\(\int \frac {\tan ^4(x)}{a+a \csc (x)} \, dx\) [1]

Optimal result

Integrand size = 13, antiderivative size = 55 \[ \int \frac {\tan ^4(x)}{a+a \csc (x)} \, dx=\frac {x}{a}-\frac {(15-8 \csc (x)) \tan (x)}{15 a}+\frac {(5-4 \csc (x)) \tan ^3(x)}{15 a}-\frac {(1-\csc (x)) \tan ^5(x)}{5 a} \]

[Out]

x/a-1/15*(15-8*csc(x))*tan(x)/a+1/15*(5-4*csc(x))*tan(x)^3/a-1/5*(1-csc(x))*tan(x)^5/a

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3973, 3967, 8} \[ \int \frac {\tan ^4(x)}{a+a \csc (x)} \, dx=\frac {x}{a}-\frac {\tan ^5(x) (1-\csc (x))}{5 a}+\frac {\tan ^3(x) (5-4 \csc (x))}{15 a}-\frac {\tan (x) (15-8 \csc (x))}{15 a} \]

[In]

Int[Tan[x]^4/(a + a*Csc[x]),x]

[Out]

x/a - ((15 - 8*Csc[x])*Tan[x])/(15*a) + ((5 - 4*Csc[x])*Tan[x]^3)/(15*a) - ((1 - Csc[x])*Tan[x]^5)/(5*a)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3967

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-(e*Cot[c
+ d*x])^(m + 1))*((a + b*Csc[c + d*x])/(d*e*(m + 1))), x] - Dist[1/(e^2*(m + 1)), Int[(e*Cot[c + d*x])^(m + 2)
*(a*(m + 1) + b*(m + 2)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[m, -1]

Rule 3973

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (-a+a \csc (x)) \tan ^6(x) \, dx}{a^2} \\ & = -\frac {(1-\csc (x)) \tan ^5(x)}{5 a}+\frac {\int (5 a-4 a \csc (x)) \tan ^4(x) \, dx}{5 a^2} \\ & = \frac {(5-4 \csc (x)) \tan ^3(x)}{15 a}-\frac {(1-\csc (x)) \tan ^5(x)}{5 a}+\frac {\int (-15 a+8 a \csc (x)) \tan ^2(x) \, dx}{15 a^2} \\ & = -\frac {(15-8 \csc (x)) \tan (x)}{15 a}+\frac {(5-4 \csc (x)) \tan ^3(x)}{15 a}-\frac {(1-\csc (x)) \tan ^5(x)}{5 a}+\frac {\int 15 a \, dx}{15 a^2} \\ & = \frac {x}{a}-\frac {(15-8 \csc (x)) \tan (x)}{15 a}+\frac {(5-4 \csc (x)) \tan ^3(x)}{15 a}-\frac {(1-\csc (x)) \tan ^5(x)}{5 a} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(111\) vs. \(2(55)=110\).

Time = 0.30 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.02 \[ \int \frac {\tan ^4(x)}{a+a \csc (x)} \, dx=\frac {200+6 (-89+120 x) \cos (x)+128 \cos (2 x)-178 \cos (3 x)+240 x \cos (3 x)+184 \cos (4 x)-64 \sin (x)-178 \sin (2 x)+240 x \sin (2 x)-128 \sin (3 x)-89 \sin (4 x)+120 x \sin (4 x)}{960 a \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^3 \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^5} \]

[In]

Integrate[Tan[x]^4/(a + a*Csc[x]),x]

[Out]

(200 + 6*(-89 + 120*x)*Cos[x] + 128*Cos[2*x] - 178*Cos[3*x] + 240*x*Cos[3*x] + 184*Cos[4*x] - 64*Sin[x] - 178*
Sin[2*x] + 240*x*Sin[2*x] - 128*Sin[3*x] - 89*Sin[4*x] + 120*x*Sin[4*x])/(960*a*(Cos[x/2] - Sin[x/2])^3*(Cos[x
/2] + Sin[x/2])^5)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.53

method	result	size
default	\(\frac {-\frac {1}{6 \left (\tan \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {5}{8 \left (\tan \left (\frac {x}{2}\right )-1\right )}+2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )+\frac {2}{5 \left (\tan \left (\frac {x}{2}\right )+1\right )^{5}}-\frac {1}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {1}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {11}{8 \left (\tan \left (\frac {x}{2}\right )+1\right )}}{a}\)	\(84\)
risch	\(\frac {x}{a}+\frac {\frac {62 i {\mathrm e}^{2 i x}}{15}+\frac {146 \,{\mathrm e}^{3 i x}}{15}+\frac {62 \,{\mathrm e}^{i x}}{15}+\frac {10 i {\mathrm e}^{4 i x}}{3}+\frac {26 \,{\mathrm e}^{5 i x}}{3}+\frac {46 i}{15}-2 i {\mathrm e}^{6 i x}+2 \,{\mathrm e}^{7 i x}}{\left (i+{\mathrm e}^{i x}\right )^{5} \left ({\mathrm e}^{i x}-i\right )^{3} a}\)	\(87\)

[In]

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

[Out]

64/a*(-1/384/(tan(1/2*x)-1)^3-1/256/(tan(1/2*x)-1)^2+5/512/(tan(1/2*x)-1)+1/32*arctan(tan(1/2*x))+1/160/(tan(1
/2*x)+1)^5-1/64/(tan(1/2*x)+1)^4+1/64/(tan(1/2*x)+1)^2+11/512/(tan(1/2*x)+1))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.05 \[ \int \frac {\tan ^4(x)}{a+a \csc (x)} \, dx=\frac {15 \, x \cos \left (x\right )^{3} + 23 \, \cos \left (x\right )^{4} - 19 \, \cos \left (x\right )^{2} + {\left (15 \, x \cos \left (x\right )^{3} - 8 \, \cos \left (x\right )^{2} + 1\right )} \sin \left (x\right ) + 4}{15 \, {\left (a \cos \left (x\right )^{3} \sin \left (x\right ) + a \cos \left (x\right )^{3}\right )}} \]

[In]

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

[Out]

1/15*(15*x*cos(x)^3 + 23*cos(x)^4 - 19*cos(x)^2 + (15*x*cos(x)^3 - 8*cos(x)^2 + 1)*sin(x) + 4)/(a*cos(x)^3*sin
(x) + a*cos(x)^3)

Sympy [F]

\[ \int \frac {\tan ^4(x)}{a+a \csc (x)} \, dx=\frac {\int \frac {\tan ^{4}{\left (x \right )}}{\csc {\left (x \right )} + 1}\, dx}{a} \]

[In]

integrate(tan(x)**4/(a+a*csc(x)),x)

[Out]

Integral(tan(x)**4/(csc(x) + 1), x)/a

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (47) = 94\).

Time = 0.33 (sec) , antiderivative size = 194, normalized size of antiderivative = 3.53 \[ \int \frac {\tan ^4(x)}{a+a \csc (x)} \, dx=\frac {2 \, {\left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac {46 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac {13 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {100 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {35 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} - \frac {30 \, \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} - \frac {15 \, \sin \left (x\right )^{7}}{{\left (\cos \left (x\right ) + 1\right )}^{7}} + 8\right )}}{15 \, {\left (a + \frac {2 \, a \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac {2 \, a \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac {6 \, a \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {6 \, a \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac {2 \, a \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} - \frac {2 \, a \sin \left (x\right )^{7}}{{\left (\cos \left (x\right ) + 1\right )}^{7}} - \frac {a \sin \left (x\right )^{8}}{{\left (\cos \left (x\right ) + 1\right )}^{8}}\right )}} + \frac {2 \, \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a} \]

[In]

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

[Out]

2/15*(sin(x)/(cos(x) + 1) - 46*sin(x)^2/(cos(x) + 1)^2 - 13*sin(x)^3/(cos(x) + 1)^3 + 100*sin(x)^4/(cos(x) + 1
)^4 + 35*sin(x)^5/(cos(x) + 1)^5 - 30*sin(x)^6/(cos(x) + 1)^6 - 15*sin(x)^7/(cos(x) + 1)^7 + 8)/(a + 2*a*sin(x
)/(cos(x) + 1) - 2*a*sin(x)^2/(cos(x) + 1)^2 - 6*a*sin(x)^3/(cos(x) + 1)^3 + 6*a*sin(x)^5/(cos(x) + 1)^5 + 2*a
*sin(x)^6/(cos(x) + 1)^6 - 2*a*sin(x)^7/(cos(x) + 1)^7 - a*sin(x)^8/(cos(x) + 1)^8) + 2*arctan(sin(x)/(cos(x)
+ 1))/a

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.45 \[ \int \frac {\tan ^4(x)}{a+a \csc (x)} \, dx=\frac {x}{a} + \frac {15 \, \tan \left (\frac {1}{2} \, x\right )^{2} - 36 \, \tan \left (\frac {1}{2} \, x\right ) + 17}{24 \, a {\left (\tan \left (\frac {1}{2} \, x\right ) - 1\right )}^{3}} + \frac {55 \, \tan \left (\frac {1}{2} \, x\right )^{4} + 260 \, \tan \left (\frac {1}{2} \, x\right )^{3} + 450 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 300 \, \tan \left (\frac {1}{2} \, x\right ) + 71}{40 \, a {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}^{5}} \]

[In]

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

[Out]

x/a + 1/24*(15*tan(1/2*x)^2 - 36*tan(1/2*x) + 17)/(a*(tan(1/2*x) - 1)^3) + 1/40*(55*tan(1/2*x)^4 + 260*tan(1/2
*x)^3 + 450*tan(1/2*x)^2 + 300*tan(1/2*x) + 71)/(a*(tan(1/2*x) + 1)^5)

Mupad [B] (verification not implemented)

Time = 18.99 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.51 \[ \int \frac {\tan ^4(x)}{a+a \csc (x)} \, dx=\frac {x}{a}-\frac {-2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^7-4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+\frac {14\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{3}+\frac {40\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{3}-\frac {26\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{15}-\frac {92\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{15}+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )}{15}+\frac {16}{15}}{a\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}^5} \]

[In]

int(tan(x)^4/(a + a/sin(x)),x)

[Out]

x/a - ((2*tan(x/2))/15 - (92*tan(x/2)^2)/15 - (26*tan(x/2)^3)/15 + (40*tan(x/2)^4)/3 + (14*tan(x/2)^5)/3 - 4*t
an(x/2)^6 - 2*tan(x/2)^7 + 16/15)/(a*(tan(x/2) - 1)^3*(tan(x/2) + 1)^5)