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

Optimal. Leaf size=130 \[ -\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}+\frac {2 b \left (a^2-b^2\right )^{3/2} \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^5}-\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{8 a^4}+\frac {x \left (3 a^4-12 a^2 b^2+8 b^4\right )}{8 a^5} \]

[Out]

1/8*(3*a^4-12*a^2*b^2+8*b^4)*x/a^5+2*b*(a^2-b^2)^(3/2)*arctanh((a+b*tan(1/2*x))/(a^2-b^2)^(1/2))/a^5-1/12*cos(
x)^3*(4*b-3*a*sin(x))/a^2-1/8*cos(x)*(8*b*(a^2-b^2)-a*(3*a^2-4*b^2)*sin(x))/a^4

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Rubi [A]  time = 0.33, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3872, 2865, 2735, 2660, 618, 206} \[ \frac {x \left (-12 a^2 b^2+3 a^4+8 b^4\right )}{8 a^5}-\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{8 a^4}+\frac {2 b \left (a^2-b^2\right )^{3/2} \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^5}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^4/(a + b*Csc[x]),x]

[Out]

((3*a^4 - 12*a^2*b^2 + 8*b^4)*x)/(8*a^5) + (2*b*(a^2 - b^2)^(3/2)*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/a
^5 - (Cos[x]^3*(4*b - 3*a*Sin[x]))/(12*a^2) - (Cos[x]*(8*b*(a^2 - b^2) - a*(3*a^2 - 4*b^2)*Sin[x]))/(8*a^4)

Rule 206

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

Rule 618

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 2660

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 2735

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 2865

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\cos ^4(x)}{a+b \csc (x)} \, dx &=\int \frac {\cos ^4(x) \sin (x)}{b+a \sin (x)} \, dx\\ &=-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}+\frac {\int \frac {\cos ^2(x) \left (-a b+\left (3 a^2-4 b^2\right ) \sin (x)\right )}{b+a \sin (x)} \, dx}{4 a^2}\\ &=-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}-\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{8 a^4}+\frac {\int \frac {-a b \left (5 a^2-4 b^2\right )+\left (3 a^4-12 a^2 b^2+8 b^4\right ) \sin (x)}{b+a \sin (x)} \, dx}{8 a^4}\\ &=\frac {\left (3 a^4-12 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}-\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{8 a^4}-\frac {\left (b \left (a^2-b^2\right )^2\right ) \int \frac {1}{b+a \sin (x)} \, dx}{a^5}\\ &=\frac {\left (3 a^4-12 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}-\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{8 a^4}-\frac {\left (2 b \left (a^2-b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{b+2 a x+b x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{a^5}\\ &=\frac {\left (3 a^4-12 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}-\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{8 a^4}+\frac {\left (4 b \left (a^2-b^2\right )^2\right ) \operatorname {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^5}\\ &=\frac {\left (3 a^4-12 a^2 b^2+8 b^4\right ) x}{8 a^5}+\frac {2 b \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 )}{a^5}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}-\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{8 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.74, size = 127, normalized size = 0.98 \[ -\frac {8 a^3 b \cos (3 x)+24 a b \left (5 a^2-4 b^2\right ) \cos (x)+192 b \left (b^2-a^2\right )^{3/2} \tan ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )-3 \left (a^4 \sin (4 x)+8 a^2 \left (a^2-b^2\right ) \sin (2 x)+4 x \left (3 a^4-12 a^2 b^2+8 b^4\right )\right )}{96 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^4/(a + b*Csc[x]),x]

[Out]

-1/96*(192*b*(-a^2 + b^2)^(3/2)*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]] + 24*a*b*(5*a^2 - 4*b^2)*Cos[x] + 8*
a^3*b*Cos[3*x] - 3*(4*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x + 8*a^2*(a^2 - b^2)*Sin[2*x] + a^4*Sin[4*x]))/a^5

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fricas [A]  time = 0.51, size = 335, normalized size = 2.58 \[ \left [-\frac {8 \, a^{3} b \cos \relax (x)^{3} + 12 \, {\left (a^{2} b - b^{3}\right )} \sqrt {a^{2} - b^{2}} \log \left (-\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2} - 2 \, {\left (b \cos \relax (x) \sin \relax (x) + a \cos \relax (x)\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) - 3 \, {\left (3 \, a^{4} - 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} x + 24 \, {\left (a^{3} b - a b^{3}\right )} \cos \relax (x) - 3 \, {\left (2 \, a^{4} \cos \relax (x)^{3} + {\left (3 \, a^{4} - 4 \, a^{2} b^{2}\right )} \cos \relax (x)\right )} \sin \relax (x)}{24 \, a^{5}}, -\frac {8 \, a^{3} b \cos \relax (x)^{3} - 24 \, {\left (a^{2} b - b^{3}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \relax (x)}\right ) - 3 \, {\left (3 \, a^{4} - 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} x + 24 \, {\left (a^{3} b - a b^{3}\right )} \cos \relax (x) - 3 \, {\left (2 \, a^{4} \cos \relax (x)^{3} + {\left (3 \, a^{4} - 4 \, a^{2} b^{2}\right )} \cos \relax (x)\right )} \sin \relax (x)}{24 \, a^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(8*a^3*b*cos(x)^3 + 12*(a^2*b - b^3)*sqrt(a^2 - b^2)*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)) - 3*(3*a^4
- 12*a^2*b^2 + 8*b^4)*x + 24*(a^3*b - a*b^3)*cos(x) - 3*(2*a^4*cos(x)^3 + (3*a^4 - 4*a^2*b^2)*cos(x))*sin(x))/
a^5, -1/24*(8*a^3*b*cos(x)^3 - 24*(a^2*b - b^3)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^2
- b^2)*cos(x))) - 3*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x + 24*(a^3*b - a*b^3)*cos(x) - 3*(2*a^4*cos(x)^3 + (3*a^4 -
4*a^2*b^2)*cos(x))*sin(x))/a^5]

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giac [B]  time = 0.64, size = 278, normalized size = 2.14 \[ \frac {{\left (3 \, a^{4} - 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} x}{8 \, a^{5}} - \frac {2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{5}} - \frac {15 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{7} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{7} + 48 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{6} - 24 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{6} - 9 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{5} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 96 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{4} - 72 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{4} + 9 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 80 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} - 72 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} - 15 \, a^{3} \tan \left (\frac {1}{2} \, x\right ) + 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right ) + 32 \, a^{2} b - 24 \, b^{3}}{12 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{4} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x/a^5 - 2*(a^4*b - 2*a^2*b^3 + b^5)*(pi*floor(1/2*x/pi + 1/2)*sgn(b) + arctan
((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2)))/(sqrt(-a^2 + b^2)*a^5) - 1/12*(15*a^3*tan(1/2*x)^7 - 12*a*b^2*tan(1/2*x
)^7 + 48*a^2*b*tan(1/2*x)^6 - 24*b^3*tan(1/2*x)^6 - 9*a^3*tan(1/2*x)^5 - 12*a*b^2*tan(1/2*x)^5 + 96*a^2*b*tan(
1/2*x)^4 - 72*b^3*tan(1/2*x)^4 + 9*a^3*tan(1/2*x)^3 + 12*a*b^2*tan(1/2*x)^3 + 80*a^2*b*tan(1/2*x)^2 - 72*b^3*t
an(1/2*x)^2 - 15*a^3*tan(1/2*x) + 12*a*b^2*tan(1/2*x) + 32*a^2*b - 24*b^3)/((tan(1/2*x)^2 + 1)^4*a^4)

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maple [B]  time = 0.32, size = 514, normalized size = 3.95 \[ -\frac {2 b \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a \sqrt {-a^{2}+b^{2}}}+\frac {4 b^{3} \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a^{3} \sqrt {-a^{2}+b^{2}}}-\frac {2 b^{5} \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a^{5} \sqrt {-a^{2}+b^{2}}}-\frac {5 \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{4 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {\left (\tan ^{7}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {4 \left (\tan ^{6}\left (\frac {x}{2}\right )\right ) b}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {2 b^{3} \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{a^{4} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {3 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{4 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {\left (\tan ^{5}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {8 \left (\tan ^{4}\left (\frac {x}{2}\right )\right ) b}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {6 \left (\tan ^{4}\left (\frac {x}{2}\right )\right ) b^{3}}{a^{4} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {3 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{4 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {\left (\tan ^{3}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {20 \left (\tan ^{2}\left (\frac {x}{2}\right )\right ) b}{3 a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {6 \left (\tan ^{2}\left (\frac {x}{2}\right )\right ) b^{3}}{a^{4} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {5 \tan \left (\frac {x}{2}\right )}{4 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {\tan \left (\frac {x}{2}\right ) b^{2}}{a^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {8 b}{3 a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {2 b^{3}}{a^{4} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {3 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{4 a}-\frac {3 \arctan \left (\tan \left (\frac {x}{2}\right )\right ) b^{2}}{a^{3}}+\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right ) b^{4}}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b/a/(-a^2+b^2)^(1/2)*arctan(1/2*(2*tan(1/2*x)*b+2*a)/(-a^2+b^2)^(1/2))+4/a^3*b^3/(-a^2+b^2)^(1/2)*arctan(1/
2*(2*tan(1/2*x)*b+2*a)/(-a^2+b^2)^(1/2))-2*b^5/a^5/(-a^2+b^2)^(1/2)*arctan(1/2*(2*tan(1/2*x)*b+2*a)/(-a^2+b^2)
^(1/2))-5/4/a/(tan(1/2*x)^2+1)^4*tan(1/2*x)^7+1/a^3/(tan(1/2*x)^2+1)^4*tan(1/2*x)^7*b^2-4/a^2/(tan(1/2*x)^2+1)
^4*tan(1/2*x)^6*b+2/a^4/(tan(1/2*x)^2+1)^4*b^3*tan(1/2*x)^6+3/4/a/(tan(1/2*x)^2+1)^4*tan(1/2*x)^5+1/a^3/(tan(1
/2*x)^2+1)^4*tan(1/2*x)^5*b^2-8/a^2/(tan(1/2*x)^2+1)^4*tan(1/2*x)^4*b+6/a^4/(tan(1/2*x)^2+1)^4*tan(1/2*x)^4*b^
3-3/4/a/(tan(1/2*x)^2+1)^4*tan(1/2*x)^3-1/a^3/(tan(1/2*x)^2+1)^4*tan(1/2*x)^3*b^2-20/3/a^2/(tan(1/2*x)^2+1)^4*
tan(1/2*x)^2*b+6/a^4/(tan(1/2*x)^2+1)^4*tan(1/2*x)^2*b^3+5/4/a/(tan(1/2*x)^2+1)^4*tan(1/2*x)-1/a^3/(tan(1/2*x)
^2+1)^4*tan(1/2*x)*b^2-8/3/a^2/(tan(1/2*x)^2+1)^4*b+2/a^4/(tan(1/2*x)^2+1)^4*b^3+3/4/a*arctan(tan(1/2*x))-3/a^
3*arctan(tan(1/2*x))*b^2+2/a^5*arctan(tan(1/2*x))*b^4

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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 details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.07, size = 2055, normalized size = 15.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^4/(a + b/sin(x)),x)

[Out]

(atan(((((32*a^4*b^10 - 96*a^6*b^8 + 96*a^8*b^6 - 36*a^10*b^4 + (9*a^12*b^2)/2)/a^11 - ((a^4*3i + b^4*8i - a^2
*b^2*12i)*((12*a^14*b + 16*a^10*b^5 - 28*a^12*b^3)/a^11 - ((32*a^3*b^2 + (tan(x/2)*(192*a^16*b - 128*a^14*b^3)
)/(2*a^12))*(a^4*3i + b^4*8i - a^2*b^2*12i))/(8*a^5) + (tan(x/2)*(128*a^10*b^6 - 256*a^12*b^4 + 128*a^14*b^2))
/(2*a^12)))/(8*a^5) + (tan(x/2)*(18*a^14*b - 128*a^4*b^11 + 576*a^6*b^9 - 960*a^8*b^7 + 712*a^10*b^5 - 217*a^1
2*b^3))/(2*a^12))*(a^4*3i + b^4*8i - a^2*b^2*12i)*1i)/(8*a^5) + (((32*a^4*b^10 - 96*a^6*b^8 + 96*a^8*b^6 - 36*
a^10*b^4 + (9*a^12*b^2)/2)/a^11 + ((a^4*3i + b^4*8i - a^2*b^2*12i)*((12*a^14*b + 16*a^10*b^5 - 28*a^12*b^3)/a^
11 + ((32*a^3*b^2 + (tan(x/2)*(192*a^16*b - 128*a^14*b^3))/(2*a^12))*(a^4*3i + b^4*8i - a^2*b^2*12i))/(8*a^5)
+ (tan(x/2)*(128*a^10*b^6 - 256*a^12*b^4 + 128*a^14*b^2))/(2*a^12)))/(8*a^5) + (tan(x/2)*(18*a^14*b - 128*a^4*
b^11 + 576*a^6*b^9 - 960*a^8*b^7 + 712*a^10*b^5 - 217*a^12*b^3))/(2*a^12))*(a^4*3i + b^4*8i - a^2*b^2*12i)*1i)
/(8*a^5))/((32*b^13 - 152*a^2*b^11 + 280*a^4*b^9 - 247*a^6*b^7 + 102*a^8*b^5 - 15*a^10*b^3)/a^11 + (tan(x/2)*(
128*b^14 - 640*a^2*b^12 + 1280*a^4*b^10 - 1296*a^6*b^8 + 690*a^8*b^6 - 180*a^10*b^4 + 18*a^12*b^2))/a^12 - (((
32*a^4*b^10 - 96*a^6*b^8 + 96*a^8*b^6 - 36*a^10*b^4 + (9*a^12*b^2)/2)/a^11 - ((a^4*3i + b^4*8i - a^2*b^2*12i)*
((12*a^14*b + 16*a^10*b^5 - 28*a^12*b^3)/a^11 - ((32*a^3*b^2 + (tan(x/2)*(192*a^16*b - 128*a^14*b^3))/(2*a^12)
)*(a^4*3i + b^4*8i - a^2*b^2*12i))/(8*a^5) + (tan(x/2)*(128*a^10*b^6 - 256*a^12*b^4 + 128*a^14*b^2))/(2*a^12))
)/(8*a^5) + (tan(x/2)*(18*a^14*b - 128*a^4*b^11 + 576*a^6*b^9 - 960*a^8*b^7 + 712*a^10*b^5 - 217*a^12*b^3))/(2
*a^12))*(a^4*3i + b^4*8i - a^2*b^2*12i))/(8*a^5) + (((32*a^4*b^10 - 96*a^6*b^8 + 96*a^8*b^6 - 36*a^10*b^4 + (9
*a^12*b^2)/2)/a^11 + ((a^4*3i + b^4*8i - a^2*b^2*12i)*((12*a^14*b + 16*a^10*b^5 - 28*a^12*b^3)/a^11 + ((32*a^3
*b^2 + (tan(x/2)*(192*a^16*b - 128*a^14*b^3))/(2*a^12))*(a^4*3i + b^4*8i - a^2*b^2*12i))/(8*a^5) + (tan(x/2)*(
128*a^10*b^6 - 256*a^12*b^4 + 128*a^14*b^2))/(2*a^12)))/(8*a^5) + (tan(x/2)*(18*a^14*b - 128*a^4*b^11 + 576*a^
6*b^9 - 960*a^8*b^7 + 712*a^10*b^5 - 217*a^12*b^3))/(2*a^12))*(a^4*3i + b^4*8i - a^2*b^2*12i))/(8*a^5)))*(a^4*
3i + b^4*8i - a^2*b^2*12i)*1i)/(4*a^5) - ((2*(4*a^2*b - 3*b^3))/(3*a^4) - (tan(x/2)*(5*a^2 - 4*b^2))/(4*a^3) +
(tan(x/2)^3*(3*a^2 + 4*b^2))/(4*a^3) - (tan(x/2)^5*(3*a^2 + 4*b^2))/(4*a^3) + (tan(x/2)^7*(5*a^2 - 4*b^2))/(4
*a^3) + (2*tan(x/2)^6*(2*a^2*b - b^3))/a^4 + (2*tan(x/2)^4*(4*a^2*b - 3*b^3))/a^4 + (2*tan(x/2)^2*(10*a^2*b -
9*b^3))/(3*a^4))/(4*tan(x/2)^2 + 6*tan(x/2)^4 + 4*tan(x/2)^6 + tan(x/2)^8 + 1) + (2*b*atanh((18*b^2*tan(x/2)*(
a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2))/(26*b^5 - 9*a^2*b^3 - (25*b^7)/a^2 + (8*b^9)/a^4 - 18*a^3*b^2*tan(x/
2) - (50*b^6*tan(x/2))/a + (16*b^8*tan(x/2))/a^3 + 52*a*b^4*tan(x/2)) - (8*b^5*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*
b^2)^(1/2))/(16*b^8*tan(x/2) - 25*a*b^7 + 26*a^3*b^5 - 9*a^5*b^3 + (8*b^9)/a - 50*a^2*b^6*tan(x/2) + 52*a^4*b^
4*tan(x/2) - 18*a^6*b^2*tan(x/2)) - (9*b^3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2))/(50*b^6*tan(x/2) - 26*a*
b^5 + 9*a^3*b^3 + (25*b^7)/a - (8*b^9)/a^3 - 52*a^2*b^4*tan(x/2) + 18*a^4*b^2*tan(x/2) - (16*b^8*tan(x/2))/a^2
) + (25*b^4*tan(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2))/(25*b^7 - 26*a^2*b^5 + 9*a^4*b^3 - (8*b^9)/a^2
- 52*a^3*b^4*tan(x/2) + 18*a^5*b^2*tan(x/2) - (16*b^8*tan(x/2))/a + 50*a*b^6*tan(x/2)) + (8*b^6*tan(x/2)*(a^6
- b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2))/(8*b^9 - 25*a^2*b^7 + 26*a^4*b^5 - 9*a^6*b^3 - 50*a^3*b^6*tan(x/2) + 52
*a^5*b^4*tan(x/2) - 18*a^7*b^2*tan(x/2) + 16*a*b^8*tan(x/2)))*((a + b)^3*(a - b)^3)^(1/2))/a^5

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{4}{\relax (x )}}{a + b \csc {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(x)**4/(a + b*csc(x)), x)

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