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

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

[Out]

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

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Rubi [A]  time = 0.13, 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} \[ \frac {a^3 \sin ^5(c+d x)}{5 d}+\frac {3 a^3 \sin ^4(c+d x)}{4 d}+\frac {a^3 \sin ^3(c+d x)}{3 d}-\frac {5 a^3 \sin ^2(c+d x)}{2 d}-\frac {5 a^3 \sin (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {3 a^3 \csc (c+d x)}{d}+\frac {a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.15, size = 86, normalized size = 0.65 \[ -\frac {a^3 \left (-12 \sin ^5(c+d x)-45 \sin ^4(c+d x)-20 \sin ^3(c+d x)+150 \sin ^2(c+d x)+300 \sin (c+d x)+30 \csc ^2(c+d x)+180 \csc (c+d x)-60 \log (\sin (c+d x))\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(a^3*(180*Csc[c + d*x] + 30*Csc[c + d*x]^2 - 60*Log[Sin[c + d*x]] + 300*Sin[c + d*x] + 150*Sin[c + d*x]^
2 - 20*Sin[c + d*x]^3 - 45*Sin[c + d*x]^4 - 12*Sin[c + d*x]^5))/d

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fricas [A]  time = 0.78, size = 145, normalized size = 1.09 \[ \frac {360 \, a^{3} \cos \left (d x + c\right )^{6} + 120 \, a^{3} \cos \left (d x + c\right )^{4} - 855 \, a^{3} \cos \left (d x + c\right )^{2} + 615 \, a^{3} + 480 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 32 \, {\left (3 \, a^{3} \cos \left (d x + c\right )^{6} - 14 \, a^{3} \cos \left (d x + c\right )^{4} - 56 \, a^{3} \cos \left (d x + c\right )^{2} + 112 \, a^{3}\right )} \sin \left (d x + c\right )}{480 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(360*a^3*cos(d*x + c)^6 + 120*a^3*cos(d*x + c)^4 - 855*a^3*cos(d*x + c)^2 + 615*a^3 + 480*(a^3*cos(d*x +
c)^2 - a^3)*log(1/2*sin(d*x + c)) + 32*(3*a^3*cos(d*x + c)^6 - 14*a^3*cos(d*x + c)^4 - 56*a^3*cos(d*x + c)^2
+ 112*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2 - d)

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giac [A]  time = 0.32, size = 120, normalized size = 0.90 \[ \frac {12 \, a^{3} \sin \left (d x + c\right )^{5} + 45 \, a^{3} \sin \left (d x + c\right )^{4} + 20 \, a^{3} \sin \left (d x + c\right )^{3} - 150 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 300 \, a^{3} \sin \left (d x + c\right ) - \frac {30 \, {\left (3 \, a^{3} \sin \left (d x + c\right )^{2} + 6 \, a^{3} \sin \left (d x + c\right ) + a^{3}\right )}}{\sin \left (d x + c\right )^{2}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*a^3*sin(d*x + c)^5 + 45*a^3*sin(d*x + c)^4 + 20*a^3*sin(d*x + c)^3 - 150*a^3*sin(d*x + c)^2 + 60*a^3*
log(abs(sin(d*x + c))) - 300*a^3*sin(d*x + c) - 30*(3*a^3*sin(d*x + c)^2 + 6*a^3*sin(d*x + c) + a^3)/sin(d*x +
c)^2)/d

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maple [A]  time = 0.46, size = 154, normalized size = 1.16 \[ -\frac {112 a^{3} \sin \left (d x +c \right )}{15 d}-\frac {14 a^{3} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{5 d}-\frac {56 a^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15 d}+\frac {\left (\cos ^{4}\left (d x +c \right )\right ) a^{3}}{4 d}+\frac {a^{3} \left (\cos ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {3 a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.61, size = 106, normalized size = 0.80 \[ \frac {12 \, a^{3} \sin \left (d x + c\right )^{5} + 45 \, a^{3} \sin \left (d x + c\right )^{4} + 20 \, a^{3} \sin \left (d x + c\right )^{3} - 150 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) - 300 \, a^{3} \sin \left (d x + c\right ) - \frac {30 \, {\left (6 \, a^{3} \sin \left (d x + c\right ) + a^{3}\right )}}{\sin \left (d x + c\right )^{2}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*a^3*sin(d*x + c)^5 + 45*a^3*sin(d*x + c)^4 + 20*a^3*sin(d*x + c)^3 - 150*a^3*sin(d*x + c)^2 + 60*a^3*
log(sin(d*x + c)) - 300*a^3*sin(d*x + c) - 30*(6*a^3*sin(d*x + c) + a^3)/sin(d*x + c)^2)/d

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mupad [B]  time = 9.01, size = 342, normalized size = 2.57 \[ \frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {46\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\frac {81\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{2}+\frac {538\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{3}+\frac {149\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+\frac {3796\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{15}+77\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {628\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+45\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+70\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+6\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^3}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

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)^3,x)

[Out]

(a^3*log(tan(c/2 + (d*x)/2)))/d - (a^3*tan(c/2 + (d*x)/2)^2)/(8*d) - (3*a^3*tan(c/2 + (d*x)/2))/(2*d) - ((5*a^
3*tan(c/2 + (d*x)/2)^2)/2 + 70*a^3*tan(c/2 + (d*x)/2)^3 + 45*a^3*tan(c/2 + (d*x)/2)^4 + (628*a^3*tan(c/2 + (d*
x)/2)^5)/3 + 77*a^3*tan(c/2 + (d*x)/2)^6 + (3796*a^3*tan(c/2 + (d*x)/2)^7)/15 + (149*a^3*tan(c/2 + (d*x)/2)^8)
/2 + (538*a^3*tan(c/2 + (d*x)/2)^9)/3 + (81*a^3*tan(c/2 + (d*x)/2)^10)/2 + 46*a^3*tan(c/2 + (d*x)/2)^11 + a^3/
2 + 6*a^3*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 + 20*tan(c/2 + (d*x)/2)^4 + 40*tan(c/2 + (d*x)/2)^6 +
40*tan(c/2 + (d*x)/2)^8 + 20*tan(c/2 + (d*x)/2)^10 + 4*tan(c/2 + (d*x)/2)^12)) - (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)**3*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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