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[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]))

________________________________________________________________________________________

Rubi [A]  time = 0.251215, antiderivative size = 91, normalized size of antiderivative = 1., 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 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 3770

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

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 8

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

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 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]

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*}

Mathematica [B]  time = 1.24768, 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)

________________________________________________________________________________________

Maple [A]  time = 0.171, size = 153, normalized size = 1.7 \begin{align*}{\frac{1}{24\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{1}{4\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+{\frac{11}{8\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-4\,{\frac{1}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-{\frac{1}{24\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{1}{4\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}-{\frac{11}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}} \end{align*}

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)-4/d/a^2/(tan(1/2*
d*x+1/2*c)+1)-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))

________________________________________________________________________________________

Maxima [B]  time = 1.19624, size = 269, normalized size = 2.96 \begin{align*} \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} \end{align*}

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

________________________________________________________________________________________

Fricas [B]  time = 1.71775, size = 801, normalized size = 8.8 \begin{align*} \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 )}} \end{align*}

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))

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.34138, size = 197, normalized size = 2.16 \begin{align*} -\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} \end{align*}

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