3.666 \(\int \cos ^2(c+d x) \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=118 \[ -\frac{a \sin ^3(c+d x)}{3 d}-\frac{a \sin ^2(c+d x)}{2 d}+\frac{3 a \sin (c+d x)}{d}-\frac{a \csc ^4(c+d x)}{4 d}-\frac{a \csc ^3(c+d x)}{3 d}+\frac{3 a \csc ^2(c+d x)}{2 d}+\frac{3 a \csc (c+d x)}{d}+\frac{3 a \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.0894501, antiderivative size = 118, 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} \[ -\frac{a \sin ^3(c+d x)}{3 d}-\frac{a \sin ^2(c+d x)}{2 d}+\frac{3 a \sin (c+d x)}{d}-\frac{a \csc ^4(c+d x)}{4 d}-\frac{a \csc ^3(c+d x)}{3 d}+\frac{3 a \csc ^2(c+d x)}{2 d}+\frac{3 a \csc (c+d x)}{d}+\frac{3 a \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Cot[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

(3*a*Csc[c + d*x])/d + (3*a*Csc[c + d*x]^2)/(2*d) - (a*Csc[c + d*x]^3)/(3*d) - (a*Csc[c + d*x]^4)/(4*d) + (3*a
*Log[Sin[c + d*x]])/d + (3*a*Sin[c + d*x])/d - (a*Sin[c + d*x]^2)/(2*d) - (a*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 ^2(c+d x) \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a^5 (a-x)^3 (a+x)^4}{x^5} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^3 (a+x)^4}{x^5} \, dx,x,a \sin (c+d x)\right )}{a^2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (3 a^2+\frac{a^7}{x^5}+\frac{a^6}{x^4}-\frac{3 a^5}{x^3}-\frac{3 a^4}{x^2}+\frac{3 a^3}{x}-a x-x^2\right ) \, dx,x,a \sin (c+d x)\right )}{a^2 d}\\ &=\frac{3 a \csc (c+d x)}{d}+\frac{3 a \csc ^2(c+d x)}{2 d}-\frac{a \csc ^3(c+d x)}{3 d}-\frac{a \csc ^4(c+d x)}{4 d}+\frac{3 a \log (\sin (c+d x))}{d}+\frac{3 a \sin (c+d x)}{d}-\frac{a \sin ^2(c+d x)}{2 d}-\frac{a \sin ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.572343, size = 105, normalized size = 0.89 \[ -\frac{a \sin ^3(c+d x)}{3 d}+\frac{3 a \sin (c+d x)}{d}-\frac{a \csc ^3(c+d x)}{3 d}+\frac{3 a \csc (c+d x)}{d}+\frac{a \left (-2 \sin ^2(c+d x)-\csc ^4(c+d x)+6 \csc ^2(c+d x)+12 \log (\sin (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Cot[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.063, size = 217, normalized size = 1.8 \begin{align*} -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{5\,a \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{3\,d\sin \left ( dx+c \right ) }}+{\frac{16\,a\sin \left ( dx+c \right ) }{3\,d}}+{\frac{5\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}a}{3\,d}}+2\,{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}a}{d}}+{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) a}{3\,d}}-{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{2\,d}}+{\frac{3\,a \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{3\,a \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+3\,{\frac{a\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)^7*csc(d*x+c)^5*(a+a*sin(d*x+c)),x)

[Out]

-1/3/d*a/sin(d*x+c)^3*cos(d*x+c)^8+5/3/d*a/sin(d*x+c)*cos(d*x+c)^8+16/3*a*sin(d*x+c)/d+5/3/d*cos(d*x+c)^6*sin(
d*x+c)*a+2/d*cos(d*x+c)^4*sin(d*x+c)*a+8/3/d*cos(d*x+c)^2*sin(d*x+c)*a-1/4/d*a/sin(d*x+c)^4*cos(d*x+c)^8+1/2/d
*a/sin(d*x+c)^2*cos(d*x+c)^8+1/2*a*cos(d*x+c)^6/d+3/4*a*cos(d*x+c)^4/d+3/2*a*cos(d*x+c)^2/d+3*a*ln(sin(d*x+c))
/d

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Maxima [A]  time = 1.0378, size = 124, normalized size = 1.05 \begin{align*} -\frac{4 \, a \sin \left (d x + c\right )^{3} + 6 \, a \sin \left (d x + c\right )^{2} - 36 \, a \log \left (\sin \left (d x + c\right )\right ) - 36 \, a \sin \left (d x + c\right ) - \frac{36 \, a \sin \left (d x + c\right )^{3} + 18 \, a \sin \left (d x + c\right )^{2} - 4 \, a \sin \left (d x + c\right ) - 3 \, a}{\sin \left (d x + c\right )^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(4*a*sin(d*x + c)^3 + 6*a*sin(d*x + c)^2 - 36*a*log(sin(d*x + c)) - 36*a*sin(d*x + c) - (36*a*sin(d*x +
c)^3 + 18*a*sin(d*x + c)^2 - 4*a*sin(d*x + c) - 3*a)/sin(d*x + c)^4)/d

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Fricas [A]  time = 1.6996, size = 375, normalized size = 3.18 \begin{align*} \frac{6 \, a \cos \left (d x + c\right )^{6} - 15 \, a \cos \left (d x + c\right )^{4} - 6 \, a \cos \left (d x + c\right )^{2} + 36 \,{\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) + 4 \,{\left (a \cos \left (d x + c\right )^{6} + 6 \, a \cos \left (d x + c\right )^{4} - 24 \, a \cos \left (d x + c\right )^{2} + 16 \, a\right )} \sin \left (d x + c\right ) + 12 \, a}{12 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, 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)^7*csc(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(6*a*cos(d*x + c)^6 - 15*a*cos(d*x + c)^4 - 6*a*cos(d*x + c)^2 + 36*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^
2 + a)*log(1/2*sin(d*x + c)) + 4*(a*cos(d*x + c)^6 + 6*a*cos(d*x + c)^4 - 24*a*cos(d*x + c)^2 + 16*a)*sin(d*x
+ c) + 12*a)/(d*cos(d*x + c)^4 - 2*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)**7*csc(d*x+c)**5*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.18705, size = 139, normalized size = 1.18 \begin{align*} -\frac{4 \, a \sin \left (d x + c\right )^{3} + 6 \, a \sin \left (d x + c\right )^{2} - 36 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 36 \, a \sin \left (d x + c\right ) + \frac{75 \, a \sin \left (d x + c\right )^{4} - 36 \, a \sin \left (d x + c\right )^{3} - 18 \, a \sin \left (d x + c\right )^{2} + 4 \, a \sin \left (d x + c\right ) + 3 \, a}{\sin \left (d x + c\right )^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/12*(4*a*sin(d*x + c)^3 + 6*a*sin(d*x + c)^2 - 36*a*log(abs(sin(d*x + c))) - 36*a*sin(d*x + c) + (75*a*sin(d
*x + c)^4 - 36*a*sin(d*x + c)^3 - 18*a*sin(d*x + c)^2 + 4*a*sin(d*x + c) + 3*a)/sin(d*x + c)^4)/d