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

Optimal. Leaf size=146 \[ \frac{5 \cos ^3(c+d x)}{6 a d}+\frac{5 \cos (c+d x)}{2 a d}-\frac{5 \cot ^3(c+d x)}{6 a d}+\frac{5 \cot (c+d x)}{2 a d}+\frac{\cos ^3(c+d x) \cot ^2(c+d x)}{2 a d}+\frac{\cos ^2(c+d x) \cot ^3(c+d x)}{2 a d}-\frac{5 \tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac{5 x}{2 a} \]

[Out]

(5*x)/(2*a) - (5*ArcTanh[Cos[c + d*x]])/(2*a*d) + (5*Cos[c + d*x])/(2*a*d) + (5*Cos[c + d*x]^3)/(6*a*d) + (5*C
ot[c + d*x])/(2*a*d) + (Cos[c + d*x]^3*Cot[c + d*x]^2)/(2*a*d) - (5*Cot[c + d*x]^3)/(6*a*d) + (Cos[c + d*x]^2*
Cot[c + d*x]^3)/(2*a*d)

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Rubi [A]  time = 0.180188, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2839, 2591, 288, 302, 203, 2592, 206} \[ \frac{5 \cos ^3(c+d x)}{6 a d}+\frac{5 \cos (c+d x)}{2 a d}-\frac{5 \cot ^3(c+d x)}{6 a d}+\frac{5 \cot (c+d x)}{2 a d}+\frac{\cos ^3(c+d x) \cot ^2(c+d x)}{2 a d}+\frac{\cos ^2(c+d x) \cot ^3(c+d x)}{2 a d}-\frac{5 \tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac{5 x}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x)/(2*a) - (5*ArcTanh[Cos[c + d*x]])/(2*a*d) + (5*Cos[c + d*x])/(2*a*d) + (5*Cos[c + d*x]^3)/(6*a*d) + (5*C
ot[c + d*x])/(2*a*d) + (Cos[c + d*x]^3*Cot[c + d*x]^2)/(2*a*d) - (5*Cot[c + d*x]^3)/(6*a*d) + (Cos[c + d*x]^2*
Cot[c + d*x]^3)/(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 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 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 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 ^4(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac{\int \cos ^3(c+d x) \cot ^3(c+d x) \, dx}{a}+\frac{\int \cos ^2(c+d x) \cot ^4(c+d x) \, dx}{a}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{a d}-\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{a d}\\ &=\frac{\cos ^3(c+d x) \cot ^2(c+d x)}{2 a d}+\frac{\cos ^2(c+d x) \cot ^3(c+d x)}{2 a d}-\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 a d}-\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 a d}\\ &=\frac{\cos ^3(c+d x) \cot ^2(c+d x)}{2 a d}+\frac{\cos ^2(c+d x) \cot ^3(c+d x)}{2 a d}-\frac{5 \operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{2 a d}-\frac{5 \operatorname{Subst}\left (\int \left (-1+x^2+\frac{1}{1+x^2}\right ) \, dx,x,\cot (c+d x)\right )}{2 a d}\\ &=\frac{5 \cos (c+d x)}{2 a d}+\frac{5 \cos ^3(c+d x)}{6 a d}+\frac{5 \cot (c+d x)}{2 a d}+\frac{\cos ^3(c+d x) \cot ^2(c+d x)}{2 a d}-\frac{5 \cot ^3(c+d x)}{6 a d}+\frac{\cos ^2(c+d x) \cot ^3(c+d x)}{2 a d}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 a d}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 a d}\\ &=\frac{5 x}{2 a}-\frac{5 \tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac{5 \cos (c+d x)}{2 a d}+\frac{5 \cos ^3(c+d x)}{6 a d}+\frac{5 \cot (c+d x)}{2 a d}+\frac{\cos ^3(c+d x) \cot ^2(c+d x)}{2 a d}-\frac{5 \cot ^3(c+d x)}{6 a d}+\frac{\cos ^2(c+d x) \cot ^3(c+d x)}{2 a d}\\ \end{align*}

Mathematica [A]  time = 0.789702, size = 197, normalized size = 1.35 \[ -\frac{\csc ^3(c+d x) \left (-180 c \sin (c+d x)-180 d x \sin (c+d x)-75 \sin (2 (c+d x))+60 c \sin (3 (c+d x))+60 d x \sin (3 (c+d x))+24 \sin (4 (c+d x))+\sin (6 (c+d x))-30 \cos (c+d x)+65 \cos (3 (c+d x))-3 \cos (5 (c+d x))-180 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+60 \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+180 \sin (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-60 \sin (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{96 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]^3*(-30*Cos[c + d*x] + 65*Cos[3*(c + d*x)] - 3*Cos[5*(c + d*x)] - 180*c*Sin[c + d*x] - 180*d*x*S
in[c + d*x] + 180*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] - 180*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] - 75*Sin[2*(c +
d*x)] + 60*c*Sin[3*(c + d*x)] + 60*d*x*Sin[3*(c + d*x)] - 60*Log[Cos[(c + d*x)/2]]*Sin[3*(c + d*x)] + 60*Log[S
in[(c + d*x)/2]]*Sin[3*(c + d*x)] + 24*Sin[4*(c + d*x)] + Sin[6*(c + d*x)]))/(96*a*d)

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Maple [B]  time = 0.148, size = 306, normalized size = 2.1 \begin{align*}{\frac{1}{24\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{9}{8\,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}}+6\,{\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}}}+8\,{\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{14}{3\,da} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+5\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{da}}-{\frac{1}{24\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{9}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{5}{2\,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)^8*csc(d*x+c)^4/(a+a*sin(d*x+c)),x)

[Out]

1/24/d/a*tan(1/2*d*x+1/2*c)^3-1/8/d/a*tan(1/2*d*x+1/2*c)^2-9/8/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+6/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^4+8/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)+14/3/d/a/(1+tan(1/2*d*x+1/2*c)
^2)^3+5/a/d*arctan(tan(1/2*d*x+1/2*c))-1/24/d/a/tan(1/2*d*x+1/2*c)^3+1/8/d/a/tan(1/2*d*x+1/2*c)^2+9/8/d/a/tan(
1/2*d*x+1/2*c)+5/2/d/a*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.54203, size = 489, normalized size = 3.35 \begin{align*} -\frac{\frac{\frac{27 \, \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{24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{121 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{102 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{201 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{80 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{147 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{3 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - 1}{\frac{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{3 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{a \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}} - \frac{120 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/24*((27*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) + 24*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 121*sin(d*x + c)
^3/(cos(d*x + c) + 1)^3 + 102*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 201*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 +
80*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 147*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 3*sin(d*x + c)^8/(cos(d*x +
c) + 1)^8 - 1)/(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 + 3*a*sin(d*x
+ c)^7/(cos(d*x + c) + 1)^7 + a*sin(d*x + c)^9/(cos(d*x + c) + 1)^9) - 120*arctan(sin(d*x + c)/(cos(d*x + c)
+ 1))/a - 60*log(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

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Fricas [A]  time = 1.18203, size = 470, normalized size = 3.22 \begin{align*} -\frac{6 \, \cos \left (d x + c\right )^{5} - 40 \, \cos \left (d x + c\right )^{3} + 15 \,{\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 ) - 15 \,{\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 ) - 2 \,{\left (2 \, \cos \left (d x + c\right )^{5} + 15 \, d x \cos \left (d x + c\right )^{2} + 10 \, \cos \left (d x + c\right )^{3} - 15 \, d x - 15 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 30 \, \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(6*cos(d*x + c)^5 - 40*cos(d*x + c)^3 + 15*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c)
- 15*(cos(d*x + c)^2 - 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 2*(2*cos(d*x + c)^5 + 15*d*x*cos(d*x +
c)^2 + 10*cos(d*x + c)^3 - 15*d*x - 15*cos(d*x + c))*sin(d*x + c) + 30*cos(d*x + c))/((a*d*cos(d*x + c)^2 - 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)**8*csc(d*x+c)**4/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.27305, size = 308, normalized size = 2.11 \begin{align*} \frac{\frac{180 \,{\left (d x + c\right )}}{a} + \frac{180 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} + \frac{3 \,{\left (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} - 27 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a^{3}} - \frac{110 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 111 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 240 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 273 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 306 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 253 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 72 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{3} a}}{72 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/72*(180*(d*x + c)/a + 180*log(abs(tan(1/2*d*x + 1/2*c)))/a + 3*(a^2*tan(1/2*d*x + 1/2*c)^3 - 3*a^2*tan(1/2*d
*x + 1/2*c)^2 - 27*a^2*tan(1/2*d*x + 1/2*c))/a^3 - (110*tan(1/2*d*x + 1/2*c)^9 - 9*tan(1/2*d*x + 1/2*c)^8 - 11
1*tan(1/2*d*x + 1/2*c)^7 - 240*tan(1/2*d*x + 1/2*c)^6 - 273*tan(1/2*d*x + 1/2*c)^5 - 306*tan(1/2*d*x + 1/2*c)^
4 - 253*tan(1/2*d*x + 1/2*c)^3 - 72*tan(1/2*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) + 3)/((tan(1/2*d*x + 1/2*c
)^3 + tan(1/2*d*x + 1/2*c))^3*a))/d