3.202 \(\int \cot ^3(e+f x) (a+b \tan ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=56 \[ -\frac{a^2 \cot ^2(e+f x)}{2 f}-\frac{a (a-2 b) \log (\tan (e+f x))}{f}-\frac{(a-b)^2 \log (\cos (e+f x))}{f} \]

[Out]

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

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Rubi [A]  time = 0.0816643, antiderivative size = 56, 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 = {3670, 446, 88} \[ -\frac{a^2 \cot ^2(e+f x)}{2 f}-\frac{a (a-2 b) \log (\tan (e+f x))}{f}-\frac{(a-b)^2 \log (\cos (e+f x))}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

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

Mathematica [A]  time = 0.243094, size = 51, normalized size = 0.91 \[ -\frac{a^2 \cot ^2(e+f x)+2 a (a-2 b) \log (\tan (e+f x))+2 (a-b)^2 \log (\cos (e+f x))}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.056, size = 62, normalized size = 1.1 \begin{align*} -{\frac{{b}^{2}\ln \left ( \cos \left ( fx+e \right ) \right ) }{f}}+2\,{\frac{ab\ln \left ( \sin \left ( fx+e \right ) \right ) }{f}}-{\frac{{a}^{2} \left ( \cot \left ( fx+e \right ) \right ) ^{2}}{2\,f}}-{\frac{{a}^{2}\ln \left ( \sin \left ( fx+e \right ) \right ) }{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.11063, size = 69, normalized size = 1.23 \begin{align*} -\frac{b^{2} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) +{\left (a^{2} - 2 \, a b\right )} \log \left (\sin \left (f x + e\right )^{2}\right ) + \frac{a^{2}}{\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*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.12393, size = 234, normalized size = 4.18 \begin{align*} -\frac{b^{2} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{2} + a^{2} \tan \left (f x + e\right )^{2} +{\left (a^{2} - 2 \, a b\right )} \log \left (\frac{\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{2} + a^{2}}{2 \, f \tan \left (f x + e\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 9.12316, size = 131, normalized size = 2.34 \begin{align*} \begin{cases} \tilde{\infty } a^{2} x & \text{for}\: \left (e = 0 \vee e = - f x\right ) \wedge \left (e = - f x \vee f = 0\right ) \\x \left (a + b \tan ^{2}{\left (e \right )}\right )^{2} \cot ^{3}{\left (e \right )} & \text{for}\: f = 0 \\\frac{a^{2} \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} - \frac{a^{2} \log{\left (\tan{\left (e + f x \right )} \right )}}{f} - \frac{a^{2}}{2 f \tan ^{2}{\left (e + f x \right )}} - \frac{a b \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{f} + \frac{2 a b \log{\left (\tan{\left (e + f x \right )} \right )}}{f} + \frac{b^{2} \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*a**2*x, (Eq(e, 0) | Eq(e, -f*x)) & (Eq(f, 0) | Eq(e, -f*x))), (x*(a + b*tan(e)**2)**2*cot(e)**3
, Eq(f, 0)), (a**2*log(tan(e + f*x)**2 + 1)/(2*f) - a**2*log(tan(e + f*x))/f - a**2/(2*f*tan(e + f*x)**2) - a*
b*log(tan(e + f*x)**2 + 1)/f + 2*a*b*log(tan(e + f*x))/f + b**2*log(tan(e + f*x)**2 + 1)/(2*f), True))

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Giac [B]  time = 1.64026, size = 223, normalized size = 3.98 \begin{align*} \frac{a^{2}{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - 4 \, b^{2} \log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2\right ) + 4 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right )}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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