3.686 \(\int \frac{\cos ^4(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=97 \[ -\frac{\sin ^3(c+d x)}{3 a d}+\frac{\sin ^2(c+d x)}{2 a d}+\frac{2 \sin (c+d x)}{a d}-\frac{\csc ^2(c+d x)}{2 a d}+\frac{\csc (c+d x)}{a d}-\frac{2 \log (\sin (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.118378, antiderivative size = 97, 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} \[ -\frac{\sin ^3(c+d x)}{3 a d}+\frac{\sin ^2(c+d x)}{2 a d}+\frac{2 \sin (c+d x)}{a d}-\frac{\csc ^2(c+d x)}{2 a d}+\frac{\csc (c+d x)}{a d}-\frac{2 \log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

Csc[c + d*x]/(a*d) - Csc[c + d*x]^2/(2*a*d) - (2*Log[Sin[c + d*x]])/(a*d) + (2*Sin[c + d*x])/(a*d) + Sin[c + d
*x]^2/(2*a*d) - Sin[c + d*x]^3/(3*a*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 \frac{\cos ^4(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a^3 (a-x)^3 (a+x)^2}{x^3} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^3 (a+x)^2}{x^3} \, dx,x,a \sin (c+d x)\right )}{a^4 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (2 a^2+\frac{a^5}{x^3}-\frac{a^4}{x^2}-\frac{2 a^3}{x}+a x-x^2\right ) \, dx,x,a \sin (c+d x)\right )}{a^4 d}\\ &=\frac{\csc (c+d x)}{a d}-\frac{\csc ^2(c+d x)}{2 a d}-\frac{2 \log (\sin (c+d x))}{a d}+\frac{2 \sin (c+d x)}{a d}+\frac{\sin ^2(c+d x)}{2 a d}-\frac{\sin ^3(c+d x)}{3 a d}\\ \end{align*}

Mathematica [A]  time = 0.108717, size = 66, normalized size = 0.68 \[ \frac{-2 \sin ^3(c+d x)+3 \sin ^2(c+d x)+12 \sin (c+d x)-3 \csc ^2(c+d x)+6 \csc (c+d x)-12 \log (\sin (c+d x))}{6 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.128, size = 94, normalized size = 1. \begin{align*} -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,da}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,da}}+2\,{\frac{\sin \left ( dx+c \right ) }{da}}+{\frac{1}{da\sin \left ( dx+c \right ) }}-2\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{da}}-{\frac{1}{2\,da \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)^7*csc(d*x+c)^3/(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.02273, size = 100, normalized size = 1.03 \begin{align*} -\frac{\frac{2 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right )}{a} + \frac{12 \, \log \left (\sin \left (d x + c\right )\right )}{a} - \frac{3 \,{\left (2 \, \sin \left (d x + c\right ) - 1\right )}}{a \sin \left (d x + c\right )^{2}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.15647, size = 244, normalized size = 2.52 \begin{align*} -\frac{6 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} + 24 \,{\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 4 \,{\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 3}{12 \,{\left (a d \cos \left (d x + c\right )^{2} - a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(6*cos(d*x + c)^4 - 9*cos(d*x + c)^2 + 24*(cos(d*x + c)^2 - 1)*log(1/2*sin(d*x + c)) - 4*(cos(d*x + c)^4
+ 4*cos(d*x + c)^2 - 8)*sin(d*x + c) - 3)/(a*d*cos(d*x + c)^2 - a*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)**7*csc(d*x+c)**3/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.33759, size = 127, normalized size = 1.31 \begin{align*} -\frac{\frac{12 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} + \frac{2 \, a^{2} \sin \left (d x + c\right )^{3} - 3 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \sin \left (d x + c\right )}{a^{3}} - \frac{3 \,{\left (6 \, \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) - 1\right )}}{a \sin \left (d x + c\right )^{2}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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