3.370 \(\int \cos ^2(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=83 \[ \frac{a \cos ^3(c+d x)}{3 d}+\frac{a \cos (c+d x)}{d}-\frac{3 a \cot (c+d x)}{2 d}+\frac{a \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac{a \tanh ^{-1}(\cos (c+d x))}{d}-\frac{3 a x}{2} \]

[Out]

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

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Rubi [A]  time = 0.113601, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2838, 2591, 288, 321, 203, 2592, 302, 206} \[ \frac{a \cos ^3(c+d x)}{3 d}+\frac{a \cos (c+d x)}{d}-\frac{3 a \cot (c+d x)}{2 d}+\frac{a \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac{a \tanh ^{-1}(\cos (c+d x))}{d}-\frac{3 a x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.414031, size = 77, normalized size = 0.93 \[ \frac{a \left (15 \cos (c+d x)+\cos (3 (c+d x))-3 \left (\sin (2 (c+d x))+4 \cot (c+d x)-4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+6 c+6 d x\right )\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(15*Cos[c + d*x] + Cos[3*(c + d*x)] - 3*(6*c + 6*d*x + 4*Cot[c + d*x] + 4*Log[Cos[(c + d*x)/2]] - 4*Log[Sin
[(c + d*x)/2]] + Sin[2*(c + d*x)])))/(12*d)

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Maple [A]  time = 0.053, size = 119, normalized size = 1.4 \begin{align*}{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{\cos \left ( dx+c \right ) a}{d}}+{\frac{a\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}-{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{d\sin \left ( dx+c \right ) }}-{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{d}}-{\frac{3\,\cos \left ( dx+c \right ) a\sin \left ( dx+c \right ) }{2\,d}}-{\frac{3\,ax}{2}}-{\frac{3\,ca}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^2*(a+a*sin(d*x+c)),x)

[Out]

1/3*a*cos(d*x+c)^3/d+a*cos(d*x+c)/d+1/d*a*ln(csc(d*x+c)-cot(d*x+c))-1/d*a/sin(d*x+c)*cos(d*x+c)^5-a*cos(d*x+c)
^3*sin(d*x+c)/d-3/2*a*cos(d*x+c)*sin(d*x+c)/d-3/2*a*x-3/2/d*c*a

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Maxima [A]  time = 1.57359, size = 122, normalized size = 1.47 \begin{align*} \frac{{\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a - 3 \,{\left (3 \, d x + 3 \, c + \frac{3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.66776, size = 300, normalized size = 3.61 \begin{align*} \frac{3 \, a \cos \left (d x + c\right )^{3} - 3 \, a \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 3 \, a \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 9 \, a \cos \left (d x + c\right ) +{\left (2 \, a \cos \left (d x + c\right )^{3} - 9 \, a d x + 6 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, 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)^4*csc(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(3*a*cos(d*x + c)^3 - 3*a*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 3*a*log(-1/2*cos(d*x + c) + 1/2)*sin(
d*x + c) - 9*a*cos(d*x + c) + (2*a*cos(d*x + c)^3 - 9*a*d*x + 6*a*cos(d*x + c))*sin(d*x + c))/(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)**4*csc(d*x+c)**2*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.38823, size = 192, normalized size = 2.31 \begin{align*} -\frac{9 \,{\left (d x + c\right )} a - 6 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{3 \,{\left (2 \, 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 )} - \frac{2 \,{\left (3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 12 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8 \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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