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

Optimal. Leaf size=131 \[ \frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {a^3 \sin (c+d x)}{d}-\frac {a^3 \csc ^4(c+d x)}{4 d}-\frac {a^3 \csc ^3(c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}+\frac {5 a^3 \csc (c+d x)}{d}-\frac {5 a^3 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2707, 88} \[ \frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {a^3 \sin (c+d x)}{d}-\frac {a^3 \csc ^4(c+d x)}{4 d}-\frac {a^3 \csc ^3(c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}+\frac {5 a^3 \csc (c+d x)}{d}-\frac {5 a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2707

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^(m - (p + 1)/2))/(a - x)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.50, size = 86, normalized size = 0.66 \[ \frac {a^3 \left (4 \sin ^3(c+d x)+18 \sin ^2(c+d x)+12 \sin (c+d x)-3 \csc ^4(c+d x)-12 \csc ^3(c+d x)-6 \csc ^2(c+d x)+60 \csc (c+d x)-60 \log (\sin (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(60*Csc[c + d*x] - 6*Csc[c + d*x]^2 - 12*Csc[c + d*x]^3 - 3*Csc[c + d*x]^4 - 60*Log[Sin[c + d*x]] + 12*Si
n[c + d*x] + 18*Sin[c + d*x]^2 + 4*Sin[c + d*x]^3))/(12*d)

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fricas [A]  time = 0.79, size = 159, normalized size = 1.21 \[ -\frac {18 \, a^{3} \cos \left (d x + c\right )^{6} - 45 \, a^{3} \cos \left (d x + c\right )^{4} + 30 \, a^{3} \cos \left (d x + c\right )^{2} + 60 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (a^{3} \cos \left (d x + c\right )^{6} - 6 \, a^{3} \cos \left (d x + c\right )^{4} + 24 \, a^{3} \cos \left (d x + c\right )^{2} - 16 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/12*(18*a^3*cos(d*x + c)^6 - 45*a^3*cos(d*x + c)^4 + 30*a^3*cos(d*x + c)^2 + 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)) + 4*(a^3*cos(d*x + c)^6 - 6*a^3*cos(d*x + c)^4 + 24*a^3*cos(d*x +
c)^2 - 16*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

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giac [A]  time = 0.36, size = 121, normalized size = 0.92 \[ \frac {4 \, a^{3} \sin \left (d x + c\right )^{3} + 18 \, a^{3} \sin \left (d x + c\right )^{2} - 60 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 12 \, a^{3} \sin \left (d x + c\right ) + \frac {125 \, a^{3} \sin \left (d x + c\right )^{4} + 60 \, a^{3} \sin \left (d x + c\right )^{3} - 6 \, a^{3} \sin \left (d x + c\right )^{2} - 12 \, a^{3} \sin \left (d x + c\right ) - 3 \, a^{3}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/12*(4*a^3*sin(d*x + c)^3 + 18*a^3*sin(d*x + c)^2 - 60*a^3*log(abs(sin(d*x + c))) + 12*a^3*sin(d*x + c) + (12
5*a^3*sin(d*x + c)^4 + 60*a^3*sin(d*x + c)^3 - 6*a^3*sin(d*x + c)^2 - 12*a^3*sin(d*x + c) - 3*a^3)/sin(d*x + c
)^4)/d

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maple [A]  time = 0.43, size = 211, normalized size = 1.61 \[ \frac {2 a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}+\frac {16 a^{3} \sin \left (d x +c \right )}{3 d}+\frac {2 a^{3} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}+\frac {8 a^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}-\frac {3 a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {3 \left (\cos ^{4}\left (d x +c \right )\right ) a^{3}}{2 d}-\frac {3 a^{3} \left (\cos ^{2}\left (d x +c \right )\right )}{d}-\frac {5 a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )^{3}}-\frac {a^{3} \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a^{3} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^5*(a+a*sin(d*x+c))^3,x)

[Out]

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

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maxima [A]  time = 0.52, size = 108, normalized size = 0.82 \[ \frac {4 \, a^{3} \sin \left (d x + c\right )^{3} + 18 \, a^{3} \sin \left (d x + c\right )^{2} - 60 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 12 \, a^{3} \sin \left (d x + c\right ) + \frac {3 \, {\left (20 \, a^{3} \sin \left (d x + c\right )^{3} - 2 \, a^{3} \sin \left (d x + c\right )^{2} - 4 \, a^{3} \sin \left (d x + c\right ) - a^{3}\right )}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/12*(4*a^3*sin(d*x + c)^3 + 18*a^3*sin(d*x + c)^2 - 60*a^3*log(sin(d*x + c)) + 12*a^3*sin(d*x + c) + 3*(20*a^
3*sin(d*x + c)^3 - 2*a^3*sin(d*x + c)^2 - 4*a^3*sin(d*x + c) - a^3)/sin(d*x + c)^4)/d

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mupad [B]  time = 8.94, size = 322, normalized size = 2.46 \[ \frac {66\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+93\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {620\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+\frac {347\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{4}+128\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {39\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{4}+28\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}-2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a^3}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,d}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {5\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {17\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {5\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x))^3)/sin(c + d*x)^5,x)

[Out]

(28*a^3*tan(c/2 + (d*x)/2)^3 - (15*a^3*tan(c/2 + (d*x)/2)^2)/4 - (39*a^3*tan(c/2 + (d*x)/2)^4)/4 + 128*a^3*tan
(c/2 + (d*x)/2)^5 + (347*a^3*tan(c/2 + (d*x)/2)^6)/4 + (620*a^3*tan(c/2 + (d*x)/2)^7)/3 + 93*a^3*tan(c/2 + (d*
x)/2)^8 + 66*a^3*tan(c/2 + (d*x)/2)^9 - a^3/4 - 2*a^3*tan(c/2 + (d*x)/2))/(d*(16*tan(c/2 + (d*x)/2)^4 + 48*tan
(c/2 + (d*x)/2)^6 + 48*tan(c/2 + (d*x)/2)^8 + 16*tan(c/2 + (d*x)/2)^10)) - (a^3*tan(c/2 + (d*x)/2)^3)/(8*d) -
(a^3*tan(c/2 + (d*x)/2)^4)/(64*d) - (3*a^3*tan(c/2 + (d*x)/2)^2)/(16*d) - (5*a^3*log(tan(c/2 + (d*x)/2)))/d +
(17*a^3*tan(c/2 + (d*x)/2))/(8*d) + (5*a^3*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)**5*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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