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

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

[Out]

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

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Rubi [A]  time = 0.120996, antiderivative size = 119, 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} \[ -\frac{a^2 \csc ^6(c+d x)}{6 d}-\frac{2 a^2 \csc ^5(c+d x)}{5 d}+\frac{a^2 \csc ^4(c+d x)}{4 d}+\frac{4 a^2 \csc ^3(c+d x)}{3 d}+\frac{a^2 \csc ^2(c+d x)}{2 d}-\frac{2 a^2 \csc (c+d x)}{d}+\frac{a^2 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Csc[c + d*x])/d + (a^2*Csc[c + d*x]^2)/(2*d) + (4*a^2*Csc[c + d*x]^3)/(3*d) + (a^2*Csc[c + d*x]^4)/(4*
d) - (2*a^2*Csc[c + d*x]^5)/(5*d) - (a^2*Csc[c + d*x]^6)/(6*d) + (a^2*Log[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 ^2(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a^7 (a-x)^2 (a+x)^4}{x^7} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{a^2 \operatorname{Subst}\left (\int \frac{(a-x)^2 (a+x)^4}{x^7} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^2 \operatorname{Subst}\left (\int \left (\frac{a^6}{x^7}+\frac{2 a^5}{x^6}-\frac{a^4}{x^5}-\frac{4 a^3}{x^4}-\frac{a^2}{x^3}+\frac{2 a}{x^2}+\frac{1}{x}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{2 a^2 \csc (c+d x)}{d}+\frac{a^2 \csc ^2(c+d x)}{2 d}+\frac{4 a^2 \csc ^3(c+d x)}{3 d}+\frac{a^2 \csc ^4(c+d x)}{4 d}-\frac{2 a^2 \csc ^5(c+d x)}{5 d}-\frac{a^2 \csc ^6(c+d x)}{6 d}+\frac{a^2 \log (\sin (c+d x))}{d}\\ \end{align*}

Mathematica [A]  time = 0.0405315, size = 102, normalized size = 0.86 \[ a^2 \left (-\frac{\csc ^6(c+d x)}{6 d}-\frac{2 \csc ^5(c+d x)}{5 d}+\frac{\csc ^4(c+d x)}{4 d}+\frac{4 \csc ^3(c+d x)}{3 d}+\frac{\csc ^2(c+d x)}{2 d}-\frac{2 \csc (c+d x)}{d}+\frac{\log (\sin (c+d x))}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.083, size = 202, normalized size = 1.7 \begin{align*} -{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{2\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}+{\frac{2\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{15\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{2\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{5\,d\sin \left ( dx+c \right ) }}-{\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{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{6\,d \left ( \sin \left ( dx+c \right ) \right ) ^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x)

[Out]

-1/4/d*a^2*cot(d*x+c)^4+1/2/d*a^2*cot(d*x+c)^2+a^2*ln(sin(d*x+c))/d-2/5/d*a^2/sin(d*x+c)^5*cos(d*x+c)^6+2/15/d
*a^2/sin(d*x+c)^3*cos(d*x+c)^6-2/5/d*a^2/sin(d*x+c)*cos(d*x+c)^6-16/15*a^2*sin(d*x+c)/d-2/5/d*sin(d*x+c)*a^2*c
os(d*x+c)^4-8/15/d*sin(d*x+c)*a^2*cos(d*x+c)^2-1/6/d*a^2/sin(d*x+c)^6*cos(d*x+c)^6

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Maxima [A]  time = 1.03126, size = 131, normalized size = 1.1 \begin{align*} \frac{60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - \frac{120 \, a^{2} \sin \left (d x + c\right )^{5} - 30 \, a^{2} \sin \left (d x + c\right )^{4} - 80 \, a^{2} \sin \left (d x + c\right )^{3} - 15 \, a^{2} \sin \left (d x + c\right )^{2} + 24 \, a^{2} \sin \left (d x + c\right ) + 10 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{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)^7*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.58512, size = 408, normalized size = 3.43 \begin{align*} -\frac{30 \, a^{2} \cos \left (d x + c\right )^{4} - 75 \, a^{2} \cos \left (d x + c\right )^{2} + 35 \, a^{2} - 60 \,{\left (a^{2} \cos \left (d x + c\right )^{6} - 3 \, a^{2} \cos \left (d x + c\right )^{4} + 3 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 8 \,{\left (15 \, a^{2} \cos \left (d x + c\right )^{4} - 20 \, a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2}\right )} \sin \left (d x + c\right )}{60 \,{\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, 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)^5*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/60*(30*a^2*cos(d*x + c)^4 - 75*a^2*cos(d*x + c)^2 + 35*a^2 - 60*(a^2*cos(d*x + c)^6 - 3*a^2*cos(d*x + c)^4
+ 3*a^2*cos(d*x + c)^2 - a^2)*log(1/2*sin(d*x + c)) - 8*(15*a^2*cos(d*x + c)^4 - 20*a^2*cos(d*x + c)^2 + 8*a^2
)*sin(d*x + c))/(d*cos(d*x + c)^6 - 3*d*cos(d*x + c)^4 + 3*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)**5*csc(d*x+c)**7*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.3006, size = 150, normalized size = 1.26 \begin{align*} \frac{60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac{147 \, a^{2} \sin \left (d x + c\right )^{6} + 120 \, a^{2} \sin \left (d x + c\right )^{5} - 30 \, a^{2} \sin \left (d x + c\right )^{4} - 80 \, a^{2} \sin \left (d x + c\right )^{3} - 15 \, a^{2} \sin \left (d x + c\right )^{2} + 24 \, a^{2} \sin \left (d x + c\right ) + 10 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{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)^7*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/60*(60*a^2*log(abs(sin(d*x + c))) - (147*a^2*sin(d*x + c)^6 + 120*a^2*sin(d*x + c)^5 - 30*a^2*sin(d*x + c)^4
- 80*a^2*sin(d*x + c)^3 - 15*a^2*sin(d*x + c)^2 + 24*a^2*sin(d*x + c) + 10*a^2)/sin(d*x + c)^6)/d