3.631 \(\int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=94 \[ \frac {3 \cos (c+d x)}{2 a d}-\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cot (c+d x)}{a d}+\frac {\cos (c+d x) \cot ^2(c+d x)}{2 a d}-\frac {3 \tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac {x}{a} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2839, 3473, 8, 2592, 288, 321, 206} \[ \frac {3 \cos (c+d x)}{2 a d}-\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cot (c+d x)}{a d}+\frac {\cos (c+d x) \cot ^2(c+d x)}{2 a d}-\frac {3 \tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac {x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

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]

Rubi steps

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

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Mathematica [A]  time = 0.91, size = 138, normalized size = 1.47 \[ \frac {\csc \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (9 \sin (2 (c+d x))-2 (3 \sin (c+d x)+4) \cos (3 (c+d x))+12 \sin ^3(c+d x) \left (3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 c+2 d x\right )\right )}{192 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^2*(12*(2*c + 2*d*x - 3*Log[Cos[(c + d
*x)/2]] + 3*Log[Sin[(c + d*x)/2]])*Sin[c + d*x]^3 - 2*Cos[3*(c + d*x)]*(4 + 3*Sin[c + d*x]) + 9*Sin[2*(c + d*x
)]))/(192*a*d*(1 + Sin[c + d*x]))

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fricas [A]  time = 0.81, size = 148, normalized size = 1.57 \[ \frac {16 \, \cos \left (d x + c\right )^{3} - 9 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 9 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 6 \, {\left (2 \, d x \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right )^{3} - 2 \, d x - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 12 \, \cos \left (d x + c\right )}{12 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(16*cos(d*x + c)^3 - 9*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 9*(cos(d*x + c)^2
- 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 6*(2*d*x*cos(d*x + c)^2 + 2*cos(d*x + c)^3 - 2*d*x - 3*cos(d*
x + c))*sin(d*x + c) - 12*cos(d*x + c))/((a*d*cos(d*x + c)^2 - a*d)*sin(d*x + c))

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giac [A]  time = 0.19, size = 157, normalized size = 1.67 \[ \frac {\frac {24 \, {\left (d x + c\right )}}{a} + \frac {36 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} + \frac {48}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a} - \frac {66 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/24*(24*(d*x + c)/a + 36*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 - 15*a^2*tan(1/2*d*x + 1/2*c))/a^3 + 48/((tan(1/2*d*x + 1/2*c)^2 + 1)*a) - (66*tan(1/2*d*x + 1/2*c)^
3 - 15*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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maple [A]  time = 0.50, size = 173, normalized size = 1.84 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 a d}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {2}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {1}{24 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{8 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {5}{8 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^4/(a+a*sin(d*x+c)),x)

[Out]

1/24/a/d*tan(1/2*d*x+1/2*c)^3-1/8/a/d*tan(1/2*d*x+1/2*c)^2-5/8/a/d*tan(1/2*d*x+1/2*c)+2/a/d/(1+tan(1/2*d*x+1/2
*c)^2)+2/a/d*arctan(tan(1/2*d*x+1/2*c))-1/24/a/d/tan(1/2*d*x+1/2*c)^3+1/8/a/d/tan(1/2*d*x+1/2*c)^2+5/8/a/d/tan
(1/2*d*x+1/2*c)+3/2/a/d*ln(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.53, size = 240, normalized size = 2.55 \[ -\frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a} - \frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {14 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {51 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 1}{\frac {a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} - \frac {48 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {36 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/24*((15*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 - (3*sin(d*x + c)/(cos(d*x + c) + 1) + 14*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 51*sin(d*x + c)^
3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 1)/(a*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 +
a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) - 48*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - 36*log(sin(d*x + c)/(c
os(d*x + c) + 1))/a)/d

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mupad [B]  time = 8.97, size = 212, normalized size = 2.26 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {2\,\mathrm {atan}\left (\frac {6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-6}+\frac {4}{4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-6}\right )}{a\,d}+\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a\,d}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {14\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{3}}{d\,\left (8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6/(sin(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) - (2*atan((6*tan(c/2 + (d*x)/2))/(4*tan(c/2 + (d*
x)/2) - 6) + 4/(4*tan(c/2 + (d*x)/2) - 6)))/(a*d) + (3*log(tan(c/2 + (d*x)/2)))/(2*a*d) + (tan(c/2 + (d*x)/2)
+ (14*tan(c/2 + (d*x)/2)^2)/3 + 17*tan(c/2 + (d*x)/2)^3 + 5*tan(c/2 + (d*x)/2)^4 - 1/3)/(d*(8*a*tan(c/2 + (d*x
)/2)^3 + 8*a*tan(c/2 + (d*x)/2)^5)) - (5*tan(c/2 + (d*x)/2))/(8*a*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)**6*csc(d*x+c)**4/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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