3.742 \(\int \frac{\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=99 \[ -\frac{\cos ^3(c+d x)}{a^3 d}+\frac{\cos (c+d x)}{a^3 d}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 a^3 d}-\frac{13 \sin (c+d x) \cos (c+d x)}{8 a^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{13 x}{8 a^3} \]

[Out]

(-13*x)/(8*a^3) - ArcTanh[Cos[c + d*x]]/(a^3*d) + Cos[c + d*x]/(a^3*d) - Cos[c + d*x]^3/(a^3*d) - (13*Cos[c +
d*x]*Sin[c + d*x])/(8*a^3*d) + (Cos[c + d*x]^3*Sin[c + d*x])/(4*a^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.242914, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.37, Rules used = {2875, 2873, 2635, 8, 2592, 321, 206, 2565, 30, 2568} \[ -\frac{\cos ^3(c+d x)}{a^3 d}+\frac{\cos (c+d x)}{a^3 d}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 a^3 d}-\frac{13 \sin (c+d x) \cos (c+d x)}{8 a^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{13 x}{8 a^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^7*Cot[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

(-13*x)/(8*a^3) - ArcTanh[Cos[c + d*x]]/(a^3*d) + Cos[c + d*x]/(a^3*d) - Cos[c + d*x]^3/(a^3*d) - (13*Cos[c +
d*x]*Sin[c + d*x])/(8*a^3*d) + (Cos[c + d*x]^3*Sin[c + d*x])/(4*a^3*d)

Rule 2875

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

Rule 2873

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

Rule 2635

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] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 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 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 30

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

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rubi steps

\begin{align*} \int \frac{\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \cos (c+d x) \cot (c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac{\int \left (-3 a^3 \cos ^2(c+d x)+a^3 \cos (c+d x) \cot (c+d x)+3 a^3 \cos ^2(c+d x) \sin (c+d x)-a^3 \cos ^2(c+d x) \sin ^2(c+d x)\right ) \, dx}{a^6}\\ &=\frac{\int \cos (c+d x) \cot (c+d x) \, dx}{a^3}-\frac{\int \cos ^2(c+d x) \sin ^2(c+d x) \, dx}{a^3}-\frac{3 \int \cos ^2(c+d x) \, dx}{a^3}+\frac{3 \int \cos ^2(c+d x) \sin (c+d x) \, dx}{a^3}\\ &=-\frac{3 \cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a^3 d}-\frac{\int \cos ^2(c+d x) \, dx}{4 a^3}-\frac{3 \int 1 \, dx}{2 a^3}-\frac{\operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int x^2 \, dx,x,\cos (c+d x)\right )}{a^3 d}\\ &=-\frac{3 x}{2 a^3}+\frac{\cos (c+d x)}{a^3 d}-\frac{\cos ^3(c+d x)}{a^3 d}-\frac{13 \cos (c+d x) \sin (c+d x)}{8 a^3 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a^3 d}-\frac{\int 1 \, dx}{8 a^3}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{a^3 d}\\ &=-\frac{13 x}{8 a^3}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{\cos (c+d x)}{a^3 d}-\frac{\cos ^3(c+d x)}{a^3 d}-\frac{13 \cos (c+d x) \sin (c+d x)}{8 a^3 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.390727, size = 80, normalized size = 0.81 \[ \frac{-24 \sin (2 (c+d x))+\sin (4 (c+d x))+8 \cos (c+d x)-8 \cos (3 (c+d x))+32 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-32 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-52 c-52 d x}{32 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^7*Cot[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

(-52*c - 52*d*x + 8*Cos[c + d*x] - 8*Cos[3*(c + d*x)] - 32*Log[Cos[(c + d*x)/2]] + 32*Log[Sin[(c + d*x)/2]] -
24*Sin[2*(c + d*x)] + Sin[4*(c + d*x)])/(32*a^3*d)

________________________________________________________________________________________

Maple [B]  time = 0.135, size = 239, normalized size = 2.4 \begin{align*}{\frac{11}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}-4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}+{\frac{19}{4\,d{a}^{3}} \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 ) ^{-4}}-{\frac{19}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}-{\frac{11}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}-{\frac{13}{4\,d{a}^{3}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{1}{d{a}^{3}}\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)/(a+a*sin(d*x+c))^3,x)

[Out]

11/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7-4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*
c)^6+19/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5-19/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*
d*x+1/2*c)^3+4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^2-11/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan
(1/2*d*x+1/2*c)-13/4/d/a^3*arctan(tan(1/2*d*x+1/2*c))+1/d/a^3*ln(tan(1/2*d*x+1/2*c))

________________________________________________________________________________________

Maxima [B]  time = 1.58602, size = 363, normalized size = 3.67 \begin{align*} -\frac{\frac{\frac{11 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{16 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{19 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{19 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{16 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{11 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{3} + \frac{4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{4 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac{13 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac{4 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/4*((11*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 19*sin(d*x + c)^3/(cos(d*
x + c) + 1)^3 - 19*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 16*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 11*sin(d*x +
c)^7/(cos(d*x + c) + 1)^7)/(a^3 + 4*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a^3*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4 + 4*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) + 13*arctan(
sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - 4*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

________________________________________________________________________________________

Fricas [A]  time = 1.17286, size = 238, normalized size = 2.4 \begin{align*} -\frac{8 \, \cos \left (d x + c\right )^{3} + 13 \, d x -{\left (2 \, \cos \left (d x + c\right )^{3} - 13 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 8 \, \cos \left (d x + c\right ) + 4 \, \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 4 \, \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{8 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/8*(8*cos(d*x + c)^3 + 13*d*x - (2*cos(d*x + c)^3 - 13*cos(d*x + c))*sin(d*x + c) - 8*cos(d*x + c) + 4*log(1
/2*cos(d*x + c) + 1/2) - 4*log(-1/2*cos(d*x + c) + 1/2))/(a^3*d)

________________________________________________________________________________________

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)/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.30986, size = 174, normalized size = 1.76 \begin{align*} -\frac{\frac{13 \,{\left (d x + c\right )}}{a^{3}} - \frac{8 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{2 \,{\left (11 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 16 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 19 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 19 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 16 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 11 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4} a^{3}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/8*(13*(d*x + c)/a^3 - 8*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 2*(11*tan(1/2*d*x + 1/2*c)^7 - 16*tan(1/2*d*x
+ 1/2*c)^6 + 19*tan(1/2*d*x + 1/2*c)^5 - 19*tan(1/2*d*x + 1/2*c)^3 + 16*tan(1/2*d*x + 1/2*c)^2 - 11*tan(1/2*d*
x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^3))/d