3.2.85 \(\int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [185]

Optimal. Leaf size=114 \[ \frac {2 b \csc (c+d x)}{a^3 d}-\frac {\csc ^2(c+d x)}{2 a^2 d}-\frac {\left (a^2-3 b^2\right ) \log (\sin (c+d x))}{a^4 d}+\frac {\left (a^2-3 b^2\right ) \log (a+b \sin (c+d x))}{a^4 d}-\frac {a^2-b^2}{a^3 d (a+b \sin (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 114, normalized size of antiderivative = 1.00, 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 = {2800, 908} \begin {gather*} \frac {2 b \csc (c+d x)}{a^3 d}-\frac {\csc ^2(c+d x)}{2 a^2 d}-\frac {\left (a^2-3 b^2\right ) \log (\sin (c+d x))}{a^4 d}+\frac {\left (a^2-3 b^2\right ) \log (a+b \sin (c+d x))}{a^4 d}-\frac {a^2-b^2}{a^3 d (a+b \sin (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

Rule 908

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]))

Rule 2800

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]

Rubi steps

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

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Mathematica [A]
time = 0.42, size = 96, normalized size = 0.84 \begin {gather*} -\frac {-4 a b \csc (c+d x)+a^2 \csc ^2(c+d x)+2 \left (a^2-3 b^2\right ) \log (\sin (c+d x))-2 \left (a^2-3 b^2\right ) \log (a+b \sin (c+d x))+\frac {2 a (a-b) (a+b)}{a+b \sin (c+d x)}}{2 a^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(-4*a*b*Csc[c + d*x] + a^2*Csc[c + d*x]^2 + 2*(a^2 - 3*b^2)*Log[Sin[c + d*x]] - 2*(a^2 - 3*b^2)*Log[a + b
*Sin[c + d*x]] + (2*a*(a - b)*(a + b))/(a + b*Sin[c + d*x]))/(a^4*d)

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Maple [A]
time = 0.36, size = 105, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {\frac {\left (a^{2}-3 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{4}}-\frac {a^{2}-b^{2}}{a^{3} \left (a +b \sin \left (d x +c \right )\right )}-\frac {1}{2 a^{2} \sin \left (d x +c \right )^{2}}+\frac {\left (-a^{2}+3 b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{4}}+\frac {2 b}{a^{3} \sin \left (d x +c \right )}}{d}\)	\(105\)
default	\(\frac {\frac {\left (a^{2}-3 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{4}}-\frac {a^{2}-b^{2}}{a^{3} \left (a +b \sin \left (d x +c \right )\right )}-\frac {1}{2 a^{2} \sin \left (d x +c \right )^{2}}+\frac {\left (-a^{2}+3 b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{4}}+\frac {2 b}{a^{3} \sin \left (d x +c \right )}}{d}\)	\(105\)
risch	\(-\frac {2 i \left (-3 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+3 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}+6 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )} b^{2}+a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+a^{2} {\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right ) d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{a^{2} d}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) b^{2}}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{a^{4} d}\)	\(282\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 116, normalized size = 1.02 \begin {gather*} \frac {\frac {3 \, a b \sin \left (d x + c\right ) - 2 \, {\left (a^{2} - 3 \, b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2}}{a^{3} b \sin \left (d x + c\right )^{3} + a^{4} \sin \left (d x + c\right )^{2}} + \frac {2 \, {\left (a^{2} - 3 \, b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4}} - \frac {2 \, {\left (a^{2} - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (112) = 224\).
time = 0.39, size = 259, normalized size = 2.27 \begin {gather*} -\frac {3 \, a^{2} b \sin \left (d x + c\right ) - 3 \, a^{3} + 6 \, a b^{2} + 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} - 3 \, a b^{2} - {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{2} b - 3 \, b^{3} - {\left (a^{2} b - 3 \, 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 (a^{3} - 3 \, a b^{2} - {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{2} b - 3 \, b^{3} - {\left (a^{2} b - 3 \, 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^{5} d \cos \left (d x + c\right )^{2} - a^{5} d + {\left (a^{4} b d \cos \left (d x + c\right )^{2} - a^{4} b d\right )} \sin \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 5.94, size = 165, normalized size = 1.45 \begin {gather*} -\frac {\frac {2 \, {\left (a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{4}} - \frac {2 \, {\left (a^{2} b - 3 \, b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b} + \frac {2 \, {\left (a^{2} b \sin \left (d x + c\right ) - 3 \, b^{3} \sin \left (d x + c\right ) + 2 \, a^{3} - 4 \, a b^{2}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} a^{4}} - \frac {3 \, 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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 6.38, size = 235, normalized size = 2.06 \begin {gather*} \frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2}{2}-8\,b^2\right )+\frac {a^2}{2}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^2\,b-2\,b^3\right )}{a}-3\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+8\,b\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-3\,b^2\right )}{a^4\,d}+\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^2-3\,b^2\right )}{a^4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*tan(c/2 + (d*x)/2))/(a^3*d) - (tan(c/2 + (d*x)/2)^2*(a^2/2 - 8*b^2) + a^2/2 - (4*tan(c/2 + (d*x)/2)^3*(3*a^
2*b - 2*b^3))/a - 3*a*b*tan(c/2 + (d*x)/2))/(d*(4*a^4*tan(c/2 + (d*x)/2)^2 + 4*a^4*tan(c/2 + (d*x)/2)^4 + 8*a^
3*b*tan(c/2 + (d*x)/2)^3)) - tan(c/2 + (d*x)/2)^2/(8*a^2*d) - (log(tan(c/2 + (d*x)/2))*(a^2 - 3*b^2))/(a^4*d)
+ (log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(a^2 - 3*b^2))/(a^4*d)

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