3.578 \(\int \cos ^3(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=134 \[ -\frac{5 a \cos ^3(c+d x)}{6 d}-\frac{5 a \cos (c+d x)}{2 d}-\frac{15 a \cot (c+d x)}{8 d}-\frac{a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac{a \cos ^4(c+d x) \cot (c+d x)}{4 d}+\frac{5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac{5 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{15 a x}{8} \]

[Out]

(-15*a*x)/8 + (5*a*ArcTanh[Cos[c + d*x]])/(2*d) - (5*a*Cos[c + d*x])/(2*d) - (5*a*Cos[c + d*x]^3)/(6*d) - (15*
a*Cot[c + d*x])/(8*d) + (5*a*Cos[c + d*x]^2*Cot[c + d*x])/(8*d) + (a*Cos[c + d*x]^4*Cot[c + d*x])/(4*d) - (a*C
os[c + d*x]^3*Cot[c + d*x]^2)/(2*d)

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Rubi [A]  time = 0.143726, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2838, 2592, 288, 302, 206, 2591, 321, 203} \[ -\frac{5 a \cos ^3(c+d x)}{6 d}-\frac{5 a \cos (c+d x)}{2 d}-\frac{15 a \cot (c+d x)}{8 d}-\frac{a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac{a \cos ^4(c+d x) \cot (c+d x)}{4 d}+\frac{5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac{5 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{15 a x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*a*x)/8 + (5*a*ArcTanh[Cos[c + d*x]])/(2*d) - (5*a*Cos[c + d*x])/(2*d) - (5*a*Cos[c + d*x]^3)/(6*d) - (15*
a*Cot[c + d*x])/(8*d) + (5*a*Cos[c + d*x]^2*Cot[c + d*x])/(8*d) + (a*Cos[c + d*x]^4*Cot[c + d*x])/(4*d) - (a*C
os[c + d*x]^3*Cot[c + d*x]^2)/(2*d)

Rule 2838

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

Rubi steps

\begin{align*} \int \cos ^3(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx &=a \int \cos ^4(c+d x) \cot ^2(c+d x) \, dx+a \int \cos ^3(c+d x) \cot ^3(c+d x) \, dx\\ &=-\frac{a \operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{d}-\frac{a \operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^3} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{a \cos ^4(c+d x) \cot (c+d x)}{4 d}-\frac{a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}-\frac{(5 a) \operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{4 d}+\frac{(5 a) \operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 d}\\ &=\frac{5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac{a \cos ^4(c+d x) \cot (c+d x)}{4 d}-\frac{a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}-\frac{(15 a) \operatorname{Subst}\left (\int \frac{x^2}{1+x^2} \, dx,x,\cot (c+d x)\right )}{8 d}+\frac{(5 a) \operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{2 d}\\ &=-\frac{5 a \cos (c+d x)}{2 d}-\frac{5 a \cos ^3(c+d x)}{6 d}-\frac{15 a \cot (c+d x)}{8 d}+\frac{5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac{a \cos ^4(c+d x) \cot (c+d x)}{4 d}-\frac{a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac{(15 a) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{8 d}+\frac{(5 a) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 d}\\ &=-\frac{15 a x}{8}+\frac{5 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{5 a \cos (c+d x)}{2 d}-\frac{5 a \cos ^3(c+d x)}{6 d}-\frac{15 a \cot (c+d x)}{8 d}+\frac{5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac{a \cos ^4(c+d x) \cot (c+d x)}{4 d}-\frac{a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 2.85452, size = 117, normalized size = 0.87 \[ -\frac{a \left (216 \cos (c+d x)+8 \cos (3 (c+d x))+3 \left (16 \sin (2 (c+d x))+\sin (4 (c+d x))+32 \cot (c+d x)+4 \csc ^2\left (\frac{1}{2} (c+d x)\right )-4 \sec ^2\left (\frac{1}{2} (c+d x)\right )+80 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-80 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+60 c+60 d x\right )\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(216*Cos[c + d*x] + 8*Cos[3*(c + d*x)] + 3*(60*c + 60*d*x + 32*Cot[c + d*x] + 4*Csc[(c + d*x)/2]^2 - 80*Lo
g[Cos[(c + d*x)/2]] + 80*Log[Sin[(c + d*x)/2]] - 4*Sec[(c + d*x)/2]^2 + 16*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]
)))/(96*d)

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Maple [A]  time = 0.062, size = 177, normalized size = 1.3 \begin{align*} -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{d\sin \left ( dx+c \right ) }}-{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) }{d}}-{\frac{5\,a \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{4\,d}}-{\frac{15\,\cos \left ( dx+c \right ) a\sin \left ( dx+c \right ) }{8\,d}}-{\frac{15\,ax}{8}}-{\frac{15\,ca}{8\,d}}-{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{2\,d}}-{\frac{5\,a \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{6\,d}}-{\frac{5\,\cos \left ( dx+c \right ) a}{2\,d}}-{\frac{5\,a\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^3*(a+a*sin(d*x+c)),x)

[Out]

-1/d*a/sin(d*x+c)*cos(d*x+c)^7-a*cos(d*x+c)^5*sin(d*x+c)/d-5/4*a*cos(d*x+c)^3*sin(d*x+c)/d-15/8*a*cos(d*x+c)*s
in(d*x+c)/d-15/8*a*x-15/8/d*c*a-1/2/d*a/sin(d*x+c)^2*cos(d*x+c)^7-1/2*a*cos(d*x+c)^5/d-5/6*a*cos(d*x+c)^3/d-5/
2*a*cos(d*x+c)/d-5/2/d*a*ln(csc(d*x+c)-cot(d*x+c))

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Maxima [A]  time = 1.62985, size = 177, normalized size = 1.32 \begin{align*} -\frac{2 \,{\left (4 \, \cos \left (d x + c\right )^{3} - \frac{6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a + 3 \,{\left (15 \, d x + 15 \, c + \frac{15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/24*(2*(4*cos(d*x + c)^3 - 6*cos(d*x + c)/(cos(d*x + c)^2 - 1) + 24*cos(d*x + c) - 15*log(cos(d*x + c) + 1)
+ 15*log(cos(d*x + c) - 1))*a + 3*(15*d*x + 15*c + (15*tan(d*x + c)^4 + 25*tan(d*x + c)^2 + 8)/(tan(d*x + c)^5
+ 2*tan(d*x + c)^3 + tan(d*x + c)))*a)/d

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Fricas [A]  time = 1.20265, size = 435, normalized size = 3.25 \begin{align*} -\frac{8 \, a \cos \left (d x + c\right )^{5} + 45 \, a d x \cos \left (d x + c\right )^{2} + 40 \, a \cos \left (d x + c\right )^{3} - 45 \, a d x - 60 \, a \cos \left (d x + c\right ) - 30 \,{\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 30 \,{\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 3 \,{\left (2 \, a \cos \left (d x + c\right )^{5} + 5 \, a \cos \left (d x + c\right )^{3} - 15 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \,{\left (d \cos \left (d x + c\right )^{2} - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/24*(8*a*cos(d*x + c)^5 + 45*a*d*x*cos(d*x + c)^2 + 40*a*cos(d*x + c)^3 - 45*a*d*x - 60*a*cos(d*x + c) - 30*
(a*cos(d*x + c)^2 - a)*log(1/2*cos(d*x + c) + 1/2) + 30*(a*cos(d*x + c)^2 - a)*log(-1/2*cos(d*x + c) + 1/2) +
3*(2*a*cos(d*x + c)^5 + 5*a*cos(d*x + c)^3 - 15*a*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 - d)

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

[Out]

Timed out

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Giac [A]  time = 1.36379, size = 289, normalized size = 2.16 \begin{align*} \frac{3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 45 \,{\left (d x + c\right )} a - 60 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 12 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{3 \,{\left (30 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}} + \frac{2 \,{\left (27 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 72 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 168 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 152 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 27 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 56 \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/24*(3*a*tan(1/2*d*x + 1/2*c)^2 - 45*(d*x + c)*a - 60*a*log(abs(tan(1/2*d*x + 1/2*c))) + 12*a*tan(1/2*d*x + 1
/2*c) + 3*(30*a*tan(1/2*d*x + 1/2*c)^2 - 4*a*tan(1/2*d*x + 1/2*c) - a)/tan(1/2*d*x + 1/2*c)^2 + 2*(27*a*tan(1/
2*d*x + 1/2*c)^7 - 72*a*tan(1/2*d*x + 1/2*c)^6 + 3*a*tan(1/2*d*x + 1/2*c)^5 - 168*a*tan(1/2*d*x + 1/2*c)^4 - 3
*a*tan(1/2*d*x + 1/2*c)^3 - 152*a*tan(1/2*d*x + 1/2*c)^2 - 27*a*tan(1/2*d*x + 1/2*c) - 56*a)/(tan(1/2*d*x + 1/
2*c)^2 + 1)^4)/d