3.1.24 \(\int \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\) [24]

Optimal. Leaf size=98 \[ -\frac {a^2 x}{2}+\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2788, 3855, 3852, 8, 3853, 2718, 2715} \begin {gather*} -\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{d}-\frac {a^2 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

-1/2*(a^2*x) + (3*a^2*ArcTanh[Cos[c + d*x]])/d - (2*a^2*Cos[c + d*x])/d - (a^2*Cot[c + d*x]^3)/(3*d) - (a^2*Co
t[c + d*x]*Csc[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 3853

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 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 ^4(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac {\int \left (-a^6-4 a^6 \csc (c+d x)-a^6 \csc ^2(c+d x)+2 a^6 \csc ^3(c+d x)+a^6 \csc ^4(c+d x)+2 a^6 \sin (c+d x)+a^6 \sin ^2(c+d x)\right ) \, dx}{a^4}\\ &=-a^2 x-a^2 \int \csc ^2(c+d x) \, dx+a^2 \int \csc ^4(c+d x) \, dx+a^2 \int \sin ^2(c+d x) \, dx+\left (2 a^2\right ) \int \csc ^3(c+d x) \, dx+\left (2 a^2\right ) \int \sin (c+d x) \, dx-\left (4 a^2\right ) \int \csc (c+d x) \, dx\\ &=-a^2 x+\frac {4 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) \csc (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+a^2 \int \csc (c+d x) \, dx+\frac {a^2 \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}-\frac {a^2 \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {a^2 x}{2}+\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x) \csc (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 = 4.01, size = 191, normalized size = 1.95 \begin {gather*} \frac {a^2 (1+\sin (c+d x))^2 \left (-12 (c+d x)-48 \cos (c+d x)+4 \cot \left (\frac {1}{2} (c+d x)\right )-6 \csc ^2\left (\frac {1}{2} (c+d x)\right )+72 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-72 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 \sec ^2\left (\frac {1}{2} (c+d x)\right )+8 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )-\frac {1}{2} \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)-6 \sin (2 (c+d x))-4 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{24 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(1 + Sin[c + d*x])^2*(-12*(c + d*x) - 48*Cos[c + d*x] + 4*Cot[(c + d*x)/2] - 6*Csc[(c + d*x)/2]^2 + 72*Lo
g[Cos[(c + d*x)/2]] - 72*Log[Sin[(c + d*x)/2]] + 6*Sec[(c + d*x)/2]^2 + 8*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 -
(Csc[(c + d*x)/2]^4*Sin[c + d*x])/2 - 6*Sin[2*(c + d*x)] - 4*Tan[(c + d*x)/2]))/(24*d*(Cos[(c + d*x)/2] + Sin[
(c + d*x)/2])^4)

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Maple [A]
time = 0.20, size = 146, normalized size = 1.49

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {\cos ^{5}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+2 a^{2} \left (-\frac {\cos ^{5}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+a^{2} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\)	\(146\)
default	\(\frac {a^{2} \left (-\frac {\cos ^{5}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+2 a^{2} \left (-\frac {\cos ^{5}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+a^{2} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\)	\(146\)
risch	\(-\frac {a^{2} x}{2}+\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{d}-\frac {i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 a^{2} \left (3 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}+i-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\)	\(176\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2*(-1/sin(d*x+c)*cos(d*x+c)^5-(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)-3/2*d*x-3/2*c)+2*a^2*(-1/2/sin(d
*x+c)^2*cos(d*x+c)^5-1/2*cos(d*x+c)^3-3/2*cos(d*x+c)-3/2*ln(csc(d*x+c)-cot(d*x+c)))+a^2*(-1/3*cot(d*x+c)^3+cot
(d*x+c)+d*x+c))

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Maxima [A]
time = 0.51, size = 139, normalized size = 1.42 \begin {gather*} -\frac {3 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2} - 2 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a^{2} - 3 \, a^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (92) = 184\).
time = 0.39, size = 192, normalized size = 1.96 \begin {gather*} \frac {3 \, a^{2} \cos \left (d x + c\right )^{5} - 4 \, a^{2} \cos \left (d x + c\right )^{3} + 3 \, a^{2} \cos \left (d x + c\right ) + 9 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 9 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, {\left (a^{2} d x \cos \left (d x + c\right )^{2} + 4 \, a^{2} \cos \left (d x + c\right )^{3} - a^{2} d x - 6 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*a^2*cos(d*x + c)^5 - 4*a^2*cos(d*x + c)^3 + 3*a^2*cos(d*x + c) + 9*(a^2*cos(d*x + c)^2 - a^2)*log(1/2*c
os(d*x + c) + 1/2)*sin(d*x + c) - 9*(a^2*cos(d*x + c)^2 - a^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*(
a^2*d*x*cos(d*x + c)^2 + 4*a^2*cos(d*x + c)^3 - a^2*d*x - 6*a^2*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^2
- 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 ^{4}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}\, dx + \int \cot ^{4}{\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)**4*(a+a*sin(d*x+c))**2,x)

[Out]

a**2*(Integral(2*sin(c + d*x)*cot(c + d*x)**4, x) + Integral(sin(c + d*x)**2*cot(c + d*x)**4, x) + Integral(co
t(c + d*x)**4, x))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (92) = 184\).
time = 6.02, size = 209, normalized size = 2.13 \begin {gather*} \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, {\left (d x + c\right )} a^{2} - 72 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {24 \, {\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}} + \frac {132 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, 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 )^{3}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*(d*x + c)*a^2 - 72*a^2*log(abs(tan(1/2*d*
x + 1/2*c))) - 3*a^2*tan(1/2*d*x + 1/2*c) + 24*(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 + (132*a^2*tan(1/2*d*x + 1/2*c)^3 + 3*a^2*tan(1
/2*d*x + 1/2*c)^2 - 6*a^2*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^3)/d

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Mupad [B]
time = 6.52, size = 293, normalized size = 2.99 \begin {gather*} \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {3\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^2\,\mathrm {atan}\left (\frac {a^4}{6\,a^4-a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {6\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{6\,a^4-a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {-9\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+34\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {19\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+36\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^2}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*tan(c/2 + (d*x)/2)^2)/(4*d) + (a^2*tan(c/2 + (d*x)/2)^3)/(24*d) - (3*a^2*log(tan(c/2 + (d*x)/2)))/d - (a^
2*atan(a^4/(6*a^4 - a^4*tan(c/2 + (d*x)/2)) + (6*a^4*tan(c/2 + (d*x)/2))/(6*a^4 - a^4*tan(c/2 + (d*x)/2))))/d
- (36*a^2*tan(c/2 + (d*x)/2)^3 - (a^2*tan(c/2 + (d*x)/2)^2)/3 + (19*a^2*tan(c/2 + (d*x)/2)^4)/3 + 34*a^2*tan(c
/2 + (d*x)/2)^5 - 9*a^2*tan(c/2 + (d*x)/2)^6 + a^2/3 + 2*a^2*tan(c/2 + (d*x)/2))/(d*(8*tan(c/2 + (d*x)/2)^3 +
16*tan(c/2 + (d*x)/2)^5 + 8*tan(c/2 + (d*x)/2)^7)) - (a^2*tan(c/2 + (d*x)/2))/(8*d)

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