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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.110436, antiderivative size = 133, 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} \[ \frac{a^3 \sin ^2(c+d x)}{2 d}+\frac{3 a^3 \sin (c+d x)}{d}-\frac{a^3 \csc ^5(c+d x)}{5 d}-\frac{3 a^3 \csc ^4(c+d x)}{4 d}-\frac{a^3 \csc ^3(c+d x)}{3 d}+\frac{5 a^3 \csc ^2(c+d x)}{2 d}+\frac{5 a^3 \csc (c+d x)}{d}+\frac{a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.16801, size = 86, normalized size = 0.65 \[ \frac{a^3 \left (30 \sin ^2(c+d x)+180 \sin (c+d x)-12 \csc ^5(c+d x)-45 \csc ^4(c+d x)-20 \csc ^3(c+d x)+150 \csc ^2(c+d x)+300 \csc (c+d x)+60 \log (\sin (c+d x))\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(300*Csc[c + d*x] + 150*Csc[c + d*x]^2 - 20*Csc[c + d*x]^3 - 45*Csc[c + d*x]^4 - 12*Csc[c + d*x]^5 + 60*L
og[Sin[c + d*x]] + 180*Sin[c + d*x] + 30*Sin[c + d*x]^2))/(60*d)

________________________________________________________________________________________

Maple [A]  time = 0.091, size = 234, normalized size = 1.8 \begin{align*} -{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}{a}^{3}}{2\,d}}-{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{14\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{15\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{14\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{5\,d\sin \left ( dx+c \right ) }}+{\frac{112\,{a}^{3}\sin \left ( dx+c \right ) }{15\,d}}+{\frac{14\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}{a}^{3}\sin \left ( dx+c \right ) }{5\,d}}+{\frac{56\,{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}}-{\frac{{a}^{3} \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))^3,x)

[Out]

-1/2/d*a^3/sin(d*x+c)^2*cos(d*x+c)^6-1/2/d*cos(d*x+c)^4*a^3-1/d*a^3*cos(d*x+c)^2+a^3*ln(sin(d*x+c))/d-14/15/d*
a^3/sin(d*x+c)^3*cos(d*x+c)^6+14/5/d*a^3/sin(d*x+c)*cos(d*x+c)^6+112/15*a^3*sin(d*x+c)/d+14/5/d*a^3*cos(d*x+c)
^4*sin(d*x+c)+56/15/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)^2-1/5/d*a^3/sin(
d*x+c)^5*cos(d*x+c)^6

________________________________________________________________________________________

Maxima [A]  time = 1.09972, size = 144, normalized size = 1.08 \begin{align*} \frac{30 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 180 \, a^{3} \sin \left (d x + c\right ) + \frac{300 \, a^{3} \sin \left (d x + c\right )^{4} + 150 \, a^{3} \sin \left (d x + c\right )^{3} - 20 \, a^{3} \sin \left (d x + c\right )^{2} - 45 \, a^{3} \sin \left (d x + c\right ) - 12 \, a^{3}}{\sin \left (d x + c\right )^{5}}}{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)^6*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(30*a^3*sin(d*x + c)^2 + 60*a^3*log(sin(d*x + c)) + 180*a^3*sin(d*x + c) + (300*a^3*sin(d*x + c)^4 + 150*
a^3*sin(d*x + c)^3 - 20*a^3*sin(d*x + c)^2 - 45*a^3*sin(d*x + c) - 12*a^3)/sin(d*x + c)^5)/d

________________________________________________________________________________________

Fricas [A]  time = 1.44621, size = 456, normalized size = 3.43 \begin{align*} -\frac{180 \, a^{3} \cos \left (d x + c\right )^{6} - 840 \, a^{3} \cos \left (d x + c\right )^{4} + 1120 \, a^{3} \cos \left (d x + c\right )^{2} - 448 \, a^{3} - 60 \,{\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 15 \,{\left (2 \, a^{3} \cos \left (d x + c\right )^{6} - 5 \, a^{3} \cos \left (d x + c\right )^{4} + 14 \, a^{3} \cos \left (d x + c\right )^{2} - 8 \, a^{3}\right )} \sin \left (d x + c\right )}{60 \,{\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))^3,x, algorithm="fricas")

[Out]

-1/60*(180*a^3*cos(d*x + c)^6 - 840*a^3*cos(d*x + c)^4 + 1120*a^3*cos(d*x + c)^2 - 448*a^3 - 60*(a^3*cos(d*x +
 c)^4 - 2*a^3*cos(d*x + c)^2 + a^3)*log(1/2*sin(d*x + c))*sin(d*x + c) + 15*(2*a^3*cos(d*x + c)^6 - 5*a^3*cos(
d*x + c)^4 + 14*a^3*cos(d*x + c)^2 - 8*a^3)*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x
 + c))

________________________________________________________________________________________

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))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.31364, size = 165, normalized size = 1.24 \begin{align*} \frac{30 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 180 \, a^{3} \sin \left (d x + c\right ) - \frac{137 \, a^{3} \sin \left (d x + c\right )^{5} - 300 \, a^{3} \sin \left (d x + c\right )^{4} - 150 \, a^{3} \sin \left (d x + c\right )^{3} + 20 \, a^{3} \sin \left (d x + c\right )^{2} + 45 \, a^{3} \sin \left (d x + c\right ) + 12 \, a^{3}}{\sin \left (d x + c\right )^{5}}}{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)^6*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*a^3*sin(d*x + c)^2 + 60*a^3*log(abs(sin(d*x + c))) + 180*a^3*sin(d*x + c) - (137*a^3*sin(d*x + c)^5 -
 300*a^3*sin(d*x + c)^4 - 150*a^3*sin(d*x + c)^3 + 20*a^3*sin(d*x + c)^2 + 45*a^3*sin(d*x + c) + 12*a^3)/sin(d
*x + c)^5)/d