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

Optimal. Leaf size=181 \[ -\frac {(a+3 b) \log (\tan (e+f x))}{a^4 f}+\frac {b^2 (3 a-2 b)}{2 a^3 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {\cot ^2(e+f x)}{2 a^3 f}+\frac {b^2}{4 a^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b^2 \left (6 a^2-8 a b+3 b^2\right ) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 f (a-b)^3}-\frac {\log (\cos (e+f x))}{f (a-b)^3} \]

[Out]

-1/2*cot(f*x+e)^2/a^3/f-ln(cos(f*x+e))/(a-b)^3/f-(a+3*b)*ln(tan(f*x+e))/a^4/f-1/2*b^2*(6*a^2-8*a*b+3*b^2)*ln(a
+b*tan(f*x+e)^2)/a^4/(a-b)^3/f+1/4*b^2/a^2/(a-b)/f/(a+b*tan(f*x+e)^2)^2+1/2*(3*a-2*b)*b^2/a^3/(a-b)^2/f/(a+b*t
an(f*x+e)^2)

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Rubi [A]  time = 0.21, antiderivative size = 181, 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} \[ \frac {b^2 (3 a-2 b)}{2 a^3 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac {b^2}{4 a^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b^2 \left (6 a^2-8 a b+3 b^2\right ) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 f (a-b)^3}-\frac {(a+3 b) \log (\tan (e+f x))}{a^4 f}-\frac {\cot ^2(e+f x)}{2 a^3 f}-\frac {\log (\cos (e+f x))}{f (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cot[e + f*x]^2/(2*a^3*f) - Log[Cos[e + f*x]]/((a - b)^3*f) - ((a + 3*b)*Log[Tan[e + f*x]])/(a^4*f) - (b^2*(6*
a^2 - 8*a*b + 3*b^2)*Log[a + b*Tan[e + f*x]^2])/(2*a^4*(a - b)^3*f) + b^2/(4*a^2*(a - b)*f*(a + b*Tan[e + f*x]
^2)^2) + ((3*a - 2*b)*b^2)/(2*a^3*(a - b)^2*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 ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^3 \left (1+x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 (1+x) (a+b x)^3} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{a^3 x^2}+\frac {-a-3 b}{a^4 x}+\frac {1}{(a-b)^3 (1+x)}-\frac {b^3}{a^2 (a-b) (a+b x)^3}-\frac {(3 a-2 b) b^3}{a^3 (a-b)^2 (a+b x)^2}-\frac {b^3 \left (6 a^2-8 a b+3 b^2\right )}{a^4 (a-b)^3 (a+b x)}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac {\cot ^2(e+f x)}{2 a^3 f}-\frac {\log (\cos (e+f x))}{(a-b)^3 f}-\frac {(a+3 b) \log (\tan (e+f x))}{a^4 f}-\frac {b^2 \left (6 a^2-8 a b+3 b^2\right ) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^4 (a-b)^3 f}+\frac {b^2}{4 a^2 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}+\frac {(3 a-2 b) b^2}{2 a^3 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 2.06, size = 144, normalized size = 0.80 \[ -\frac {-\frac {b^4}{2 a^4 (a-b) \left (a \cot ^2(e+f x)+b\right )^2}+\frac {b^3 (4 a-3 b)}{a^4 (a-b)^2 \left (a \cot ^2(e+f x)+b\right )}+\frac {\cot ^2(e+f x)}{a^3}+\frac {b^2 \left (6 a^2-8 a b+3 b^2\right ) \log \left (a \cot ^2(e+f x)+b\right )}{a^4 (a-b)^3}+\frac {2 \log (\sin (e+f x))}{(a-b)^3}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Cot[e + f*x]^2/a^3 - b^4/(2*a^4*(a - b)*(b + a*Cot[e + f*x]^2)^2) + ((4*a - 3*b)*b^3)/(a^4*(a - b)^2*(b
+ a*Cot[e + f*x]^2)) + (b^2*(6*a^2 - 8*a*b + 3*b^2)*Log[b + a*Cot[e + f*x]^2])/(a^4*(a - b)^3) + (2*Log[Sin[e
+ f*x]])/(a - b)^3)/f

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fricas [B]  time = 0.58, size = 545, normalized size = 3.01 \[ -\frac {{\left (2 \, a^{4} b^{2} - 6 \, a^{3} b^{3} + 13 \, a^{2} b^{4} - 6 \, a b^{5}\right )} \tan \left (f x + e\right )^{6} + 2 \, a^{6} - 6 \, a^{5} b + 6 \, a^{4} b^{2} - 2 \, a^{3} b^{3} + 2 \, {\left (2 \, a^{5} b - 5 \, a^{4} b^{2} + 7 \, a^{3} b^{3} + 2 \, a^{2} b^{4} - 3 \, a b^{5}\right )} \tan \left (f x + e\right )^{4} + {\left (2 \, a^{6} - 2 \, a^{5} b - 6 \, a^{4} b^{2} + 18 \, a^{3} b^{3} - 9 \, a^{2} b^{4}\right )} \tan \left (f x + e\right )^{2} + 2 \, {\left ({\left (a^{4} b^{2} - 6 \, a^{2} b^{4} + 8 \, a b^{5} - 3 \, b^{6}\right )} \tan \left (f x + e\right )^{6} + 2 \, {\left (a^{5} b - 6 \, a^{3} b^{3} + 8 \, a^{2} b^{4} - 3 \, a b^{5}\right )} \tan \left (f x + e\right )^{4} + {\left (a^{6} - 6 \, a^{4} b^{2} + 8 \, a^{3} b^{3} - 3 \, a^{2} b^{4}\right )} \tan \left (f x + e\right )^{2}\right )} \log \left (\frac {\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left ({\left (6 \, a^{2} b^{4} - 8 \, a b^{5} + 3 \, b^{6}\right )} \tan \left (f x + e\right )^{6} + 2 \, {\left (6 \, a^{3} b^{3} - 8 \, a^{2} b^{4} + 3 \, a b^{5}\right )} \tan \left (f x + e\right )^{4} + {\left (6 \, a^{4} b^{2} - 8 \, a^{3} b^{3} + 3 \, a^{2} b^{4}\right )} \tan \left (f x + e\right )^{2}\right )} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \, {\left ({\left (a^{7} b^{2} - 3 \, a^{6} b^{3} + 3 \, a^{5} b^{4} - a^{4} b^{5}\right )} f \tan \left (f x + e\right )^{6} + 2 \, {\left (a^{8} b - 3 \, a^{7} b^{2} + 3 \, a^{6} b^{3} - a^{5} b^{4}\right )} f \tan \left (f x + e\right )^{4} + {\left (a^{9} - 3 \, a^{8} b + 3 \, a^{7} b^{2} - a^{6} b^{3}\right )} f \tan \left (f x + e\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((2*a^4*b^2 - 6*a^3*b^3 + 13*a^2*b^4 - 6*a*b^5)*tan(f*x + e)^6 + 2*a^6 - 6*a^5*b + 6*a^4*b^2 - 2*a^3*b^3
+ 2*(2*a^5*b - 5*a^4*b^2 + 7*a^3*b^3 + 2*a^2*b^4 - 3*a*b^5)*tan(f*x + e)^4 + (2*a^6 - 2*a^5*b - 6*a^4*b^2 + 18
*a^3*b^3 - 9*a^2*b^4)*tan(f*x + e)^2 + 2*((a^4*b^2 - 6*a^2*b^4 + 8*a*b^5 - 3*b^6)*tan(f*x + e)^6 + 2*(a^5*b -
6*a^3*b^3 + 8*a^2*b^4 - 3*a*b^5)*tan(f*x + e)^4 + (a^6 - 6*a^4*b^2 + 8*a^3*b^3 - 3*a^2*b^4)*tan(f*x + e)^2)*lo
g(tan(f*x + e)^2/(tan(f*x + e)^2 + 1)) + 2*((6*a^2*b^4 - 8*a*b^5 + 3*b^6)*tan(f*x + e)^6 + 2*(6*a^3*b^3 - 8*a^
2*b^4 + 3*a*b^5)*tan(f*x + e)^4 + (6*a^4*b^2 - 8*a^3*b^3 + 3*a^2*b^4)*tan(f*x + e)^2)*log((b*tan(f*x + e)^2 +
a)/(tan(f*x + e)^2 + 1)))/((a^7*b^2 - 3*a^6*b^3 + 3*a^5*b^4 - a^4*b^5)*f*tan(f*x + e)^6 + 2*(a^8*b - 3*a^7*b^2
+ 3*a^6*b^3 - a^5*b^4)*f*tan(f*x + e)^4 + (a^9 - 3*a^8*b + 3*a^7*b^2 - a^6*b^3)*f*tan(f*x + e)^2)

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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)^3/(a+b*tan(f*x+e)^2)^3,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)))*a+12*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b-a)*1/16/a^4/(1-cos(f*x+exp(1)))*(
1+cos(f*x+exp(1)))+(18*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^4*a^4*b^2-24*((1-cos(f*x+exp(1)))/(1+cos(f*x+
exp(1))))^4*a^3*b^3+9*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^4*a^2*b^4-72*((1-cos(f*x+exp(1)))/(1+cos(f*x+e
xp(1))))^3*a^4*b^2+208*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a^3*b^3-172*((1-cos(f*x+exp(1)))/(1+cos(f*x
+exp(1))))^3*a^2*b^4+48*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a*b^5+108*((1-cos(f*x+exp(1)))/(1+cos(f*x+
exp(1))))^2*a^4*b^2-368*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^3*b^3+502*((1-cos(f*x+exp(1)))/(1+cos(f*
x+exp(1))))^2*a^2*b^4-288*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a*b^5+64*((1-cos(f*x+exp(1)))/(1+cos(f*x
+exp(1))))^2*b^6-72*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^4*b^2+208*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))
)*a^3*b^3-172*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^2*b^4+48*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a*b^5
+18*a^4*b^2-24*a^3*b^3+9*a^2*b^4)/(8*a^7-24*a^6*b+24*a^5*b^2-8*a^4*b^3)/(((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)^2-(1-cos(f
*x+exp(1)))/(1+cos(f*x+exp(1)))*1/16/a^3+1/(2*a^3-6*a^2*b+6*a*b^2-2*b^3)*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x
+exp(1)))+1))+(-a-3*b)*1/4/a^4*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1))))+(-6*a^2*b^2+8*a*b^3-3*b^4)/(4
*a^7-12*a^6*b+12*a^5*b^2-4*a^4*b^3)*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))

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maple [B]  time = 1.00, size = 362, normalized size = 2.00 \[ -\frac {2 b^{3}}{f \,a^{2} \left (a -b \right )^{3} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}+\frac {b^{4}}{f \,a^{3} \left (a -b \right )^{3} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}-\frac {3 b^{2} \ln \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}{f \,a^{2} \left (a -b \right )^{3}}+\frac {4 b^{3} \ln \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}{f \,a^{3} \left (a -b \right )^{3}}-\frac {3 b^{4} \ln \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}{2 f \,a^{4} \left (a -b \right )^{3}}+\frac {b^{4}}{4 f \,a^{2} \left (a -b \right )^{3} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}+\frac {1}{4 f \,a^{3} \left (-1+\cos \left (f x +e \right )\right )}-\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{2 f \,a^{3}}-\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right ) b}{2 f \,a^{4}}-\frac {1}{4 f \,a^{3} \left (1+\cos \left (f x +e \right )\right )}-\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{2 f \,a^{3}}-\frac {3 \ln \left (1+\cos \left (f x +e \right )\right ) b}{2 f \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.97, size = 345, normalized size = 1.91 \[ -\frac {\frac {2 \, {\left (6 \, a^{2} b^{2} - 8 \, a b^{3} + 3 \, b^{4}\right )} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}} + \frac {2 \, a^{5} - 6 \, a^{4} b + 6 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + 2 \, {\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 14 \, a^{2} b^{3} + 11 \, a b^{4} - 3 \, b^{5}\right )} \sin \left (f x + e\right )^{4} - {\left (4 \, a^{5} - 16 \, a^{4} b + 24 \, a^{3} b^{2} - 24 \, a^{2} b^{3} + 9 \, a b^{4}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{8} - 5 \, a^{7} b + 10 \, a^{6} b^{2} - 10 \, a^{5} b^{3} + 5 \, a^{4} b^{4} - a^{3} b^{5}\right )} \sin \left (f x + e\right )^{6} - 2 \, {\left (a^{8} - 4 \, a^{7} b + 6 \, a^{6} b^{2} - 4 \, a^{5} b^{3} + a^{4} b^{4}\right )} \sin \left (f x + e\right )^{4} + {\left (a^{8} - 3 \, a^{7} b + 3 \, a^{6} b^{2} - a^{5} b^{3}\right )} \sin \left (f x + e\right )^{2}} + \frac {2 \, {\left (a + 3 \, b\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{4}}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*(6*a^2*b^2 - 8*a*b^3 + 3*b^4)*log(-(a - b)*sin(f*x + e)^2 + a)/(a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3) +
(2*a^5 - 6*a^4*b + 6*a^3*b^2 - 2*a^2*b^3 + 2*(a^5 - 5*a^4*b + 10*a^3*b^2 - 14*a^2*b^3 + 11*a*b^4 - 3*b^5)*sin
(f*x + e)^4 - (4*a^5 - 16*a^4*b + 24*a^3*b^2 - 24*a^2*b^3 + 9*a*b^4)*sin(f*x + e)^2)/((a^8 - 5*a^7*b + 10*a^6*
b^2 - 10*a^5*b^3 + 5*a^4*b^4 - a^3*b^5)*sin(f*x + e)^6 - 2*(a^8 - 4*a^7*b + 6*a^6*b^2 - 4*a^5*b^3 + a^4*b^4)*s
in(f*x + e)^4 + (a^8 - 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*sin(f*x + e)^2) + 2*(a + 3*b)*log(sin(f*x + e)^2)/a^4)/f

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mupad [B]  time = 13.46, size = 229, normalized size = 1.27 \[ \frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f\,{\left (a-b\right )}^3}-\frac {\frac {1}{2\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (a^2\,b^2-5\,a\,b^3+3\,b^4\right )}{2\,a^3\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (4\,a^2\,b-15\,a\,b^2+9\,b^3\right )}{4\,a^2\,\left (a^2-2\,a\,b+b^2\right )}}{f\,\left (a^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,a\,b\,{\mathrm {tan}\left (e+f\,x\right )}^4+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^6\right )}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a+3\,b\right )}{a^4\,f}-\frac {b^2\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )\,\left (6\,a^2-8\,a\,b+3\,b^2\right )}{2\,a^4\,f\,{\left (a-b\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(tan(e + f*x)^2 + 1)/(2*f*(a - b)^3) - (1/(2*a) + (tan(e + f*x)^4*(3*b^4 - 5*a*b^3 + a^2*b^2))/(2*a^3*(a^2
- 2*a*b + b^2)) + (tan(e + f*x)^2*(4*a^2*b - 15*a*b^2 + 9*b^3))/(4*a^2*(a^2 - 2*a*b + b^2)))/(f*(a^2*tan(e + f
*x)^2 + b^2*tan(e + f*x)^6 + 2*a*b*tan(e + f*x)^4)) - (log(tan(e + f*x))*(a + 3*b))/(a^4*f) - (b^2*log(a + b*t
an(e + f*x)^2)*(6*a^2 - 8*a*b + 3*b^2))/(2*a^4*f*(a - b)^3)

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

[Out]

Timed out

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