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

Optimal. Leaf size=86 \[ \frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^2(c+d x)}{2 d}-\frac {2 a \sin (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \csc (c+d x)}{d}-\frac {2 a \log (\sin (c+d x))}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 86, 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} \[ \frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^2(c+d x)}{2 d}-\frac {2 a \sin (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \csc (c+d x)}{d}-\frac {2 a \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 77, normalized size = 0.90 \[ \frac {a \sin ^3(c+d x)}{3 d}-\frac {2 a \sin (c+d x)}{d}-\frac {a \csc (c+d x)}{d}-\frac {a \left (-\sin ^2(c+d x)+\csc ^2(c+d x)+4 \log (\sin (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.69, size = 102, normalized size = 1.19 \[ -\frac {6 \, a \cos \left (d x + c\right )^{4} - 9 \, a \cos \left (d x + c\right )^{2} + 24 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} - 8 \, a\right )} \sin \left (d x + c\right ) - 3 \, a}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

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)),x, algorithm="fricas")

[Out]

-1/12*(6*a*cos(d*x + c)^4 - 9*a*cos(d*x + c)^2 + 24*(a*cos(d*x + c)^2 - a)*log(1/2*sin(d*x + c)) + 4*(a*cos(d*
x + c)^4 + 4*a*cos(d*x + c)^2 - 8*a)*sin(d*x + c) - 3*a)/(d*cos(d*x + c)^2 - d)

________________________________________________________________________________________

giac [A]  time = 0.20, size = 82, normalized size = 0.95 \[ \frac {2 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 12 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 12 \, a \sin \left (d x + c\right ) + \frac {3 \, {\left (6 \, a \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right ) - a\right )}}{\sin \left (d x + c\right )^{2}}}{6 \, d} \]

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)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.38, size = 139, normalized size = 1.62 \[ -\frac {a \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {8 a \sin \left (d x +c \right )}{3 d}-\frac {\left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a}{d}-\frac {4 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a}{3 d}-\frac {a \left (\cos ^{6}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {a \left (\cos ^{4}\left (d x +c \right )\right )}{2 d}-\frac {a \left (\cos ^{2}\left (d x +c \right )\right )}{d}-\frac {2 a \ln \left (\sin \left (d x +c \right )\right )}{d} \]

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)),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.60, size = 68, normalized size = 0.79 \[ \frac {2 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 12 \, a \log \left (\sin \left (d x + c\right )\right ) - 12 \, a \sin \left (d x + c\right ) - \frac {3 \, {\left (2 \, a \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )^{2}}}{6 \, d} \]

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)),x, algorithm="maxima")

[Out]

1/6*(2*a*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - 12*a*log(sin(d*x + c)) - 12*a*sin(d*x + c) - 3*(2*a*sin(d*x + c
) + a)/sin(d*x + c)^2)/d

________________________________________________________________________________________

mupad [B]  time = 8.86, size = 229, normalized size = 2.66 \[ \frac {2\,a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {18\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {82\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-\frac {13\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+22\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {2\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (a/2 + 2*a*tan(c/2 + (d*x)/2) + (3*a*tan(c/2 + (d*x)/2)^2)/2 + 22*a*ta
n(c/2 + (d*x)/2)^3 - (13*a*tan(c/2 + (d*x)/2)^4)/2 + (82*a*tan(c/2 + (d*x)/2)^5)/3 - (15*a*tan(c/2 + (d*x)/2)^
6)/2 + 18*a*tan(c/2 + (d*x)/2)^7)/(d*(4*tan(c/2 + (d*x)/2)^2 + 12*tan(c/2 + (d*x)/2)^4 + 12*tan(c/2 + (d*x)/2)
^6 + 4*tan(c/2 + (d*x)/2)^8)) - (a*tan(c/2 + (d*x)/2))/(2*d) - (a*tan(c/2 + (d*x)/2)^2)/(8*d) - (2*a*log(tan(c
/2 + (d*x)/2)))/d

________________________________________________________________________________________

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)),x)

[Out]

Timed out

________________________________________________________________________________________