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]

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

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Rubi [A]  time = 0.24, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, 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 8

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

Rule 30

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

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 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 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]

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 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 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 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]

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*}

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Mathematica [A]  time = 0.41, 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)

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fricas [A]  time = 0.48, size = 84, normalized size = 0.85 \[ -\frac {8 \, \cos \left (d x + c\right )^{3} + 13 \, d x - {\left (2 \, \cos \left (d x + c\right )^{3} - 13 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 8 \, \cos \left (d x + c\right ) + 4 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4 \, \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{8 \, a^{3} d} \]

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)

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giac [A]  time = 0.25, size = 129, normalized size = 1.30 \[ -\frac {\frac {13 \, {\left (d x + c\right )}}{a^{3}} - \frac {8 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {2 \, {\left (11 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 16 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 19 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 19 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{3}}}{8 \, d} \]

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

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maple [B]  time = 0.57, size = 239, normalized size = 2.41 \[ \frac {11 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {4 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {19 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {19 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {13 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3} d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}} \]

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

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maxima [B]  time = 0.60, size = 269, normalized size = 2.72 \[ -\frac {\frac {\frac {11 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {16 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {19 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {19 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {16 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {11 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{3} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac {13 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {4 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{4 \, d} \]

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

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mupad [B]  time = 11.00, size = 222, normalized size = 2.24 \[ \frac {13\,\mathrm {atan}\left (\frac {169}{16\,\left (\frac {169\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}+\frac {13}{2}\right )}-\frac {13\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {169\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}+\frac {13}{2}\right )}\right )}{4\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {-\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {11\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^8/(sin(c + d*x)*(a + a*sin(c + d*x))^3),x)

[Out]

(13*atan(169/(16*((169*tan(c/2 + (d*x)/2))/16 + 13/2)) - (13*tan(c/2 + (d*x)/2))/(2*((169*tan(c/2 + (d*x)/2))/
16 + 13/2))))/(4*a^3*d) + log(tan(c/2 + (d*x)/2))/(a^3*d) - ((11*tan(c/2 + (d*x)/2))/4 - 4*tan(c/2 + (d*x)/2)^
2 + (19*tan(c/2 + (d*x)/2)^3)/4 - (19*tan(c/2 + (d*x)/2)^5)/4 + 4*tan(c/2 + (d*x)/2)^6 - (11*tan(c/2 + (d*x)/2
)^7)/4)/(d*(4*a^3*tan(c/2 + (d*x)/2)^2 + 6*a^3*tan(c/2 + (d*x)/2)^4 + 4*a^3*tan(c/2 + (d*x)/2)^6 + a^3*tan(c/2
+ (d*x)/2)^8 + a^3))

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

[Out]

Timed out

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