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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.325443, antiderivative size = 130, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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])

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*}

Mathematica [A]  time = 0.749434, size = 127, normalized size = 0.98 \[ -\frac{-3 \left (4 x \left (-12 a^2 b^2+3 a^4+8 b^4\right )+8 a^2 \left (a^2-b^2\right ) \sin (2 x)+a^4 \sin (4 x)\right )+24 a b \left (5 a^2-4 b^2\right ) \cos (x)+192 b \left (b^2-a^2\right )^{3/2} \tan ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )+8 a^3 b \cos (3 x)}{96 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^4/(a + b*Csc[x]),x]

[Out]

-(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]))/(96*a^5)

________________________________________________________________________________________

Maple [B]  time = 0.065, size = 514, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*arcta
n(tan(1/2*x))*b^2+2/a^5*arctan(tan(1/2*x))*b^4-2*b/a/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*x)+2*a)/(-a^2+b^
2)^(1/2))+4*b^3/a^3/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*x)+2*a)/(-a^2+b^2)^(1/2))-2*b^5/a^5/(-a^2+b^2)^(1
/2)*arctan(1/2*(2*b*tan(1/2*x)+2*a)/(-a^2+b^2)^(1/2))

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 0.563399, size = 791, normalized size = 6.08 \begin{align*} \left [-\frac{8 \, a^{3} b \cos \left (x\right )^{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 \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 ) - 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 \left (x\right ) - 3 \,{\left (2 \, a^{4} \cos \left (x\right )^{3} +{\left (3 \, a^{4} - 4 \, a^{2} b^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{24 \, a^{5}}, -\frac{8 \, a^{3} b \cos \left (x\right )^{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 \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (x\right )}\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 \left (x\right ) - 3 \,{\left (2 \, a^{4} \cos \left (x\right )^{3} +{\left (3 \, a^{4} - 4 \, a^{2} b^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{24 \, a^{5}}\right ] \end{align*}

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]

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Giac [B]  time = 1.24925, size = 375, normalized size = 2.88 \begin{align*} \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}\left (b\right ) + \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}} \end{align*}

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)