3.10 \(\int \frac{\tan ^4(x)}{a+b \cos (x)} \, dx\)

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

[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)

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Rubi [A]  time = 0.418255, antiderivative size = 113, normalized size of antiderivative = 1., 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 = {2725, 3055, 3001, 3770, 2659, 205} \[ -\frac{\left (4 a^2-3 b^2\right ) \tan (x)}{3 a^3}+\frac{b \left (3 a^2-2 b^2\right ) \tanh ^{-1}(\sin (x))}{2 a^4}+\frac{2 (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^4}-\frac{b \tan (x) \sec (x)}{2 a^2}+\frac{\tan (x) \sec ^2(x)}{3 a} \]

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 2725

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*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])/Sin[e + f*x]^2, 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 3055

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] + Dis
t[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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

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 3770

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

Rule 2659

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 205

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

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 ) \operatorname{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*}

Mathematica [A]  time = 1.18297, size = 190, normalized size = 1.68 \[ -\frac{48 \left (b^2-a^2\right )^{3/2} \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )+\sec ^3(x) \left (2 a \left (\left (4 a^2-3 b^2\right ) \sin (3 x)+3 a b \sin (2 x)-3 b^2 \sin (x)\right )+9 b \left (3 a^2-2 b^2\right ) \cos (x) \left (\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )+3 b \left (3 a^2-2 b^2\right ) \cos (3 x) \left (\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )\right )}{24 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^4/(a + b*Cos[x]),x]

[Out]

-(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]*(L
og[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])))/(24*a^4)

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Maple [B]  time = 0.094, size = 338, normalized size = 3. \begin{align*} -{\frac{1}{3\,a} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{2\,a} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}-{\frac{b}{2\,{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{a} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{b}{2\,{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{{b}^{2}}{{a}^{3}} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{3\,b}{2\,{a}^{2}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{{b}^{3}}{{a}^{4}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{1}{3\,a} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{2\,a} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{\frac{b}{2\,{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{a} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{b}{2\,{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{{b}^{2}}{{a}^{3}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{3\,b}{2\,{a}^{2}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{{b}^{3}}{{a}^{4}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) }+2\,{\frac{1}{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( x/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }-4\,{\frac{{b}^{2}}{{a}^{2}\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( x/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }+2\,{\frac{{b}^{4}}{{a}^{4}\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( x/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/a/(tan(1/2*x)-1)^3-1/2/a/(tan(1/2*x)-1)^2-1/2/a^2/(tan(1/2*x)-1)^2*b+1/a/(tan(1/2*x)-1)-1/2/a^2/(tan(1/2*
x)-1)*b-1/a^3/(tan(1/2*x)-1)*b^2-3/2*b/a^2*ln(tan(1/2*x)-1)+b^3/a^4*ln(tan(1/2*x)-1)-1/3/a/(tan(1/2*x)+1)^3+1/
2/a/(tan(1/2*x)+1)^2+1/2/a^2/(tan(1/2*x)+1)^2*b+1/a/(tan(1/2*x)+1)-1/2/a^2/(tan(1/2*x)+1)*b-1/a^3/(tan(1/2*x)+
1)*b^2+3/2*b/a^2*ln(tan(1/2*x)+1)-b^3/a^4*ln(tan(1/2*x)+1)+2/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*x)*(a-b)/((a-b
)*(a+b))^(1/2))-4/a^2/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*x)*(a-b)/((a-b)*(a+b))^(1/2))*b^2+2/a^4/((a-b)*(a+b))
^(1/2)*arctan(tan(1/2*x)*(a-b)/((a-b)*(a+b))^(1/2))*b^4

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

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

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Fricas [A]  time = 2.47612, size = 853, normalized size = 7.55 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\left (x \right )}}{a + b \cos{\left (x \right )}}\, dx \end{align*}

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]  time = 1.47192, size = 305, normalized size = 2.7 \begin{align*} \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{align*}

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)