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

Optimal. Leaf size=114 \[ \frac{a^2 \sin ^5(c+d x)}{5 d}+\frac{a^2 \sin ^4(c+d x)}{2 d}-\frac{a^2 \sin ^3(c+d x)}{3 d}-\frac{2 a^2 \sin ^2(c+d x)}{d}-\frac{a^2 \sin (c+d x)}{d}-\frac{a^2 \csc (c+d x)}{d}+\frac{2 a^2 \log (\sin (c+d x))}{d} \]

[Out]

-((a^2*Csc[c + d*x])/d) + (2*a^2*Log[Sin[c + d*x]])/d - (a^2*Sin[c + d*x])/d - (2*a^2*Sin[c + d*x]^2)/d - (a^2
*Sin[c + d*x]^3)/(3*d) + (a^2*Sin[c + d*x]^4)/(2*d) + (a^2*Sin[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.118494, antiderivative size = 114, 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} \[ \frac{a^2 \sin ^5(c+d x)}{5 d}+\frac{a^2 \sin ^4(c+d x)}{2 d}-\frac{a^2 \sin ^3(c+d x)}{3 d}-\frac{2 a^2 \sin ^2(c+d x)}{d}-\frac{a^2 \sin (c+d x)}{d}-\frac{a^2 \csc (c+d x)}{d}+\frac{2 a^2 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0546058, size = 114, normalized size = 1. \[ \frac{a^2 \sin ^5(c+d x)}{5 d}+\frac{a^2 \sin ^4(c+d x)}{2 d}-\frac{a^2 \sin ^3(c+d x)}{3 d}-\frac{2 a^2 \sin ^2(c+d x)}{d}-\frac{a^2 \sin (c+d x)}{d}-\frac{a^2 \csc (c+d x)}{d}+\frac{2 a^2 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*Csc[c + d*x])/d) + (2*a^2*Log[Sin[c + d*x]])/d - (a^2*Sin[c + d*x])/d - (2*a^2*Sin[c + d*x]^2)/d - (a^2
*Sin[c + d*x]^3)/(3*d) + (a^2*Sin[c + d*x]^4)/(2*d) + (a^2*Sin[c + d*x]^5)/(5*d)

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Maple [A]  time = 0.069, size = 130, normalized size = 1.1 \begin{align*} -{\frac{32\,{a}^{2}\sin \left ( dx+c \right ) }{15\,d}}-{\frac{4\,{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{16\,{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}{a}^{2}}{2\,d}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{d}}+2\,{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-32/15*a^2*sin(d*x+c)/d-4/5/d*sin(d*x+c)*a^2*cos(d*x+c)^4-16/15/d*sin(d*x+c)*a^2*cos(d*x+c)^2+1/2/d*cos(d*x+c)
^4*a^2+1/d*a^2*cos(d*x+c)^2+2*a^2*ln(sin(d*x+c))/d-1/d*a^2/sin(d*x+c)*cos(d*x+c)^6

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Maxima [A]  time = 1.12375, size = 127, normalized size = 1.11 \begin{align*} \frac{6 \, a^{2} \sin \left (d x + c\right )^{5} + 15 \, a^{2} \sin \left (d x + c\right )^{4} - 10 \, a^{2} \sin \left (d x + c\right )^{3} - 60 \, a^{2} \sin \left (d x + c\right )^{2} + 60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - 30 \, a^{2} \sin \left (d x + c\right ) - \frac{30 \, a^{2}}{\sin \left (d x + c\right )}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(6*a^2*sin(d*x + c)^5 + 15*a^2*sin(d*x + c)^4 - 10*a^2*sin(d*x + c)^3 - 60*a^2*sin(d*x + c)^2 + 60*a^2*lo
g(sin(d*x + c)) - 30*a^2*sin(d*x + c) - 30*a^2/sin(d*x + c))/d

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Fricas [A]  time = 1.16981, size = 306, normalized size = 2.68 \begin{align*} -\frac{48 \, a^{2} \cos \left (d x + c\right )^{6} - 64 \, a^{2} \cos \left (d x + c\right )^{4} - 256 \, a^{2} \cos \left (d x + c\right )^{2} - 480 \, a^{2} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 512 \, a^{2} - 15 \,{\left (8 \, a^{2} \cos \left (d x + c\right )^{4} + 16 \, a^{2} \cos \left (d x + c\right )^{2} - 11 \, a^{2}\right )} \sin \left (d x + c\right )}{240 \, 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)^5*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/240*(48*a^2*cos(d*x + c)^6 - 64*a^2*cos(d*x + c)^4 - 256*a^2*cos(d*x + c)^2 - 480*a^2*log(1/2*sin(d*x + c))
*sin(d*x + c) + 512*a^2 - 15*(8*a^2*cos(d*x + c)^4 + 16*a^2*cos(d*x + c)^2 - 11*a^2)*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)**5*csc(d*x+c)**2*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.36175, size = 144, normalized size = 1.26 \begin{align*} \frac{6 \, a^{2} \sin \left (d x + c\right )^{5} + 15 \, a^{2} \sin \left (d x + c\right )^{4} - 10 \, a^{2} \sin \left (d x + c\right )^{3} - 60 \, a^{2} \sin \left (d x + c\right )^{2} + 60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 30 \, a^{2} \sin \left (d x + c\right ) - \frac{30 \,{\left (2 \, a^{2} \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(6*a^2*sin(d*x + c)^5 + 15*a^2*sin(d*x + c)^4 - 10*a^2*sin(d*x + c)^3 - 60*a^2*sin(d*x + c)^2 + 60*a^2*lo
g(abs(sin(d*x + c))) - 30*a^2*sin(d*x + c) - 30*(2*a^2*sin(d*x + c) + a^2)/sin(d*x + c))/d