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

Optimal. Leaf size=53 \[ -\frac {a \sin ^2(c+d x)}{2 d}-\frac {a \sin (c+d x)}{d}-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 53, 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} \[ -\frac {a \sin ^2(c+d x)}{2 d}-\frac {a \sin (c+d x)}{d}-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 53, normalized size = 1.00 \[ -\frac {a \sin ^2(c+d x)}{2 d}-\frac {a \sin (c+d x)}{d}-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 68, normalized size = 1.28 \[ \frac {4 \, a \cos \left (d x + c\right )^{2} + 4 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + {\left (2 \, a \cos \left (d x + c\right )^{2} - a\right )} \sin \left (d x + c\right ) - 8 \, a}{4 \, d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 47, normalized size = 0.89 \[ -\frac {a \sin \left (d x + c\right )^{2} - 2 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 2 \, a \sin \left (d x + c\right ) + \frac {2 \, a}{\sin \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.28, size = 82, normalized size = 1.55 \[ \frac {a \left (\cos ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a \left (\cos ^{4}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a}{d}-\frac {2 a \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 46, normalized size = 0.87 \[ -\frac {a \sin \left (d x + c\right )^{2} - 2 \, a \log \left (\sin \left (d x + c\right )\right ) + 2 \, a \sin \left (d x + c\right ) + \frac {2 \, a}{\sin \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.70, size = 140, normalized size = 2.64 \[ \frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]

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)^2,x)

[Out]

(a*log(tan(c/2 + (d*x)/2)))/d - (a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (a*tan(c/2 + (d*x)/2))/(2*d) - (a + 6*a*
tan(c/2 + (d*x)/2)^2 + 4*a*tan(c/2 + (d*x)/2)^3 + 5*a*tan(c/2 + (d*x)/2)^4)/(d*(2*tan(c/2 + (d*x)/2) + 4*tan(c
/2 + (d*x)/2)^3 + 2*tan(c/2 + (d*x)/2)^5))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \cos ^{3}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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