3.5.46 \(\int \frac {\cot ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) [446]

Optimal. Leaf size=63 \[ -\frac {\csc ^2(c+d x)}{2 a d}-\frac {(a+b) \log (\sin (c+d x))}{a^2 d}+\frac {(a+b) \log \left (a+b \sin ^2(c+d x)\right )}{2 a^2 d} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3273, 78} \begin {gather*} \frac {(a+b) \log \left (a+b \sin ^2(c+d x)\right )}{2 a^2 d}-\frac {(a+b) \log (\sin (c+d x))}{a^2 d}-\frac {\csc ^2(c+d x)}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 3273

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

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 50, normalized size = 0.79 \begin {gather*} -\frac {a \csc ^2(c+d x)+(a+b) \left (2 \log (\sin (c+d x))-\log \left (a+b \sin ^2(c+d x)\right )\right )}{2 a^2 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*(a*Csc[c + d*x]^2 + (a + b)*(2*Log[Sin[c + d*x]] - Log[a + b*Sin[c + d*x]^2]))/(a^2*d)

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Maple [A]
time = 0.53, size = 101, normalized size = 1.60

method	result	size
derivativedivides	\(\frac {\frac {1}{4 a \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (-a -b \right ) \ln \left (\cos \left (d x +c \right )-1\right )}{2 a^{2}}-\frac {1}{4 a \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (-a -b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{2 a^{2}}+\frac {\left (a +b \right ) \ln \left (a +b -b \left (\cos ^{2}\left (d x +c \right )\right )\right )}{2 a^{2}}}{d}\)	\(101\)
default	\(\frac {\frac {1}{4 a \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (-a -b \right ) \ln \left (\cos \left (d x +c \right )-1\right )}{2 a^{2}}-\frac {1}{4 a \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (-a -b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{2 a^{2}}+\frac {\left (a +b \right ) \ln \left (a +b -b \left (\cos ^{2}\left (d x +c \right )\right )\right )}{2 a^{2}}}{d}\)	\(101\)
risch	\(\frac {2 \,{\mathrm e}^{2 i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{4 i \left (d x +c \right )}-\frac {2 \left (2 a +b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{b}+1\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{4 i \left (d x +c \right )}-\frac {2 \left (2 a +b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{b}+1\right ) b}{2 a^{2} d}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 56, normalized size = 0.89 \begin {gather*} \frac {\frac {{\left (a + b\right )} \log \left (b \sin \left (d x + c\right )^{2} + a\right )}{a^{2}} - \frac {{\left (a + b\right )} \log \left (\sin \left (d x + c\right )^{2}\right )}{a^{2}} - \frac {1}{a \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="maxima")

[Out]

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

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Fricas [A]
time = 0.44, size = 91, normalized size = 1.44 \begin {gather*} \frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \log \left (-b \cos \left (d x + c\right )^{2} + a + b\right ) - 2 \, {\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + a}{2 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\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*(((a + b)*cos(d*x + c)^2 - a - b)*log(-b*cos(d*x + c)^2 + a + b) - 2*((a + b)*cos(d*x + c)^2 - a - b)*log(
1/2*sin(d*x + c)) + a)/(a^2*d*cos(d*x + c)^2 - a^2*d)

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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 )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, 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 = 0.55, size = 108, normalized size = 1.71 \begin {gather*} \frac {\frac {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}}{a} + \frac {4 \, {\left (a + b\right )} \log \left ({\left | -a {\left (\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )} + 2 \, a + 4 \, b \right |}\right )}{a^{2}}}{8 \, 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]

1/8*(((cos(d*x + c) + 1)/(cos(d*x + c) - 1) + (cos(d*x + c) - 1)/(cos(d*x + c) + 1))/a + 4*(a + b)*log(abs(-a*
((cos(d*x + c) + 1)/(cos(d*x + c) - 1) + (cos(d*x + c) - 1)/(cos(d*x + c) + 1)) + 2*a + 4*b))/a^2)/d

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Mupad [B]
time = 14.46, size = 69, normalized size = 1.10 \begin {gather*} \frac {\ln \left (a+a\,{\mathrm {tan}\left (c+d\,x\right )}^2+b\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )\,\left (a+b\right )}{2\,a^2\,d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2}{2\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a+b\right )}{a^2\,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]

(log(a + a*tan(c + d*x)^2 + b*tan(c + d*x)^2)*(a + b))/(2*a^2*d) - cot(c + d*x)^2/(2*a*d) - (log(tan(c + d*x))
*(a + b))/(a^2*d)

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