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

Optimal. Leaf size=108 \[ \frac{\left (a^2+3 a b+3 b^2\right ) \log (\sin (e+f x))}{f (a+b)^3}+\frac{b^3 \log \left (a \cos ^2(e+f x)+b\right )}{2 a f (a+b)^3}-\frac{\csc ^4(e+f x)}{4 f (a+b)}+\frac{(2 a+3 b) \csc ^2(e+f x)}{2 f (a+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.148455, antiderivative size = 108, 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} \[ \frac{\left (a^2+3 a b+3 b^2\right ) \log (\sin (e+f x))}{f (a+b)^3}+\frac{b^3 \log \left (a \cos ^2(e+f x)+b\right )}{2 a f (a+b)^3}-\frac{\csc ^4(e+f x)}{4 f (a+b)}+\frac{(2 a+3 b) \csc ^2(e+f x)}{2 f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a + 3*b)*Csc[e + f*x]^2)/(2*(a + b)^2*f) - Csc[e + f*x]^4/(4*(a + b)*f) + (b^3*Log[b + a*Cos[e + f*x]^2])/
(2*a*(a + b)^3*f) + ((a^2 + 3*a*b + 3*b^2)*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 ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^7}{\left (1-x^2\right )^3 \left (b+a x^2\right )} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^3}{(1-x)^3 (b+a x)} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{(a+b) (-1+x)^3}+\frac{-2 a-3 b}{(a+b)^2 (-1+x)^2}+\frac{-a^2-3 a b-3 b^2}{(a+b)^3 (-1+x)}-\frac{b^3}{(a+b)^3 (b+a x)}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac{(2 a+3 b) \csc ^2(e+f x)}{2 (a+b)^2 f}-\frac{\csc ^4(e+f x)}{4 (a+b) f}+\frac{b^3 \log \left (b+a \cos ^2(e+f x)\right )}{2 a (a+b)^3 f}+\frac{\left (a^2+3 a b+3 b^2\right ) \log (\sin (e+f x))}{(a+b)^3 f}\\ \end{align*}

Mathematica [A]  time = 0.653677, size = 138, normalized size = 1.28 \[ \frac{\sec ^2(e+f x) (a \cos (2 e+2 f x)+a+2 b) \left (\frac{4 \left (a^2+3 a b+3 b^2\right ) \log (\sin (e+f x))}{(a+b)^3}+\frac{2 b^3 \log \left (-a \sin ^2(e+f x)+a+b\right )}{a (a+b)^3}-\frac{\csc ^4(e+f x)}{a+b}+\frac{2 (2 a+3 b) \csc ^2(e+f x)}{(a+b)^2}\right )}{8 f \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.089, size = 293, normalized size = 2.7 \begin{align*}{\frac{{b}^{3}\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,a \left ( a+b \right ) ^{3}f}}-{\frac{1}{2\,f \left ( 8\,a+8\,b \right ) \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}+{\frac{7\,a}{16\,f \left ( a+b \right ) ^{2} \left ( 1+\cos \left ( fx+e \right ) \right ) }}+{\frac{11\,b}{16\,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}}{2\,f \left ( a+b \right ) ^{3}}}+{\frac{3\,\ln \left ( 1+\cos \left ( fx+e \right ) \right ) ab}{2\,f \left ( a+b \right ) ^{3}}}+{\frac{3\,\ln \left ( 1+\cos \left ( fx+e \right ) \right ){b}^{2}}{2\,f \left ( a+b \right ) ^{3}}}-{\frac{1}{2\,f \left ( 8\,a+8\,b \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}}}-{\frac{7\,a}{16\,f \left ( a+b \right ) ^{2} \left ( -1+\cos \left ( fx+e \right ) \right ) }}-{\frac{11\,b}{16\,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}}{2\,f \left ( a+b \right ) ^{3}}}+{\frac{3\,\ln \left ( -1+\cos \left ( fx+e \right ) \right ) ab}{2\,f \left ( a+b \right ) ^{3}}}+{\frac{3\,\ln \left ( -1+\cos \left ( fx+e \right ) \right ){b}^{2}}{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)^5/(a+b*sec(f*x+e)^2),x)

[Out]

1/2*b^3*ln(b+a*cos(f*x+e)^2)/a/(a+b)^3/f-1/2/f/(8*a+8*b)/(1+cos(f*x+e))^2+7/16/f/(a+b)^2/(1+cos(f*x+e))*a+11/1
6/f/(a+b)^2/(1+cos(f*x+e))*b+1/2/f/(a+b)^3*ln(1+cos(f*x+e))*a^2+3/2/f/(a+b)^3*ln(1+cos(f*x+e))*a*b+3/2/f/(a+b)
^3*ln(1+cos(f*x+e))*b^2-1/2/f/(8*a+8*b)/(-1+cos(f*x+e))^2-7/16/f/(a+b)^2/(-1+cos(f*x+e))*a-11/16/f/(a+b)^2/(-1
+cos(f*x+e))*b+1/2/f/(a+b)^3*ln(-1+cos(f*x+e))*a^2+3/2/f/(a+b)^3*ln(-1+cos(f*x+e))*a*b+3/2/f/(a+b)^3*ln(-1+cos
(f*x+e))*b^2

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Maxima [A]  time = 0.994865, size = 196, normalized size = 1.81 \begin{align*} \frac{\frac{2 \, b^{3} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}} + \frac{2 \,{\left (a^{2} + 3 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac{2 \,{\left (2 \, a + 3 \, b\right )} \sin \left (f x + e\right )^{2} - a - b}{{\left (a^{2} + 2 \, a b + b^{2}\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),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.32138, size = 610, normalized size = 5.65 \begin{align*} \frac{3 \, a^{3} + 8 \, a^{2} b + 5 \, a b^{2} - 2 \,{\left (2 \, a^{3} + 5 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left (b^{3} \cos \left (f x + e\right )^{4} - 2 \, b^{3} \cos \left (f x + e\right )^{2} + b^{3}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) + 4 \,{\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} - 2 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2}\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^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \,{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\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),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.42843, size = 733, normalized size = 6.79 \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),x, algorithm="giac")

[Out]

1/64*(32*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^4 + 3*a^3*b +
3*a^2*b^2 + a*b^3) + 32*(a^2 + 3*a*b + 3*b^2)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/(a^3 + 3*a^2*b + 3*
a*b^2 + b^3) - (12*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 20*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(c
os(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^2 + 2*a*b + b^2) - (
a^2 + 2*a*b + b^2 + 12*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 32*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1
) + 20*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 + 144*a*b*
(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 144*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) +
1)^2/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(cos(f*x + e) - 1)^2) - 64*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) +
1)/a)/f