3.1.58 \(\int \frac {\cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx\) [58]

Optimal. Leaf size=58 \[ -\frac {\tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2785, 2687, 30, 2691, 3855} \begin {gather*} -\frac {\cot ^3(c+d x)}{3 a d}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2691

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 2785

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(124\) vs. \(2(58)=116\).
time = 0.36, size = 124, normalized size = 2.14 \begin {gather*} -\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (\cos (3 (c+d x))+\cos (c+d x) (3-6 \sin (c+d x))+6 \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 ) \sin ^3(c+d x)\right )}{96 a d (1+\sin (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/96*(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^2*(Cos[3*(c + d*x)] + Cos[c + d
*x]*(3 - 6*Sin[c + d*x]) + 6*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]])*Sin[c + d*x]^3))/(a*d*(1 + Sin[c
+ d*x]))

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Maple [A]
time = 0.21, size = 94, normalized size = 1.62

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{8 d a}\)	\(94\)
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{8 d a}\)	\(94\)
risch	\(-\frac {-6 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}-2 i-3 \,{\mathrm e}^{i \left (d x +c \right )}}{3 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d a}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d/a*(1/3*tan(1/2*d*x+1/2*c)^3-tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)-1/3/tan(1/2*d*x+1/2*c)^3+4*ln(tan(1/
2*d*x+1/2*c))+1/tan(1/2*d*x+1/2*c)+1/tan(1/2*d*x+1/2*c)^2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (52) = 104\).
time = 0.29, size = 155, normalized size = 2.67 \begin {gather*} -\frac {\frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {{\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a \sin \left (d x + 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)),x, algorithm="maxima")

[Out]

-1/24*((3*sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a - 12*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - (3*sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)
^2/(cos(d*x + c) + 1)^2 - 1)*(cos(d*x + c) + 1)^3/(a*sin(d*x + c)^3))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (52) = 104\).
time = 0.36, size = 111, normalized size = 1.91 \begin {gather*} \frac {4 \, \cos \left (d x + c\right )^{3} - 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 ) \sin \left (d x + c\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 ) \sin \left (d x + c\right ) - 6 \, \cos \left (d x + c\right ) \sin \left (d x + c\right )}{12 \, {\left (a d \cos \left (d x + c\right )^{2} - a 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)),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\cot ^{4}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \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)),x)

[Out]

Integral(cot(c + d*x)**4/(sin(c + d*x) + 1), x)/a

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (52) = 104\).
time = 8.76, size = 127, normalized size = 2.19 \begin {gather*} \frac {\frac {12 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {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} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} - \frac {22 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a \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)),x, algorithm="giac")

[Out]

1/24*(12*log(abs(tan(1/2*d*x + 1/2*c)))/a + (a^2*tan(1/2*d*x + 1/2*c)^3 - 3*a^2*tan(1/2*d*x + 1/2*c)^2 - 3*a^2
*tan(1/2*d*x + 1/2*c))/a^3 - (22*tan(1/2*d*x + 1/2*c)^3 - 3*tan(1/2*d*x + 1/2*c)^2 - 3*tan(1/2*d*x + 1/2*c) +
1)/(a*tan(1/2*d*x + 1/2*c)^3))/d

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Mupad [B]
time = 6.63, size = 115, normalized size = 1.98 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{3}\right )}{8\,a\,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)),x)

[Out]

tan(c/2 + (d*x)/2)^3/(24*a*d) - tan(c/2 + (d*x)/2)^2/(8*a*d) + log(tan(c/2 + (d*x)/2))/(2*a*d) - tan(c/2 + (d*
x)/2)/(8*a*d) + (cot(c/2 + (d*x)/2)^3*(tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^2 - 1/3))/(8*a*d)

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