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

Optimal. Leaf size=161 \[ \frac {b^3 (4 a-3 b) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 f (a-b)^2}-\frac {b^3}{2 a^3 f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac {(a+2 b) \cot ^2(e+f x)}{2 a^3 f}-\frac {\cot ^4(e+f x)}{4 a^2 f}+\frac {\left (a^2+2 a b+3 b^2\right ) \log (\tan (e+f x))}{a^4 f}+\frac {\log (\cos (e+f x))}{f (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3670, 446, 88} \[ -\frac {b^3}{2 a^3 f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac {b^3 (4 a-3 b) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 f (a-b)^2}+\frac {\left (a^2+2 a b+3 b^2\right ) \log (\tan (e+f x))}{a^4 f}+\frac {(a+2 b) \cot ^2(e+f x)}{2 a^3 f}-\frac {\cot ^4(e+f x)}{4 a^2 f}+\frac {\log (\cos (e+f x))}{f (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

\begin {align*} \int \frac {\cot ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^5 \left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^3 (1+x) (a+b x)^2} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{a^2 x^3}+\frac {-a-2 b}{a^3 x^2}+\frac {a^2+2 a b+3 b^2}{a^4 x}-\frac {1}{(a-b)^2 (1+x)}+\frac {b^4}{a^3 (a-b) (a+b x)^2}+\frac {(4 a-3 b) b^4}{a^4 (a-b)^2 (a+b x)}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {(a+2 b) \cot ^2(e+f x)}{2 a^3 f}-\frac {\cot ^4(e+f x)}{4 a^2 f}+\frac {\log (\cos (e+f x))}{(a-b)^2 f}+\frac {\left (a^2+2 a b+3 b^2\right ) \log (\tan (e+f x))}{a^4 f}+\frac {(4 a-3 b) b^3 \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 (a-b)^2 f}-\frac {b^3}{2 a^3 (a-b) f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.09, size = 121, normalized size = 0.75 \[ -\frac {-\frac {b^4}{a^4 (a-b) \left (a \cot ^2(e+f x)+b\right )}-\frac {b^3 (4 a-3 b) \log \left (a \cot ^2(e+f x)+b\right )}{a^4 (a-b)^2}-\frac {(a+2 b) \cot ^2(e+f x)}{a^3}+\frac {\cot ^4(e+f x)}{2 a^2}-\frac {2 \log (\sin (e+f x))}{(a-b)^2}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(-(((a + 2*b)*Cot[e + f*x]^2)/a^3) + Cot[e + f*x]^4/(2*a^2) - b^4/(a^4*(a - b)*(b + a*Cot[e + f*x]^2)) -
((4*a - 3*b)*b^3*Log[b + a*Cot[e + f*x]^2])/(a^4*(a - b)^2) - (2*Log[Sin[e + f*x]])/(a - b)^2)/f

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fricas [B]  time = 0.52, size = 347, normalized size = 2.16 \[ \frac {{\left (3 \, a^{4} b - 2 \, a^{3} b^{2} - 5 \, a^{2} b^{3} + 6 \, a b^{4}\right )} \tan \left (f x + e\right )^{6} - a^{5} + 2 \, a^{4} b - a^{3} b^{2} + {\left (3 \, a^{5} - 5 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + 6 \, a b^{4}\right )} \tan \left (f x + e\right )^{4} + {\left (2 \, a^{5} - a^{4} b - 4 \, a^{3} b^{2} + 3 \, a^{2} b^{3}\right )} \tan \left (f x + e\right )^{2} + 2 \, {\left ({\left (a^{4} b - 4 \, a b^{4} + 3 \, b^{5}\right )} \tan \left (f x + e\right )^{6} + {\left (a^{5} - 4 \, a^{2} b^{3} + 3 \, a b^{4}\right )} \tan \left (f x + e\right )^{4}\right )} \log \left (\frac {\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left ({\left (4 \, a b^{4} - 3 \, b^{5}\right )} \tan \left (f x + e\right )^{6} + {\left (4 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \tan \left (f x + e\right )^{4}\right )} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \, {\left ({\left (a^{6} b - 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \tan \left (f x + e\right )^{6} + {\left (a^{7} - 2 \, a^{6} b + a^{5} b^{2}\right )} f \tan \left (f x + e\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^5/(a+b*tan(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*((-4*((1-cos(f
*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^2*b^3+3*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a*b^4+8*(1-cos(f*x+ex
p(1)))/(1+cos(f*x+exp(1)))*a^2*b^3-18*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a*b^4+8*(1-cos(f*x+exp(1)))/(1+c
os(f*x+exp(1)))*b^5-4*a^2*b^3+3*a*b^4)/(4*a^6-8*a^5*b+4*a^4*b^2)/(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*
a-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+4*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b+a)+(-48*((1-cos(f*x+
exp(1)))/(1+cos(f*x+exp(1))))^2*a^2-96*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a*b-144*((1-cos(f*x+exp(1))
)/(1+cos(f*x+exp(1))))^2*b^2+12*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^2+16*(1-cos(f*x+exp(1)))/(1+cos(f*x+
exp(1)))*a*b-a^2)*1/128/a^4/((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2+(-32*((1-cos(f*x+exp(1)))/(1+cos(f*x+e
xp(1))))^2*a^2+384*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^2+512*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a*b
)*1/4096/a^4-1/(2*a^2-4*a*b+2*b^2)*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1))+(4*a*b^3-3*b^4)/(4*a^6-8
*a^5*b+4*a^4*b^2)*ln(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a
+4*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b+a)+(a^2+2*a*b+3*b^2)*1/4/a^4*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(
f*x+exp(1)))))

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maple [B]  time = 0.99, size = 347, normalized size = 2.16 \[ \frac {b^{4}}{2 f \,a^{3} \left (a -b \right )^{2} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}+\frac {2 b^{3} \ln \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}{f \,a^{3} \left (a -b \right )^{2}}-\frac {3 b^{4} \ln \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}{2 f \,a^{4} \left (a -b \right )^{2}}-\frac {1}{16 f \,a^{2} \left (-1+\cos \left (f x +e \right )\right )^{2}}-\frac {7}{16 f \,a^{2} \left (-1+\cos \left (f x +e \right )\right )}-\frac {b}{2 f \,a^{3} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{2 f \,a^{2}}+\frac {\ln \left (-1+\cos \left (f x +e \right )\right ) b}{f \,a^{3}}+\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right ) b^{2}}{2 f \,a^{4}}-\frac {1}{16 f \,a^{2} \left (1+\cos \left (f x +e \right )\right )^{2}}+\frac {7}{16 f \,a^{2} \left (1+\cos \left (f x +e \right )\right )}+\frac {b}{2 f \,a^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{2 f \,a^{2}}+\frac {\ln \left (1+\cos \left (f x +e \right )\right ) b}{f \,a^{3}}+\frac {3 \ln \left (1+\cos \left (f x +e \right )\right ) b^{2}}{2 f \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.19, size = 236, normalized size = 1.47 \[ \frac {\frac {2 \, {\left (4 \, a b^{3} - 3 \, b^{4}\right )} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{6} - 2 \, a^{5} b + a^{4} b^{2}} + \frac {2 \, {\left (2 \, a^{4} - 4 \, a^{3} b + 4 \, a b^{3} - 3 \, b^{4}\right )} \sin \left (f x + e\right )^{4} + a^{4} - 2 \, a^{3} b + a^{2} b^{2} - {\left (5 \, a^{4} - 7 \, a^{3} b - a^{2} b^{2} + 3 \, a b^{3}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}\right )} \sin \left (f x + e\right )^{6} - {\left (a^{6} - 2 \, a^{5} b + a^{4} b^{2}\right )} \sin \left (f x + e\right )^{4}} + \frac {2 \, {\left (a^{2} + 2 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{4}}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 12.37, size = 191, normalized size = 1.19 \[ \frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,a+3\,b\right )}{4\,a^2}-\frac {1}{4\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (a^2\,b+a\,b^2-3\,b^3\right )}{2\,a^3\,\left (a-b\right )}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^6+a\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f\,{\left (a-b\right )}^2}+\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^2+2\,a\,b+3\,b^2\right )}{a^4\,f}+\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )\,\left (4\,a\,b^3-3\,b^4\right )}{f\,\left (2\,a^6-4\,a^5\,b+2\,a^4\,b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**5/(a+b*tan(f*x+e)**2)**2,x)

[Out]

Timed out

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