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

Optimal. Leaf size=95 \[ -\frac{\cos ^3(c+d x)}{3 a d}-\frac{\cos (c+d x)}{a d}-\frac{3 \cot (c+d x)}{2 a d}+\frac{\cos ^2(c+d x) \cot (c+d x)}{2 a d}+\frac{\tanh ^{-1}(\cos (c+d x))}{a d}-\frac{3 x}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.156155, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2839, 2591, 288, 321, 203, 2592, 302, 206} \[ -\frac{\cos ^3(c+d x)}{3 a d}-\frac{\cos (c+d x)}{a d}-\frac{3 \cot (c+d x)}{2 a d}+\frac{\cos ^2(c+d x) \cot (c+d x)}{2 a d}+\frac{\tanh ^{-1}(\cos (c+d x))}{a d}-\frac{3 x}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2591

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

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 203

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

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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])

Rubi steps

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

Mathematica [A]  time = 0.780923, size = 122, normalized size = 1.28 \[ -\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \left (\cot \left (\frac{1}{2} (c+d x)\right )+1\right )^2 \left (27 \cos (c+d x)+(2 \sin (c+d x)-3) \cos (3 (c+d x))+6 \sin (c+d x) \left (5 \cos (c+d x)+4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+6 c+6 d x\right )\right )}{48 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 + Cot[(c + d*x)/2])^2*(27*Cos[c + d*x] + 6*(6*c + 6*d*x + 5*Cos[c + d*x] - 4*Log[Cos[(c + d*x)/2]] + 4*Lo
g[Sin[(c + d*x)/2]])*Sin[c + d*x] + Cos[3*(c + d*x)]*(-3 + 2*Sin[c + d*x]))*Tan[(c + d*x)/2])/(48*a*d*(1 + Sin
[c + d*x]))

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Maple [B]  time = 0.119, size = 230, normalized size = 2.4 \begin{align*}{\frac{1}{2\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}-4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}-{\frac{8}{3\,da} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}-3\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{da}}-{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{1}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^2/(a+a*sin(d*x+c)),x)

[Out]

1/2/d/a*tan(1/2*d*x+1/2*c)+1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5-4/d/a/(1+tan(1/2*d*x+1/2*c)^2
)^3*tan(1/2*d*x+1/2*c)^4-4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2-1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^
3*tan(1/2*d*x+1/2*c)-8/3/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3-3/a/d*arctan(tan(1/2*d*x+1/2*c))-1/2/d/a/tan(1/2*d*x+1
/2*c)-1/d/a*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.54743, size = 374, normalized size = 3.94 \begin{align*} -\frac{\frac{\frac{16 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{15 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{24 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{9 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{24 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{3 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 3}{\frac{a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{3 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} + \frac{18 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac{3 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/6*((16*sin(d*x + c)/(cos(d*x + c) + 1) + 15*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 24*sin(d*x + c)^3/(cos(d*
x + c) + 1)^3 + 9*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 24*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 3*sin(d*x + c
)^6/(cos(d*x + c) + 1)^6 + 3)/(a*sin(d*x + c)/(cos(d*x + c) + 1) + 3*a*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3
*a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) + 18*arctan(sin(d*x + c)/(cos(
d*x + c) + 1))/a + 6*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - 3*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.1328, size = 284, normalized size = 2.99 \begin{align*} \frac{3 \, \cos \left (d x + c\right )^{3} -{\left (2 \, \cos \left (d x + c\right )^{3} + 9 \, d x + 6 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 3 \, \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 3 \, \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 9 \, \cos \left (d x + c\right )}{6 \, a d \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*csc(d*x+c)**2/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.34769, size = 198, normalized size = 2.08 \begin{align*} -\frac{\frac{9 \,{\left (d x + c\right )}}{a} + \frac{6 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} - \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} - \frac{3 \,{\left (2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}}{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} - \frac{2 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 12 \, \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 ) - 8\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} a}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(9*(d*x + c)/a + 6*log(abs(tan(1/2*d*x + 1/2*c)))/a - 3*tan(1/2*d*x + 1/2*c)/a - 3*(2*tan(1/2*d*x + 1/2*c
) - 1)/(a*tan(1/2*d*x + 1/2*c)) - 2*(3*tan(1/2*d*x + 1/2*c)^5 - 12*tan(1/2*d*x + 1/2*c)^4 - 12*tan(1/2*d*x + 1
/2*c)^2 - 3*tan(1/2*d*x + 1/2*c) - 8)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a))/d