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

Optimal. Leaf size=56 \[ -\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \sin ^2(c+d x)}{2 d}+\frac {a \sin (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2836, 12, 75} \[ -\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \sin ^2(c+d x)}{2 d}+\frac {a \sin (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Log[Sin[c + d*x]])/d + (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 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

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 (c+d x) (a+a \sin (c+d x)) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a (a-x) (a+x)^2}{x} \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a-x) (a+x)^2}{x} \, dx,x,a \sin (c+d x)\right )}{a^2 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (a^2+\frac {a^3}{x}-a x-x^2\right ) \, dx,x,a \sin (c+d x)\right )}{a^2 d}\\ &=\frac {a \log (\sin (c+d x))}{d}+\frac {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*}

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Mathematica [A]  time = 0.03, size = 56, normalized size = 1.00 \[ -\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \sin ^2(c+d x)}{2 d}+\frac {a \sin (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 51, normalized size = 0.91 \[ \frac {3 \, a \cos \left (d x + c\right )^{2} + 6 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 2 \, {\left (a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 48, normalized size = 0.86 \[ -\frac {2 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 6 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 6 \, a \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*csc(d*x+c)*(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 - 6*a*log(abs(sin(d*x + c))) - 6*a*sin(d*x + c))/d

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maple [A]  time = 0.33, size = 60, normalized size = 1.07 \[ \frac {\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a}{3 d}+\frac {2 a \sin \left (d x +c \right )}{3 d}+\frac {a \left (\cos ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a \ln \left (\sin \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 47, normalized size = 0.84 \[ -\frac {2 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 6 \, a \log \left (\sin \left (d x + c\right )\right ) - 6 \, a \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*csc(d*x+c)*(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 - 6*a*log(sin(d*x + c)) - 6*a*sin(d*x + c))/d

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mupad [B]  time = 8.73, size = 92, normalized size = 1.64 \[ \frac {a\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a\,{\cos \left (c+d\,x\right )}^2}{2\,d}+\frac {2\,a\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \cos ^{3}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*csc(d*x+c)*(a+a*sin(d*x+c)),x)

[Out]

a*(Integral(cos(c + d*x)**3*csc(c + d*x), x) + Integral(sin(c + d*x)*cos(c + d*x)**3*csc(c + d*x), x))

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