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

Optimal. Leaf size=85 \[ \frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac {\csc ^2(c+d x)}{2 a^2 d}+\frac {2 \csc (c+d x)}{a^2 d}+\frac {3 \log (\sin (c+d x))}{a^2 d}-\frac {3 \log (\sin (c+d x)+1)}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2833, 12, 44} \[ \frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac {\csc ^2(c+d x)}{2 a^2 d}+\frac {2 \csc (c+d x)}{a^2 d}+\frac {3 \log (\sin (c+d x))}{a^2 d}-\frac {3 \log (\sin (c+d x)+1)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2833

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 61, normalized size = 0.72 \[ \frac {\frac {2}{\sin (c+d x)+1}-\csc ^2(c+d x)+4 \csc (c+d x)+6 \log (\sin (c+d x))-6 \log (\sin (c+d x)+1)}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Csc[c + d*x] - Csc[c + d*x]^2 + 6*Log[Sin[c + d*x]] - 6*Log[1 + Sin[c + d*x]] + 2/(1 + Sin[c + d*x]))/(2*a^
2*d)

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fricas [A]  time = 0.50, size = 147, normalized size = 1.73 \[ \frac {6 \, \cos \left (d x + c\right )^{2} + 6 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 6 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \sin \left (d x + c\right ) - 5}{2 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d + {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*cos(d*x + c)^2 + 6*(cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - 1)*log(1/2*sin(d*x + c)) - 6*(
cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - 1)*log(sin(d*x + c) + 1) - 3*sin(d*x + c) - 5)/(a^2*d*cos
(d*x + c)^2 - 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 = 87, normalized size = 1.02 \[ \frac {\frac {6 \, \log \left ({\left | -\frac {a}{a \sin \left (d x + c\right ) + a} + 1 \right |}\right )}{a^{2}} + \frac {2}{{\left (a \sin \left (d x + c\right ) + a\right )} a} - \frac {\frac {6 \, a}{a \sin \left (d x + c\right ) + a} - 5}{a^{2} {\left (\frac {a}{a \sin \left (d x + c\right ) + a} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*log(abs(-a/(a*sin(d*x + c) + a) + 1))/a^2 + 2/((a*sin(d*x + c) + a)*a) - (6*a/(a*sin(d*x + c) + a) - 5)
/(a^2*(a/(a*sin(d*x + c) + a) - 1)^2))/d

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maple [A]  time = 0.30, size = 83, normalized size = 0.98 \[ -\frac {1}{2 a^{2} d \sin \left (d x +c \right )^{2}}+\frac {2}{a^{2} d \sin \left (d x +c \right )}+\frac {3 \ln \left (\sin \left (d x +c \right )\right )}{a^{2} d}+\frac {1}{d \,a^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a^2/d/sin(d*x+c)^2+2/a^2/d/sin(d*x+c)+3*ln(sin(d*x+c))/a^2/d+1/d/a^2/(1+sin(d*x+c))-3*ln(1+sin(d*x+c))/a^
2/d

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maxima [A]  time = 0.30, size = 80, normalized size = 0.94 \[ \frac {\frac {6 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 1}{a^{2} \sin \left (d x + c\right )^{3} + a^{2} \sin \left (d x + c\right )^{2}} - \frac {6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {6 \, \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((6*sin(d*x + c)^2 + 3*sin(d*x + c) - 1)/(a^2*sin(d*x + c)^3 + a^2*sin(d*x + c)^2) - 6*log(sin(d*x + c) +
1)/a^2 + 6*log(sin(d*x + c))/a^2)/d

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mupad [B]  time = 8.61, size = 168, normalized size = 1.98 \[ \frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}-\frac {6\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^2\,d}+\frac {-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}}{d\,\left (4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*log(tan(c/2 + (d*x)/2)))/(a^2*d) - tan(c/2 + (d*x)/2)^2/(8*a^2*d) - (6*log(tan(c/2 + (d*x)/2) + 1))/(a^2*d)
+ (3*tan(c/2 + (d*x)/2) + (15*tan(c/2 + (d*x)/2)^2)/2 - 4*tan(c/2 + (d*x)/2)^3 - 1/2)/(d*(4*a^2*tan(c/2 + (d*
x)/2)^2 + 8*a^2*tan(c/2 + (d*x)/2)^3 + 4*a^2*tan(c/2 + (d*x)/2)^4)) + tan(c/2 + (d*x)/2)/(a^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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