3.517 \(\int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=110 \[ \frac{a^2 \sin ^3(c+d x)}{3 d}+\frac{a^2 \sin ^2(c+d x)}{d}-\frac{a^2 \sin (c+d x)}{d}-\frac{a^2 \csc ^3(c+d x)}{3 d}-\frac{a^2 \csc ^2(c+d x)}{d}+\frac{a^2 \csc (c+d x)}{d}-\frac{4 a^2 \log (\sin (c+d x))}{d} \]

[Out]

(a^2*Csc[c + d*x])/d - (a^2*Csc[c + d*x]^2)/d - (a^2*Csc[c + d*x]^3)/(3*d) - (4*a^2*Log[Sin[c + d*x]])/d - (a^
2*Sin[c + d*x])/d + (a^2*Sin[c + d*x]^2)/d + (a^2*Sin[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.101402, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2836, 12, 88} \[ \frac{a^2 \sin ^3(c+d x)}{3 d}+\frac{a^2 \sin ^2(c+d x)}{d}-\frac{a^2 \sin (c+d x)}{d}-\frac{a^2 \csc ^3(c+d x)}{3 d}-\frac{a^2 \csc ^2(c+d x)}{d}+\frac{a^2 \csc (c+d x)}{d}-\frac{4 a^2 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*Cot[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*Csc[c + d*x])/d - (a^2*Csc[c + d*x]^2)/d - (a^2*Csc[c + d*x]^3)/(3*d) - (4*a^2*Log[Sin[c + d*x]])/d - (a^
2*Sin[c + d*x])/d + (a^2*Sin[c + d*x]^2)/d + (a^2*Sin[c + d*x]^3)/(3*d)

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

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

Mathematica [A]  time = 0.176152, size = 74, normalized size = 0.67 \[ \frac{a^2 \left (\sin ^3(c+d x)+3 \sin ^2(c+d x)-3 \sin (c+d x)-\csc ^3(c+d x)-3 \csc ^2(c+d x)+3 \csc (c+d x)-12 \log (\sin (c+d x))\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*Cot[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(3*Csc[c + d*x] - 3*Csc[c + d*x]^2 - Csc[c + d*x]^3 - 12*Log[Sin[c + d*x]] - 3*Sin[c + d*x] + 3*Sin[c + d
*x]^2 + Sin[c + d*x]^3))/(3*d)

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Maple [A]  time = 0.08, size = 97, normalized size = 0.9 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}{a}^{2}}{d}}-2\,{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{d}}-4\,{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x)

[Out]

-1/d*a^2/sin(d*x+c)^2*cos(d*x+c)^6-1/d*cos(d*x+c)^4*a^2-2/d*a^2*cos(d*x+c)^2-4*a^2*ln(sin(d*x+c))/d-1/3/d*a^2/
sin(d*x+c)^3*cos(d*x+c)^6

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Maxima [A]  time = 1.09278, size = 126, normalized size = 1.15 \begin{align*} \frac{a^{2} \sin \left (d x + c\right )^{3} + 3 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - 3 \, a^{2} \sin \left (d x + c\right ) + \frac{3 \, a^{2} \sin \left (d x + c\right )^{2} - 3 \, a^{2} \sin \left (d x + c\right ) - a^{2}}{\sin \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/3*(a^2*sin(d*x + c)^3 + 3*a^2*sin(d*x + c)^2 - 12*a^2*log(sin(d*x + c)) - 3*a^2*sin(d*x + c) + (3*a^2*sin(d*
x + c)^2 - 3*a^2*sin(d*x + c) - a^2)/sin(d*x + c)^3)/d

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Fricas [A]  time = 1.4313, size = 273, normalized size = 2.48 \begin{align*} \frac{2 \, a^{2} \cos \left (d x + c\right )^{6} - 24 \,{\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 3 \,{\left (2 \, a^{2} \cos \left (d x + c\right )^{4} - 3 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sin \left (d x + c\right )}{6 \,{\left (d \cos \left (d x + c\right )^{2} - 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)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(2*a^2*cos(d*x + c)^6 - 24*(a^2*cos(d*x + c)^2 - a^2)*log(1/2*sin(d*x + c))*sin(d*x + c) - 3*(2*a^2*cos(d*
x + c)^4 - 3*a^2*cos(d*x + c)^2 - a^2)*sin(d*x + c))/((d*cos(d*x + c)^2 - 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)**5*csc(d*x+c)**4*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.2978, size = 144, normalized size = 1.31 \begin{align*} \frac{a^{2} \sin \left (d x + c\right )^{3} + 3 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 3 \, a^{2} \sin \left (d x + c\right ) + \frac{22 \, a^{2} \sin \left (d x + c\right )^{3} + 3 \, a^{2} \sin \left (d x + c\right )^{2} - 3 \, a^{2} \sin \left (d x + c\right ) - a^{2}}{\sin \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/3*(a^2*sin(d*x + c)^3 + 3*a^2*sin(d*x + c)^2 - 12*a^2*log(abs(sin(d*x + c))) - 3*a^2*sin(d*x + c) + (22*a^2*
sin(d*x + c)^3 + 3*a^2*sin(d*x + c)^2 - 3*a^2*sin(d*x + c) - a^2)/sin(d*x + c)^3)/d