3.1231 \(\int \frac{\cos ^2(c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=131 \[ \frac{\left (a^2-b^2\right )^2}{a^3 b^2 d (a+b \sin (c+d x))}-\frac{\left (2 a^2-3 b^2\right ) \log (\sin (c+d x))}{a^4 d}+\frac{\left (2 a^2 b^2+a^4-3 b^4\right ) \log (a+b \sin (c+d x))}{a^4 b^2 d}+\frac{2 b \csc (c+d x)}{a^3 d}-\frac{\csc ^2(c+d x)}{2 a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.199803, antiderivative size = 131, 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} \[ \frac{\left (a^2-b^2\right )^2}{a^3 b^2 d (a+b \sin (c+d x))}-\frac{\left (2 a^2-3 b^2\right ) \log (\sin (c+d x))}{a^4 d}+\frac{\left (2 a^2 b^2+a^4-3 b^4\right ) \log (a+b \sin (c+d x))}{a^4 b^2 d}+\frac{2 b \csc (c+d x)}{a^3 d}-\frac{\csc ^2(c+d x)}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.716844, size = 116, normalized size = 0.89 \[ \frac{\frac{2 a \left (a^2-b^2\right )^2}{b^2 (a+b \sin (c+d x))}-2 \left (2 a^2-3 b^2\right ) \log (\sin (c+d x))+\frac{2 \left (2 a^2 b^2+a^4-3 b^4\right ) \log (a+b \sin (c+d x))}{b^2}-a^2 \csc ^2(c+d x)+4 a b \csc (c+d x)}{2 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.154, size = 189, normalized size = 1.4 \begin{align*}{\frac{a}{d{b}^{2} \left ( a+b\sin \left ( dx+c \right ) \right ) }}-2\,{\frac{1}{da \left ( a+b\sin \left ( dx+c \right ) \right ) }}+{\frac{{b}^{2}}{d{a}^{3} \left ( a+b\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{b}^{2}}}+2\,{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{a}^{2}}}-3\,{\frac{{b}^{2}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{a}^{4}}}-{\frac{1}{2\,d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-2\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{d{a}^{2}}}+3\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ){b}^{2}}{d{a}^{4}}}+2\,{\frac{b}{d{a}^{3}\sin \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^3/(a+b*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.01919, size = 198, normalized size = 1.51 \begin{align*} \frac{\frac{3 \, a b^{3} \sin \left (d x + c\right ) - a^{2} b^{2} + 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + 3 \, b^{4}\right )} \sin \left (d x + c\right )^{2}}{a^{3} b^{3} \sin \left (d x + c\right )^{3} + a^{4} b^{2} \sin \left (d x + c\right )^{2}} - \frac{2 \,{\left (2 \, a^{2} - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{4}} + \frac{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} - 3 \, b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.19302, size = 743, normalized size = 5.67 \begin{align*} -\frac{3 \, a^{2} b^{3} \sin \left (d x + c\right ) + 2 \, a^{5} - 5 \, a^{3} b^{2} + 6 \, a b^{4} - 2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} - 3 \, a b^{4} -{\left (a^{5} + 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} +{\left (a^{4} b + 2 \, a^{2} b^{3} - 3 \, b^{5} -{\left (a^{4} b + 2 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \,{\left (2 \, a^{3} b^{2} - 3 \, a b^{4} -{\left (2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} +{\left (2 \, a^{2} b^{3} - 3 \, b^{5} -{\left (2 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \sin \left (d x + c\right )\right )}{2 \,{\left (a^{5} b^{2} d \cos \left (d x + c\right )^{2} - a^{5} b^{2} d +{\left (a^{4} b^{3} d \cos \left (d x + c\right )^{2} - a^{4} b^{3} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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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)**3/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.22335, size = 257, normalized size = 1.96 \begin{align*} -\frac{\frac{2 \,{\left (2 \, a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{4}} - \frac{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} - 3 \, b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b^{2}} + \frac{2 \,{\left (a^{4} \sin \left (d x + c\right ) + 2 \, a^{2} b^{2} \sin \left (d x + c\right ) - 3 \, b^{4} \sin \left (d x + c\right ) + 4 \, a^{3} b - 4 \, a b^{3}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} a^{4} b} - \frac{6 \, a^{2} \sin \left (d x + c\right )^{2} - 9 \, b^{2} \sin \left (d x + c\right )^{2} + 4 \, a b \sin \left (d x + c\right ) - a^{2}}{a^{4} \sin \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*(2*(2*a^2 - 3*b^2)*log(abs(sin(d*x + c)))/a^4 - 2*(a^4 + 2*a^2*b^2 - 3*b^4)*log(abs(b*sin(d*x + c) + a))/
(a^4*b^2) + 2*(a^4*sin(d*x + c) + 2*a^2*b^2*sin(d*x + c) - 3*b^4*sin(d*x + c) + 4*a^3*b - 4*a*b^3)/((b*sin(d*x
+ c) + a)*a^4*b) - (6*a^2*sin(d*x + c)^2 - 9*b^2*sin(d*x + c)^2 + 4*a*b*sin(d*x + c) - a^2)/(a^4*sin(d*x + c)
^2))/d