3.1.23 \(\int \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\) [23]

Optimal. Leaf size=74 \[ -\frac {a^2 x}{2}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2788, 3855, 3852, 8, 2718, 2715} \begin {gather*} \frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {a^2 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a^2*x) - (2*a^2*ArcTanh[Cos[c + d*x]])/d + (2*a^2*Cos[c + d*x])/d - (a^2*Cot[c + d*x])/d + (a^2*Cos[c +
d*x]*Sin[c + d*x])/(2*d)

Rule 8

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 3852

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 3855

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

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Mathematica [A]
time = 0.40, size = 94, normalized size = 1.27 \begin {gather*} -\frac {a^2 \csc \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (7 \cos (c+d x)+\cos (3 (c+d x))+4 \left (c+d x-4 \cos (c+d x)+4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin (c+d x)\right )}{16 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(a^2*Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(7*Cos[c + d*x] + Cos[3*(c + d*x)] + 4*(c + d*x - 4*Cos[c + d*x]
+ 4*Log[Cos[(c + d*x)/2]] - 4*Log[Sin[(c + d*x)/2]])*Sin[c + d*x]))/d

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Maple [A]
time = 0.13, size = 80, normalized size = 1.08

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a^{2} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(80\)
default	\(\frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a^{2} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(80\)
risch	\(-\frac {a^{2} x}{2}-\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 79, normalized size = 1.07 \begin {gather*} \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 4 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a^{2} + 4 \, a^{2} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 105, normalized size = 1.42 \begin {gather*} -\frac {a^{2} \cos \left (d x + c\right )^{3} + 2 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + a^{2} \cos \left (d x + c\right ) + {\left (a^{2} d x - 4 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \sin \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int 2 \sin {\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx + \int \cot ^{2}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2*(a+a*sin(d*x+c))**2,x)

[Out]

a**2*(Integral(2*sin(c + d*x)*cot(c + d*x)**2, x) + Integral(sin(c + d*x)**2*cot(c + d*x)**2, x) + Integral(co
t(c + d*x)**2, x))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (70) = 140\).
time = 7.51, size = 143, normalized size = 1.93 \begin {gather*} -\frac {{\left (d x + c\right )} a^{2} - 4 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 6.52, size = 201, normalized size = 2.72 \begin {gather*} \frac {2\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {-3\,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+8\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a^2}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {a^4}{4\,a^4+a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {4\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^4+a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^2*(a + a*sin(c + d*x))^2,x)

[Out]

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

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