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

Optimal. Leaf size=111 \[ -\frac{b^3}{2 a^2 f (a+b)^2 \left (a \cos ^2(e+f x)+b\right )}-\frac{b^2 (3 a+b) \log \left (a \cos ^2(e+f x)+b\right )}{2 a^2 f (a+b)^3}-\frac{\csc ^2(e+f x)}{2 f (a+b)^2}-\frac{(a+3 b) \log (\sin (e+f x))}{f (a+b)^3} \]

[Out]

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

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Rubi [A]  time = 0.157109, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4138, 446, 88} \[ -\frac{b^3}{2 a^2 f (a+b)^2 \left (a \cos ^2(e+f x)+b\right )}-\frac{b^2 (3 a+b) \log \left (a \cos ^2(e+f x)+b\right )}{2 a^2 f (a+b)^3}-\frac{\csc ^2(e+f x)}{2 f (a+b)^2}-\frac{(a+3 b) \log (\sin (e+f x))}{f (a+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-b^3/(2*a^2*(a + b)^2*f*(b + a*Cos[e + f*x]^2)) - Csc[e + f*x]^2/(2*(a + b)^2*f) - (b^2*(3*a + b)*Log[b + a*Co
s[e + f*x]^2])/(2*a^2*(a + b)^3*f) - ((a + 3*b)*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 ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^7}{\left (1-x^2\right )^2 \left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^3}{(1-x)^2 (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)^2}+\frac{a+3 b}{(a+b)^3 (-1+x)}-\frac{b^3}{a (a+b)^2 (b+a x)^2}+\frac{b^2 (3 a+b)}{a (a+b)^3 (b+a x)}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{b^3}{2 a^2 (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )}-\frac{\csc ^2(e+f x)}{2 (a+b)^2 f}-\frac{b^2 (3 a+b) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^2 (a+b)^3 f}-\frac{(a+3 b) \log (\sin (e+f x))}{(a+b)^3 f}\\ \end{align*}

Mathematica [A]  time = 1.33795, size = 130, normalized size = 1.17 \[ -\frac{\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b)^2 \left (\frac{b^2 \left (\frac{2 b (a+b)}{a \cos (2 (e+f x))+a+2 b}+(3 a+b) \log \left (-a \sin ^2(e+f x)+a+b\right )\right )}{a^2}+(a+b) \csc ^2(e+f x)+2 (a+3 b) \log (\sin (e+f x))\right )}{8 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])^2*((a + b)*Csc[e + f*x]^2 + 2*(a + 3*b)*Log[Sin[e + f*x]] + (b^2*((2*b*(a + b
))/(a + 2*b + a*Cos[2*(e + f*x)]) + (3*a + b)*Log[a + b - a*Sin[e + f*x]^2]))/a^2)*Sec[e + f*x]^4)/(8*(a + b)^
3*f*(a + b*Sec[e + f*x]^2)^2)

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Maple [B]  time = 0.109, size = 240, normalized size = 2.2 \begin{align*} -{\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}}-{\frac{{b}^{3}\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}}{2\,f \left ( a+b \right ) ^{3}a \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{{b}^{4}}{2\,f \left ( a+b \right ) ^{3}{a}^{2} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{1}{4\,f \left ( a+b \right ) ^{2} \left ( 1+\cos \left ( fx+e \right ) \right ) }}-{\frac{\ln \left ( 1+\cos \left ( fx+e \right ) \right ) a}{2\,f \left ( a+b \right ) ^{3}}}-{\frac{3\,\ln \left ( 1+\cos \left ( fx+e \right ) \right ) b}{2\,f \left ( a+b \right ) ^{3}}}+{\frac{1}{4\,f \left ( a+b \right ) ^{2} \left ( -1+\cos \left ( fx+e \right ) \right ) }}-{\frac{\ln \left ( -1+\cos \left ( fx+e \right ) \right ) a}{2\,f \left ( a+b \right ) ^{3}}}-{\frac{3\,\ln \left ( -1+\cos \left ( fx+e \right ) \right ) b}{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)^3/(a+b*sec(f*x+e)^2)^2,x)

[Out]

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

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Maxima [A]  time = 1.02126, size = 259, normalized size = 2.33 \begin{align*} -\frac{\frac{{\left (3 \, a b^{2} + b^{3}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}} + \frac{{\left (a + 3 \, b\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{a^{3} + a^{2} b -{\left (a^{3} - b^{3}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sin \left (f x + e\right )^{4} -{\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} \sin \left (f x + e\right )^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.53509, size = 664, normalized size = 5.98 \begin{align*} \frac{a^{3} b + a^{2} b^{2} + a b^{3} + b^{4} +{\left (a^{4} + a^{3} b - a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2} -{\left ({\left (3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{4} - 3 \, a b^{3} - b^{4} -{\left (3 \, a^{2} b^{2} - 2 \, a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 2 \,{\left ({\left (a^{4} + 3 \, a^{3} b\right )} \cos \left (f x + e\right )^{4} - a^{3} b - 3 \, a^{2} b^{2} -{\left (a^{4} + 2 \, a^{3} b - 3 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (f x + e\right )\right )}{2 \,{\left ({\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} f \cos \left (f x + e\right )^{4} -{\left (a^{6} + 2 \, a^{5} b - 2 \, a^{3} b^{3} - a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{2} -{\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^3*b + a^2*b^2 + a*b^3 + b^4 + (a^4 + a^3*b - a*b^3 - b^4)*cos(f*x + e)^2 - ((3*a^2*b^2 + a*b^3)*cos(f*x
+ e)^4 - 3*a*b^3 - b^4 - (3*a^2*b^2 - 2*a*b^3 - b^4)*cos(f*x + e)^2)*log(a*cos(f*x + e)^2 + b) - 2*((a^4 + 3*
a^3*b)*cos(f*x + e)^4 - a^3*b - 3*a^2*b^2 - (a^4 + 2*a^3*b - 3*a^2*b^2)*cos(f*x + e)^2)*log(1/2*sin(f*x + e)))
/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*f*cos(f*x + e)^4 - (a^6 + 2*a^5*b - 2*a^3*b^3 - a^2*b^4)*f*cos(f*x + e
)^2 - (a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*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)**3/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.94323, size = 1165, normalized size = 10.5 \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)^3/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/24*(12*(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^
5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3) + 12*(a + 3*b)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/(a^3 + 3*a^2*b +
3*a*b^2 + b^3) - (3*a^4 + 6*a^3*b + 3*a^2*b^2 + 10*a^4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 16*a^3*b*(cos(
f*x + e) - 1)/(cos(f*x + e) + 1) + 30*a^2*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 32*a*b^3*(cos(f*x + e) -
1)/(cos(f*x + e) + 1) + 8*b^4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 11*a^4*(cos(f*x + e) - 1)^2/(cos(f*x +
e) + 1)^2 + 22*a^3*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 27*a^2*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e)
+ 1)^2 + 16*a*b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 16*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^
2 + 4*a^4*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 16*a^3*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 36*
a^2*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 32*a*b^3*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 8*b^4
*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3)/((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*(a*(cos(f*x + e) - 1)/(cos(
f*x + e) + 1) + b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 2*a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 2*b*
(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + a*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + b*(cos(f*x + e) - 1)
^3/(cos(f*x + e) + 1)^3)) - 3*(cos(f*x + e) - 1)/((a^2 + 2*a*b + b^2)*(cos(f*x + e) + 1)) - 24*log(-(cos(f*x +
e) - 1)/(cos(f*x + e) + 1) + 1)/a^2)/f