3.664 \(\int \cos ^4(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=115 \[ -\frac{a \sin ^5(c+d x)}{5 d}-\frac{a \sin ^4(c+d x)}{4 d}+\frac{a \sin ^3(c+d x)}{d}+\frac{3 a \sin ^2(c+d x)}{2 d}-\frac{3 a \sin (c+d x)}{d}-\frac{a \csc ^2(c+d x)}{2 d}-\frac{a \csc (c+d x)}{d}-\frac{3 a \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.0869306, antiderivative size = 115, 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} \[ -\frac{a \sin ^5(c+d x)}{5 d}-\frac{a \sin ^4(c+d x)}{4 d}+\frac{a \sin ^3(c+d x)}{d}+\frac{3 a \sin ^2(c+d x)}{2 d}-\frac{3 a \sin (c+d x)}{d}-\frac{a \csc ^2(c+d x)}{2 d}-\frac{a \csc (c+d x)}{d}-\frac{3 a \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.125145, size = 100, normalized size = 0.87 \[ -\frac{a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^3(c+d x)}{d}-\frac{3 a \sin (c+d x)}{d}-\frac{a \csc (c+d x)}{d}-\frac{a \left (\sin ^4(c+d x)-6 \sin ^2(c+d x)+2 \csc ^2(c+d x)+12 \log (\sin (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Csc[c + d*x])/d) - (3*a*Sin[c + d*x])/d + (a*Sin[c + d*x]^3)/d - (a*Sin[c + d*x]^5)/(5*d) - (a*(2*Csc[c +
d*x]^2 + 12*Log[Sin[c + d*x]] - 6*Sin[c + d*x]^2 + Sin[c + d*x]^4))/(4*d)

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Maple [A]  time = 0.061, size = 173, normalized size = 1.5 \begin{align*} -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{d\sin \left ( dx+c \right ) }}-{\frac{16\,a\sin \left ( dx+c \right ) }{5\,d}}-{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}a}{d}}-{\frac{6\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}a}{5\,d}}-{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) a}{5\,d}}-{\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)^3*(a+a*sin(d*x+c)),x)

[Out]

-1/d*a/sin(d*x+c)*cos(d*x+c)^8-16/5*a*sin(d*x+c)/d-1/d*cos(d*x+c)^6*sin(d*x+c)*a-6/5/d*cos(d*x+c)^4*sin(d*x+c)
*a-8/5/d*cos(d*x+c)^2*sin(d*x+c)*a-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 = 1.02327, size = 122, normalized size = 1.06 \begin{align*} -\frac{4 \, a \sin \left (d x + c\right )^{5} + 5 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} - 30 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left (\sin \left (d x + c\right )\right ) + 60 \, a \sin \left (d x + c\right ) + \frac{10 \,{\left (2 \, a \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )^{2}}}{20 \, 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]

-1/20*(4*a*sin(d*x + c)^5 + 5*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3 - 30*a*sin(d*x + c)^2 + 60*a*log(sin(d*x
+ c)) + 60*a*sin(d*x + c) + 10*(2*a*sin(d*x + c) + a)/sin(d*x + c)^2)/d

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Fricas [A]  time = 1.67183, size = 328, normalized size = 2.85 \begin{align*} -\frac{40 \, a \cos \left (d x + c\right )^{6} + 120 \, a \cos \left (d x + c\right )^{4} - 255 \, a \cos \left (d x + c\right )^{2} + 480 \,{\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) + 32 \,{\left (a \cos \left (d x + c\right )^{6} + 2 \, a \cos \left (d x + c\right )^{4} + 8 \, a \cos \left (d x + c\right )^{2} - 16 \, a\right )} \sin \left (d x + c\right ) + 15 \, a}{160 \,{\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)^7*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/160*(40*a*cos(d*x + c)^6 + 120*a*cos(d*x + c)^4 - 255*a*cos(d*x + c)^2 + 480*(a*cos(d*x + c)^2 - a)*log(1/2
*sin(d*x + c)) + 32*(a*cos(d*x + c)^6 + 2*a*cos(d*x + c)^4 + 8*a*cos(d*x + c)^2 - 16*a)*sin(d*x + c) + 15*a)/(
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)**3*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.21013, size = 140, normalized size = 1.22 \begin{align*} -\frac{4 \, a \sin \left (d x + c\right )^{5} + 5 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} - 30 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, a \sin \left (d x + c\right ) - \frac{10 \,{\left (9 \, a \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right ) - a\right )}}{\sin \left (d x + c\right )^{2}}}{20 \, 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]

-1/20*(4*a*sin(d*x + c)^5 + 5*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3 - 30*a*sin(d*x + c)^2 + 60*a*log(abs(sin(
d*x + c))) + 60*a*sin(d*x + c) - 10*(9*a*sin(d*x + c)^2 - 2*a*sin(d*x + c) - a)/sin(d*x + c)^2)/d