3.524 \(\int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\)

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

[Out]

-a^3*csc(d*x+c)/d+3*a^3*ln(sin(d*x+c))/d+a^3*sin(d*x+c)/d-5/2*a^3*sin(d*x+c)^2/d-5/3*a^3*sin(d*x+c)^3/d+1/4*a^
3*sin(d*x+c)^4/d+3/5*a^3*sin(d*x+c)^5/d+1/6*a^3*sin(d*x+c)^6/d

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Rubi [A]  time = 0.12, antiderivative size = 133, normalized size of antiderivative = 1.00, 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} \[ \frac {a^3 \sin ^6(c+d x)}{6 d}+\frac {3 a^3 \sin ^5(c+d x)}{5 d}+\frac {a^3 \sin ^4(c+d x)}{4 d}-\frac {5 a^3 \sin ^3(c+d x)}{3 d}-\frac {5 a^3 \sin ^2(c+d x)}{2 d}+\frac {a^3 \sin (c+d x)}{d}-\frac {a^3 \csc (c+d x)}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.17, size = 86, normalized size = 0.65 \[ -\frac {a^3 \left (-10 \sin ^6(c+d x)-36 \sin ^5(c+d x)-15 \sin ^4(c+d x)+100 \sin ^3(c+d x)+150 \sin ^2(c+d x)-60 \sin (c+d x)+60 \csc (c+d x)-180 \log (\sin (c+d x))\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(a^3*(60*Csc[c + d*x] - 180*Log[Sin[c + d*x]] - 60*Sin[c + d*x] + 150*Sin[c + d*x]^2 + 100*Sin[c + d*x]^
3 - 15*Sin[c + d*x]^4 - 36*Sin[c + d*x]^5 - 10*Sin[c + d*x]^6))/d

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fricas [A]  time = 0.71, size = 131, normalized size = 0.98 \[ -\frac {144 \, a^{3} \cos \left (d x + c\right )^{6} - 32 \, a^{3} \cos \left (d x + c\right )^{4} - 128 \, a^{3} \cos \left (d x + c\right )^{2} - 720 \, a^{3} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 256 \, a^{3} + 5 \, {\left (8 \, a^{3} \cos \left (d x + c\right )^{6} - 36 \, a^{3} \cos \left (d x + c\right )^{4} - 72 \, a^{3} \cos \left (d x + c\right )^{2} + 47 \, a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/240*(144*a^3*cos(d*x + c)^6 - 32*a^3*cos(d*x + c)^4 - 128*a^3*cos(d*x + c)^2 - 720*a^3*log(1/2*sin(d*x + c)
)*sin(d*x + c) + 256*a^3 + 5*(8*a^3*cos(d*x + c)^6 - 36*a^3*cos(d*x + c)^4 - 72*a^3*cos(d*x + c)^2 + 47*a^3)*s
in(d*x + c))/(d*sin(d*x + c))

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giac [A]  time = 0.31, size = 120, normalized size = 0.90 \[ \frac {10 \, a^{3} \sin \left (d x + c\right )^{6} + 36 \, a^{3} \sin \left (d x + c\right )^{5} + 15 \, a^{3} \sin \left (d x + c\right )^{4} - 100 \, a^{3} \sin \left (d x + c\right )^{3} - 150 \, a^{3} \sin \left (d x + c\right )^{2} + 180 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, a^{3} \sin \left (d x + c\right ) - \frac {60 \, {\left (3 \, a^{3} \sin \left (d x + c\right ) + a^{3}\right )}}{\sin \left (d x + c\right )}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.40, size = 147, normalized size = 1.11 \[ -\frac {a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{6 d}-\frac {16 a^{3} \sin \left (d x +c \right )}{15 d}-\frac {2 a^{3} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{5 d}-\frac {8 a^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15 d}+\frac {3 \left (\cos ^{4}\left (d x +c \right )\right ) a^{3}}{4 d}+\frac {3 a^{3} \left (\cos ^{2}\left (d x +c \right )\right )}{2 d}+\frac {3 a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x)

[Out]

-1/6/d*a^3*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*cos(d*x+c)^4*a^3+3/2/d*a^3*cos(d*x+c)^2+3*a^3*ln(sin(d*x+c))/d-1/d*a^3/sin(d*x+c)*cos(d*x+c)^6

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maxima [A]  time = 0.51, size = 107, normalized size = 0.80 \[ \frac {10 \, a^{3} \sin \left (d x + c\right )^{6} + 36 \, a^{3} \sin \left (d x + c\right )^{5} + 15 \, a^{3} \sin \left (d x + c\right )^{4} - 100 \, a^{3} \sin \left (d x + c\right )^{3} - 150 \, a^{3} \sin \left (d x + c\right )^{2} + 180 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 60 \, a^{3} \sin \left (d x + c\right ) - \frac {60 \, a^{3}}{\sin \left (d x + c\right )}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.18, size = 371, normalized size = 2.79 \[ \frac {14\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}-\frac {10\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d}+\frac {8\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3\,d}-\frac {28\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d}+\frac {32\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{d}-\frac {32\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{3\,d}-\frac {3\,a^3\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {3\,a^3\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {46\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {688\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {1064\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{15\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {288\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {96\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {3\,a^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {a^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \]

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)^2,x)

[Out]

(14*a^3*cos(c/2 + (d*x)/2)^4)/d - (10*a^3*cos(c/2 + (d*x)/2)^2)/d + (8*a^3*cos(c/2 + (d*x)/2)^6)/(3*d) - (28*a
^3*cos(c/2 + (d*x)/2)^8)/d + (32*a^3*cos(c/2 + (d*x)/2)^10)/d - (32*a^3*cos(c/2 + (d*x)/2)^12)/(3*d) - (3*a^3*
log(1/cos(c/2 + (d*x)/2)^2))/d + (3*a^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (46*a^3*cos(c/2 + (d*x
)/2)^3)/(3*d*sin(c/2 + (d*x)/2)) + (688*a^3*cos(c/2 + (d*x)/2)^5)/(15*d*sin(c/2 + (d*x)/2)) - (1064*a^3*cos(c/
2 + (d*x)/2)^7)/(15*d*sin(c/2 + (d*x)/2)) + (288*a^3*cos(c/2 + (d*x)/2)^9)/(5*d*sin(c/2 + (d*x)/2)) - (96*a^3*
cos(c/2 + (d*x)/2)^11)/(5*d*sin(c/2 + (d*x)/2)) + (3*a^3*cos(c/2 + (d*x)/2))/(2*d*sin(c/2 + (d*x)/2)) - (a^3*s
in(c/2 + (d*x)/2))/(2*d*cos(c/2 + (d*x)/2))

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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)**2*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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