3.366 \(\int \frac{\cot (e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=130 \[ -\frac{b^3}{4 a^3 f (a+b) \left (a \cos ^2(e+f x)+b\right )^2}+\frac{b^2 (3 a+2 b)}{2 a^3 f (a+b)^2 \left (a \cos ^2(e+f x)+b\right )}+\frac{b \left (3 a^2+3 a b+b^2\right ) \log \left (a \cos ^2(e+f x)+b\right )}{2 a^3 f (a+b)^3}+\frac{\log (\sin (e+f x))}{f (a+b)^3} \]

[Out]

-b^3/(4*a^3*(a + b)*f*(b + a*Cos[e + f*x]^2)^2) + (b^2*(3*a + 2*b))/(2*a^3*(a + b)^2*f*(b + a*Cos[e + f*x]^2))
+ (b*(3*a^2 + 3*a*b + b^2)*Log[b + a*Cos[e + f*x]^2])/(2*a^3*(a + b)^3*f) + Log[Sin[e + f*x]]/((a + b)^3*f)

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Rubi [A]  time = 0.165065, antiderivative size = 130, normalized size of antiderivative = 1., 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} \[ -\frac{b^3}{4 a^3 f (a+b) \left (a \cos ^2(e+f x)+b\right )^2}+\frac{b^2 (3 a+2 b)}{2 a^3 f (a+b)^2 \left (a \cos ^2(e+f x)+b\right )}+\frac{b \left (3 a^2+3 a b+b^2\right ) \log \left (a \cos ^2(e+f x)+b\right )}{2 a^3 f (a+b)^3}+\frac{\log (\sin (e+f x))}{f (a+b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

-b^3/(4*a^3*(a + b)*f*(b + a*Cos[e + f*x]^2)^2) + (b^2*(3*a + 2*b))/(2*a^3*(a + b)^2*f*(b + a*Cos[e + f*x]^2))
+ (b*(3*a^2 + 3*a*b + b^2)*Log[b + a*Cos[e + f*x]^2])/(2*a^3*(a + b)^3*f) + Log[Sin[e + f*x]]/((a + b)^3*f)

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]

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 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]))

Rubi steps

\begin{align*} \int \frac{\cot (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^7}{\left (1-x^2\right ) \left (b+a x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^3}{(1-x) (b+a x)^3} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{(a+b)^3 (-1+x)}-\frac{b^3}{a^2 (a+b) (b+a x)^3}+\frac{b^2 (3 a+2 b)}{a^2 (a+b)^2 (b+a x)^2}-\frac{b \left (3 a^2+3 a b+b^2\right )}{a^2 (a+b)^3 (b+a x)}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{b^3}{4 a^3 (a+b) f \left (b+a \cos ^2(e+f x)\right )^2}+\frac{b^2 (3 a+2 b)}{2 a^3 (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )}+\frac{b \left (3 a^2+3 a b+b^2\right ) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^3 (a+b)^3 f}+\frac{\log (\sin (e+f x))}{(a+b)^3 f}\\ \end{align*}

Mathematica [A]  time = 1.09815, size = 158, normalized size = 1.22 \[ \frac{\sec ^6(e+f x) (a \cos (2 (e+f x))+a+2 b)^3 \left (-\frac{b^3 (a+b)^2}{a^3 \left (-a \sin ^2(e+f x)+a+b\right )^2}+\frac{2 b^2 (a+b) (3 a+2 b)}{a^3 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{2 b \left (3 a^2+3 a b+b^2\right ) \log \left (-a \sin ^2(e+f x)+a+b\right )}{a^3}+4 \log (\sin (e+f x))\right )}{32 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^6*(4*Log[Sin[e + f*x]] + (2*b*(3*a^2 + 3*a*b + b^2)*Log[a + b -
a*Sin[e + f*x]^2])/a^3 - (b^3*(a + b)^2)/(a^3*(a + b - a*Sin[e + f*x]^2)^2) + (2*b^2*(a + b)*(3*a + 2*b))/(a^
3*(a + b - a*Sin[e + f*x]^2))))/(32*(a + b)^3*f*(a + b*Sec[e + f*x]^2)^3)

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Maple [B]  time = 0.102, size = 304, normalized size = 2.3 \begin{align*}{\frac{3\,b\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f \left ( a+b \right ) ^{3}a}}+{\frac{3\,{b}^{2}\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f \left ( a+b \right ) ^{3}{a}^{2}}}+{\frac{{b}^{3}\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f \left ( a+b \right ) ^{3}{a}^{3}}}+{\frac{3\,{b}^{2}}{2\,f \left ( a+b \right ) ^{3}a \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{5\,{b}^{3}}{2\,f \left ( a+b \right ) ^{3}{a}^{2} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{{b}^{4}}{f \left ( a+b \right ) ^{3}{a}^{3} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{{b}^{3}}{4\,f \left ( a+b \right ) ^{3}a \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{b}^{4}}{2\,f \left ( a+b \right ) ^{3}{a}^{2} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{b}^{5}}{4\,f \left ( a+b \right ) ^{3}{a}^{3} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{\ln \left ( 1+\cos \left ( fx+e \right ) \right ) }{2\,f \left ( a+b \right ) ^{3}}}+{\frac{\ln \left ( -1+\cos \left ( fx+e \right ) \right ) }{2\,f \left ( a+b \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)/(a+b*sec(f*x+e)^2)^3,x)

[Out]

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

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Maxima [A]  time = 1.01423, size = 328, normalized size = 2.52 \begin{align*} \frac{\frac{2 \,{\left (3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}} + \frac{6 \, a^{2} b^{2} + 9 \, a b^{3} + 3 \, b^{4} - 2 \,{\left (3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \sin \left (f x + e\right )^{2}}{a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4} +{\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} \sin \left (f x + e\right )^{4} - 2 \,{\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} \sin \left (f x + e\right )^{2}} + \frac{2 \, \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

1/4*(2*(3*a^2*b + 3*a*b^2 + b^3)*log(a*sin(f*x + e)^2 - a - b)/(a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3) + (6*a^2*
b^2 + 9*a*b^3 + 3*b^4 - 2*(3*a^2*b^2 + 2*a*b^3)*sin(f*x + e)^2)/(a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b
^4 + (a^7 + 2*a^6*b + a^5*b^2)*sin(f*x + e)^4 - 2*(a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*sin(f*x + e)^2) + 2*lo
g(sin(f*x + e)^2)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3))/f

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Fricas [B]  time = 1.92186, size = 670, normalized size = 5.15 \begin{align*} \frac{5 \, a^{2} b^{3} + 8 \, a b^{4} + 3 \, b^{5} + 2 \,{\left (3 \, a^{3} b^{2} + 5 \, a^{2} b^{3} + 2 \, a b^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left (3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5} +{\left (3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) + 4 \,{\left (a^{5} \cos \left (f x + e\right )^{4} + 2 \, a^{4} b \cos \left (f x + e\right )^{2} + a^{3} b^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (f x + e\right )\right )}{4 \,{\left ({\left (a^{8} + 3 \, a^{7} b + 3 \, a^{6} b^{2} + a^{5} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{7} b + 3 \, a^{6} b^{2} + 3 \, a^{5} b^{3} + a^{4} b^{4}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{6} b^{2} + 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} + a^{3} b^{5}\right )} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

1/4*(5*a^2*b^3 + 8*a*b^4 + 3*b^5 + 2*(3*a^3*b^2 + 5*a^2*b^3 + 2*a*b^4)*cos(f*x + e)^2 + 2*(3*a^2*b^3 + 3*a*b^4
+ b^5 + (3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cos(f*x + e)^4 + 2*(3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(f*x + e)^2)*lo
g(a*cos(f*x + e)^2 + b) + 4*(a^5*cos(f*x + e)^4 + 2*a^4*b*cos(f*x + e)^2 + a^3*b^2)*log(1/2*sin(f*x + e)))/((a
^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*f*cos(f*x + e)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*f*cos(f*x +
e)^2 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*f)

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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(cot(f*x+e)/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.53223, size = 1100, normalized size = 8.46 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

1/4*(2*(3*a^2*b + 3*a*b^2 + b^3)*log(a + b + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1
)/(cos(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)
^2)/(a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3) + 2*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/(a^3 + 3*a^2*b + 3*a
*b^2 + b^3) - (9*a^3*b + 18*a^2*b^2 + 12*a*b^3 + 3*b^4 + 36*a^3*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 24*a
^2*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 16*a*b^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 12*b^4*(cos(f*
x + e) - 1)/(cos(f*x + e) + 1) + 54*a^3*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 12*a^2*b^2*(cos(f*x + e)
- 1)^2/(cos(f*x + e) + 1)^2 + 8*a*b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 18*b^4*(cos(f*x + e) - 1)^2
/(cos(f*x + e) + 1)^2 + 36*a^3*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 24*a^2*b^2*(cos(f*x + e) - 1)^3/(
cos(f*x + e) + 1)^3 - 16*a*b^3*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 12*b^4*(cos(f*x + e) - 1)^3/(cos(f*
x + e) + 1)^3 + 9*a^3*b*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 18*a^2*b^2*(cos(f*x + e) - 1)^4/(cos(f*x +
e) + 1)^4 + 12*a*b^3*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 3*b^4*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1
)^4)/((a^5 + 2*a^4*b + a^3*b^2)*(a + b + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1)/(c
os(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)^
2) - 4*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1)/a^3)/f