3.529 \(\int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=133 \[ \frac{a^3 \sin (c+d x)}{d}-\frac{a^3 \csc ^6(c+d x)}{6 d}-\frac{3 a^3 \csc ^5(c+d x)}{5 d}-\frac{a^3 \csc ^4(c+d x)}{4 d}+\frac{5 a^3 \csc ^3(c+d x)}{3 d}+\frac{5 a^3 \csc ^2(c+d x)}{2 d}-\frac{a^3 \csc (c+d x)}{d}+\frac{3 a^3 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.126247, antiderivative size = 133, 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} \[ \frac{a^3 \sin (c+d x)}{d}-\frac{a^3 \csc ^6(c+d x)}{6 d}-\frac{3 a^3 \csc ^5(c+d x)}{5 d}-\frac{a^3 \csc ^4(c+d x)}{4 d}+\frac{5 a^3 \csc ^3(c+d x)}{3 d}+\frac{5 a^3 \csc ^2(c+d x)}{2 d}-\frac{a^3 \csc (c+d x)}{d}+\frac{3 a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*Csc[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

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

Mathematica [A]  time = 0.0396926, size = 113, normalized size = 0.85 \[ a^3 \left (\frac{\sin (c+d x)}{d}-\frac{\csc ^6(c+d x)}{6 d}-\frac{3 \csc ^5(c+d x)}{5 d}-\frac{\csc ^4(c+d x)}{4 d}+\frac{5 \csc ^3(c+d x)}{3 d}+\frac{5 \csc ^2(c+d x)}{2 d}-\frac{\csc (c+d x)}{d}+\frac{3 \log (\sin (c+d x))}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*Csc[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

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

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Maple [A]  time = 0.093, size = 203, normalized size = 1.5 \begin{align*} -{\frac{2\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{15\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{2\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{5\,d\sin \left ( dx+c \right ) }}+{\frac{16\,{a}^{3}\sin \left ( dx+c \right ) }{15\,d}}+{\frac{2\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sin \left ( dx+c \right ) }{5\,d}}+{\frac{8\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) }{15\,d}}-{\frac{3\,{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{3\,{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+3\,{\frac{{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{3\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{{a}^{3} \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))^3,x)

[Out]

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

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Maxima [A]  time = 1.0321, size = 146, normalized size = 1.1 \begin{align*} \frac{180 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 60 \, a^{3} \sin \left (d x + c\right ) - \frac{60 \, a^{3} \sin \left (d x + c\right )^{5} - 150 \, a^{3} \sin \left (d x + c\right )^{4} - 100 \, a^{3} \sin \left (d x + c\right )^{3} + 15 \, a^{3} \sin \left (d x + c\right )^{2} + 36 \, a^{3} \sin \left (d x + c\right ) + 10 \, a^{3}}{\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))^3,x, algorithm="maxima")

[Out]

1/60*(180*a^3*log(sin(d*x + c)) + 60*a^3*sin(d*x + c) - (60*a^3*sin(d*x + c)^5 - 150*a^3*sin(d*x + c)^4 - 100*
a^3*sin(d*x + c)^3 + 15*a^3*sin(d*x + c)^2 + 36*a^3*sin(d*x + c) + 10*a^3)/sin(d*x + c)^6)/d

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Fricas [A]  time = 1.56869, size = 447, normalized size = 3.36 \begin{align*} -\frac{150 \, a^{3} \cos \left (d x + c\right )^{4} - 285 \, a^{3} \cos \left (d x + c\right )^{2} + 125 \, a^{3} - 180 \,{\left (a^{3} \cos \left (d x + c\right )^{6} - 3 \, a^{3} \cos \left (d x + c\right )^{4} + 3 \, a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 4 \,{\left (15 \, a^{3} \cos \left (d x + c\right )^{6} - 30 \, a^{3} \cos \left (d x + c\right )^{4} + 40 \, a^{3} \cos \left (d x + c\right )^{2} - 16 \, a^{3}\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))^3,x, algorithm="fricas")

[Out]

-1/60*(150*a^3*cos(d*x + c)^4 - 285*a^3*cos(d*x + c)^2 + 125*a^3 - 180*(a^3*cos(d*x + c)^6 - 3*a^3*cos(d*x + c
)^4 + 3*a^3*cos(d*x + c)^2 - a^3)*log(1/2*sin(d*x + c)) - 4*(15*a^3*cos(d*x + c)^6 - 30*a^3*cos(d*x + c)^4 + 4
0*a^3*cos(d*x + c)^2 - 16*a^3)*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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.4107, size = 165, normalized size = 1.24 \begin{align*} \frac{180 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, a^{3} \sin \left (d x + c\right ) - \frac{441 \, a^{3} \sin \left (d x + c\right )^{6} + 60 \, a^{3} \sin \left (d x + c\right )^{5} - 150 \, a^{3} \sin \left (d x + c\right )^{4} - 100 \, a^{3} \sin \left (d x + c\right )^{3} + 15 \, a^{3} \sin \left (d x + c\right )^{2} + 36 \, a^{3} \sin \left (d x + c\right ) + 10 \, a^{3}}{\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))^3,x, algorithm="giac")

[Out]

1/60*(180*a^3*log(abs(sin(d*x + c))) + 60*a^3*sin(d*x + c) - (441*a^3*sin(d*x + c)^6 + 60*a^3*sin(d*x + c)^5 -
150*a^3*sin(d*x + c)^4 - 100*a^3*sin(d*x + c)^3 + 15*a^3*sin(d*x + c)^2 + 36*a^3*sin(d*x + c) + 10*a^3)/sin(d
*x + c)^6)/d