3.355 \(\int \frac{\cot ^5(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=140 \[ \frac{b^4}{2 a^2 f (a+b)^3 \left (a \cos ^2(e+f x)+b\right )}+\frac{\left (a^2+4 a b+6 b^2\right ) \log (\sin (e+f x))}{f (a+b)^4}+\frac{b^3 (4 a+b) \log \left (a \cos ^2(e+f x)+b\right )}{2 a^2 f (a+b)^4}-\frac{\csc ^4(e+f x)}{4 f (a+b)^2}+\frac{(a+2 b) \csc ^2(e+f x)}{f (a+b)^3} \]

[Out]

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

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Rubi [A]  time = 0.197253, antiderivative size = 140, 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} \[ \frac{b^4}{2 a^2 f (a+b)^3 \left (a \cos ^2(e+f x)+b\right )}+\frac{\left (a^2+4 a b+6 b^2\right ) \log (\sin (e+f x))}{f (a+b)^4}+\frac{b^3 (4 a+b) \log \left (a \cos ^2(e+f x)+b\right )}{2 a^2 f (a+b)^4}-\frac{\csc ^4(e+f x)}{4 f (a+b)^2}+\frac{(a+2 b) \csc ^2(e+f x)}{f (a+b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^5/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

b^4/(2*a^2*(a + b)^3*f*(b + a*Cos[e + f*x]^2)) + ((a + 2*b)*Csc[e + f*x]^2)/((a + b)^3*f) - Csc[e + f*x]^4/(4*
(a + b)^2*f) + (b^3*(4*a + b)*Log[b + a*Cos[e + f*x]^2])/(2*a^2*(a + b)^4*f) + ((a^2 + 4*a*b + 6*b^2)*Log[Sin[
e + f*x]])/((a + b)^4*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 ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^9}{\left (1-x^2\right )^3 \left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^4}{(1-x)^3 (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)^3}-\frac{2 (a+2 b)}{(a+b)^3 (-1+x)^2}+\frac{-a^2-4 a b-6 b^2}{(a+b)^4 (-1+x)}+\frac{b^4}{a (a+b)^3 (b+a x)^2}-\frac{b^3 (4 a+b)}{a (a+b)^4 (b+a x)}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac{b^4}{2 a^2 (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )}+\frac{(a+2 b) \csc ^2(e+f x)}{(a+b)^3 f}-\frac{\csc ^4(e+f x)}{4 (a+b)^2 f}+\frac{b^3 (4 a+b) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^2 (a+b)^4 f}+\frac{\left (a^2+4 a b+6 b^2\right ) \log (\sin (e+f x))}{(a+b)^4 f}\\ \end{align*}

Mathematica [A]  time = 2.00365, size = 162, normalized size = 1.16 \[ \frac{\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b)^2 \left (\frac{2 b^4 (a+b)}{a^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{2 b^3 (4 a+b) \log \left (-a \sin ^2(e+f x)+a+b\right )}{a^2}+4 \left (a^2+4 a b+6 b^2\right ) \log (\sin (e+f x))-(a+b)^2 \csc ^4(e+f x)+4 (a+b) (a+2 b) \csc ^2(e+f x)\right )}{16 f (a+b)^4 \left (a+b \sec ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^5/(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*(4*(a + b)*(a + 2*b)*Csc[e + f*x]^2 - (a + b)^2*Csc[e + f*x]^
4 + 4*(a^2 + 4*a*b + 6*b^2)*Log[Sin[e + f*x]] + (2*b^3*(4*a + b)*Log[a + b - a*Sin[e + f*x]^2])/a^2 + (2*b^4*(
a + b))/(a^2*(a + b - a*Sin[e + f*x]^2))))/(16*(a + b)^4*f*(a + b*Sec[e + f*x]^2)^2)

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Maple [B]  time = 0.126, size = 374, normalized size = 2.7 \begin{align*} 2\,{\frac{{b}^{3}\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{f \left ( a+b \right ) ^{4}a}}+{\frac{{b}^{4}\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f \left ( a+b \right ) ^{4}{a}^{2}}}+{\frac{{b}^{4}}{2\,f \left ( a+b \right ) ^{4}a \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{{b}^{5}}{2\,f \left ( a+b \right ) ^{4}{a}^{2} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{1}{16\,f \left ( a+b \right ) ^{2} \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}+{\frac{7\,a}{16\,f \left ( a+b \right ) ^{3} \left ( 1+\cos \left ( fx+e \right ) \right ) }}+{\frac{15\,b}{16\,f \left ( a+b \right ) ^{3} \left ( 1+\cos \left ( fx+e \right ) \right ) }}+{\frac{\ln \left ( 1+\cos \left ( fx+e \right ) \right ){a}^{2}}{2\,f \left ( a+b \right ) ^{4}}}+2\,{\frac{\ln \left ( 1+\cos \left ( fx+e \right ) \right ) ab}{f \left ( a+b \right ) ^{4}}}+3\,{\frac{\ln \left ( 1+\cos \left ( fx+e \right ) \right ){b}^{2}}{f \left ( a+b \right ) ^{4}}}-{\frac{1}{16\,f \left ( a+b \right ) ^{2} \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}}}-{\frac{7\,a}{16\,f \left ( a+b \right ) ^{3} \left ( -1+\cos \left ( fx+e \right ) \right ) }}-{\frac{15\,b}{16\,f \left ( a+b \right ) ^{3} \left ( -1+\cos \left ( fx+e \right ) \right ) }}+{\frac{\ln \left ( -1+\cos \left ( fx+e \right ) \right ){a}^{2}}{2\,f \left ( a+b \right ) ^{4}}}+2\,{\frac{\ln \left ( -1+\cos \left ( fx+e \right ) \right ) ab}{f \left ( a+b \right ) ^{4}}}+3\,{\frac{\ln \left ( -1+\cos \left ( fx+e \right ) \right ){b}^{2}}{f \left ( a+b \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x)

[Out]

2/f*b^3/(a+b)^4/a*ln(b+a*cos(f*x+e)^2)+1/2/f*b^4/(a+b)^4/a^2*ln(b+a*cos(f*x+e)^2)+1/2/f*b^4/(a+b)^4/a/(b+a*cos
(f*x+e)^2)+1/2/f*b^5/(a+b)^4/a^2/(b+a*cos(f*x+e)^2)-1/16/f/(a+b)^2/(1+cos(f*x+e))^2+7/16/f/(a+b)^3/(1+cos(f*x+
e))*a+15/16/f/(a+b)^3/(1+cos(f*x+e))*b+1/2/f/(a+b)^4*ln(1+cos(f*x+e))*a^2+2/f/(a+b)^4*ln(1+cos(f*x+e))*a*b+3/f
/(a+b)^4*ln(1+cos(f*x+e))*b^2-1/16/f/(a+b)^2/(-1+cos(f*x+e))^2-7/16/f/(a+b)^3/(-1+cos(f*x+e))*a-15/16/f/(a+b)^
3/(-1+cos(f*x+e))*b+1/2/f/(a+b)^4*ln(-1+cos(f*x+e))*a^2+2/f/(a+b)^4*ln(-1+cos(f*x+e))*a*b+3/f/(a+b)^4*ln(-1+co
s(f*x+e))*b^2

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Maxima [B]  time = 1.01132, size = 377, normalized size = 2.69 \begin{align*} \frac{\frac{2 \,{\left (4 \, a b^{3} + b^{4}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}} + \frac{2 \,{\left (a^{2} + 4 \, a b + 6 \, b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac{2 \,{\left (2 \, a^{4} + 4 \, a^{3} b - b^{4}\right )} \sin \left (f x + e\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} -{\left (5 \, a^{4} + 13 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \sin \left (f x + e\right )^{6} -{\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \sin \left (f x + e\right )^{4}}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

1/4*(2*(4*a*b^3 + b^4)*log(a*sin(f*x + e)^2 - a - b)/(a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4) + 2*(a^
2 + 4*a*b + 6*b^2)*log(sin(f*x + e)^2)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) + (2*(2*a^4 + 4*a^3*b - b^4
)*sin(f*x + e)^4 + a^4 + 2*a^3*b + a^2*b^2 - (5*a^4 + 13*a^3*b + 8*a^2*b^2)*sin(f*x + e)^2)/((a^6 + 3*a^5*b +
3*a^4*b^2 + a^3*b^3)*sin(f*x + e)^6 - (a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*sin(f*x + e)^4))/f

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Fricas [B]  time = 2.69687, size = 1196, normalized size = 8.54 \begin{align*} \frac{3 \, a^{4} b + 10 \, a^{3} b^{2} + 7 \, a^{2} b^{3} + 2 \, a b^{4} + 2 \, b^{5} - 2 \,{\left (2 \, a^{5} + 6 \, a^{4} b + 4 \, a^{3} b^{2} - a b^{4} - b^{5}\right )} \cos \left (f x + e\right )^{4} +{\left (3 \, a^{5} + 6 \, a^{4} b - 5 \, a^{3} b^{2} - 8 \, a^{2} b^{3} - 4 \, a b^{4} - 4 \, b^{5}\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left ({\left (4 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (f x + e\right )^{6} + 4 \, a b^{4} + b^{5} -{\left (8 \, a^{2} b^{3} - 2 \, a b^{4} - b^{5}\right )} \cos \left (f x + e\right )^{4} +{\left (4 \, a^{2} b^{3} - 7 \, a b^{4} - 2 \, b^{5}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) + 4 \,{\left ({\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2}\right )} \cos \left (f x + e\right )^{6} + a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} -{\left (2 \, a^{5} + 7 \, a^{4} b + 8 \, a^{3} b^{2} - 6 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{4} +{\left (a^{5} + 2 \, a^{4} b - 2 \, a^{3} b^{2} - 12 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (f x + e\right )\right )}{4 \,{\left ({\left (a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{6} -{\left (2 \, a^{7} + 7 \, a^{6} b + 8 \, a^{5} b^{2} + 2 \, a^{4} b^{3} - 2 \, a^{3} b^{4} - a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{4} +{\left (a^{7} + 2 \, a^{6} b - 2 \, a^{5} b^{2} - 8 \, a^{4} b^{3} - 7 \, a^{3} b^{4} - 2 \, a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{6} b + 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/4*(3*a^4*b + 10*a^3*b^2 + 7*a^2*b^3 + 2*a*b^4 + 2*b^5 - 2*(2*a^5 + 6*a^4*b + 4*a^3*b^2 - a*b^4 - b^5)*cos(f*
x + e)^4 + (3*a^5 + 6*a^4*b - 5*a^3*b^2 - 8*a^2*b^3 - 4*a*b^4 - 4*b^5)*cos(f*x + e)^2 + 2*((4*a^2*b^3 + a*b^4)
*cos(f*x + e)^6 + 4*a*b^4 + b^5 - (8*a^2*b^3 - 2*a*b^4 - b^5)*cos(f*x + e)^4 + (4*a^2*b^3 - 7*a*b^4 - 2*b^5)*c
os(f*x + e)^2)*log(a*cos(f*x + e)^2 + b) + 4*((a^5 + 4*a^4*b + 6*a^3*b^2)*cos(f*x + e)^6 + a^4*b + 4*a^3*b^2 +
6*a^2*b^3 - (2*a^5 + 7*a^4*b + 8*a^3*b^2 - 6*a^2*b^3)*cos(f*x + e)^4 + (a^5 + 2*a^4*b - 2*a^3*b^2 - 12*a^2*b^
3)*cos(f*x + e)^2)*log(1/2*sin(f*x + e)))/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^6
- (2*a^7 + 7*a^6*b + 8*a^5*b^2 + 2*a^4*b^3 - 2*a^3*b^4 - a^2*b^5)*f*cos(f*x + e)^4 + (a^7 + 2*a^6*b - 2*a^5*b^
2 - 8*a^4*b^3 - 7*a^3*b^4 - 2*a^2*b^5)*f*cos(f*x + e)^2 + (a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*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)**5/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.57041, size = 1305, normalized size = 9.32 \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)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/64*(32*(4*a*b^3 + b^4)*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
+ 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4) + 32*(a^2 + 4*a*b + 6*b^2)*log(-(cos(f*x + e) - 1)/(cos(f*x + e)
+ 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - (12*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 40*a*b*(c
os(f*x + e) - 1)/(cos(f*x + e) + 1) + 28*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a^2*(cos(f*x + e) - 1)^2/
(cos(f*x + e) + 1)^2 + 2*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e
) + 1)^2)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - (a^2 + 2*a*b + b^2 + 12*a^2*(cos(f*x + e) - 1)/(cos(f*
x + e) + 1) + 40*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 28*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 48
*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 192*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 288*b^2*(
cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1)^2/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*(c
os(f*x + e) - 1)^2) - 32*(4*a^2*b^3 + 5*a*b^4 + b^5 + 8*a^2*b^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*a*b^
4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*b^5*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 4*a^2*b^3*(cos(f*x + e
) - 1)^2/(cos(f*x + e) + 1)^2 + 5*a*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b^5*(cos(f*x + e) - 1)^2/(
cos(f*x + e) + 1)^2)/((a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*(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*(co
s(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)) - 64*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1)/a^2)/f