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

Optimal. Leaf size=145 \[ -\frac{a^2-3 b^2}{a^4 d (a+b \sin (c+d x))}-\frac{a^2-b^2}{2 a^3 d (a+b \sin (c+d x))^2}-\frac{\left (a^2-6 b^2\right ) \log (\sin (c+d x))}{a^5 d}+\frac{\left (a^2-6 b^2\right ) \log (a+b \sin (c+d x))}{a^5 d}+\frac{3 b \csc (c+d x)}{a^4 d}-\frac{\csc ^2(c+d x)}{2 a^3 d} \]

[Out]

(3*b*Csc[c + d*x])/(a^4*d) - Csc[c + d*x]^2/(2*a^3*d) - ((a^2 - 6*b^2)*Log[Sin[c + d*x]])/(a^5*d) + ((a^2 - 6*
b^2)*Log[a + b*Sin[c + d*x]])/(a^5*d) - (a^2 - b^2)/(2*a^3*d*(a + b*Sin[c + d*x])^2) - (a^2 - 3*b^2)/(a^4*d*(a
+ b*Sin[c + d*x]))

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Rubi [A]  time = 0.133294, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2721, 894} \[ -\frac{a^2-3 b^2}{a^4 d (a+b \sin (c+d x))}-\frac{a^2-b^2}{2 a^3 d (a+b \sin (c+d x))^2}-\frac{\left (a^2-6 b^2\right ) \log (\sin (c+d x))}{a^5 d}+\frac{\left (a^2-6 b^2\right ) \log (a+b \sin (c+d x))}{a^5 d}+\frac{3 b \csc (c+d x)}{a^4 d}-\frac{\csc ^2(c+d x)}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*Csc[c + d*x])/(a^4*d) - Csc[c + d*x]^2/(2*a^3*d) - ((a^2 - 6*b^2)*Log[Sin[c + d*x]])/(a^5*d) + ((a^2 - 6*
b^2)*Log[a + b*Sin[c + d*x]])/(a^5*d) - (a^2 - b^2)/(2*a^3*d*(a + b*Sin[c + d*x])^2) - (a^2 - 3*b^2)/(a^4*d*(a
+ b*Sin[c + d*x]))

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

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{\cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b^2-x^2}{x^3 (a+x)^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{b^2}{a^3 x^3}-\frac{3 b^2}{a^4 x^2}+\frac{-a^2+6 b^2}{a^5 x}+\frac{a^2-b^2}{a^3 (a+x)^3}+\frac{a^2-3 b^2}{a^4 (a+x)^2}+\frac{a^2-6 b^2}{a^5 (a+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{3 b \csc (c+d x)}{a^4 d}-\frac{\csc ^2(c+d x)}{2 a^3 d}-\frac{\left (a^2-6 b^2\right ) \log (\sin (c+d x))}{a^5 d}+\frac{\left (a^2-6 b^2\right ) \log (a+b \sin (c+d x))}{a^5 d}-\frac{a^2-b^2}{2 a^3 d (a+b \sin (c+d x))^2}-\frac{a^2-3 b^2}{a^4 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.931238, size = 121, normalized size = 0.83 \[ -\frac{\frac{2 a \left (a^2-3 b^2\right )}{a+b \sin (c+d x)}+2 \left (a^2-6 b^2\right ) \log (\sin (c+d x))-2 \left (a^2-6 b^2\right ) \log (a+b \sin (c+d x))+\frac{a^2 (a-b) (a+b)}{(a+b \sin (c+d x))^2}+a^2 \csc ^2(c+d x)-6 a b \csc (c+d x)}{2 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-6*a*b*Csc[c + d*x] + a^2*Csc[c + d*x]^2 + 2*(a^2 - 6*b^2)*Log[Sin[c + d*x]] - 2*(a^2 - 6*b^2)*Log[a + b*Sin
[c + d*x]] + (a^2*(a - b)*(a + b))/(a + b*Sin[c + d*x])^2 + (2*a*(a^2 - 3*b^2))/(a + b*Sin[c + d*x]))/(2*a^5*d
)

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Maple [A]  time = 0.111, size = 194, normalized size = 1.3 \begin{align*}{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{{a}^{3}d}}-6\,{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ){b}^{2}}{d{a}^{5}}}-{\frac{1}{{a}^{2}d \left ( a+b\sin \left ( dx+c \right ) \right ) }}+3\,{\frac{{b}^{2}}{d{a}^{4} \left ( a+b\sin \left ( dx+c \right ) \right ) }}-{\frac{1}{2\,da \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{2}}{2\,{a}^{3}d \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{1}{2\,{a}^{3}d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{{a}^{3}d}}+6\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ){b}^{2}}{d{a}^{5}}}+3\,{\frac{b}{d{a}^{4}\sin \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.92069, size = 211, normalized size = 1.46 \begin{align*} \frac{\frac{4 \, a^{2} b \sin \left (d x + c\right ) - 2 \,{\left (a^{2} b - 6 \, b^{3}\right )} \sin \left (d x + c\right )^{3} - a^{3} - 3 \,{\left (a^{3} - 6 \, a b^{2}\right )} \sin \left (d x + c\right )^{2}}{a^{4} b^{2} \sin \left (d x + c\right )^{4} + 2 \, a^{5} b \sin \left (d x + c\right )^{3} + a^{6} \sin \left (d x + c\right )^{2}} + \frac{2 \,{\left (a^{2} - 6 \, b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5}} - \frac{2 \,{\left (a^{2} - 6 \, b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{5}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.3421, size = 911, normalized size = 6.28 \begin{align*} -\frac{4 \, a^{4} - 18 \, a^{2} b^{2} - 3 \,{\left (a^{4} - 6 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left ({\left (a^{2} b^{2} - 6 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 5 \, a^{2} b^{2} - 6 \, b^{4} -{\left (a^{4} - 4 \, a^{2} b^{2} - 12 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{3} b - 6 \, a b^{3} -{\left (a^{3} b - 6 \, a b^{3}\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 ({\left (a^{2} b^{2} - 6 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 5 \, a^{2} b^{2} - 6 \, b^{4} -{\left (a^{4} - 4 \, a^{2} b^{2} - 12 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{3} b - 6 \, a b^{3} -{\left (a^{3} b - 6 \, a b^{3}\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^{3} b + 6 \, a b^{3} +{\left (a^{3} b - 6 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{5} b^{2} d \cos \left (d x + c\right )^{4} -{\left (a^{7} + 2 \, a^{5} b^{2}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{7} + a^{5} b^{2}\right )} d - 2 \,{\left (a^{6} b d \cos \left (d x + c\right )^{2} - a^{6} b d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*(4*a^4 - 18*a^2*b^2 - 3*(a^4 - 6*a^2*b^2)*cos(d*x + c)^2 - 2*((a^2*b^2 - 6*b^4)*cos(d*x + c)^4 + a^4 - 5*
a^2*b^2 - 6*b^4 - (a^4 - 4*a^2*b^2 - 12*b^4)*cos(d*x + c)^2 + 2*(a^3*b - 6*a*b^3 - (a^3*b - 6*a*b^3)*cos(d*x +
c)^2)*sin(d*x + c))*log(b*sin(d*x + c) + a) + 2*((a^2*b^2 - 6*b^4)*cos(d*x + c)^4 + a^4 - 5*a^2*b^2 - 6*b^4 -
(a^4 - 4*a^2*b^2 - 12*b^4)*cos(d*x + c)^2 + 2*(a^3*b - 6*a*b^3 - (a^3*b - 6*a*b^3)*cos(d*x + c)^2)*sin(d*x +
c))*log(-1/2*sin(d*x + c)) - 2*(a^3*b + 6*a*b^3 + (a^3*b - 6*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a^5*b^2*d*c
os(d*x + c)^4 - (a^7 + 2*a^5*b^2)*d*cos(d*x + c)^2 + (a^7 + a^5*b^2)*d - 2*(a^6*b*d*cos(d*x + c)^2 - a^6*b*d)*
sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{3}{\left (c + d x \right )}}{\left (a + b \sin{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.80037, size = 208, normalized size = 1.43 \begin{align*} -\frac{\frac{2 \,{\left (a^{2} - 6 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{5}} - \frac{2 \,{\left (a^{2} b - 6 \, b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b} + \frac{2 \, a^{2} b \sin \left (d x + c\right )^{3} - 12 \, b^{3} \sin \left (d x + c\right )^{3} + 3 \, a^{3} \sin \left (d x + c\right )^{2} - 18 \, a b^{2} \sin \left (d x + c\right )^{2} - 4 \, a^{2} b \sin \left (d x + c\right ) + a^{3}}{{\left (b \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right )\right )}^{2} a^{4}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(2*(a^2 - 6*b^2)*log(abs(sin(d*x + c)))/a^5 - 2*(a^2*b - 6*b^3)*log(abs(b*sin(d*x + c) + a))/(a^5*b) + (2
*a^2*b*sin(d*x + c)^3 - 12*b^3*sin(d*x + c)^3 + 3*a^3*sin(d*x + c)^2 - 18*a*b^2*sin(d*x + c)^2 - 4*a^2*b*sin(d
*x + c) + a^3)/((b*sin(d*x + c)^2 + a*sin(d*x + c))^2*a^4))/d