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

Optimal. Leaf size=54 \[ \frac {2 \cot (c+d x)}{a^2 d}-\frac {3 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d} \]

[Out]

-3/2*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

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Rubi [A]  time = 0.15, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2869, 2757, 3770, 3767, 8, 3768} \[ \frac {2 \cot (c+d x)}{a^2 d}-\frac {3 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*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)

Rule 8

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

Rule 2757

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

Rule 2869

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

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

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Mathematica [A]  time = 0.53, size = 86, normalized size = 1.59 \[ -\frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (\cot (c+d x) (\csc (c+d x)-4)+3 \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{2 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 93, normalized size = 1.72 \[ -\frac {3 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 8 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 98, normalized size = 1.81 \[ \frac {\frac {12 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} - \frac {18 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*(12*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 - (
18*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 = 0.60, size = 93, normalized size = 1.72 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{2} d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2}}+\frac {1}{d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}-\frac {1}{8 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x)

[Out]

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

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maxima [B]  time = 0.34, size = 115, normalized size = 2.13 \[ -\frac {\frac {\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {{\left (\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a^{2} \sin \left (d x + c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/8*((8*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^2 - 12*log(sin(d*x + c)/(cos
(d*x + c) + 1))/a^2 - (8*sin(d*x + c)/(cos(d*x + c) + 1) - 1)*(cos(d*x + c) + 1)^2/(a^2*sin(d*x + c)^2))/d

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mupad [B]  time = 8.69, size = 84, normalized size = 1.56 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}+\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{8}\right )}{a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(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) + (3*log(tan(c/2 + (d*x)/2)))/(2*a^2*d) - tan(c/2 + (d*x)/2)/(a^2*d) + (cot(c/2
+ (d*x)/2)^2*(tan(c/2 + (d*x)/2) - 1/8))/(a^2*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**3/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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