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

Optimal. Leaf size=44 \[ -\frac {x}{8 a}-\frac {\cos ^3(x)}{3 a}+\frac {\sin (x) \cos ^3(x)}{4 a}-\frac {\sin (x) \cos (x)}{8 a} \]

[Out]

-1/8*x/a-1/3*cos(x)^3/a-1/8*cos(x)*sin(x)/a+1/4*cos(x)^3*sin(x)/a

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Rubi [A]  time = 0.12, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3872, 2839, 2565, 30, 2568, 2635, 8} \[ -\frac {x}{8 a}-\frac {\cos ^3(x)}{3 a}+\frac {\sin (x) \cos ^3(x)}{4 a}-\frac {\sin (x) \cos (x)}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^4/(a + a*Csc[x]),x]

[Out]

-x/(8*a) - Cos[x]^3/(3*a) - (Cos[x]*Sin[x])/(8*a) + (Cos[x]^3*Sin[x])/(4*a)

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2839

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

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

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Mathematica [A]  time = 0.04, size = 40, normalized size = 0.91 \[ -\frac {x}{8 a}+\frac {\sin (4 x)}{32 a}-\frac {\cos (x)}{4 a}-\frac {\cos (3 x)}{12 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^4/(a + a*Csc[x]),x]

[Out]

-1/8*x/a - Cos[x]/(4*a) - Cos[3*x]/(12*a) + Sin[4*x]/(32*a)

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fricas [A]  time = 0.78, size = 30, normalized size = 0.68 \[ -\frac {8 \, \cos \relax (x)^{3} - 3 \, {\left (2 \, \cos \relax (x)^{3} - \cos \relax (x)\right )} \sin \relax (x) + 3 \, x}{24 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(8*cos(x)^3 - 3*(2*cos(x)^3 - cos(x))*sin(x) + 3*x)/a

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giac [B]  time = 0.29, size = 78, normalized size = 1.77 \[ -\frac {x}{8 \, a} - \frac {3 \, \tan \left (\frac {1}{2} \, x\right )^{7} + 24 \, \tan \left (\frac {1}{2} \, x\right )^{6} - 21 \, \tan \left (\frac {1}{2} \, x\right )^{5} + 24 \, \tan \left (\frac {1}{2} \, x\right )^{4} + 21 \, \tan \left (\frac {1}{2} \, x\right )^{3} + 8 \, \tan \left (\frac {1}{2} \, x\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, x\right ) + 8}{12 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{4} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*x/a - 1/12*(3*tan(1/2*x)^7 + 24*tan(1/2*x)^6 - 21*tan(1/2*x)^5 + 24*tan(1/2*x)^4 + 21*tan(1/2*x)^3 + 8*ta
n(1/2*x)^2 - 3*tan(1/2*x) + 8)/((tan(1/2*x)^2 + 1)^4*a)

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maple [B]  time = 0.25, size = 172, normalized size = 3.91 \[ -\frac {\tan ^{7}\left (\frac {x}{2}\right )}{4 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {2 \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {7 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{4 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {2 \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {7 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{4 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {2 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{3 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {\tan \left (\frac {x}{2}\right )}{4 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {2}{3 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {\arctan \left (\tan \left (\frac {x}{2}\right )\right )}{4 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.42, size = 157, normalized size = 3.57 \[ \frac {\frac {3 \, \sin \relax (x)}{\cos \relax (x) + 1} - \frac {8 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {21 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - \frac {24 \, \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {21 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} - \frac {24 \, \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} - \frac {3 \, \sin \relax (x)^{7}}{{\left (\cos \relax (x) + 1\right )}^{7}} - 8}{12 \, {\left (a + \frac {4 \, a \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {6 \, a \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {4 \, a \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + \frac {a \sin \relax (x)^{8}}{{\left (\cos \relax (x) + 1\right )}^{8}}\right )}} - \frac {\arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{4 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*sin(x)/(cos(x) + 1) - 8*sin(x)^2/(cos(x) + 1)^2 - 21*sin(x)^3/(cos(x) + 1)^3 - 24*sin(x)^4/(cos(x) + 1
)^4 + 21*sin(x)^5/(cos(x) + 1)^5 - 24*sin(x)^6/(cos(x) + 1)^6 - 3*sin(x)^7/(cos(x) + 1)^7 - 8)/(a + 4*a*sin(x)
^2/(cos(x) + 1)^2 + 6*a*sin(x)^4/(cos(x) + 1)^4 + 4*a*sin(x)^6/(cos(x) + 1)^6 + a*sin(x)^8/(cos(x) + 1)^8) - 1
/4*arctan(sin(x)/(cos(x) + 1))/a

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mupad [B]  time = 0.31, size = 25, normalized size = 0.57 \[ -\frac {3\,x+2\,\cos \left (3\,x\right )-\frac {3\,\sin \left (4\,x\right )}{4}+6\,\cos \relax (x)}{24\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^4/(a + a/sin(x)),x)

[Out]

-(3*x + 2*cos(3*x) - (3*sin(4*x))/4 + 6*cos(x))/(24*a)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cos ^{4}{\relax (x )}}{\csc {\relax (x )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(x)**4/(csc(x) + 1), x)/a

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