3.239 \(\int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=101 \[ -\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac {\csc ^3(c+d x)}{3 a^2 d}+\frac {\csc ^2(c+d x)}{a^2 d}-\frac {3 \csc (c+d x)}{a^2 d}-\frac {4 \log (\sin (c+d x))}{a^2 d}+\frac {4 \log (\sin (c+d x)+1)}{a^2 d} \]

[Out]

-3*csc(d*x+c)/a^2/d+csc(d*x+c)^2/a^2/d-1/3*csc(d*x+c)^3/a^2/d-4*ln(sin(d*x+c))/a^2/d+4*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 = 0.10, antiderivative size = 101, 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} \[ -\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac {\csc ^3(c+d x)}{3 a^2 d}+\frac {\csc ^2(c+d x)}{a^2 d}-\frac {3 \csc (c+d x)}{a^2 d}-\frac {4 \log (\sin (c+d x))}{a^2 d}+\frac {4 \log (\sin (c+d x)+1)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]*Csc[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-3*Csc[c + d*x])/(a^2*d) + Csc[c + d*x]^2/(a^2*d) - Csc[c + d*x]^3/(3*a^2*d) - (4*Log[Sin[c + d*x]])/(a^2*d)
+ (4*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 ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a^4}{x^4 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{x^4 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^3 \operatorname {Subst}\left (\int \left (\frac {1}{a^2 x^4}-\frac {2}{a^3 x^3}+\frac {3}{a^4 x^2}-\frac {4}{a^5 x}+\frac {1}{a^4 (a+x)^2}+\frac {4}{a^5 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {3 \csc (c+d x)}{a^2 d}+\frac {\csc ^2(c+d x)}{a^2 d}-\frac {\csc ^3(c+d x)}{3 a^2 d}-\frac {4 \log (\sin (c+d x))}{a^2 d}+\frac {4 \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 = 2.46, size = 98, normalized size = 0.97 \[ -\frac {1}{a^2 d (\sin (c+d x)+1)}-\frac {\csc ^3(c+d x)}{3 a^2 d}+\frac {\csc ^2(c+d x)}{a^2 d}-\frac {3 \csc (c+d x)}{a^2 d}-\frac {4 \log (\sin (c+d x))}{a^2 d}+\frac {4 \log (\sin (c+d x)+1)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]*Csc[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-3*Csc[c + d*x])/(a^2*d) + Csc[c + d*x]^2/(a^2*d) - Csc[c + d*x]^3/(3*a^2*d) - (4*Log[Sin[c + d*x]])/(a^2*d)
+ (4*Log[1 + Sin[c + d*x]])/(a^2*d) - 1/(a^2*d*(1 + Sin[c + d*x]))

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fricas [A]  time = 0.52, size = 195, normalized size = 1.93 \[ \frac {6 \, \cos \left (d x + c\right )^{2} - 12 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 12 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) - 7}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d - {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(6*cos(d*x + c)^2 - 12*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^2 - 1)*sin(d*x + c) + 1)*log(1/2
*sin(d*x + c)) + 12*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^2 - 1)*sin(d*x + c) + 1)*log(sin(d*x +
c) + 1) + 2*(6*cos(d*x + c)^2 - 5)*sin(d*x + c) - 7)/(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)^2 - a^2*d)*sin(d*x + c))

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giac [A]  time = 0.18, size = 103, normalized size = 1.02 \[ -\frac {\frac {12 \, \log \left ({\left | -\frac {a}{a \sin \left (d x + c\right ) + a} + 1 \right |}\right )}{a^{2}} + \frac {3}{{\left (a \sin \left (d x + c\right ) + a\right )} a} + \frac {\frac {30 \, a}{a \sin \left (d x + c\right ) + a} - \frac {18 \, a^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} - 13}{a^{2} {\left (\frac {a}{a \sin \left (d x + c\right ) + a} - 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(12*log(abs(-a/(a*sin(d*x + c) + a) + 1))/a^2 + 3/((a*sin(d*x + c) + a)*a) + (30*a/(a*sin(d*x + c) + a) -
18*a^2/(a*sin(d*x + c) + a)^2 - 13)/(a^2*(a/(a*sin(d*x + c) + a) - 1)^3))/d

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maple [A]  time = 0.29, size = 99, normalized size = 0.98 \[ -\frac {1}{3 a^{2} d \sin \left (d x +c \right )^{3}}+\frac {1}{a^{2} d \sin \left (d x +c \right )^{2}}-\frac {3}{a^{2} d \sin \left (d x +c \right )}-\frac {4 \ln \left (\sin \left (d x +c \right )\right )}{a^{2} d}-\frac {1}{d \,a^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {4 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/a^2/d/sin(d*x+c)^3+1/a^2/d/sin(d*x+c)^2-3/a^2/d/sin(d*x+c)-4*ln(sin(d*x+c))/a^2/d-1/d/a^2/(1+sin(d*x+c))+
4*ln(1+sin(d*x+c))/a^2/d

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maxima [A]  time = 0.32, size = 90, normalized size = 0.89 \[ -\frac {\frac {12 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}{a^{2} \sin \left (d x + c\right )^{4} + a^{2} \sin \left (d x + c\right )^{3}} - \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {12 \, \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((12*sin(d*x + c)^3 + 6*sin(d*x + c)^2 - 2*sin(d*x + c) + 1)/(a^2*sin(d*x + c)^4 + a^2*sin(d*x + c)^3) -
12*log(sin(d*x + c) + 1)/a^2 + 12*log(sin(d*x + c))/a^2)/d

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mupad [B]  time = 8.63, size = 202, normalized size = 2.00 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}-\frac {-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {1}{3}}{d\,\left (8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^2\,d}-\frac {13\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)^2/(4*a^2*d) - tan(c/2 + (d*x)/2)^3/(24*a^2*d) - ((28*tan(c/2 + (d*x)/2)^2)/3 - (4*tan(c/2 +
(d*x)/2))/3 + 24*tan(c/2 + (d*x)/2)^3 - 3*tan(c/2 + (d*x)/2)^4 + 1/3)/(d*(8*a^2*tan(c/2 + (d*x)/2)^3 + 16*a^2
*tan(c/2 + (d*x)/2)^4 + 8*a^2*tan(c/2 + (d*x)/2)^5)) - (4*log(tan(c/2 + (d*x)/2)))/(a^2*d) + (8*log(tan(c/2 +
(d*x)/2) + 1))/(a^2*d) - (13*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 \[ \frac {\int \frac {\cos {\left (c + d x \right )} \csc ^{4}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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