\(\int \frac {\tan ^4(x)}{a+b \csc (x)} \, dx\) [19]

Optimal result

Integrand size = 13, antiderivative size = 159 \[ \int \frac {\tan ^4(x)}{a+b \csc (x)} \, dx=\frac {x}{a}+\frac {2 b^5 \text {arctanh}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}-\frac {b^3 \sec (x)}{\left (a^2-b^2\right )^2}+\frac {b \sec (x)}{a^2-b^2}-\frac {b \sec ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a b^2 \tan (x)}{\left (a^2-b^2\right )^2}-\frac {a \tan (x)}{a^2-b^2}+\frac {a \tan ^3(x)}{3 \left (a^2-b^2\right )} \]

[Out]

x/a+2*b^5*arctanh((a+b*tan(1/2*x))/(a^2-b^2)^(1/2))/a/(a^2-b^2)^(5/2)-b^3*sec(x)/(a^2-b^2)^2+b*sec(x)/(a^2-b^2
)-1/3*b*sec(x)^3/(a^2-b^2)+a*b^2*tan(x)/(a^2-b^2)^2-a*tan(x)/(a^2-b^2)+1/3*a*tan(x)^3/(a^2-b^2)

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.29, number of steps used = 16, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {3983, 2981, 2686, 3554, 8, 2814, 2739, 632, 212} \[ \int \frac {\tan ^4(x)}{a+b \csc (x)} \, dx=\frac {2 b^5 \text {arctanh}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}-\frac {a b^2 x}{\left (a^2-b^2\right )^2}+\frac {a x}{a^2-b^2}+\frac {a \tan ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a b^2 \tan (x)}{\left (a^2-b^2\right )^2}-\frac {a \tan (x)}{a^2-b^2}-\frac {b \sec ^3(x)}{3 \left (a^2-b^2\right )}+\frac {b \sec (x)}{a^2-b^2}+\frac {b^4 x}{a \left (a^2-b^2\right )^2}-\frac {b^3 \sec (x)}{\left (a^2-b^2\right )^2} \]

[In]

Int[Tan[x]^4/(a + b*Csc[x]),x]

[Out]

-((a*b^2*x)/(a^2 - b^2)^2) + (b^4*x)/(a*(a^2 - b^2)^2) + (a*x)/(a^2 - b^2) + (2*b^5*ArcTanh[(a + b*Tan[x/2])/S
qrt[a^2 - b^2]])/(a*(a^2 - b^2)^(5/2)) - (b^3*Sec[x])/(a^2 - b^2)^2 + (b*Sec[x])/(a^2 - b^2) - (b*Sec[x]^3)/(3
*(a^2 - b^2)) + (a*b^2*Tan[x])/(a^2 - b^2)^2 - (a*Tan[x])/(a^2 - b^2) + (a*Tan[x]^3)/(3*(a^2 - b^2))

Rule 8

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*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 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2981

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[a*(d^2/(a^2 - b^2)), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[b*(d/(a^2 - b^2)), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[a^2*(d^2/(g^2*(a^2 - b^2
))), Int[(g*Cos[e + f*x])^(p + 2)*((d*Sin[e + f*x])^(n - 2)/(a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3983

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[Cos[c + d*x]^m
*((b + a*Sin[c + d*x])^n/Sin[c + d*x]^(m + n)), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[
n] && IntegerQ[m] && (IntegerQ[m/2] || LeQ[m, 1])

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sin (x) \tan ^4(x)}{b+a \sin (x)} \, dx \\ & = \frac {a \int \tan ^4(x) \, dx}{a^2-b^2}-\frac {b \int \sec (x) \tan ^3(x) \, dx}{a^2-b^2}+\frac {b^2 \int \frac {\sin (x) \tan ^2(x)}{b+a \sin (x)} \, dx}{a^2-b^2} \\ & = \frac {a \tan ^3(x)}{3 \left (a^2-b^2\right )}+\frac {\left (a b^2\right ) \int \tan ^2(x) \, dx}{\left (a^2-b^2\right )^2}-\frac {b^3 \int \sec (x) \tan (x) \, dx}{\left (a^2-b^2\right )^2}+\frac {b^4 \int \frac {\sin (x)}{b+a \sin (x)} \, dx}{\left (a^2-b^2\right )^2}-\frac {a \int \tan ^2(x) \, dx}{a^2-b^2}-\frac {b \text {Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\sec (x)\right )}{a^2-b^2} \\ & = \frac {b^4 x}{a \left (a^2-b^2\right )^2}+\frac {b \sec (x)}{a^2-b^2}-\frac {b \sec ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a b^2 \tan (x)}{\left (a^2-b^2\right )^2}-\frac {a \tan (x)}{a^2-b^2}+\frac {a \tan ^3(x)}{3 \left (a^2-b^2\right )}-\frac {\left (a b^2\right ) \int 1 \, dx}{\left (a^2-b^2\right )^2}-\frac {b^3 \text {Subst}(\int 1 \, dx,x,\sec (x))}{\left (a^2-b^2\right )^2}-\frac {b^5 \int \frac {1}{b+a \sin (x)} \, dx}{a \left (a^2-b^2\right )^2}+\frac {a \int 1 \, dx}{a^2-b^2} \\ & = -\frac {a b^2 x}{\left (a^2-b^2\right )^2}+\frac {b^4 x}{a \left (a^2-b^2\right )^2}+\frac {a x}{a^2-b^2}-\frac {b^3 \sec (x)}{\left (a^2-b^2\right )^2}+\frac {b \sec (x)}{a^2-b^2}-\frac {b \sec ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a b^2 \tan (x)}{\left (a^2-b^2\right )^2}-\frac {a \tan (x)}{a^2-b^2}+\frac {a \tan ^3(x)}{3 \left (a^2-b^2\right )}-\frac {\left (2 b^5\right ) \text {Subst}\left (\int \frac {1}{b+2 a x+b x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )^2} \\ & = -\frac {a b^2 x}{\left (a^2-b^2\right )^2}+\frac {b^4 x}{a \left (a^2-b^2\right )^2}+\frac {a x}{a^2-b^2}-\frac {b^3 \sec (x)}{\left (a^2-b^2\right )^2}+\frac {b \sec (x)}{a^2-b^2}-\frac {b \sec ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a b^2 \tan (x)}{\left (a^2-b^2\right )^2}-\frac {a \tan (x)}{a^2-b^2}+\frac {a \tan ^3(x)}{3 \left (a^2-b^2\right )}+\frac {\left (4 b^5\right ) \text {Subst}\left (\int \frac {1}{4 \left (a^2-b^2\right )-x^2} \, dx,x,2 a+2 b \tan \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )^2} \\ & = -\frac {a b^2 x}{\left (a^2-b^2\right )^2}+\frac {b^4 x}{a \left (a^2-b^2\right )^2}+\frac {a x}{a^2-b^2}+\frac {2 b^5 \text {arctanh}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}-\frac {b^3 \sec (x)}{\left (a^2-b^2\right )^2}+\frac {b \sec (x)}{a^2-b^2}-\frac {b \sec ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a b^2 \tan (x)}{\left (a^2-b^2\right )^2}-\frac {a \tan (x)}{a^2-b^2}+\frac {a \tan ^3(x)}{3 \left (a^2-b^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.30 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.22 \[ \int \frac {\tan ^4(x)}{a+b \csc (x)} \, dx=\frac {\csc (x) (b+a \sin (x)) \left (-\frac {24 b^5 \arctan \left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+\frac {\sec ^3(x) \left (2 a^3 b-8 a b^3+9 \left (a^2-b^2\right )^2 x \cos (x)+6 a b \left (a^2-2 b^2\right ) \cos (2 x)+3 a^4 x \cos (3 x)-6 a^2 b^2 x \cos (3 x)+3 b^4 x \cos (3 x)+3 a^2 b^2 \sin (x)-4 a^4 \sin (3 x)+7 a^2 b^2 \sin (3 x)\right )}{(a-b)^2 (a+b)^2}\right )}{12 a (a+b \csc (x))} \]

[In]

Integrate[Tan[x]^4/(a + b*Csc[x]),x]

[Out]

(Csc[x]*(b + a*Sin[x])*((-24*b^5*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + (Sec[x]^3*(2*
a^3*b - 8*a*b^3 + 9*(a^2 - b^2)^2*x*Cos[x] + 6*a*b*(a^2 - 2*b^2)*Cos[2*x] + 3*a^4*x*Cos[3*x] - 6*a^2*b^2*x*Cos
[3*x] + 3*b^4*x*Cos[3*x] + 3*a^2*b^2*Sin[x] - 4*a^4*Sin[3*x] + 7*a^2*b^2*Sin[3*x]))/((a - b)^2*(a + b)^2)))/(1
2*a*(a + b*Csc[x]))

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.19

method	result	size
default	\(\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a}-\frac {2 b^{5} \arctan \left (\frac {2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a \sqrt {-a^{2}+b^{2}}}-\frac {64}{3 \left (\tan \left (\frac {x}{2}\right )-1\right )^{3} \left (64 a +64 b \right )}-\frac {32}{\left (64 a +64 b \right ) \left (\tan \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-2 a -3 b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {x}{2}\right )-1\right )}-\frac {64}{3 \left (\tan \left (\frac {x}{2}\right )+1\right )^{3} \left (64 a -64 b \right )}+\frac {32}{\left (64 a -64 b \right ) \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-2 a +3 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )}\)	\(190\)
risch	\(\frac {x}{a}-\frac {2 \left (6 i a^{3} {\mathrm e}^{4 i x}-9 i a \,b^{2} {\mathrm e}^{4 i x}-3 a^{2} b \,{\mathrm e}^{5 i x}+6 b^{3} {\mathrm e}^{5 i x}+6 i a^{3} {\mathrm e}^{2 i x}-12 i a \,b^{2} {\mathrm e}^{2 i x}-2 a^{2} b \,{\mathrm e}^{3 i x}+8 b^{3} {\mathrm e}^{3 i x}+4 i a^{3}-7 i a \,b^{2}-3 a^{2} b \,{\mathrm e}^{i x}+6 \,{\mathrm e}^{i x} b^{3}\right )}{3 \left (-a^{2}+b^{2}\right )^{2} \left ({\mathrm e}^{2 i x}+1\right )^{3}}+\frac {b^{5} \ln \left ({\mathrm e}^{i x}+\frac {i b \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} a}-\frac {b^{5} \ln \left ({\mathrm e}^{i x}+\frac {i b \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} a}\)	\(305\)

[In]

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

[Out]

2/a*arctan(tan(1/2*x))-2/(a-b)^2/(a+b)^2*b^5/a/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*x)+2*a)/(-a^2+b^2)^(1/
2))-64/3/(tan(1/2*x)-1)^3/(64*a+64*b)-32/(64*a+64*b)/(tan(1/2*x)-1)^2-1/2/(a+b)^2*(-2*a-3*b)/(tan(1/2*x)-1)-64
/3/(tan(1/2*x)+1)^3/(64*a-64*b)+32/(64*a-64*b)/(tan(1/2*x)+1)^2-1/2/(a-b)^2*(-2*a+3*b)/(tan(1/2*x)+1)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 483, normalized size of antiderivative = 3.04 \[ \int \frac {\tan ^4(x)}{a+b \csc (x)} \, dx=\left [\frac {3 \, \sqrt {a^{2} - b^{2}} b^{5} \cos \left (x\right )^{3} \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2} + 2 \, {\left (b \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right )\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - 2 \, a^{5} b + 4 \, a^{3} b^{3} - 2 \, a b^{5} + 6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} x \cos \left (x\right )^{3} + 6 \, {\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (x\right )^{2} + 2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} - {\left (4 \, a^{6} - 11 \, a^{4} b^{2} + 7 \, a^{2} b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (x\right )^{3}}, \frac {3 \, \sqrt {-a^{2} + b^{2}} b^{5} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (x\right )}\right ) \cos \left (x\right )^{3} - a^{5} b + 2 \, a^{3} b^{3} - a b^{5} + 3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} x \cos \left (x\right )^{3} + 3 \, {\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (x\right )^{2} + {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} - {\left (4 \, a^{6} - 11 \, a^{4} b^{2} + 7 \, a^{2} b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{3 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (x\right )^{3}}\right ] \]

[In]

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

[Out]

[1/6*(3*sqrt(a^2 - b^2)*b^5*cos(x)^3*log(((a^2 - 2*b^2)*cos(x)^2 + 2*a*b*sin(x) + a^2 + b^2 + 2*(b*cos(x)*sin(
x) + a*cos(x))*sqrt(a^2 - b^2))/(a^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) - 2*a^5*b + 4*a^3*b^3 - 2*a*b^5 + 6
*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*cos(x)^3 + 6*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos(x)^2 + 2*(a^6 - 2*a^4*b^
2 + a^2*b^4 - (4*a^6 - 11*a^4*b^2 + 7*a^2*b^4)*cos(x)^2)*sin(x))/((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cos(x)
^3), 1/3*(3*sqrt(-a^2 + b^2)*b^5*arctan(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^2 - b^2)*cos(x)))*cos(x)^3 - a^5*
b + 2*a^3*b^3 - a*b^5 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*cos(x)^3 + 3*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos
(x)^2 + (a^6 - 2*a^4*b^2 + a^2*b^4 - (4*a^6 - 11*a^4*b^2 + 7*a^2*b^4)*cos(x)^2)*sin(x))/((a^7 - 3*a^5*b^2 + 3*
a^3*b^4 - a*b^6)*cos(x)^3)]

Sympy [F]

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

[In]

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

[Out]

Integral(tan(x)**4/(a + b*csc(x)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^4(x)}{a+b \csc (x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(tan(x)^4/(a+b*csc(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*a^2-4*b^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.36 \[ \int \frac {\tan ^4(x)}{a+b \csc (x)} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b^{5}}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {x}{a} + \frac {2 \, {\left (3 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{5} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 3 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{4} - 10 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 16 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 6 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} - 12 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, a^{3} \tan \left (\frac {1}{2} \, x\right ) - 6 \, a b^{2} \tan \left (\frac {1}{2} \, x\right ) - 2 \, a^{2} b + 5 \, b^{3}\right )}}{3 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} - 1\right )}^{3}} \]

[In]

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

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2)))*b^5/((a^5 - 2*a^3*b^2 + a*b
^4)*sqrt(-a^2 + b^2)) + x/a + 2/3*(3*a^3*tan(1/2*x)^5 - 6*a*b^2*tan(1/2*x)^5 + 3*b^3*tan(1/2*x)^4 - 10*a^3*tan
(1/2*x)^3 + 16*a*b^2*tan(1/2*x)^3 + 6*a^2*b*tan(1/2*x)^2 - 12*b^3*tan(1/2*x)^2 + 3*a^3*tan(1/2*x) - 6*a*b^2*ta
n(1/2*x) - 2*a^2*b + 5*b^3)/((a^4 - 2*a^2*b^2 + b^4)*(tan(1/2*x)^2 - 1)^3)

Mupad [B] (verification not implemented)

Time = 27.54 (sec) , antiderivative size = 4468, normalized size of antiderivative = 28.10 \[ \int \frac {\tan ^4(x)}{a+b \csc (x)} \, dx=\text {Too large to display} \]

[In]

int(tan(x)^4/(a + b/sin(x)),x)

[Out]

((4*tan(x/2)^2*(a^2*b - 2*b^3))/(a^4 + b^4 - 2*a^2*b^2) - (2*(2*a^2*b - 5*b^3))/(3*(a^4 + b^4 - 2*a^2*b^2)) +
(2*b^3*tan(x/2)^4)/(a^4 + b^4 - 2*a^2*b^2) - (4*a*tan(x/2)^3*(5*a^2 - 8*b^2))/(3*(a^4 + b^4 - 2*a^2*b^2)) + (2
*a*tan(x/2)*(a^2 - 2*b^2))/(a^4 + b^4 - 2*a^2*b^2) + (2*a*tan(x/2)^5*(a^2 - 2*b^2))/(a^4 + b^4 - 2*a^2*b^2))/(
3*tan(x/2)^2 - 3*tan(x/2)^4 + tan(x/2)^6 - 1) + (2*atan((320*a*b^21*tan(x/2))/(320*a*b^21 + 64*a^21*b - 2560*a
^3*b^19 + 9280*a^5*b^17 - 20160*a^7*b^15 + 29184*a^9*b^13 - 29504*a^11*b^11 + 21120*a^13*b^9 - 10560*a^15*b^7
+ 3520*a^17*b^5 - 704*a^19*b^3) + (64*a^21*b*tan(x/2))/(320*a*b^21 + 64*a^21*b - 2560*a^3*b^19 + 9280*a^5*b^17
- 20160*a^7*b^15 + 29184*a^9*b^13 - 29504*a^11*b^11 + 21120*a^13*b^9 - 10560*a^15*b^7 + 3520*a^17*b^5 - 704*a
^19*b^3) - (2560*a^3*b^19*tan(x/2))/(320*a*b^21 + 64*a^21*b - 2560*a^3*b^19 + 9280*a^5*b^17 - 20160*a^7*b^15 +
29184*a^9*b^13 - 29504*a^11*b^11 + 21120*a^13*b^9 - 10560*a^15*b^7 + 3520*a^17*b^5 - 704*a^19*b^3) + (9280*a^
5*b^17*tan(x/2))/(320*a*b^21 + 64*a^21*b - 2560*a^3*b^19 + 9280*a^5*b^17 - 20160*a^7*b^15 + 29184*a^9*b^13 - 2
9504*a^11*b^11 + 21120*a^13*b^9 - 10560*a^15*b^7 + 3520*a^17*b^5 - 704*a^19*b^3) - (20160*a^7*b^15*tan(x/2))/(
320*a*b^21 + 64*a^21*b - 2560*a^3*b^19 + 9280*a^5*b^17 - 20160*a^7*b^15 + 29184*a^9*b^13 - 29504*a^11*b^11 + 2
1120*a^13*b^9 - 10560*a^15*b^7 + 3520*a^17*b^5 - 704*a^19*b^3) + (29184*a^9*b^13*tan(x/2))/(320*a*b^21 + 64*a^
21*b - 2560*a^3*b^19 + 9280*a^5*b^17 - 20160*a^7*b^15 + 29184*a^9*b^13 - 29504*a^11*b^11 + 21120*a^13*b^9 - 10
560*a^15*b^7 + 3520*a^17*b^5 - 704*a^19*b^3) - (29504*a^11*b^11*tan(x/2))/(320*a*b^21 + 64*a^21*b - 2560*a^3*b
^19 + 9280*a^5*b^17 - 20160*a^7*b^15 + 29184*a^9*b^13 - 29504*a^11*b^11 + 21120*a^13*b^9 - 10560*a^15*b^7 + 35
20*a^17*b^5 - 704*a^19*b^3) + (21120*a^13*b^9*tan(x/2))/(320*a*b^21 + 64*a^21*b - 2560*a^3*b^19 + 9280*a^5*b^1
7 - 20160*a^7*b^15 + 29184*a^9*b^13 - 29504*a^11*b^11 + 21120*a^13*b^9 - 10560*a^15*b^7 + 3520*a^17*b^5 - 704*
a^19*b^3) - (10560*a^15*b^7*tan(x/2))/(320*a*b^21 + 64*a^21*b - 2560*a^3*b^19 + 9280*a^5*b^17 - 20160*a^7*b^15
+ 29184*a^9*b^13 - 29504*a^11*b^11 + 21120*a^13*b^9 - 10560*a^15*b^7 + 3520*a^17*b^5 - 704*a^19*b^3) + (3520*
a^17*b^5*tan(x/2))/(320*a*b^21 + 64*a^21*b - 2560*a^3*b^19 + 9280*a^5*b^17 - 20160*a^7*b^15 + 29184*a^9*b^13 -
29504*a^11*b^11 + 21120*a^13*b^9 - 10560*a^15*b^7 + 3520*a^17*b^5 - 704*a^19*b^3) - (704*a^19*b^3*tan(x/2))/(
320*a*b^21 + 64*a^21*b - 2560*a^3*b^19 + 9280*a^5*b^17 - 20160*a^7*b^15 + 29184*a^9*b^13 - 29504*a^11*b^11 + 2
1120*a^13*b^9 - 10560*a^15*b^7 + 3520*a^17*b^5 - 704*a^19*b^3)))/a + (b^5*atan(((b^5*((a + b)^5*(a - b)^5)^(1/
2)*(tan(x/2)*(64*a^22*b - 64*b^23 + 576*a^2*b^21 - 2560*a^4*b^19 + 7360*a^6*b^17 - 14880*a^8*b^15 + 21696*a^10
*b^13 - 22880*a^12*b^11 + 17280*a^14*b^9 - 9120*a^16*b^7 + 3200*a^18*b^5 - 672*a^20*b^3) + 32*a*b^22 - 320*a^3
*b^20 + 1440*a^5*b^18 - 3840*a^7*b^16 + 6720*a^9*b^14 - 8064*a^11*b^12 + 6720*a^13*b^10 - 3840*a^15*b^8 + 1440
*a^17*b^6 - 320*a^19*b^4 + 32*a^21*b^2 + (b^5*((a + b)^5*(a - b)^5)^(1/2)*(32*a^23*b + 96*a^3*b^21 - 864*a^5*b
^19 + 3488*a^7*b^17 - 8320*a^9*b^15 + 12992*a^11*b^13 - 13888*a^13*b^11 + 10304*a^15*b^9 - 5248*a^17*b^7 + 176
0*a^19*b^5 - 352*a^21*b^3 + tan(x/2)*(64*a^2*b^22 - 512*a^4*b^20 + 1792*a^6*b^18 - 3584*a^8*b^16 + 4480*a^10*b
^14 - 3584*a^12*b^12 + 1792*a^14*b^10 - 512*a^16*b^8 + 64*a^18*b^6) + (b^5*((a + b)^5*(a - b)^5)^(1/2)*(tan(x/
2)*(96*a^24*b - 64*a^2*b^23 + 736*a^4*b^21 - 3840*a^6*b^19 + 12000*a^8*b^17 - 24960*a^10*b^15 + 36288*a^12*b^1
3 - 37632*a^14*b^11 + 27840*a^16*b^9 - 14400*a^18*b^7 + 4960*a^20*b^5 - 1024*a^22*b^3) + 32*a^3*b^22 - 320*a^5
*b^20 + 1440*a^7*b^18 - 3840*a^9*b^16 + 6720*a^11*b^14 - 8064*a^13*b^12 + 6720*a^15*b^10 - 3840*a^17*b^8 + 144
0*a^19*b^6 - 320*a^21*b^4 + 32*a^23*b^2))/(a*b^10 - a^11 - 5*a^3*b^8 + 10*a^5*b^6 - 10*a^7*b^4 + 5*a^9*b^2)))/
(a*b^10 - a^11 - 5*a^3*b^8 + 10*a^5*b^6 - 10*a^7*b^4 + 5*a^9*b^2))*1i)/(a*b^10 - a^11 - 5*a^3*b^8 + 10*a^5*b^6
- 10*a^7*b^4 + 5*a^9*b^2) + (b^5*((a + b)^5*(a - b)^5)^(1/2)*(tan(x/2)*(64*a^22*b - 64*b^23 + 576*a^2*b^21 -
2560*a^4*b^19 + 7360*a^6*b^17 - 14880*a^8*b^15 + 21696*a^10*b^13 - 22880*a^12*b^11 + 17280*a^14*b^9 - 9120*a^1
6*b^7 + 3200*a^18*b^5 - 672*a^20*b^3) + 32*a*b^22 - 320*a^3*b^20 + 1440*a^5*b^18 - 3840*a^7*b^16 + 6720*a^9*b^
14 - 8064*a^11*b^12 + 6720*a^13*b^10 - 3840*a^15*b^8 + 1440*a^17*b^6 - 320*a^19*b^4 + 32*a^21*b^2 - (b^5*((a +
b)^5*(a - b)^5)^(1/2)*(32*a^23*b + 96*a^3*b^21 - 864*a^5*b^19 + 3488*a^7*b^17 - 8320*a^9*b^15 + 12992*a^11*b^
13 - 13888*a^13*b^11 + 10304*a^15*b^9 - 5248*a^17*b^7 + 1760*a^19*b^5 - 352*a^21*b^3 + tan(x/2)*(64*a^2*b^22 -
512*a^4*b^20 + 1792*a^6*b^18 - 3584*a^8*b^16 + 4480*a^10*b^14 - 3584*a^12*b^12 + 1792*a^14*b^10 - 512*a^16*b^
8 + 64*a^18*b^6) - (b^5*((a + b)^5*(a - b)^5)^(1/2)*(tan(x/2)*(96*a^24*b - 64*a^2*b^23 + 736*a^4*b^21 - 3840*a
^6*b^19 + 12000*a^8*b^17 - 24960*a^10*b^15 + 36288*a^12*b^13 - 37632*a^14*b^11 + 27840*a^16*b^9 - 14400*a^18*b
^7 + 4960*a^20*b^5 - 1024*a^22*b^3) + 32*a^3*b^22 - 320*a^5*b^20 + 1440*a^7*b^18 - 3840*a^9*b^16 + 6720*a^11*b
^14 - 8064*a^13*b^12 + 6720*a^15*b^10 - 3840*a^17*b^8 + 1440*a^19*b^6 - 320*a^21*b^4 + 32*a^23*b^2))/(a*b^10 -
a^11 - 5*a^3*b^8 + 10*a^5*b^6 - 10*a^7*b^4 + 5*a^9*b^2)))/(a*b^10 - a^11 - 5*a^3*b^8 + 10*a^5*b^6 - 10*a^7*b^
4 + 5*a^9*b^2))*1i)/(a*b^10 - a^11 - 5*a^3*b^8 + 10*a^5*b^6 - 10*a^7*b^4 + 5*a^9*b^2))/(2*tan(x/2)*(64*b^22 -
512*a^2*b^20 + 1792*a^4*b^18 - 3584*a^6*b^16 + 4480*a^8*b^14 - 3584*a^10*b^12 + 1792*a^12*b^10 - 512*a^14*b^8
+ 64*a^16*b^6) - 128*a*b^21 + 832*a^3*b^19 - 2304*a^5*b^17 + 3520*a^7*b^15 - 3200*a^9*b^13 + 1728*a^11*b^11 -
512*a^13*b^9 + 64*a^15*b^7 + (b^5*((a + b)^5*(a - b)^5)^(1/2)*(tan(x/2)*(64*a^22*b - 64*b^23 + 576*a^2*b^21 -
2560*a^4*b^19 + 7360*a^6*b^17 - 14880*a^8*b^15 + 21696*a^10*b^13 - 22880*a^12*b^11 + 17280*a^14*b^9 - 9120*a^1
6*b^7 + 3200*a^18*b^5 - 672*a^20*b^3) + 32*a*b^22 - 320*a^3*b^20 + 1440*a^5*b^18 - 3840*a^7*b^16 + 6720*a^9*b^
14 - 8064*a^11*b^12 + 6720*a^13*b^10 - 3840*a^15*b^8 + 1440*a^17*b^6 - 320*a^19*b^4 + 32*a^21*b^2 + (b^5*((a +
b)^5*(a - b)^5)^(1/2)*(32*a^23*b + 96*a^3*b^21 - 864*a^5*b^19 + 3488*a^7*b^17 - 8320*a^9*b^15 + 12992*a^11*b^
13 - 13888*a^13*b^11 + 10304*a^15*b^9 - 5248*a^17*b^7 + 1760*a^19*b^5 - 352*a^21*b^3 + tan(x/2)*(64*a^2*b^22 -
512*a^4*b^20 + 1792*a^6*b^18 - 3584*a^8*b^16 + 4480*a^10*b^14 - 3584*a^12*b^12 + 1792*a^14*b^10 - 512*a^16*b^
8 + 64*a^18*b^6) + (b^5*((a + b)^5*(a - b)^5)^(1/2)*(tan(x/2)*(96*a^24*b - 64*a^2*b^23 + 736*a^4*b^21 - 3840*a
^6*b^19 + 12000*a^8*b^17 - 24960*a^10*b^15 + 36288*a^12*b^13 - 37632*a^14*b^11 + 27840*a^16*b^9 - 14400*a^18*b
^7 + 4960*a^20*b^5 - 1024*a^22*b^3) + 32*a^3*b^22 - 320*a^5*b^20 + 1440*a^7*b^18 - 3840*a^9*b^16 + 6720*a^11*b
^14 - 8064*a^13*b^12 + 6720*a^15*b^10 - 3840*a^17*b^8 + 1440*a^19*b^6 - 320*a^21*b^4 + 32*a^23*b^2))/(a*b^10 -
a^11 - 5*a^3*b^8 + 10*a^5*b^6 - 10*a^7*b^4 + 5*a^9*b^2)))/(a*b^10 - a^11 - 5*a^3*b^8 + 10*a^5*b^6 - 10*a^7*b^
4 + 5*a^9*b^2)))/(a*b^10 - a^11 - 5*a^3*b^8 + 10*a^5*b^6 - 10*a^7*b^4 + 5*a^9*b^2) - (b^5*((a + b)^5*(a - b)^5
)^(1/2)*(tan(x/2)*(64*a^22*b - 64*b^23 + 576*a^2*b^21 - 2560*a^4*b^19 + 7360*a^6*b^17 - 14880*a^8*b^15 + 21696
*a^10*b^13 - 22880*a^12*b^11 + 17280*a^14*b^9 - 9120*a^16*b^7 + 3200*a^18*b^5 - 672*a^20*b^3) + 32*a*b^22 - 32
0*a^3*b^20 + 1440*a^5*b^18 - 3840*a^7*b^16 + 6720*a^9*b^14 - 8064*a^11*b^12 + 6720*a^13*b^10 - 3840*a^15*b^8 +
1440*a^17*b^6 - 320*a^19*b^4 + 32*a^21*b^2 - (b^5*((a + b)^5*(a - b)^5)^(1/2)*(32*a^23*b + 96*a^3*b^21 - 864*
a^5*b^19 + 3488*a^7*b^17 - 8320*a^9*b^15 + 12992*a^11*b^13 - 13888*a^13*b^11 + 10304*a^15*b^9 - 5248*a^17*b^7
+ 1760*a^19*b^5 - 352*a^21*b^3 + tan(x/2)*(64*a^2*b^22 - 512*a^4*b^20 + 1792*a^6*b^18 - 3584*a^8*b^16 + 4480*a
^10*b^14 - 3584*a^12*b^12 + 1792*a^14*b^10 - 512*a^16*b^8 + 64*a^18*b^6) - (b^5*((a + b)^5*(a - b)^5)^(1/2)*(t
an(x/2)*(96*a^24*b - 64*a^2*b^23 + 736*a^4*b^21 - 3840*a^6*b^19 + 12000*a^8*b^17 - 24960*a^10*b^15 + 36288*a^1
2*b^13 - 37632*a^14*b^11 + 27840*a^16*b^9 - 14400*a^18*b^7 + 4960*a^20*b^5 - 1024*a^22*b^3) + 32*a^3*b^22 - 32
0*a^5*b^20 + 1440*a^7*b^18 - 3840*a^9*b^16 + 6720*a^11*b^14 - 8064*a^13*b^12 + 6720*a^15*b^10 - 3840*a^17*b^8
+ 1440*a^19*b^6 - 320*a^21*b^4 + 32*a^23*b^2))/(a*b^10 - a^11 - 5*a^3*b^8 + 10*a^5*b^6 - 10*a^7*b^4 + 5*a^9*b^
2)))/(a*b^10 - a^11 - 5*a^3*b^8 + 10*a^5*b^6 - 10*a^7*b^4 + 5*a^9*b^2)))/(a*b^10 - a^11 - 5*a^3*b^8 + 10*a^5*b
^6 - 10*a^7*b^4 + 5*a^9*b^2)))*((a + b)^5*(a - b)^5)^(1/2)*2i)/(a*b^10 - a^11 - 5*a^3*b^8 + 10*a^5*b^6 - 10*a^
7*b^4 + 5*a^9*b^2)