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

Optimal. Leaf size=78 \[ \frac {2 \cot (c+d x)}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d} \]

[Out]

-5/2*arctanh(cos(d*x+c))/a^2/d+2*cot(d*x+c)/a^2/d-1/2*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 = 0.22, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2874, 2966, 3770, 3767, 8, 3768, 2648} \[ \frac {2 \cot (c+d x)}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*ArcTanh[Cos[c + d*x]])/(2*a^2*d) + (2*Cot[c + d*x])/(a^2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a^2*d) + (2*C
os[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 (c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \frac {\csc ^3(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)+\csc ^3(c+d x)-\frac {2}{1+\sin (c+d x)}\right ) \, dx}{a^2}\\ &=\frac {\int \csc ^3(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 \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)}{2 a^2 d}+\frac {2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))}+\frac {\int \csc (c+d x) \, dx}{2 a^2}+\frac {2 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d}\\ &=-\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 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 = 0.78, size = 364, normalized size = 4.67 \[ -\frac {4 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}{d (a \sin (c+d x)+a)^2}-\frac {5 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}{2 d (a \sin (c+d x)+a)^2}+\frac {5 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}{2 d (a \sin (c+d x)+a)^2}-\frac {\tan \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}{d (a \sin (c+d x)+a)^2}+\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}{d (a \sin (c+d x)+a)^2}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}{8 d (a \sin (c+d x)+a)^2}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}{8 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)/(d*(a + a*Sin[c + d*x])^2) + (Cot[(c + d*x)/2]*(
Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)/(d*(a + a*Sin[c + d*x])^2) - (Csc[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + S
in[(c + d*x)/2])^4)/(8*d*(a + a*Sin[c + d*x])^2) - (5*Log[Cos[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2])^4)/(2*d*(a + a*Sin[c + d*x])^2) + (5*Log[Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)/(2*d*(
a + a*Sin[c + d*x])^2) + (Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)/(8*d*(a + a*Sin[c + d*x]
)^2) - ((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*Tan[(c + d*x)/2])/(d*(a + a*Sin[c + d*x])^2)

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fricas [B]  time = 0.50, size = 246, normalized size = 3.15 \[ \frac {16 \, \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{2} - 5 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 5 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (8 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 14 \, \cos \left (d x + c\right ) - 8}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - a^{2} d + {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/4*(16*cos(d*x + c)^3 + 10*cos(d*x + c)^2 - 5*(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x
+ c) - cos(d*x + c) - 1)*log(1/2*cos(d*x + c) + 1/2) + 5*(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 -
1)*sin(d*x + c) - cos(d*x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) - 2*(8*cos(d*x + c)^2 + 3*cos(d*x + c) - 4)*
sin(d*x + c) - 14*cos(d*x + c) - 8)/(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
+ (a^2*d*cos(d*x + c)^2 - a^2*d)*sin(d*x + c))

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giac [A]  time = 0.19, size = 116, normalized size = 1.49 \[ \frac {\frac {20 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} + \frac {32}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}} - \frac {30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*(20*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + (a^2*tan(1/2*d*x + 1/2*c)^2 - 8*a^2*tan(1/2*d*x + 1/2*c))/a^4 + 3
2/(a^2*(tan(1/2*d*x + 1/2*c) + 1)) - (30*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 1)/(a^2*tan(1/2*d*x
+ 1/2*c)^2))/d

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maple [A]  time = 0.76, size = 114, normalized size = 1.46 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{2}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2}}-\frac {1}{8 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {1}{d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}+\frac {4}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x)

[Out]

1/8/d/a^2*tan(1/2*d*x+1/2*c)^2-1/d/a^2*tan(1/2*d*x+1/2*c)-1/8/d/a^2/tan(1/2*d*x+1/2*c)^2+1/d/a^2/tan(1/2*d*x+1
/2*c)+5/2/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 = 0.34, size = 161, normalized size = 2.06 \[ \frac {\frac {\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}{\frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} - \frac {\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} + \frac {20 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/8*((7*sin(d*x + c)/(cos(d*x + c) + 1) + 40*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)/(a^2*sin(d*x + c)^2/(cos
(d*x + c) + 1)^2 + a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3) - (8*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c
)^2/(cos(d*x + c) + 1)^2)/a^2 + 20*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 8.66, size = 120, normalized size = 1.54 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}+\frac {5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^2\,d}+\frac {20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {1}{2}}{d\,\left (4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2/(sin(c + d*x)^3*(a + a*sin(c + d*x))^2),x)

[Out]

tan(c/2 + (d*x)/2)^2/(8*a^2*d) + (5*log(tan(c/2 + (d*x)/2)))/(2*a^2*d) + ((7*tan(c/2 + (d*x)/2))/2 + 20*tan(c/
2 + (d*x)/2)^2 - 1/2)/(d*(4*a^2*tan(c/2 + (d*x)/2)^2 + 4*a^2*tan(c/2 + (d*x)/2)^3)) - tan(c/2 + (d*x)/2)/(a^2*
d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cos ^{2}{\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**3/(a+a*sin(d*x+c))**2,x)

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)**3/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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