3.1.10 \(\int \frac {\sin ^4(x)}{a+a \csc (x)} \, dx\) [10]

Optimal. Leaf size=66 \[ \frac {15 x}{8 a}+\frac {4 \cos (x)}{a}-\frac {4 \cos ^3(x)}{3 a}-\frac {15 \cos (x) \sin (x)}{8 a}-\frac {5 \cos (x) \sin ^3(x)}{4 a}+\frac {\cos (x) \sin ^3(x)}{a+a \csc (x)} \]

[Out]

15/8*x/a+4*cos(x)/a-4/3*cos(x)^3/a-15/8*cos(x)*sin(x)/a-5/4*cos(x)*sin(x)^3/a+cos(x)*sin(x)^3/(a+a*csc(x))

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Rubi [A]
time = 0.06, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3904, 3872, 2715, 8, 2713} \begin {gather*} \frac {15 x}{8 a}-\frac {4 \cos ^3(x)}{3 a}+\frac {4 \cos (x)}{a}-\frac {5 \sin ^3(x) \cos (x)}{4 a}-\frac {15 \sin (x) \cos (x)}{8 a}+\frac {\sin ^3(x) \cos (x)}{a \csc (x)+a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[x]^4/(a + a*Csc[x]),x]

[Out]

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

Rule 8

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

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

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] && Integ
erQ[2*n]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3904

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[Cot[e + f*x
]*((d*Csc[e + f*x])^n/(f*(a + b*Csc[e + f*x]))), x] - Dist[1/a^2, Int[(d*Csc[e + f*x])^n*(a*(n - 1) - b*n*Csc[
e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\sin ^4(x)}{a+a \csc (x)} \, dx &=\frac {\cos (x) \sin ^3(x)}{a+a \csc (x)}-\frac {\int (-5 a+4 a \csc (x)) \sin ^4(x) \, dx}{a^2}\\ &=\frac {\cos (x) \sin ^3(x)}{a+a \csc (x)}-\frac {4 \int \sin ^3(x) \, dx}{a}+\frac {5 \int \sin ^4(x) \, dx}{a}\\ &=-\frac {5 \cos (x) \sin ^3(x)}{4 a}+\frac {\cos (x) \sin ^3(x)}{a+a \csc (x)}+\frac {15 \int \sin ^2(x) \, dx}{4 a}+\frac {4 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (x)\right )}{a}\\ &=\frac {4 \cos (x)}{a}-\frac {4 \cos ^3(x)}{3 a}-\frac {15 \cos (x) \sin (x)}{8 a}-\frac {5 \cos (x) \sin ^3(x)}{4 a}+\frac {\cos (x) \sin ^3(x)}{a+a \csc (x)}+\frac {15 \int 1 \, dx}{8 a}\\ &=\frac {15 x}{8 a}+\frac {4 \cos (x)}{a}-\frac {4 \cos ^3(x)}{3 a}-\frac {15 \cos (x) \sin (x)}{8 a}-\frac {5 \cos (x) \sin ^3(x)}{4 a}+\frac {\cos (x) \sin ^3(x)}{a+a \csc (x)}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 57, normalized size = 0.86 \begin {gather*} \frac {168 \cos (x)-8 \cos (3 x)+3 \left (60 x-\frac {64 \sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )}-16 \sin (2 x)+\sin (4 x)\right )}{96 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^4/(a + a*Csc[x]),x]

[Out]

(168*Cos[x] - 8*Cos[3*x] + 3*(60*x - (64*Sin[x/2])/(Cos[x/2] + Sin[x/2]) - 16*Sin[2*x] + Sin[4*x]))/(96*a)

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Maple [A]
time = 0.08, size = 90, normalized size = 1.36

method	result	size
risch	\(\frac {15 x}{8 a}+\frac {7 \,{\mathrm e}^{i x}}{8 a}+\frac {7 \,{\mathrm e}^{-i x}}{8 a}+\frac {2}{\left (i+{\mathrm e}^{i x}\right ) a}+\frac {\sin \left (4 x \right )}{32 a}-\frac {\cos \left (3 x \right )}{12 a}-\frac {\sin \left (2 x \right )}{2 a}\)	\(70\)
default	\(\frac {\frac {64}{32 \tan \left (\frac {x}{2}\right )+32}+\frac {2 \left (\frac {7 \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{8}+\tan ^{6}\left (\frac {x}{2}\right )+\frac {15 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{8}+5 \left (\tan ^{4}\left (\frac {x}{2}\right )\right )-\frac {15 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{8}+\frac {17 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{3}-\frac {7 \tan \left (\frac {x}{2}\right )}{8}+\frac {5}{3}\right )}{\left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {15 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{4}}{a}\)	\(90\)
norman	\(\frac {\frac {15 x}{8 a}+\frac {15}{4 a}+\frac {15 x \tan \left (\frac {x}{2}\right )}{8 a}+\frac {15 x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2 a}+\frac {15 x \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2 a}+\frac {45 x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{4 a}+\frac {45 x \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{4 a}+\frac {15 x \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{2 a}+\frac {15 x \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{2 a}+\frac {15 x \left (\tan ^{8}\left (\frac {x}{2}\right )\right )}{8 a}+\frac {15 x \left (\tan ^{9}\left (\frac {x}{2}\right )\right )}{8 a}+\frac {13 \left (\tan ^{8}\left (\frac {x}{2}\right )\right )}{6 a}-\frac {19 \left (\tan ^{9}\left (\frac {x}{2}\right )\right )}{12 a}+\frac {5 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{4 a}+\frac {45 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{4 a}+\frac {17 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{4 a}+\frac {35 \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{4 a}-\frac {31 \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{12 a}+\frac {89 \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{12 a}}{\left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4} \left (\tan \left (\frac {x}{2}\right )+1\right )}\)	\(226\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

64/a*(1/32/(tan(1/2*x)+1)+1/32*(7/8*tan(1/2*x)^7+tan(1/2*x)^6+15/8*tan(1/2*x)^5+5*tan(1/2*x)^4-15/8*tan(1/2*x)
^3+17/3*tan(1/2*x)^2-7/8*tan(1/2*x)+5/3)/(tan(1/2*x)^2+1)^4+15/256*arctan(tan(1/2*x)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (58) = 116\).
time = 0.48, size = 230, normalized size = 3.48 \begin {gather*} \frac {\frac {19 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {211 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {91 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {219 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {165 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac {165 \, \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} + \frac {45 \, \sin \left (x\right )^{7}}{{\left (\cos \left (x\right ) + 1\right )}^{7}} + \frac {45 \, \sin \left (x\right )^{8}}{{\left (\cos \left (x\right ) + 1\right )}^{8}} + 64}{12 \, {\left (a + \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {4 \, a \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {4 \, a \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {6 \, a \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {6 \, a \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac {4 \, a \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} + \frac {4 \, 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}} + \frac {a \sin \left (x\right )^{9}}{{\left (\cos \left (x\right ) + 1\right )}^{9}}\right )}} + \frac {15 \, \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{4 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(19*sin(x)/(cos(x) + 1) + 211*sin(x)^2/(cos(x) + 1)^2 + 91*sin(x)^3/(cos(x) + 1)^3 + 219*sin(x)^4/(cos(x)
+ 1)^4 + 165*sin(x)^5/(cos(x) + 1)^5 + 165*sin(x)^6/(cos(x) + 1)^6 + 45*sin(x)^7/(cos(x) + 1)^7 + 45*sin(x)^8
/(cos(x) + 1)^8 + 64)/(a + a*sin(x)/(cos(x) + 1) + 4*a*sin(x)^2/(cos(x) + 1)^2 + 4*a*sin(x)^3/(cos(x) + 1)^3 +
6*a*sin(x)^4/(cos(x) + 1)^4 + 6*a*sin(x)^5/(cos(x) + 1)^5 + 4*a*sin(x)^6/(cos(x) + 1)^6 + 4*a*sin(x)^7/(cos(x
) + 1)^7 + a*sin(x)^8/(cos(x) + 1)^8 + a*sin(x)^9/(cos(x) + 1)^9) + 15/4*arctan(sin(x)/(cos(x) + 1))/a

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Fricas [A]
time = 3.89, size = 81, normalized size = 1.23 \begin {gather*} -\frac {6 \, \cos \left (x\right )^{5} + 8 \, \cos \left (x\right )^{4} - 25 \, \cos \left (x\right )^{3} - 45 \, {\left (x + 1\right )} \cos \left (x\right ) - 48 \, \cos \left (x\right )^{2} - {\left (6 \, \cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{3} - 27 \, \cos \left (x\right )^{2} + 45 \, x + 21 \, \cos \left (x\right ) - 24\right )} \sin \left (x\right ) - 45 \, x - 24}{24 \, {\left (a \cos \left (x\right ) + a \sin \left (x\right ) + a\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(6*cos(x)^5 + 8*cos(x)^4 - 25*cos(x)^3 - 45*(x + 1)*cos(x) - 48*cos(x)^2 - (6*cos(x)^4 - 2*cos(x)^3 - 27
*cos(x)^2 + 45*x + 21*cos(x) - 24)*sin(x) - 45*x - 24)/(a*cos(x) + a*sin(x) + a)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sin ^{4}{\left (x \right )}}{\csc {\left (x \right )} + 1}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(x)**4/(csc(x) + 1), x)/a

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Giac [A]
time = 0.40, size = 91, normalized size = 1.38 \begin {gather*} \frac {15 \, x}{8 \, a} + \frac {2}{a {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}} + \frac {21 \, \tan \left (\frac {1}{2} \, x\right )^{7} + 24 \, \tan \left (\frac {1}{2} \, x\right )^{6} + 45 \, \tan \left (\frac {1}{2} \, x\right )^{5} + 120 \, \tan \left (\frac {1}{2} \, x\right )^{4} - 45 \, \tan \left (\frac {1}{2} \, x\right )^{3} + 136 \, \tan \left (\frac {1}{2} \, x\right )^{2} - 21 \, \tan \left (\frac {1}{2} \, x\right ) + 40}{12 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{4} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/8*x/a + 2/(a*(tan(1/2*x) + 1)) + 1/12*(21*tan(1/2*x)^7 + 24*tan(1/2*x)^6 + 45*tan(1/2*x)^5 + 120*tan(1/2*x)
^4 - 45*tan(1/2*x)^3 + 136*tan(1/2*x)^2 - 21*tan(1/2*x) + 40)/((tan(1/2*x)^2 + 1)^4*a)

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Mupad [B]
time = 0.40, size = 93, normalized size = 1.41 \begin {gather*} \frac {15\,x}{8\,a}+\frac {\frac {15\,{\mathrm {tan}\left (\frac {x}{2}\right )}^8}{4}+\frac {15\,{\mathrm {tan}\left (\frac {x}{2}\right )}^7}{4}+\frac {55\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6}{4}+\frac {55\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{4}+\frac {73\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{4}+\frac {91\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{12}+\frac {211\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{12}+\frac {19\,\mathrm {tan}\left (\frac {x}{2}\right )}{12}+\frac {16}{3}}{a\,{\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}^4\,\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a + a/sin(x)),x)

[Out]

(15*x)/(8*a) + ((19*tan(x/2))/12 + (211*tan(x/2)^2)/12 + (91*tan(x/2)^3)/12 + (73*tan(x/2)^4)/4 + (55*tan(x/2)
^5)/4 + (55*tan(x/2)^6)/4 + (15*tan(x/2)^7)/4 + (15*tan(x/2)^8)/4 + 16/3)/(a*(tan(x/2)^2 + 1)^4*(tan(x/2) + 1)
)

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