3.340 \(\int \frac{\cos (c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=60 \[ -\frac{\left (1-\frac{b^2}{a^2}\right ) \log (a+b \sin (c+d x))}{b d}-\frac{b \log (\sin (c+d x))}{a^2 d}-\frac{\csc (c+d x)}{a d} \]

[Out]

-(Csc[c + d*x]/(a*d)) - (b*Log[Sin[c + d*x]])/(a^2*d) - ((1 - b^2/a^2)*Log[a + b*Sin[c + d*x]])/(b*d)

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Rubi [A]  time = 0.123461, antiderivative size = 60, normalized size of antiderivative = 1., 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 = {2837, 12, 894} \[ -\frac{\left (1-\frac{b^2}{a^2}\right ) \log (a+b \sin (c+d x))}{b d}-\frac{b \log (\sin (c+d x))}{a^2 d}-\frac{\csc (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]/(a*d)) - (b*Log[Sin[c + d*x]])/(a^2*d) - ((1 - b^2/a^2)*Log[a + b*Sin[c + d*x]])/(b*d)

Rule 2837

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*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

\begin{align*} \int \frac{\cos (c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b^2 \left (b^2-x^2\right )}{x^2 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{b^2-x^2}{x^2 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{b^2}{a x^2}-\frac{b^2}{a^2 x}+\frac{-a^2+b^2}{a^2 (a+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{b d}\\ &=-\frac{\csc (c+d x)}{a d}-\frac{b \log (\sin (c+d x))}{a^2 d}-\frac{\left (1-\frac{b^2}{a^2}\right ) \log (a+b \sin (c+d x))}{b d}\\ \end{align*}

Mathematica [A]  time = 0.0937522, size = 54, normalized size = 0.9 \[ \frac{\left (b^2-a^2\right ) \log (a+b \sin (c+d x))-a b \csc (c+d x)+b^2 (-\log (\sin (c+d x)))}{a^2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 72, normalized size = 1.2 \begin{align*} -{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{bd}}+{\frac{b\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{a}^{2}}}-{\frac{1}{da\sin \left ( dx+c \right ) }}-{\frac{b\ln \left ( \sin \left ( dx+c \right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*cot(d*x+c)^2/(a+b*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.973981, size = 77, normalized size = 1.28 \begin{align*} -\frac{\frac{b \log \left (\sin \left (d x + c\right )\right )}{a^{2}} + \frac{{\left (a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} b} + \frac{1}{a \sin \left (d x + c\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.95147, size = 166, normalized size = 2.77 \begin{align*} -\frac{b^{2} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) + a b}{a^{2} b d \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}}{a + b \sin{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*cot(d*x+c)**2/(a+b*sin(d*x+c)),x)

[Out]

Integral(cos(c + d*x)*cot(c + d*x)**2/(a + b*sin(c + d*x)), x)

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Giac [A]  time = 1.15671, size = 97, normalized size = 1.62 \begin{align*} -\frac{\frac{b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac{{\left (a^{2} - b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b} - \frac{b \sin \left (d x + c\right ) - a}{a^{2} \sin \left (d x + c\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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