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

Optimal. Leaf size=112 \[ \frac{a^2 \sin (c+d x)}{d}-\frac{a^2 \csc ^5(c+d x)}{5 d}-\frac{a^2 \csc ^4(c+d x)}{2 d}+\frac{a^2 \csc ^3(c+d x)}{3 d}+\frac{2 a^2 \csc ^2(c+d x)}{d}+\frac{a^2 \csc (c+d x)}{d}+\frac{2 a^2 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.101829, antiderivative size = 112, 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} \[ \frac{a^2 \sin (c+d x)}{d}-\frac{a^2 \csc ^5(c+d x)}{5 d}-\frac{a^2 \csc ^4(c+d x)}{2 d}+\frac{a^2 \csc ^3(c+d x)}{3 d}+\frac{2 a^2 \csc ^2(c+d x)}{d}+\frac{a^2 \csc (c+d x)}{d}+\frac{2 a^2 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Csc[c + d*x])/d + (2*a^2*Csc[c + d*x]^2)/d + (a^2*Csc[c + d*x]^3)/(3*d) - (a^2*Csc[c + d*x]^4)/(2*d) - (a
^2*Csc[c + d*x]^5)/(5*d) + (2*a^2*Log[Sin[c + d*x]])/d + (a^2*Sin[c + d*x])/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 \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a^6 (a-x)^2 (a+x)^4}{x^6} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{a \operatorname{Subst}\left (\int \frac{(a-x)^2 (a+x)^4}{x^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a \operatorname{Subst}\left (\int \left (1+\frac{a^6}{x^6}+\frac{2 a^5}{x^5}-\frac{a^4}{x^4}-\frac{4 a^3}{x^3}-\frac{a^2}{x^2}+\frac{2 a}{x}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^2 \csc (c+d x)}{d}+\frac{2 a^2 \csc ^2(c+d x)}{d}+\frac{a^2 \csc ^3(c+d x)}{3 d}-\frac{a^2 \csc ^4(c+d x)}{2 d}-\frac{a^2 \csc ^5(c+d x)}{5 d}+\frac{2 a^2 \log (\sin (c+d x))}{d}+\frac{a^2 \sin (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.127337, size = 76, normalized size = 0.68 \[ \frac{a^2 \left (30 \sin (c+d x)-6 \csc ^5(c+d x)-15 \csc ^4(c+d x)+10 \csc ^3(c+d x)+60 \csc ^2(c+d x)+30 \csc (c+d x)+60 \log (\sin (c+d x))\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.081, size = 178, normalized size = 1.6 \begin{align*} -{\frac{4\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{15\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{4\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{5\,d\sin \left ( dx+c \right ) }}+{\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{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{2\,d}}+{\frac{{a}^{2} \left ( \cot \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}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x)

[Out]

-4/15/d*a^2/sin(d*x+c)^3*cos(d*x+c)^6+4/5/d*a^2/sin(d*x+c)*cos(d*x+c)^6+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*a^2*cot(d*x+c)^4+1/d*a^2*cot(d*x+c)^2+2*a^2*ln(si
n(d*x+c))/d-1/5/d*a^2/sin(d*x+c)^5*cos(d*x+c)^6

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Maxima [A]  time = 1.13529, size = 127, normalized size = 1.13 \begin{align*} \frac{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 )^{4} + 60 \, a^{2} \sin \left (d x + c\right )^{3} + 10 \, a^{2} \sin \left (d x + c\right )^{2} - 15 \, a^{2} \sin \left (d x + c\right ) - 6 \, a^{2}}{\sin \left (d x + c\right )^{5}}}{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)^6*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.53737, size = 389, normalized size = 3.47 \begin{align*} -\frac{30 \, a^{2} \cos \left (d x + c\right )^{6} - 120 \, a^{2} \cos \left (d x + c\right )^{4} + 160 \, a^{2} \cos \left (d x + c\right )^{2} - 60 \,{\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 64 \, a^{2} + 15 \,{\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2}\right )} \sin \left (d x + c\right )}{30 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \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)^6*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/30*(30*a^2*cos(d*x + c)^6 - 120*a^2*cos(d*x + c)^4 + 160*a^2*cos(d*x + c)^2 - 60*(a^2*cos(d*x + c)^4 - 2*a^
2*cos(d*x + c)^2 + a^2)*log(1/2*sin(d*x + c))*sin(d*x + c) - 64*a^2 + 15*(4*a^2*cos(d*x + c)^2 - 3*a^2)*sin(d*
x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + 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)**6*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.33154, size = 147, normalized size = 1.31 \begin{align*} \frac{60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 30 \, a^{2} \sin \left (d x + c\right ) - \frac{137 \, a^{2} \sin \left (d x + c\right )^{5} - 30 \, a^{2} \sin \left (d x + c\right )^{4} - 60 \, a^{2} \sin \left (d x + c\right )^{3} - 10 \, a^{2} \sin \left (d x + c\right )^{2} + 15 \, a^{2} \sin \left (d x + c\right ) + 6 \, a^{2}}{\sin \left (d x + c\right )^{5}}}{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)^6*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/30*(60*a^2*log(abs(sin(d*x + c))) + 30*a^2*sin(d*x + c) - (137*a^2*sin(d*x + c)^5 - 30*a^2*sin(d*x + c)^4 -
60*a^2*sin(d*x + c)^3 - 10*a^2*sin(d*x + c)^2 + 15*a^2*sin(d*x + c) + 6*a^2)/sin(d*x + c)^5)/d