3.516 \(\int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=116 \[ \frac{a^2 \sin ^4(c+d x)}{4 d}+\frac{2 a^2 \sin ^3(c+d x)}{3 d}-\frac{a^2 \sin ^2(c+d x)}{2 d}-\frac{4 a^2 \sin (c+d x)}{d}-\frac{a^2 \csc ^2(c+d x)}{2 d}-\frac{2 a^2 \csc (c+d x)}{d}-\frac{a^2 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.117751, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2836, 12, 88} \[ \frac{a^2 \sin ^4(c+d x)}{4 d}+\frac{2 a^2 \sin ^3(c+d x)}{3 d}-\frac{a^2 \sin ^2(c+d x)}{2 d}-\frac{4 a^2 \sin (c+d x)}{d}-\frac{a^2 \csc ^2(c+d x)}{2 d}-\frac{2 a^2 \csc (c+d x)}{d}-\frac{a^2 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Cot[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

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

Mathematica [A]  time = 0.167488, size = 76, normalized size = 0.66 \[ -\frac{a^2 \left (-3 \sin ^4(c+d x)-8 \sin ^3(c+d x)+6 \sin ^2(c+d x)+48 \sin (c+d x)+6 \csc ^2(c+d x)+24 \csc (c+d x)+12 \log (\sin (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Cot[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

-(a^2*(24*Csc[c + d*x] + 6*Csc[c + d*x]^2 + 12*Log[Sin[c + d*x]] + 48*Sin[c + d*x] + 6*Sin[c + d*x]^2 - 8*Sin[
c + d*x]^3 - 3*Sin[c + d*x]^4))/(12*d)

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Maple [A]  time = 0.079, size = 155, normalized size = 1.3 \begin{align*} -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}{a}^{2}}{4\,d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }}-{\frac{16\,{a}^{2}\sin \left ( dx+c \right ) }{3\,d}}-2\,{\frac{{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{d}}-{\frac{8\,{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x)

[Out]

-1/4/d*cos(d*x+c)^4*a^2-1/2/d*a^2*cos(d*x+c)^2-a^2*ln(sin(d*x+c))/d-2/d*a^2/sin(d*x+c)*cos(d*x+c)^6-16/3*a^2*s
in(d*x+c)/d-2/d*sin(d*x+c)*a^2*cos(d*x+c)^4-8/3/d*sin(d*x+c)*a^2*cos(d*x+c)^2-1/2/d*a^2/sin(d*x+c)^2*cos(d*x+c
)^6

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Maxima [A]  time = 1.06963, size = 126, normalized size = 1.09 \begin{align*} \frac{3 \, a^{2} \sin \left (d x + c\right )^{4} + 8 \, a^{2} \sin \left (d x + c\right )^{3} - 6 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - 48 \, a^{2} \sin \left (d x + c\right ) - \frac{6 \,{\left (4 \, a^{2} \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )^{2}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(3*a^2*sin(d*x + c)^4 + 8*a^2*sin(d*x + c)^3 - 6*a^2*sin(d*x + c)^2 - 12*a^2*log(sin(d*x + c)) - 48*a^2*s
in(d*x + c) - 6*(4*a^2*sin(d*x + c) + a^2)/sin(d*x + c)^2)/d

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Fricas [A]  time = 1.31846, size = 315, normalized size = 2.72 \begin{align*} \frac{24 \, a^{2} \cos \left (d x + c\right )^{6} - 24 \, a^{2} \cos \left (d x + c\right )^{4} - 9 \, a^{2} \cos \left (d x + c\right )^{2} + 57 \, a^{2} - 96 \,{\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 ) - 64 \,{\left (a^{2} \cos \left (d x + c\right )^{4} + 4 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a^{2}\right )} \sin \left (d x + c\right )}{96 \,{\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)^5*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/96*(24*a^2*cos(d*x + c)^6 - 24*a^2*cos(d*x + c)^4 - 9*a^2*cos(d*x + c)^2 + 57*a^2 - 96*(a^2*cos(d*x + c)^2 -
a^2)*log(1/2*sin(d*x + c)) - 64*(a^2*cos(d*x + c)^4 + 4*a^2*cos(d*x + c)^2 - 8*a^2)*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)**5*csc(d*x+c)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.28033, size = 147, normalized size = 1.27 \begin{align*} \frac{3 \, a^{2} \sin \left (d x + c\right )^{4} + 8 \, a^{2} \sin \left (d x + c\right )^{3} - 6 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 48 \, a^{2} \sin \left (d x + c\right ) + \frac{6 \,{\left (3 \, a^{2} \sin \left (d x + c\right )^{2} - 4 \, a^{2} \sin \left (d x + c\right ) - a^{2}\right )}}{\sin \left (d x + c\right )^{2}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/12*(3*a^2*sin(d*x + c)^4 + 8*a^2*sin(d*x + c)^3 - 6*a^2*sin(d*x + c)^2 - 12*a^2*log(abs(sin(d*x + c))) - 48*
a^2*sin(d*x + c) + 6*(3*a^2*sin(d*x + c)^2 - 4*a^2*sin(d*x + c) - a^2)/sin(d*x + c)^2)/d