Integrand size = 13, antiderivative size = 113 \[ \int \frac {\tan ^4(x)}{a+b \cos (x)} \, dx=\frac {2 (a-b)^{3/2} (a+b)^{3/2} \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4}+\frac {b \left (3 a^2-2 b^2\right ) \text {arctanh}(\sin (x))}{2 a^4}-\frac {\left (4 a^2-3 b^2\right ) \tan (x)}{3 a^3}-\frac {b \sec (x) \tan (x)}{2 a^2}+\frac {\sec ^2(x) \tan (x)}{3 a} \] Output:
2*(a-b)^(3/2)*(a+b)^(3/2)*arctan((a-b)^(1/2)*tan(1/2*x)/(a+b)^(1/2))/a^4+1 /2*b*(3*a^2-2*b^2)*arctanh(sin(x))/a^4-1/3*(4*a^2-3*b^2)*tan(x)/a^3-1/2*b* sec(x)*tan(x)/a^2+1/3*sec(x)^2*tan(x)/a
Time = 1.35 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.68 \[ \int \frac {\tan ^4(x)}{a+b \cos (x)} \, dx=-\frac {48 \left (-a^2+b^2\right )^{3/2} \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )+\sec ^3(x) \left (9 b \left (3 a^2-2 b^2\right ) \cos (x) \left (\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )\right )+3 b \left (3 a^2-2 b^2\right ) \cos (3 x) \left (\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )\right )+2 a \left (-3 b^2 \sin (x)+3 a b \sin (2 x)+\left (4 a^2-3 b^2\right ) \sin (3 x)\right )\right )}{24 a^4} \] Input:
Integrate[Tan[x]^4/(a + b*Cos[x]),x]
Output:
-1/24*(48*(-a^2 + b^2)^(3/2)*ArcTanh[((a - b)*Tan[x/2])/Sqrt[-a^2 + b^2]] + Sec[x]^3*(9*b*(3*a^2 - 2*b^2)*Cos[x]*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos [x/2] + Sin[x/2]]) + 3*b*(3*a^2 - 2*b^2)*Cos[3*x]*(Log[Cos[x/2] - Sin[x/2] ] - Log[Cos[x/2] + Sin[x/2]]) + 2*a*(-3*b^2*Sin[x] + 3*a*b*Sin[2*x] + (4*a ^2 - 3*b^2)*Sin[3*x])))/a^4
Time = 0.87 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.18, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {3042, 3204, 3042, 3534, 27, 3042, 3480, 3042, 3138, 218, 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.
| \(\displaystyle \int \frac {\tan ^4(x)}{a+b \cos (x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan \left (x-\frac {\pi }{2}\right )^4 \left (a-b \sin \left (x-\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3204 |
| \(\displaystyle -\frac {\int \frac {\left (-3 \left (2 a^2-b^2\right ) \cos ^2(x)-a b \cos (x)+2 \left (4 a^2-3 b^2\right )\right ) \sec ^2(x)}{a+b \cos (x)}dx}{6 a^2}-\frac {b \tan (x) \sec (x)}{2 a^2}+\frac {\tan (x) \sec ^2(x)}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {-3 \left (2 a^2-b^2\right ) \sin \left (x+\frac {\pi }{2}\right )^2-a b \sin \left (x+\frac {\pi }{2}\right )+2 \left (4 a^2-3 b^2\right )}{\sin \left (x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (x+\frac {\pi }{2}\right )\right )}dx}{6 a^2}-\frac {b \tan (x) \sec (x)}{2 a^2}+\frac {\tan (x) \sec ^2(x)}{3 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle -\frac {\frac {\int -\frac {3 \left (b \left (3 a^2-2 b^2\right )+a \left (2 a^2-b^2\right ) \cos (x)\right ) \sec (x)}{a+b \cos (x)}dx}{a}+\frac {2 \left (4 a^2-3 b^2\right ) \tan (x)}{a}}{6 a^2}-\frac {b \tan (x) \sec (x)}{2 a^2}+\frac {\tan (x) \sec ^2(x)}{3 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {2 \left (4 a^2-3 b^2\right ) \tan (x)}{a}-\frac {3 \int \frac {\left (b \left (3 a^2-2 b^2\right )+a \left (2 a^2-b^2\right ) \cos (x)\right ) \sec (x)}{a+b \cos (x)}dx}{a}}{6 a^2}-\frac {b \tan (x) \sec (x)}{2 a^2}+\frac {\tan (x) \sec ^2(x)}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {2 \left (4 a^2-3 b^2\right ) \tan (x)}{a}-\frac {3 \int \frac {b \left (3 a^2-2 b^2\right )+a \left (2 a^2-b^2\right ) \sin \left (x+\frac {\pi }{2}\right )}{\sin \left (x+\frac {\pi }{2}\right ) \left (a+b \sin \left (x+\frac {\pi }{2}\right )\right )}dx}{a}}{6 a^2}-\frac {b \tan (x) \sec (x)}{2 a^2}+\frac {\tan (x) \sec ^2(x)}{3 a}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle -\frac {\frac {2 \left (4 a^2-3 b^2\right ) \tan (x)}{a}-\frac {3 \left (\frac {2 \left (a^2-b^2\right )^2 \int \frac {1}{a+b \cos (x)}dx}{a}+\frac {b \left (3 a^2-2 b^2\right ) \int \sec (x)dx}{a}\right )}{a}}{6 a^2}-\frac {b \tan (x) \sec (x)}{2 a^2}+\frac {\tan (x) \sec ^2(x)}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {2 \left (4 a^2-3 b^2\right ) \tan (x)}{a}-\frac {3 \left (\frac {2 \left (a^2-b^2\right )^2 \int \frac {1}{a+b \sin \left (x+\frac {\pi }{2}\right )}dx}{a}+\frac {b \left (3 a^2-2 b^2\right ) \int \csc \left (x+\frac {\pi }{2}\right )dx}{a}\right )}{a}}{6 a^2}-\frac {b \tan (x) \sec (x)}{2 a^2}+\frac {\tan (x) \sec ^2(x)}{3 a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle -\frac {\frac {2 \left (4 a^2-3 b^2\right ) \tan (x)}{a}-\frac {3 \left (\frac {4 \left (a^2-b^2\right )^2 \int \frac {1}{(a-b) \tan ^2\left (\frac {x}{2}\right )+a+b}d\tan \left (\frac {x}{2}\right )}{a}+\frac {b \left (3 a^2-2 b^2\right ) \int \csc \left (x+\frac {\pi }{2}\right )dx}{a}\right )}{a}}{6 a^2}-\frac {b \tan (x) \sec (x)}{2 a^2}+\frac {\tan (x) \sec ^2(x)}{3 a}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {\frac {2 \left (4 a^2-3 b^2\right ) \tan (x)}{a}-\frac {3 \left (\frac {b \left (3 a^2-2 b^2\right ) \int \csc \left (x+\frac {\pi }{2}\right )dx}{a}+\frac {4 \left (a^2-b^2\right )^2 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{6 a^2}-\frac {b \tan (x) \sec (x)}{2 a^2}+\frac {\tan (x) \sec ^2(x)}{3 a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\frac {2 \left (4 a^2-3 b^2\right ) \tan (x)}{a}-\frac {3 \left (\frac {4 \left (a^2-b^2\right )^2 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}}+\frac {b \left (3 a^2-2 b^2\right ) \text {arctanh}(\sin (x))}{a}\right )}{a}}{6 a^2}-\frac {b \tan (x) \sec (x)}{2 a^2}+\frac {\tan (x) \sec ^2(x)}{3 a}\) |
Input:
Int[Tan[x]^4/(a + b*Cos[x]),x]
Output:
-1/2*(b*Sec[x]*Tan[x])/a^2 + (Sec[x]^2*Tan[x])/(3*a) - ((-3*((4*(a^2 - b^2 )^2*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b] ) + (b*(3*a^2 - 2*b^2)*ArcTanh[Sin[x]])/a))/a + (2*(4*a^2 - 3*b^2)*Tan[x]) /a)/(6*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> Simp[(-Cos[e + f*x])*((a + b*Sin[e + f*x])^(m + 1)/(3*a*f*Sin[ e + f*x]^3)), x] + (-Simp[b*(m - 2)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(6*a^2*f*Sin[e + f*x]^2)), x] - Simp[1/(6*a^2) Int[((a + b*Sin[e + f* x])^m/Sin[e + f*x]^2)*Simp[8*a^2 - b^2*(m - 1)*(m - 2) + a*b*m*Sin[e + f*x] - (6*a^2 - b^2*m*(m - 2))*Sin[e + f*x]^2, x], x], x]) /; FreeQ[{a, b, e, f , m}, x] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1] && IntegerQ[2*m]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(218\) vs. \(2(95)=190\).
Time = 0.91 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.94
| method | result | size |
| default | \(-\frac {1}{3 a \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {-a -b}{2 a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-2 a^{2}+a b +2 b^{2}}{2 a^{3} \left (\tan \left (\frac {x}{2}\right )+1\right )}+\frac {b \left (3 a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{2 a^{4}}+\frac {2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{3 a \left (\tan \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a +b}{2 a^{2} \left (\tan \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-2 a^{2}+a b +2 b^{2}}{2 a^{3} \left (\tan \left (\frac {x}{2}\right )-1\right )}-\frac {b \left (3 a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{2 a^{4}}\) | \(219\) |
| risch | \(\frac {i \left (3 a b \,{\mathrm e}^{5 i x}-12 a^{2} {\mathrm e}^{4 i x}+6 b^{2} {\mathrm e}^{4 i x}-12 \,{\mathrm e}^{2 i x} a^{2}+12 \,{\mathrm e}^{2 i x} b^{2}-3 b \,{\mathrm e}^{i x} a -8 a^{2}+6 b^{2}\right )}{3 a^{3} \left ({\mathrm e}^{2 i x}+1\right )^{3}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right )}{a^{2}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right ) b^{2}}{a^{4}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right )}{a^{2}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right ) b^{2}}{a^{4}}-\frac {3 b \ln \left ({\mathrm e}^{i x}-i\right )}{2 a^{2}}+\frac {b^{3} \ln \left ({\mathrm e}^{i x}-i\right )}{a^{4}}+\frac {3 b \ln \left ({\mathrm e}^{i x}+i\right )}{2 a^{2}}-\frac {b^{3} \ln \left ({\mathrm e}^{i x}+i\right )}{a^{4}}\) | \(331\) |
Input:
int(tan(x)^4/(a+b*cos(x)),x,method=_RETURNVERBOSE)
Output:
-1/3/a/(tan(1/2*x)+1)^3-1/2*(-a-b)/a^2/(tan(1/2*x)+1)^2-1/2*(-2*a^2+a*b+2* b^2)/a^3/(tan(1/2*x)+1)+1/2*b*(3*a^2-2*b^2)/a^4*ln(tan(1/2*x)+1)+2*(a^4-2* a^2*b^2+b^4)/a^4/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*x)/((a-b)*(a+b)) ^(1/2))-1/3/a/(tan(1/2*x)-1)^3-1/2*(a+b)/a^2/(tan(1/2*x)-1)^2-1/2*(-2*a^2+ a*b+2*b^2)/a^3/(tan(1/2*x)-1)-1/2*b*(3*a^2-2*b^2)/a^4*ln(tan(1/2*x)-1)
Time = 0.20 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.94 \[ \int \frac {\tan ^4(x)}{a+b \cos (x)} \, dx=\left [-\frac {6 \, {\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (x\right )^{3} \log \left (\frac {2 \, a b \cos \left (x\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) - 3 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (x\right )^{3} \log \left (\sin \left (x\right ) + 1\right ) + 3 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (x\right )^{3} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, {\left (3 \, a^{2} b \cos \left (x\right ) - 2 \, a^{3} + 2 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{12 \, a^{4} \cos \left (x\right )^{3}}, \frac {12 \, {\left (a^{2} - b^{2}\right )}^{\frac {3}{2}} \arctan \left (-\frac {a \cos \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (x\right )}\right ) \cos \left (x\right )^{3} + 3 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (x\right )^{3} \log \left (\sin \left (x\right ) + 1\right ) - 3 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (x\right )^{3} \log \left (-\sin \left (x\right ) + 1\right ) - 2 \, {\left (3 \, a^{2} b \cos \left (x\right ) - 2 \, a^{3} + 2 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{12 \, a^{4} \cos \left (x\right )^{3}}\right ] \] Input:
integrate(tan(x)^4/(a+b*cos(x)),x, algorithm="fricas")
Output:
[-1/12*(6*(a^2 - b^2)*sqrt(-a^2 + b^2)*cos(x)^3*log((2*a*b*cos(x) + (2*a^2 - b^2)*cos(x)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(x) + b)*sin(x) - a^2 + 2*b^2) /(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2)) - 3*(3*a^2*b - 2*b^3)*cos(x)^3*log(s in(x) + 1) + 3*(3*a^2*b - 2*b^3)*cos(x)^3*log(-sin(x) + 1) + 2*(3*a^2*b*co s(x) - 2*a^3 + 2*(4*a^3 - 3*a*b^2)*cos(x)^2)*sin(x))/(a^4*cos(x)^3), 1/12* (12*(a^2 - b^2)^(3/2)*arctan(-(a*cos(x) + b)/(sqrt(a^2 - b^2)*sin(x)))*cos (x)^3 + 3*(3*a^2*b - 2*b^3)*cos(x)^3*log(sin(x) + 1) - 3*(3*a^2*b - 2*b^3) *cos(x)^3*log(-sin(x) + 1) - 2*(3*a^2*b*cos(x) - 2*a^3 + 2*(4*a^3 - 3*a*b^ 2)*cos(x)^2)*sin(x))/(a^4*cos(x)^3)]
\[ \int \frac {\tan ^4(x)}{a+b \cos (x)} \, dx=\int \frac {\tan ^{4}{\left (x \right )}}{a + b \cos {\left (x \right )}}\, dx \] Input:
integrate(tan(x)**4/(a+b*cos(x)),x)
Output:
Integral(tan(x)**4/(a + b*cos(x)), x)
Exception generated. \[ \int \frac {\tan ^4(x)}{a+b \cos (x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(tan(x)^4/(a+b*cos(x)),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (95) = 190\).
Time = 0.39 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.00 \[ \int \frac {\tan ^4(x)}{a+b \cos (x)} \, dx=\frac {{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{2 \, a^{4}} - \frac {{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{2 \, a^{4}} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{4}} + \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{5} - 3 \, a b \tan \left (\frac {1}{2} \, x\right )^{5} - 6 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} - 20 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 12 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, x\right ) + 3 \, a b \tan \left (\frac {1}{2} \, x\right ) - 6 \, b^{2} \tan \left (\frac {1}{2} \, x\right )}{3 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} - 1\right )}^{3} a^{3}} \] Input:
integrate(tan(x)^4/(a+b*cos(x)),x, algorithm="giac")
Output:
1/2*(3*a^2*b - 2*b^3)*log(abs(tan(1/2*x) + 1))/a^4 - 1/2*(3*a^2*b - 2*b^3) *log(abs(tan(1/2*x) - 1))/a^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*(pi*floor(1/2*x/ pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*tan(1/2*x))/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^4) + 1/3*(6*a^2*tan(1/2*x)^5 - 3*a*b*tan(1/2* x)^5 - 6*b^2*tan(1/2*x)^5 - 20*a^2*tan(1/2*x)^3 + 12*b^2*tan(1/2*x)^3 + 6* a^2*tan(1/2*x) + 3*a*b*tan(1/2*x) - 6*b^2*tan(1/2*x))/((tan(1/2*x)^2 - 1)^ 3*a^3)
Time = 41.06 (sec) , antiderivative size = 1666, normalized size of antiderivative = 14.74 \[ \int \frac {\tan ^4(x)}{a+b \cos (x)} \, dx=\text {Too large to display} \] Input:
int(tan(x)^4/(a + b*cos(x)),x)
Output:
(2*atanh((64*tan(x/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(112*a*b^ 2 + 128*a^2*b - 64*a^3 - 352*b^3 + (16*b^4)/a + (320*b^5)/a^2 - (112*b^6)/ a^3 - (96*b^7)/a^4 + (48*b^8)/a^5) + (144*b^2*tan(x/2)*(b^6 - a^6 - 3*a^2* b^4 + 3*a^4*b^2)^(1/2))/(16*a*b^4 + 128*a^4*b - 64*a^5 + 320*b^5 - 352*a^2 *b^3 + 112*a^3*b^2 - (112*b^6)/a - (96*b^7)/a^2 + (48*b^8)/a^3) + (80*b^3* tan(x/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(320*a*b^5 + 128*a^5*b - 64*a^6 - 112*b^6 + 16*a^2*b^4 - 352*a^3*b^3 + 112*a^4*b^2 - (96*b^7)/a + (48*b^8)/a^2) - (144*b^4*tan(x/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1 /2))/(128*a^6*b - 112*a*b^6 - 64*a^7 - 96*b^7 + 320*a^2*b^5 + 16*a^3*b^4 - 352*a^4*b^3 + 112*a^5*b^2 + (48*b^8)/a) + (48*b^5*tan(x/2)*(b^6 - a^6 - 3 *a^2*b^4 + 3*a^4*b^2)^(1/2))/(128*a^7*b - 96*a*b^7 - 64*a^8 + 48*b^8 - 112 *a^2*b^6 + 320*a^3*b^5 + 16*a^4*b^4 - 352*a^5*b^3 + 112*a^6*b^2) - (192*b* tan(x/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(128*a^3*b - 352*a*b^3 - 64*a^4 + 16*b^4 + 112*a^2*b^2 + (320*b^5)/a - (112*b^6)/a^2 - (96*b^7)/ a^3 + (48*b^8)/a^4))*(-(a + b)^3*(a - b)^3)^(1/2))/a^4 - ((4*tan(x/2)^3*(5 *a^2 - 3*b^2))/(3*a^3) - (tan(x/2)*(a*b + 2*a^2 - 2*b^2))/a^3 + (tan(x/2)^ 5*(a*b - 2*a^2 + 2*b^2))/a^3)/(3*tan(x/2)^2 - 3*tan(x/2)^4 + tan(x/2)^6 - 1) - (atan(((((3*a^2*b)/2 - b^3)*((((3*a^2*b)/2 - b^3)*((8*(2*a^12*b - 4*a ^13 + 4*a^8*b^5 - 6*a^9*b^4 - 6*a^10*b^3 + 10*a^11*b^2))/a^9 - (8*tan(x/2) *((3*a^2*b)/2 - b^3)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/a^10))/a^4 - ...
Time = 0.17 (sec) , antiderivative size = 368, normalized size of antiderivative = 3.26 \[ \int \frac {\tan ^4(x)}{a+b \cos (x)} \, dx=\frac {12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right )^{2} a^{2}-12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right )^{2} b^{2}-12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2}+12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) b^{2}-6 \cos \left (x \right ) \sin \left (x \right ) a \,b^{2}-9 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{2} a^{2} b +6 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{2} b^{3}+9 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) a^{2} b -6 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) b^{3}+9 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{2} a^{2} b -6 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{2} b^{3}-9 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) a^{2} b +6 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) b^{3}+2 \sin \left (x \right )^{2} \tan \left (x \right )^{3} a^{3}-6 \sin \left (x \right )^{2} \tan \left (x \right ) a^{3}+3 \sin \left (x \right ) a^{2} b -2 \tan \left (x \right )^{3} a^{3}+6 \tan \left (x \right ) a^{3}}{6 a^{4} \left (\sin \left (x \right )^{2}-1\right )} \] Input:
int(tan(x)^4/(a+b*cos(x)),x)
Output:
(12*sqrt(a**2 - b**2)*atan((tan(x/2)*a - tan(x/2)*b)/sqrt(a**2 - b**2))*si n(x)**2*a**2 - 12*sqrt(a**2 - b**2)*atan((tan(x/2)*a - tan(x/2)*b)/sqrt(a* *2 - b**2))*sin(x)**2*b**2 - 12*sqrt(a**2 - b**2)*atan((tan(x/2)*a - tan(x /2)*b)/sqrt(a**2 - b**2))*a**2 + 12*sqrt(a**2 - b**2)*atan((tan(x/2)*a - t an(x/2)*b)/sqrt(a**2 - b**2))*b**2 - 6*cos(x)*sin(x)*a*b**2 - 9*log(tan(x/ 2) - 1)*sin(x)**2*a**2*b + 6*log(tan(x/2) - 1)*sin(x)**2*b**3 + 9*log(tan( x/2) - 1)*a**2*b - 6*log(tan(x/2) - 1)*b**3 + 9*log(tan(x/2) + 1)*sin(x)** 2*a**2*b - 6*log(tan(x/2) + 1)*sin(x)**2*b**3 - 9*log(tan(x/2) + 1)*a**2*b + 6*log(tan(x/2) + 1)*b**3 + 2*sin(x)**2*tan(x)**3*a**3 - 6*sin(x)**2*tan (x)*a**3 + 3*sin(x)*a**2*b - 2*tan(x)**3*a**3 + 6*tan(x)*a**3)/(6*a**4*(si n(x)**2 - 1))