - 3*cos(x)^3*log(-sin(x) + 1) - 2*(2*cos( x)^2 + 3*cos(x) - 2)*sin(x))/(a*cos(x)^3)
\[ \int \frac {\tan ^4(x)}{a+a \cos (x)} \, dx=\frac {\int \frac {\tan ^{4}{\left (x \right )}}{\cos {\left (x \right )} + 1}\, dx}{a} \] Input:
integrate(tan(x)**4/(a+a*cos(x)),x)
Output:
Integral(tan(x)**4/(cos(x) + 1), x)/a
Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (27) = 54\).
Time = 0.03 (sec) , antiderivative size = 115, normalized size of antiderivative = 3.48 \[ \int \frac {\tan ^4(x)}{a+a \cos (x)} \, dx=-\frac {\frac {3 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac {8 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} - \frac {3 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}}}{3 \, {\left (a - \frac {3 \, a \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} - \frac {a \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}}\right )}} + \frac {\log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}{2 \, a} - \frac {\log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right )}{2 \, a} \] Input:
integrate(tan(x)^4/(a+a*cos(x)),x, algorithm="maxima")
Output:
-1/3*(3*sin(x)/(cos(x) + 1) - 8*sin(x)^3/(cos(x) + 1)^3 - 3*sin(x)^5/(cos( x) + 1)^5)/(a - 3*a*sin(x)^2/(cos(x) + 1)^2 + 3*a*sin(x)^4/(cos(x) + 1)^4 - a*sin(x)^6/(cos(x) + 1)^6) + 1/2*log(sin(x)/(cos(x) + 1) + 1)/a - 1/2*lo g(sin(x)/(cos(x) + 1) - 1)/a
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (27) = 54\).
Time = 0.34 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.97 \[ \int \frac {\tan ^4(x)}{a+a \cos (x)} \, dx=\frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{2 \, a} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{2 \, a} - \frac {3 \, \tan \left (\frac {1}{2} \, x\right )^{5} + 8 \, \tan \left (\frac {1}{2} \, x\right )^{3} - 3 \, \tan \left (\frac {1}{2} \, x\right )}{3 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} - 1\right )}^{3} a} \] Input:
integrate(tan(x)^4/(a+a*cos(x)),x, algorithm="giac")
Output:
1/2*log(abs(tan(1/2*x) + 1))/a - 1/2*log(abs(tan(1/2*x) - 1))/a - 1/3*(3*t an(1/2*x)^5 + 8*tan(1/2*x)^3 - 3*tan(1/2*x))/((tan(1/2*x)^2 - 1)^3*a)
Time = 40.81 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39 \[ \int \frac {\tan ^4(x)}{a+a \cos (x)} \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^5+\frac {8\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{3}-\mathrm {tan}\left (\frac {x}{2}\right )}{a\,{\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2-1\right )}^3} \] Input:
int(tan(x)^4/(a + a*cos(x)),x)
Output:
atanh(tan(x/2))/a - ((8*tan(x/2)^3)/3 - tan(x/2) + tan(x/2)^5)/(a*(tan(x/2 )^2 - 1)^3)
Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.91 \[ \int \frac {\tan ^4(x)}{a+a \cos (x)} \, dx=\frac {-6 \cos \left (x \right ) \sin \left (x \right )-3 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right )+3 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right )+2 \sin \left (x \right )^{2} \tan \left (x \right )^{3}-6 \sin \left (x \right )^{2} \tan \left (x \right )+3 \sin \left (x \right )-2 \tan \left (x \right )^{3}+6 \tan \left (x \right )}{6 a \left (\sin \left (x \right )^{2}-1\right )} \] Input:
int(tan(x)^4/(a+a*cos(x)),x)
Output:
( - 6*cos(x)*sin(x) - 3*log(tan(x/2) - 1)*sin(x)**2 + 3*log(tan(x/2) - 1) + 3*log(tan(x/2) + 1)*sin(x)**2 - 3*log(tan(x/2) + 1) + 2*sin(x)**2*tan(x) **3 - 6*sin(x)**2*tan(x) + 3*sin(x) - 2*tan(x)**3 + 6*tan(x))/(6*a*(sin(x) **2 - 1))