3.271 \(\int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=52 \[ -\frac {a \cot (c+d x)}{d}+\frac {a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a \cot (c+d x) \csc (c+d x)}{2 d}-a x \]

[Out]

-a*x+1/2*a*arctanh(cos(d*x+c))/d-a*cot(d*x+c)/d-1/2*a*cot(d*x+c)*csc(d*x+c)/d

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Rubi [A]  time = 0.08, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2838, 2611, 3770, 3473, 8} \[ -\frac {a \cot (c+d x)}{d}+\frac {a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a \cot (c+d x) \csc (c+d x)}{2 d}-a x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*x) + (a*ArcTanh[Cos[c + d*x]])/(2*d) - (a*Cot[c + d*x])/d - (a*Cot[c + d*x]*Csc[c + d*x])/(2*d)

Rule 8

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

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

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

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Mathematica [C]  time = 0.04, size = 109, normalized size = 2.10 \[ -\frac {a \cot (c+d x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\tan ^2(c+d x)\right )}{d}-\frac {a \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(a*Csc[(c + d*x)/2]^2)/d - (a*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2])/d + (a*Log[C
os[(c + d*x)/2]])/(2*d) - (a*Log[Sin[(c + d*x)/2]])/(2*d) + (a*Sec[(c + d*x)/2]^2)/(8*d)

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fricas [B]  time = 0.53, size = 114, normalized size = 2.19 \[ -\frac {4 \, a d x \cos \left (d x + c\right )^{2} - 4 \, a d x - 4 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

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)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 95, normalized size = 1.83 \[ \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, {\left (d x + c\right )} a - 4 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {6 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

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)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.25, size = 81, normalized size = 1.56 \[ -a x -\frac {a \cot \left (d x +c \right )}{d}-\frac {c a}{d}-\frac {a \left (\cos ^{3}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {a \cos \left (d x +c \right )}{2 d}-\frac {a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d} \]

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)),x)

[Out]

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

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maxima [A]  time = 0.45, size = 66, normalized size = 1.27 \[ -\frac {4 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a - a {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \]

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)),x, algorithm="maxima")

[Out]

-1/4*(4*(d*x + c + 1/tan(d*x + c))*a - a*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) + log(cos(d*x + c) + 1) - log(co
s(d*x + c) - 1)))/d

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mupad [B]  time = 8.68, size = 145, normalized size = 2.79 \[ \frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {a\,\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {2\,a\,\mathrm {atan}\left (\frac {2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2\,d}-\frac {a\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*tan(c/2 + (d*x)/2))/(2*d) - (a*cot(c/2 + (d*x)/2))/(2*d) - (2*a*atan((2*cos(c/2 + (d*x)/2) + sin(c/2 + (d*x
)/2))/(cos(c/2 + (d*x)/2) - 2*sin(c/2 + (d*x)/2))))/d - (a*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(2*d) -
(a*cot(c/2 + (d*x)/2)^2)/(8*d) + (a*tan(c/2 + (d*x)/2)^2)/(8*d)

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

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)),x)

[Out]

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

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