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

Optimal. Leaf size=91 \[ -\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {3 \cot (c+d x)}{a^2 d}-\frac {2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2874, 2966, 3770, 3767, 8, 3768, 2648} \[ -\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {3 \cot (c+d x)}{a^2 d}-\frac {2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*ArcTanh[Cos[c + d*x]])/(a^2*d) - (3*Cot[c + d*x])/(a^2*d) - Cot[c + d*x]^3/(3*a^2*d) + (Cot[c + d*x]*Csc[c
+ d*x])/(a^2*d) - (2*Cos[c + d*x])/(a^2*d*(1 + Sin[c + d*x]))

Rule 8

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2874

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 0])

Rule 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

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

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Mathematica [B]  time = 1.29, size = 472, normalized size = 5.19 \[ \frac {\left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (12 \sin \left (\frac {1}{2} (c+d x)\right )-6 \sin \left (\frac {3}{2} (c+d x)\right )-2 \sin \left (\frac {5}{2} (c+d x)\right )+8 \sin \left (\frac {7}{2} (c+d x)\right )-10 \cos \left (\frac {5}{2} (c+d x)\right )+20 \cos \left (\frac {7}{2} (c+d x)\right )-27 \sin \left (\frac {1}{2} (c+d x)\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-27 \sin \left (\frac {3}{2} (c+d x)\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 \sin \left (\frac {5}{2} (c+d x)\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 \sin \left (\frac {7}{2} (c+d x)\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-9 \cos \left (\frac {5}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+9 \cos \left (\frac {7}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+9 \cos \left (\frac {5}{2} (c+d x)\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 \cos \left (\frac {1}{2} (c+d x)\right ) \left (-9 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+8\right )-3 \cos \left (\frac {3}{2} (c+d x)\right ) \left (-9 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+14\right )-9 \cos \left (\frac {7}{2} (c+d x)\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+27 \sin \left (\frac {1}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+27 \sin \left (\frac {3}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-9 \sin \left (\frac {5}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-9 \sin \left (\frac {7}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{192 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^3*(-10*Cos[(5*(c + d*x))/2] + 20*Cos[(7*(c + d*x))/2] - 9*Cos[(5*(c + d
*x))/2]*Log[Cos[(c + d*x)/2]] + 9*Cos[(7*(c + d*x))/2]*Log[Cos[(c + d*x)/2]] + 3*Cos[(c + d*x)/2]*(8 + 9*Log[C
os[(c + d*x)/2]] - 9*Log[Sin[(c + d*x)/2]]) - 3*Cos[(3*(c + d*x))/2]*(14 + 9*Log[Cos[(c + d*x)/2]] - 9*Log[Sin
[(c + d*x)/2]]) + 9*Cos[(5*(c + d*x))/2]*Log[Sin[(c + d*x)/2]] - 9*Cos[(7*(c + d*x))/2]*Log[Sin[(c + d*x)/2]]
+ 12*Sin[(c + d*x)/2] + 27*Log[Cos[(c + d*x)/2]]*Sin[(c + d*x)/2] - 27*Log[Sin[(c + d*x)/2]]*Sin[(c + d*x)/2]
- 6*Sin[(3*(c + d*x))/2] + 27*Log[Cos[(c + d*x)/2]]*Sin[(3*(c + d*x))/2] - 27*Log[Sin[(c + d*x)/2]]*Sin[(3*(c
+ d*x))/2] - 2*Sin[(5*(c + d*x))/2] - 9*Log[Cos[(c + d*x)/2]]*Sin[(5*(c + d*x))/2] + 9*Log[Sin[(c + d*x)/2]]*S
in[(5*(c + d*x))/2] + 8*Sin[(7*(c + d*x))/2] - 9*Log[Cos[(c + d*x)/2]]*Sin[(7*(c + d*x))/2] + 9*Log[Sin[(c + d
*x)/2]]*Sin[(7*(c + d*x))/2]))/(192*a^2*d*(1 + Sin[c + d*x])^2)

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fricas [B]  time = 0.53, size = 302, normalized size = 3.32 \[ \frac {28 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} - 42 \, \cos \left (d x + c\right )^{2} + 9 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 9 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (14 \, \cos \left (d x + c\right )^{3} + 9 \, \cos \left (d x + c\right )^{2} - 12 \, \cos \left (d x + c\right ) - 6\right )} \sin \left (d x + c\right ) - 12 \, \cos \left (d x + c\right ) + 12}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d - {\left (a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(28*cos(d*x + c)^4 + 10*cos(d*x + c)^3 - 42*cos(d*x + c)^2 + 9*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d
*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*sin(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) - 9*(cos(d*x + c)
^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*sin(d*x + c) + 1)*log(-1/2*cos(d*
x + c) + 1/2) + 2*(14*cos(d*x + c)^3 + 9*cos(d*x + c)^2 - 12*cos(d*x + c) - 6)*sin(d*x + c) - 12*cos(d*x + c)
+ 12)/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d - (a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2 -
a^2*d*cos(d*x + c) - a^2*d)*sin(d*x + c))

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giac [A]  time = 0.20, size = 146, normalized size = 1.60 \[ -\frac {\frac {72 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {96}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}} - \frac {132 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 33 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 33 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/24*(72*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 96/(a^2*(tan(1/2*d*x + 1/2*c) + 1)) - (132*tan(1/2*d*x + 1/2*c)
^3 - 33*tan(1/2*d*x + 1/2*c)^2 + 6*tan(1/2*d*x + 1/2*c) - 1)/(a^2*tan(1/2*d*x + 1/2*c)^3) - (a^4*tan(1/2*d*x +
1/2*c)^3 - 6*a^4*tan(1/2*d*x + 1/2*c)^2 + 33*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [A]  time = 0.58, size = 153, normalized size = 1.68 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d \,a^{2}}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{2}}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{2}}-\frac {1}{24 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{4 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {11}{8 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {4}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x)

[Out]

1/24/d/a^2*tan(1/2*d*x+1/2*c)^3-1/4/d/a^2*tan(1/2*d*x+1/2*c)^2+11/8/d/a^2*tan(1/2*d*x+1/2*c)-1/24/d/a^2/tan(1/
2*d*x+1/2*c)^3+1/4/d/a^2/tan(1/2*d*x+1/2*c)^2-11/8/d/a^2/tan(1/2*d*x+1/2*c)-3/d/a^2*ln(tan(1/2*d*x+1/2*c))-4/d
/a^2/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.33, size = 199, normalized size = 2.19 \[ \frac {\frac {\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {27 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {129 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 1}{\frac {a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {33 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {6 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {72 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/24*((5*sin(d*x + c)/(cos(d*x + c) + 1) - 27*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 129*sin(d*x + c)^3/(cos(d*
x + c) + 1)^3 - 1)/(a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (33*s
in(d*x + c)/(cos(d*x + c) + 1) - 6*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/
a^2 - 72*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 8.65, size = 153, normalized size = 1.68 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^2\,d}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {43\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {1}{3}}{d\,\left (8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}+\frac {11\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2/(sin(c + d*x)^4*(a + a*sin(c + d*x))^2),x)

[Out]

tan(c/2 + (d*x)/2)^3/(24*a^2*d) - tan(c/2 + (d*x)/2)^2/(4*a^2*d) - (3*log(tan(c/2 + (d*x)/2)))/(a^2*d) - (9*ta
n(c/2 + (d*x)/2)^2 - (5*tan(c/2 + (d*x)/2))/3 + 43*tan(c/2 + (d*x)/2)^3 + 1/3)/(d*(8*a^2*tan(c/2 + (d*x)/2)^3
+ 8*a^2*tan(c/2 + (d*x)/2)^4)) + (11*tan(c/2 + (d*x)/2))/(8*a^2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cos ^{2}{\left (c + d x \right )} \csc ^{4}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**4/(a+a*sin(d*x+c))**2,x)

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)**4/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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