3.353 \(\int \frac {\cot (e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=83 \[ \frac {b^2}{2 a^2 f (a+b) \left (a \cos ^2(e+f x)+b\right )}+\frac {b (2 a+b) \log \left (a \cos ^2(e+f x)+b\right )}{2 a^2 f (a+b)^2}+\frac {\log (\sin (e+f x))}{f (a+b)^2} \]

[Out]

1/2*b^2/a^2/(a+b)/f/(b+a*cos(f*x+e)^2)+1/2*b*(2*a+b)*ln(b+a*cos(f*x+e)^2)/a^2/(a+b)^2/f+ln(sin(f*x+e))/(a+b)^2
/f

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Rubi [A]  time = 0.11, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4138, 446, 88} \[ \frac {b^2}{2 a^2 f (a+b) \left (a \cos ^2(e+f x)+b\right )}+\frac {b (2 a+b) \log \left (a \cos ^2(e+f x)+b\right )}{2 a^2 f (a+b)^2}+\frac {\log (\sin (e+f x))}{f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

b^2/(2*a^2*(a + b)*f*(b + a*Cos[e + f*x]^2)) + (b*(2*a + b)*Log[b + a*Cos[e + f*x]^2])/(2*a^2*(a + b)^2*f) + L
og[Sin[e + f*x]]/((a + b)^2*f)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4138

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\cot (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x^5}{\left (1-x^2\right ) \left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x^2}{(1-x) (b+a x)^2} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {\operatorname {Subst}\left (\int \left (-\frac {1}{(a+b)^2 (-1+x)}+\frac {b^2}{a (a+b) (b+a x)^2}-\frac {b (2 a+b)}{a (a+b)^2 (b+a x)}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac {b^2}{2 a^2 (a+b) f \left (b+a \cos ^2(e+f x)\right )}+\frac {b (2 a+b) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^2 (a+b)^2 f}+\frac {\log (\sin (e+f x))}{(a+b)^2 f}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 112, normalized size = 1.35 \[ \frac {\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b)^2 \left (\frac {b^2 (a+b)}{a^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac {b (2 a+b) \log \left (-a \sin ^2(e+f x)+a+b\right )}{a^2}+2 \log (\sin (e+f x))\right )}{8 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])^2*Sec[e + f*x]^4*(2*Log[Sin[e + f*x]] + (b*(2*a + b)*Log[a + b - a*Sin[e + f*x
]^2])/a^2 + (b^2*(a + b))/(a^2*(a + b - a*Sin[e + f*x]^2))))/(8*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^2)

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fricas [A]  time = 0.72, size = 138, normalized size = 1.66 \[ \frac {a b^{2} + b^{3} + {\left (2 \, a b^{2} + b^{3} + {\left (2 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) + 2 \, {\left (a^{3} \cos \left (f x + e\right )^{2} + a^{2} b\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right )}{2 \, {\left ({\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/2*(a*b^2 + b^3 + (2*a*b^2 + b^3 + (2*a^2*b + a*b^2)*cos(f*x + e)^2)*log(a*cos(f*x + e)^2 + b) + 2*(a^3*cos(f
*x + e)^2 + a^2*b)*log(1/2*sin(f*x + e)))/((a^5 + 2*a^4*b + a^3*b^2)*f*cos(f*x + e)^2 + (a^4*b + 2*a^3*b^2 + a
^2*b^3)*f)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2/f*(1/(4*a^2+8*a*
b+4*b^2)*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1))))-1/2/a^2*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1
)))+1))+(2*a*b+b^2)/(4*a^4+8*a^3*b+4*a^2*b^2)*ln(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a+((1-cos(f*x+exp
(1)))/(1+cos(f*x+exp(1))))^2*b-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+2*(1-cos(f*x+exp(1)))/(1+cos(f*x+ex
p(1)))*b+a+b)+(-2*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a*b-((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*
b^2+4*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a*b-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b^2-2*a*b-b^2)/(4*
a^3+4*a^2*b)/(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a+((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b-2*(1
-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b+a+b))

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maple [A]  time = 1.09, size = 155, normalized size = 1.87 \[ \frac {b \ln \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{f \left (a +b \right )^{2} a}+\frac {b^{2} \ln \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{2 f \left (a +b \right )^{2} a^{2}}+\frac {b^{2}}{2 f \left (a +b \right )^{2} a \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}+\frac {b^{3}}{2 a^{2} \left (a +b \right )^{2} f \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}+\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{2 f \left (a +b \right )^{2}}+\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{2 f \left (a +b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)/(a+b*sec(f*x+e)^2)^2,x)

[Out]

1/f*b/(a+b)^2/a*ln(b+a*cos(f*x+e)^2)+1/2/f*b^2/(a+b)^2/a^2*ln(b+a*cos(f*x+e)^2)+1/2/f*b^2/(a+b)^2/a/(b+a*cos(f
*x+e)^2)+1/2*b^3/a^2/(a+b)^2/f/(b+a*cos(f*x+e)^2)+1/2/f/(a+b)^2*ln(-1+cos(f*x+e))+1/2/f/(a+b)^2*ln(1+cos(f*x+e
))

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maxima [A]  time = 0.35, size = 117, normalized size = 1.41 \[ \frac {\frac {b^{2}}{a^{4} + 2 \, a^{3} b + a^{2} b^{2} - {\left (a^{4} + a^{3} b\right )} \sin \left (f x + e\right )^{2}} + \frac {{\left (2 \, a b + b^{2}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{4} + 2 \, a^{3} b + a^{2} b^{2}} + \frac {\log \left (\sin \left (f x + e\right )^{2}\right )}{a^{2} + 2 \, a b + b^{2}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.85, size = 106, normalized size = 1.28 \[ \frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )}{f\,\left (a^2+2\,a\,b+b^2\right )}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a^2\,f}-\frac {b}{2\,a\,f\,\left (a+b\right )\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}+\frac {b\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )\,\left (2\,a+b\right )}{2\,a^2\,f\,{\left (a+b\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)/(a + b/cos(e + f*x)^2)^2,x)

[Out]

log(tan(e + f*x))/(f*(2*a*b + a^2 + b^2)) - log(tan(e + f*x)^2 + 1)/(2*a^2*f) - b/(2*a*f*(a + b)*(a + b + b*ta
n(e + f*x)^2)) + (b*log(a + b + b*tan(e + f*x)^2)*(2*a + b))/(2*a^2*f*(a + b)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot {\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Integral(cot(e + f*x)/(a + b*sec(e + f*x)**2)**2, x)

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