3.306 \(\int \frac {\cot ^2(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=114 \[ -\frac {\cot ^5(c+d x)}{5 a d}-\frac {2 \cot ^3(c+d x)}{3 a d}-\frac {\cot (c+d x)}{a d}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d} \]

[Out]

3/8*arctanh(cos(d*x+c))/a/d-cot(d*x+c)/a/d-2/3*cot(d*x+c)^3/a/d-1/5*cot(d*x+c)^5/a/d+3/8*cot(d*x+c)*csc(d*x+c)
/a/d+1/4*cot(d*x+c)*csc(d*x+c)^3/a/d

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Rubi [A]  time = 0.14, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2839, 3767, 3768, 3770} \[ -\frac {\cot ^5(c+d x)}{5 a d}-\frac {2 \cot ^3(c+d x)}{3 a d}-\frac {\cot (c+d x)}{a d}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^2*Csc[c + d*x]^4)/(a + a*Sin[c + d*x]),x]

[Out]

(3*ArcTanh[Cos[c + d*x]])/(8*a*d) - Cot[c + d*x]/(a*d) - (2*Cot[c + d*x]^3)/(3*a*d) - Cot[c + d*x]^5/(5*a*d) +
(3*Cot[c + d*x]*Csc[c + d*x])/(8*a*d) + (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a*d)

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 3767

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

Rule 3768

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\cot ^2(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac {\int \csc ^5(c+d x) \, dx}{a}+\frac {\int \csc ^6(c+d x) \, dx}{a}\\ &=\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {3 \int \csc ^3(c+d x) \, dx}{4 a}-\frac {\operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (c+d x)\right )}{a d}\\ &=-\frac {\cot (c+d x)}{a d}-\frac {2 \cot ^3(c+d x)}{3 a d}-\frac {\cot ^5(c+d x)}{5 a d}+\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {3 \int \csc (c+d x) \, dx}{8 a}\\ &=\frac {3 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac {\cot (c+d x)}{a d}-\frac {2 \cot ^3(c+d x)}{3 a d}-\frac {\cot ^5(c+d x)}{5 a d}+\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}\\ \end {align*}

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Mathematica [A]  time = 0.66, size = 189, normalized size = 1.66 \[ \frac {\csc ^5(c+d x) \left (420 \sin (2 (c+d x))-90 \sin (4 (c+d x))-640 \cos (c+d x)+320 \cos (3 (c+d x))-64 \cos (5 (c+d x))-450 \sin (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+225 \sin (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-45 \sin (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+450 \sin (c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-225 \sin (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+45 \sin (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{1920 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^2*Csc[c + d*x]^4)/(a + a*Sin[c + d*x]),x]

[Out]

(Csc[c + d*x]^5*(-640*Cos[c + d*x] + 320*Cos[3*(c + d*x)] - 64*Cos[5*(c + d*x)] + 450*Log[Cos[(c + d*x)/2]]*Si
n[c + d*x] - 450*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] + 420*Sin[2*(c + d*x)] - 225*Log[Cos[(c + d*x)/2]]*Sin[3*(
c + d*x)] + 225*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 90*Sin[4*(c + d*x)] + 45*Log[Cos[(c + d*x)/2]]*Sin[5*
(c + d*x)] - 45*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/(1920*a*d)

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fricas [A]  time = 0.50, size = 173, normalized size = 1.52 \[ -\frac {128 \, \cos \left (d x + c\right )^{5} - 320 \, \cos \left (d x + c\right )^{3} - 45 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 45 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 30 \, {\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 240 \, \cos \left (d x + c\right )}{240 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/240*(128*cos(d*x + c)^5 - 320*cos(d*x + c)^3 - 45*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x +
c) + 1/2)*sin(d*x + c) + 45*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c)
+ 30*(3*cos(d*x + c)^3 - 5*cos(d*x + c))*sin(d*x + c) + 240*cos(d*x + c))/((a*d*cos(d*x + c)^4 - 2*a*d*cos(d*
x + c)^2 + a*d)*sin(d*x + c))

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giac [A]  time = 0.19, size = 187, normalized size = 1.64 \[ -\frac {\frac {360 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 50 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 300 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{5}} - \frac {822 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 300 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 50 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/960*(360*log(abs(tan(1/2*d*x + 1/2*c)))/a - (6*a^4*tan(1/2*d*x + 1/2*c)^5 - 15*a^4*tan(1/2*d*x + 1/2*c)^4 +
50*a^4*tan(1/2*d*x + 1/2*c)^3 - 120*a^4*tan(1/2*d*x + 1/2*c)^2 + 300*a^4*tan(1/2*d*x + 1/2*c))/a^5 - (822*tan
(1/2*d*x + 1/2*c)^5 - 300*tan(1/2*d*x + 1/2*c)^4 + 120*tan(1/2*d*x + 1/2*c)^3 - 50*tan(1/2*d*x + 1/2*c)^2 + 15
*tan(1/2*d*x + 1/2*c) - 6)/(a*tan(1/2*d*x + 1/2*c)^5))/d

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maple [A]  time = 0.46, size = 208, normalized size = 1.82 \[ \frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 a d}-\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 a d}+\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 a d}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a d}-\frac {5}{16 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}-\frac {1}{160 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {1}{8 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {1}{64 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {5}{96 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^6/(a+a*sin(d*x+c)),x)

[Out]

1/160/a/d*tan(1/2*d*x+1/2*c)^5-1/64/a/d*tan(1/2*d*x+1/2*c)^4+5/96/a/d*tan(1/2*d*x+1/2*c)^3-1/8/a/d*tan(1/2*d*x
+1/2*c)^2+5/16/a/d*tan(1/2*d*x+1/2*c)-5/16/a/d/tan(1/2*d*x+1/2*c)-3/8/a/d*ln(tan(1/2*d*x+1/2*c))-1/160/a/d/tan
(1/2*d*x+1/2*c)^5+1/8/a/d/tan(1/2*d*x+1/2*c)^2+1/64/a/d/tan(1/2*d*x+1/2*c)^4-5/96/a/d/tan(1/2*d*x+1/2*c)^3

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maxima [B]  time = 0.34, size = 234, normalized size = 2.05 \[ \frac {\frac {\frac {300 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {120 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {50 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {6 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a} - \frac {360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {{\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {50 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {120 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {300 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 6\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a \sin \left (d x + c\right )^{5}}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/960*((300*sin(d*x + c)/(cos(d*x + c) + 1) - 120*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 50*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 - 15*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 6*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a - 360*log
(sin(d*x + c)/(cos(d*x + c) + 1))/a + (15*sin(d*x + c)/(cos(d*x + c) + 1) - 50*sin(d*x + c)^2/(cos(d*x + c) +
1)^2 + 120*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 300*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 6)*(cos(d*x + c) +
1)^5/(a*sin(d*x + c)^5))/d

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mupad [B]  time = 9.25, size = 291, normalized size = 2.55 \[ -\frac {6\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-50\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-300\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+300\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+50\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+360\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{960\,a\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2/(sin(c + d*x)^6*(a + a*sin(c + d*x))),x)

[Out]

-(6*cos(c/2 + (d*x)/2)^10 - 6*sin(c/2 + (d*x)/2)^10 + 15*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^9 - 15*cos(c/2
+ (d*x)/2)^9*sin(c/2 + (d*x)/2) - 50*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 + 120*cos(c/2 + (d*x)/2)^3*sin(
c/2 + (d*x)/2)^7 - 300*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 + 300*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)
^4 - 120*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^3 + 50*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2 + 360*log(si
n(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5)/(960*a*d*cos(c/2 + (d*x)/2)^5*
sin(c/2 + (d*x)/2)^5)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**6/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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