3.1230 \(\int \frac{\cos ^3(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=109 \[ -\frac{\left (a^2-b^2\right )^2}{a^2 b^3 d (a+b \sin (c+d x))}-\frac{2 \left (a^4-b^4\right ) \log (a+b \sin (c+d x))}{a^3 b^3 d}-\frac{2 b \log (\sin (c+d x))}{a^3 d}-\frac{\csc (c+d x)}{a^2 d}+\frac{\sin (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.170333, antiderivative size = 109, normalized size of antiderivative = 1., 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 = {2837, 12, 894} \[ -\frac{\left (a^2-b^2\right )^2}{a^2 b^3 d (a+b \sin (c+d x))}-\frac{2 \left (a^4-b^4\right ) \log (a+b \sin (c+d x))}{a^3 b^3 d}-\frac{2 b \log (\sin (c+d x))}{a^3 d}-\frac{\csc (c+d x)}{a^2 d}+\frac{\sin (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 ^3(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b^2 \left (b^2-x^2\right )^2}{x^2 (a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (b^2-x^2\right )^2}{x^2 (a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{b^4}{a^2 x^2}-\frac{2 b^4}{a^3 x}+\frac{\left (a^2-b^2\right )^2}{a^2 (a+x)^2}-\frac{2 \left (a^4-b^4\right )}{a^3 (a+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=-\frac{\csc (c+d x)}{a^2 d}-\frac{2 b \log (\sin (c+d x))}{a^3 d}-\frac{2 \left (a^4-b^4\right ) \log (a+b \sin (c+d x))}{a^3 b^3 d}+\frac{\sin (c+d x)}{b^2 d}-\frac{\left (a^2-b^2\right )^2}{a^2 b^3 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.592501, size = 95, normalized size = 0.87 \[ -\frac{\frac{\left (a^2-b^2\right )^2}{a^2 b^3 (a+b \sin (c+d x))}+2 \left (\frac{a}{b^3}-\frac{b}{a^3}\right ) \log (a+b \sin (c+d x))+\frac{2 b \log (\sin (c+d x))}{a^3}+\frac{\csc (c+d x)}{a^2}-\frac{\sin (c+d x)}{b^2}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.131, size = 151, normalized size = 1.4 \begin{align*}{\frac{\sin \left ( dx+c \right ) }{{b}^{2}d}}-2\,{\frac{a\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{{b}^{3}d}}+2\,{\frac{b\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{a}^{3}}}-{\frac{{a}^{2}}{{b}^{3}d \left ( a+b\sin \left ( dx+c \right ) \right ) }}+2\,{\frac{1}{bd \left ( a+b\sin \left ( dx+c \right ) \right ) }}-{\frac{b}{d{a}^{2} \left ( a+b\sin \left ( dx+c \right ) \right ) }}-{\frac{1}{d{a}^{2}\sin \left ( dx+c \right ) }}-2\,{\frac{b\ln \left ( \sin \left ( dx+c \right ) \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 0.99743, size = 162, normalized size = 1.49 \begin{align*} -\frac{\frac{a b^{3} +{\left (a^{4} - 2 \, a^{2} b^{2} + 2 \, b^{4}\right )} \sin \left (d x + c\right )}{a^{2} b^{4} \sin \left (d x + c\right )^{2} + a^{3} b^{3} \sin \left (d x + c\right )} + \frac{2 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac{\sin \left (d x + c\right )}{b^{2}} + \frac{2 \,{\left (a^{4} - b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{3} b^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-((a*b^3 + (a^4 - 2*a^2*b^2 + 2*b^4)*sin(d*x + c))/(a^2*b^4*sin(d*x + c)^2 + a^3*b^3*sin(d*x + c)) + 2*b*log(s
in(d*x + c))/a^3 - sin(d*x + c)/b^2 + 2*(a^4 - b^4)*log(b*sin(d*x + c) + a)/(a^3*b^3))/d

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Fricas [A]  time = 2.13863, size = 473, normalized size = 4.34 \begin{align*} \frac{a^{4} b \cos \left (d x + c\right )^{2} - a^{4} b + a^{2} b^{3} + 2 \,{\left (a^{4} b - b^{5} -{\left (a^{4} b - b^{5}\right )} \cos \left (d x + c\right )^{2} +{\left (a^{5} - a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \,{\left (b^{5} \cos \left (d x + c\right )^{2} - a b^{4} \sin \left (d x + c\right ) - b^{5}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) +{\left (a^{3} b^{2} \cos \left (d x + c\right )^{2} + a^{5} - 3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \sin \left (d x + c\right )}{a^{3} b^{4} d \cos \left (d x + c\right )^{2} - a^{4} b^{3} d \sin \left (d x + c\right ) - a^{3} b^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

(a^4*b*cos(d*x + c)^2 - a^4*b + a^2*b^3 + 2*(a^4*b - b^5 - (a^4*b - b^5)*cos(d*x + c)^2 + (a^5 - a*b^4)*sin(d*
x + c))*log(b*sin(d*x + c) + a) - 2*(b^5*cos(d*x + c)^2 - a*b^4*sin(d*x + c) - b^5)*log(1/2*sin(d*x + c)) + (a
^3*b^2*cos(d*x + c)^2 + a^5 - 3*a^3*b^2 + 2*a*b^4)*sin(d*x + c))/(a^3*b^4*d*cos(d*x + c)^2 - a^4*b^3*d*sin(d*x
+ c) - a^3*b^4*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**2/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.21541, size = 177, normalized size = 1.62 \begin{align*} -\frac{\frac{2 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac{\sin \left (d x + c\right )}{b^{2}} - \frac{a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{2} b \sin \left (d x + c\right ) - 2 \, b^{3} \sin \left (d x + c\right ) - a b^{2}}{{\left (b \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right )\right )} a^{2} b^{2}} + \frac{2 \,{\left (a^{4} - b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{3} b^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-(2*b*log(abs(sin(d*x + c)))/a^3 - sin(d*x + c)/b^2 - (a^3*sin(d*x + c)^2 + 2*a^2*b*sin(d*x + c) - 2*b^3*sin(d
*x + c) - a*b^2)/((b*sin(d*x + c)^2 + a*sin(d*x + c))*a^2*b^2) + 2*(a^4 - b^4)*log(abs(b*sin(d*x + c) + a))/(a
^3*b^3))/d