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(d*x+c)/d+2*a^2*csc(d*x+c)^2/d+1/3*a^2*csc(d*x+c)^3/d-1/2*a^2*csc(d*x+c)^4/d-1/5*a^2*csc(d*x+c)^5/d+2*a
^2*ln(sin(d*x+c))/d+a^2*sin(d*x+c)/d

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Rubi [A]  time = 0.10, antiderivative size = 112, normalized size of antiderivative = 1.00, 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 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]))

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]

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*}

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Mathematica [A]  time = 0.13, 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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fricas [A]  time = 0.83, size = 153, normalized size = 1.37 \[ -\frac {30 \, a^{2} \cos \left (d x + c\right )^{6} - 120 \, a^{2} \cos \left (d x + c\right )^{4} + 160 \, a^{2} \cos \left (d x + c\right )^{2} - 60 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 64 \, a^{2} + 15 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2}\right )} \sin \left (d x + c\right )}{30 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

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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giac [A]  time = 0.27, size = 109, normalized size = 0.97 \[ \frac {60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 30 \, a^{2} \sin \left (d x + c\right ) - \frac {137 \, a^{2} \sin \left (d x + c\right )^{5} - 30 \, a^{2} \sin \left (d x + c\right )^{4} - 60 \, a^{2} \sin \left (d x + c\right )^{3} - 10 \, a^{2} \sin \left (d x + c\right )^{2} + 15 \, a^{2} \sin \left (d x + c\right ) + 6 \, a^{2}}{\sin \left (d x + c\right )^{5}}}{30 \, d} \]

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

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maple [A]  time = 0.42, size = 178, normalized size = 1.59 \[ -\frac {4 a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{15 d \sin \left (d x +c \right )^{3}}+\frac {4 a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )}+\frac {32 a^{2} \sin \left (d x +c \right )}{15 d}+\frac {4 \sin \left (d x +c \right ) a^{2} \left (\cos ^{4}\left (d x +c \right )\right )}{5 d}+\frac {16 \sin \left (d x +c \right ) a^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{15 d}-\frac {a^{2} \left (\cot ^{4}\left (d x +c \right )\right )}{2 d}+\frac {a^{2} \left (\cot ^{2}\left (d x +c \right )\right )}{d}+\frac {2 a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )^{5}} \]

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 = 0.56, size = 94, normalized size = 0.84 \[ \frac {60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 30 \, a^{2} \sin \left (d x + c\right ) + \frac {30 \, a^{2} \sin \left (d x + c\right )^{4} + 60 \, a^{2} \sin \left (d x + c\right )^{3} + 10 \, a^{2} \sin \left (d x + c\right )^{2} - 15 \, a^{2} \sin \left (d x + c\right ) - 6 \, a^{2}}{\sin \left (d x + c\right )^{5}}}{30 \, d} \]

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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mupad [B]  time = 8.79, size = 267, normalized size = 2.38 \[ \frac {82\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {55\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a^2}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {2\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {9\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}-\frac {2\,a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*a^2*tan(c/2 + (d*x)/2)^2)/15 + 11*a^2*tan(c/2 + (d*x)/2)^3 + (55*a^2*tan(c/2 + (d*x)/2)^4)/3 + 12*a^2*tan(
c/2 + (d*x)/2)^5 + 82*a^2*tan(c/2 + (d*x)/2)^6 - a^2/5 - a^2*tan(c/2 + (d*x)/2))/(d*(32*tan(c/2 + (d*x)/2)^5 +
32*tan(c/2 + (d*x)/2)^7)) + (3*a^2*tan(c/2 + (d*x)/2)^2)/(8*d) + (a^2*tan(c/2 + (d*x)/2)^3)/(96*d) - (a^2*tan
(c/2 + (d*x)/2)^4)/(32*d) - (a^2*tan(c/2 + (d*x)/2)^5)/(160*d) + (2*a^2*log(tan(c/2 + (d*x)/2)))/d + (9*a^2*ta
n(c/2 + (d*x)/2))/(16*d) - (2*a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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