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

Optimal. Leaf size=119 \[ \frac{a^2 \sin ^6(c+d x)}{6 d}+\frac{2 a^2 \sin ^5(c+d x)}{5 d}-\frac{a^2 \sin ^4(c+d x)}{4 d}-\frac{4 a^2 \sin ^3(c+d x)}{3 d}-\frac{a^2 \sin ^2(c+d x)}{2 d}+\frac{2 a^2 \sin (c+d x)}{d}+\frac{a^2 \log (\sin (c+d x))}{d} \]

[Out]

(a^2*Log[Sin[c + d*x]])/d + (2*a^2*Sin[c + d*x])/d - (a^2*Sin[c + d*x]^2)/(2*d) - (4*a^2*Sin[c + d*x]^3)/(3*d)
- (a^2*Sin[c + d*x]^4)/(4*d) + (2*a^2*Sin[c + d*x]^5)/(5*d) + (a^2*Sin[c + d*x]^6)/(6*d)

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Rubi [A]  time = 0.100646, antiderivative size = 119, 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} \[ \frac{a^2 \sin ^6(c+d x)}{6 d}+\frac{2 a^2 \sin ^5(c+d x)}{5 d}-\frac{a^2 \sin ^4(c+d x)}{4 d}-\frac{4 a^2 \sin ^3(c+d x)}{3 d}-\frac{a^2 \sin ^2(c+d x)}{2 d}+\frac{2 a^2 \sin (c+d x)}{d}+\frac{a^2 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0839573, size = 78, normalized size = 0.66 \[ \frac{a^2 \left (10 \sin ^6(c+d x)+24 \sin ^5(c+d x)-15 \sin ^4(c+d x)-80 \sin ^3(c+d x)-30 \sin ^2(c+d x)+120 \sin (c+d x)+60 \log (\sin (c+d x))\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(60*Log[Sin[c + d*x]] + 120*Sin[c + d*x] - 30*Sin[c + d*x]^2 - 80*Sin[c + d*x]^3 - 15*Sin[c + d*x]^4 + 24
*Sin[c + d*x]^5 + 10*Sin[c + d*x]^6))/(60*d)

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Maple [A]  time = 0.07, size = 122, normalized size = 1. \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{6\,d}}+{\frac{16\,{a}^{2}\sin \left ( dx+c \right ) }{15\,d}}+{\frac{2\,{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{8\,{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}}{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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d*a^2*cos(d*x+c)^6+16/15*a^2*sin(d*x+c)/d+2/5/d*sin(d*x+c)*a^2*cos(d*x+c)^4+8/15/d*sin(d*x+c)*a^2*cos(d*x
+c)^2+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

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Maxima [A]  time = 1.05009, size = 127, normalized size = 1.07 \begin{align*} \frac{10 \, a^{2} \sin \left (d x + c\right )^{6} + 24 \, a^{2} \sin \left (d x + c\right )^{5} - 15 \, a^{2} \sin \left (d x + c\right )^{4} - 80 \, a^{2} \sin \left (d x + c\right )^{3} - 30 \, a^{2} \sin \left (d x + c\right )^{2} + 60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 120 \, a^{2} \sin \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(10*a^2*sin(d*x + c)^6 + 24*a^2*sin(d*x + c)^5 - 15*a^2*sin(d*x + c)^4 - 80*a^2*sin(d*x + c)^3 - 30*a^2*s
in(d*x + c)^2 + 60*a^2*log(sin(d*x + c)) + 120*a^2*sin(d*x + c))/d

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Fricas [A]  time = 1.16847, size = 247, normalized size = 2.08 \begin{align*} -\frac{10 \, a^{2} \cos \left (d x + c\right )^{6} - 15 \, a^{2} \cos \left (d x + c\right )^{4} - 30 \, a^{2} \cos \left (d x + c\right )^{2} - 60 \, a^{2} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 8 \,{\left (3 \, 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 )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(10*a^2*cos(d*x + c)^6 - 15*a^2*cos(d*x + c)^4 - 30*a^2*cos(d*x + c)^2 - 60*a^2*log(1/2*sin(d*x + c)) -
8*(3*a^2*cos(d*x + c)^4 + 4*a^2*cos(d*x + c)^2 + 8*a^2)*sin(d*x + c))/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)*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.30226, size = 128, normalized size = 1.08 \begin{align*} \frac{10 \, a^{2} \sin \left (d x + c\right )^{6} + 24 \, a^{2} \sin \left (d x + c\right )^{5} - 15 \, a^{2} \sin \left (d x + c\right )^{4} - 80 \, a^{2} \sin \left (d x + c\right )^{3} - 30 \, a^{2} \sin \left (d x + c\right )^{2} + 60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 120 \, a^{2} \sin \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(10*a^2*sin(d*x + c)^6 + 24*a^2*sin(d*x + c)^5 - 15*a^2*sin(d*x + c)^4 - 80*a^2*sin(d*x + c)^3 - 30*a^2*s
in(d*x + c)^2 + 60*a^2*log(abs(sin(d*x + c))) + 120*a^2*sin(d*x + c))/d