3.1.10 \(\int \frac {\tan ^4(x)}{a+b \cos (x)} \, dx\) [10]

Optimal. Leaf size=113 \[ \frac {2 (a-b)^{3/2} (a+b)^{3/2} \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4}+\frac {b \left (3 a^2-2 b^2\right ) \tanh ^{-1}(\sin (x))}{2 a^4}-\frac {\left (4 a^2-3 b^2\right ) \tan (x)}{3 a^3}-\frac {b \sec (x) \tan (x)}{2 a^2}+\frac {\sec ^2(x) \tan (x)}{3 a} \]

[Out]

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

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Rubi [A]
time = 0.28, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2804, 3134, 3080, 3855, 2738, 211} \begin {gather*} \frac {2 (a-b)^{3/2} (a+b)^{3/2} \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4}-\frac {b \tan (x) \sec (x)}{2 a^2}+\frac {b \left (3 a^2-2 b^2\right ) \tanh ^{-1}(\sin (x))}{2 a^4}-\frac {\left (4 a^2-3 b^2\right ) \tan (x)}{3 a^3}+\frac {\tan (x) \sec ^2(x)}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[x]^4/(a + b*Cos[x]),x]

[Out]

(2*(a - b)^(3/2)*(a + b)^(3/2)*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/a^4 + (b*(3*a^2 - 2*b^2)*ArcTanh[Si
n[x]])/(2*a^4) - ((4*a^2 - 3*b^2)*Tan[x])/(3*a^3) - (b*Sec[x]*Tan[x])/(2*a^2) + (Sec[x]^2*Tan[x])/(3*a)

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[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]

Rule 2804

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] + (-Dist[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] - 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]) /; FreeQ[{a
, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1] && IntegerQ[2*m]

Rule 3080

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(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]

Rule 3134

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(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] + D
ist[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])))

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\tan ^4(x)}{a+b \cos (x)} \, dx &=-\frac {b \sec (x) \tan (x)}{2 a^2}+\frac {\sec ^2(x) \tan (x)}{3 a}-\frac {\int \frac {\left (2 \left (4 a^2-3 b^2\right )-a b \cos (x)-3 \left (2 a^2-b^2\right ) \cos ^2(x)\right ) \sec ^2(x)}{a+b \cos (x)} \, dx}{6 a^2}\\ &=-\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}-\frac {\int \frac {\left (-3 b \left (3 a^2-2 b^2\right )-3 a \left (2 a^2-b^2\right ) \cos (x)\right ) \sec (x)}{a+b \cos (x)} \, dx}{6 a^3}\\ &=-\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}+\frac {\left (b \left (3 a^2-2 b^2\right )\right ) \int \sec (x) \, dx}{2 a^4}+\frac {\left (a^2-b^2\right )^2 \int \frac {1}{a+b \cos (x)} \, dx}{a^4}\\ &=\frac {b \left (3 a^2-2 b^2\right ) \tanh ^{-1}(\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}+\frac {\left (2 \left (a^2-b^2\right )^2\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{a^4}\\ &=\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\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 ) \tanh ^{-1}(\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}\\ \end {align*}

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Mathematica [A]
time = 1.24, size = 190, normalized size = 1.68 \begin {gather*} -\frac {48 \left (-a^2+b^2\right )^{3/2} \tanh ^{-1}\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^4/(a + b*Cos[x]),x]

[Out]

-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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(218\) vs. \(2(95)=190\).
time = 0.29, size = 219, normalized size = 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	\(\frac {i \left (3 a b \,{\mathrm e}^{5 i x}-12 a^{2} {\mathrm e}^{4 i x}+6 b^{2} {\mathrm e}^{4 i x}-12 a^{2} {\mathrm e}^{2 i x}+12 b^{2} {\mathrm e}^{2 i x}-3 b \,{\mathrm e}^{i x} a -8 a^{2}+6 b^{2}\right )}{3 a^{3} \left ({\mathrm e}^{2 i x}+1\right )^{3}}-\frac {3 b \ln \left ({\mathrm e}^{i x}-i\right )}{2 a^{2}}+\frac {b^{3} \ln \left ({\mathrm e}^{i x}-i\right )}{a^{4}}+\frac {3 b \ln \left ({\mathrm e}^{i x}+i\right )}{2 a^{2}}-\frac {b^{3} \ln \left ({\mathrm e}^{i x}+i\right )}{a^{4}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right )}{a^{2}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right ) b^{2}}{a^{4}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right )}{a^{2}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right ) b^{2}}{a^{4}}\)	\(331\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^4/(a+b*cos(x)),x,method=_RETURNVERBOSE)

[Out]

-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)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^4/(a+b*cos(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

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Fricas [A]
time = 0.53, size = 332, normalized size = 2.94 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^4/(a+b*cos(x)),x, algorithm="fricas")

[Out]

[-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
(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), 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)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{4}{\left (x \right )}}{a + b \cos {\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)**4/(a+b*cos(x)),x)

[Out]

Integral(tan(x)**4/(a + b*cos(x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (95) = 190\).
time = 0.47, size = 226, normalized size = 2.00 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^4/(a+b*cos(x)),x, algorithm="giac")

[Out]

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)

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Mupad [B]
time = 1.10, size = 1666, normalized size = 14.74 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\frac {64\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{112\,a\,b^2+128\,a^2\,b-64\,a^3-352\,b^3+\frac {16\,b^4}{a}+\frac {320\,b^5}{a^2}-\frac {112\,b^6}{a^3}-\frac {96\,b^7}{a^4}+\frac {48\,b^8}{a^5}}+\frac {144\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{16\,a\,b^4+128\,a^4\,b-64\,a^5+320\,b^5-352\,a^2\,b^3+112\,a^3\,b^2-\frac {112\,b^6}{a}-\frac {96\,b^7}{a^2}+\frac {48\,b^8}{a^3}}+\frac {80\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{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-\frac {96\,b^7}{a}+\frac {48\,b^8}{a^2}}-\frac {144\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{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+\frac {48\,b^8}{a}}+\frac {48\,b^5\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{-64\,a^8+128\,a^7\,b+112\,a^6\,b^2-352\,a^5\,b^3+16\,a^4\,b^4+320\,a^3\,b^5-112\,a^2\,b^6-96\,a\,b^7+48\,b^8}-\frac {192\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{128\,a^3\,b-352\,a\,b^3-64\,a^4+16\,b^4+112\,a^2\,b^2+\frac {320\,b^5}{a}-\frac {112\,b^6}{a^2}-\frac {96\,b^7}{a^3}+\frac {48\,b^8}{a^4}}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{a^4}-\frac {\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (5\,a^2-3\,b^2\right )}{3\,a^3}-\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a^2+a\,b-2\,b^2\right )}{a^3}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (-2\,a^2+a\,b+2\,b^2\right )}{a^3}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-1}-\frac {\mathrm {atan}\left (\frac {\frac {\left (\frac {3\,a^2\,b}{2}-b^3\right )\,\left (\frac {\left (\frac {3\,a^2\,b}{2}-b^3\right )\,\left (\frac {8\,\left (-4\,a^{13}+2\,a^{12}\,b+10\,a^{11}\,b^2-6\,a^{10}\,b^3-6\,a^9\,b^4+4\,a^8\,b^5\right )}{a^9}-\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {3\,a^2\,b}{2}-b^3\right )\,\left (8\,a^{10}\,b-16\,a^9\,b^2+8\,a^8\,b^3\right )}{a^{10}}\right )}{a^4}-\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (4\,a^9-4\,a^8\,b-7\,a^7\,b^2-11\,a^6\,b^3+39\,a^5\,b^4+3\,a^4\,b^5-48\,a^3\,b^6+16\,a^2\,b^7+16\,a\,b^8-8\,b^9\right )}{a^6}\right )\,1{}\mathrm {i}}{a^4}-\frac {\left (\frac {3\,a^2\,b}{2}-b^3\right )\,\left (\frac {\left (\frac {3\,a^2\,b}{2}-b^3\right )\,\left (\frac {8\,\left (-4\,a^{13}+2\,a^{12}\,b+10\,a^{11}\,b^2-6\,a^{10}\,b^3-6\,a^9\,b^4+4\,a^8\,b^5\right )}{a^9}+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {3\,a^2\,b}{2}-b^3\right )\,\left (8\,a^{10}\,b-16\,a^9\,b^2+8\,a^8\,b^3\right )}{a^{10}}\right )}{a^4}+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (4\,a^9-4\,a^8\,b-7\,a^7\,b^2-11\,a^6\,b^3+39\,a^5\,b^4+3\,a^4\,b^5-48\,a^3\,b^6+16\,a^2\,b^7+16\,a\,b^8-8\,b^9\right )}{a^6}\right )\,1{}\mathrm {i}}{a^4}}{\frac {16\,\left (-6\,a^{10}\,b+15\,a^9\,b^2+10\,a^8\,b^3-49\,a^7\,b^4+8\,a^6\,b^5+59\,a^5\,b^6-26\,a^4\,b^7-31\,a^3\,b^8+18\,a^2\,b^9+6\,a\,b^{10}-4\,b^{11}\right )}{a^9}+\frac {\left (\frac {3\,a^2\,b}{2}-b^3\right )\,\left (\frac {\left (\frac {3\,a^2\,b}{2}-b^3\right )\,\left (\frac {8\,\left (-4\,a^{13}+2\,a^{12}\,b+10\,a^{11}\,b^2-6\,a^{10}\,b^3-6\,a^9\,b^4+4\,a^8\,b^5\right )}{a^9}-\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {3\,a^2\,b}{2}-b^3\right )\,\left (8\,a^{10}\,b-16\,a^9\,b^2+8\,a^8\,b^3\right )}{a^{10}}\right )}{a^4}-\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (4\,a^9-4\,a^8\,b-7\,a^7\,b^2-11\,a^6\,b^3+39\,a^5\,b^4+3\,a^4\,b^5-48\,a^3\,b^6+16\,a^2\,b^7+16\,a\,b^8-8\,b^9\right )}{a^6}\right )}{a^4}+\frac {\left (\frac {3\,a^2\,b}{2}-b^3\right )\,\left (\frac {\left (\frac {3\,a^2\,b}{2}-b^3\right )\,\left (\frac {8\,\left (-4\,a^{13}+2\,a^{12}\,b+10\,a^{11}\,b^2-6\,a^{10}\,b^3-6\,a^9\,b^4+4\,a^8\,b^5\right )}{a^9}+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {3\,a^2\,b}{2}-b^3\right )\,\left (8\,a^{10}\,b-16\,a^9\,b^2+8\,a^8\,b^3\right )}{a^{10}}\right )}{a^4}+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (4\,a^9-4\,a^8\,b-7\,a^7\,b^2-11\,a^6\,b^3+39\,a^5\,b^4+3\,a^4\,b^5-48\,a^3\,b^6+16\,a^2\,b^7+16\,a\,b^8-8\,b^9\right )}{a^6}\right )}{a^4}}\right )\,\left (\frac {3\,a^2\,b}{2}-b^3\right )\,2{}\mathrm {i}}{a^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^4/(a + b*cos(x)),x)

[Out]

(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 - (8*tan(x/2)*(16*a*b^8 - 4*a^8*b + 4*a^9 - 8*b^9 + 16*a^2*b^7 -
48*a^3*b^6 + 3*a^4*b^5 + 39*a^5*b^4 - 11*a^6*b^3 - 7*a^7*b^2))/a^6)*1i)/a^4 - (((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 + (8*tan(x/2)*(16*a*b^8 - 4*a^8*b + 4*a^9 - 8*b
^9 + 16*a^2*b^7 - 48*a^3*b^6 + 3*a^4*b^5 + 39*a^5*b^4 - 11*a^6*b^3 - 7*a^7*b^2))/a^6)*1i)/a^4)/((16*(6*a*b^10
- 6*a^10*b - 4*b^11 + 18*a^2*b^9 - 31*a^3*b^8 - 26*a^4*b^7 + 59*a^5*b^6 + 8*a^6*b^5 - 49*a^7*b^4 + 10*a^8*b^3
+ 15*a^9*b^2))/a^9 + (((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 - (8*tan(x/2)*(16*a*b^8 - 4*a^8*b + 4*a^9 - 8*b^9 + 16*a^2*b^7 - 48*a^3*b^6 + 3*a^4*b^5 + 39*a^5*b^4 - 1
1*a^6*b^3 - 7*a^7*b^2))/a^6))/a^4 + (((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 + (8*tan(x/2)*(16*a*b^8 - 4*a^8*b + 4*a^9 - 8*b^9 + 16*a^2*b^7 - 48*a^3*b^6 + 3*a^4*b^5 +
39*a^5*b^4 - 11*a^6*b^3 - 7*a^7*b^2))/a^6))/a^4))*((3*a^2*b)/2 - b^3)*2i)/a^4

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