3.196 \(\int \cot ^4(e+f x) (a+b \tan ^2(e+f x)) \, dx\)

Optimal. Leaf size=39 \[ \frac{(a-b) \cot (e+f x)}{f}+x (a-b)-\frac{a \cot ^3(e+f x)}{3 f} \]

[Out]

(a - b)*x + ((a - b)*Cot[e + f*x])/f - (a*Cot[e + f*x]^3)/(3*f)

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Rubi [A]  time = 0.0370697, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3629, 12, 3473, 8} \[ \frac{(a-b) \cot (e+f x)}{f}+x (a-b)-\frac{a \cot ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^4*(a + b*Tan[e + f*x]^2),x]

[Out]

(a - b)*x + ((a - b)*Cot[e + f*x])/f - (a*Cot[e + f*x]^3)/(3*f)

Rule 3629

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[
((A*b^2 + a^2*C)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*
Tan[e + f*x])^(m + 1)*Simp[a*(A - C) - (A*b - b*C)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] &&
NeQ[A*b^2 + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [C]  time = 0.0371482, size = 65, normalized size = 1.67 \[ -\frac{a \cot ^3(e+f x) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-\tan ^2(e+f x)\right )}{3 f}-\frac{b \cot (e+f x) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-\tan ^2(e+f x)\right )}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^4*(a + b*Tan[e + f*x]^2),x]

[Out]

-(a*Cot[e + f*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[e + f*x]^2])/(3*f) - (b*Cot[e + f*x]*Hypergeometric2F
1[-1/2, 1, 1/2, -Tan[e + f*x]^2])/f

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Maple [A]  time = 0.04, size = 47, normalized size = 1.2 \begin{align*}{\frac{1}{f} \left ( b \left ( -\cot \left ( fx+e \right ) -fx-e \right ) +a \left ( -{\frac{ \left ( \cot \left ( fx+e \right ) \right ) ^{3}}{3}}+\cot \left ( fx+e \right ) +fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^4*(a+b*tan(f*x+e)^2),x)

[Out]

1/f*(b*(-cot(f*x+e)-f*x-e)+a*(-1/3*cot(f*x+e)^3+cot(f*x+e)+f*x+e))

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Maxima [A]  time = 1.6165, size = 62, normalized size = 1.59 \begin{align*} \frac{3 \,{\left (f x + e\right )}{\left (a - b\right )} + \frac{3 \,{\left (a - b\right )} \tan \left (f x + e\right )^{2} - a}{\tan \left (f x + e\right )^{3}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4*(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.06863, size = 116, normalized size = 2.97 \begin{align*} \frac{3 \,{\left (a - b\right )} f x \tan \left (f x + e\right )^{3} + 3 \,{\left (a - b\right )} \tan \left (f x + e\right )^{2} - a}{3 \, f \tan \left (f x + e\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4*(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.62876, size = 70, normalized size = 1.79 \begin{align*} \begin{cases} \tilde{\infty } a 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 ) \cot ^{4}{\left (e \right )} & \text{for}\: f = 0 \\a x + \frac{a}{f \tan{\left (e + f x \right )}} - \frac{a}{3 f \tan ^{3}{\left (e + f x \right )}} - b x - \frac{b}{f \tan{\left (e + f x \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**4*(a+b*tan(f*x+e)**2),x)

[Out]

Piecewise((zoo*a*x, (Eq(e, 0) | Eq(e, -f*x)) & (Eq(f, 0) | Eq(e, -f*x))), (x*(a + b*tan(e)**2)*cot(e)**4, Eq(f
, 0)), (a*x + a/(f*tan(e + f*x)) - a/(3*f*tan(e + f*x)**3) - b*x - b/(f*tan(e + f*x)), True))

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Giac [B]  time = 1.32586, size = 143, normalized size = 3.67 \begin{align*} \frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 24 \,{\left (f x + e\right )}{\left (a - b\right )} - 15 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 12 \, b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{15 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 12 \, b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3}}}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4*(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

1/24*(a*tan(1/2*f*x + 1/2*e)^3 + 24*(f*x + e)*(a - b) - 15*a*tan(1/2*f*x + 1/2*e) + 12*b*tan(1/2*f*x + 1/2*e)
+ (15*a*tan(1/2*f*x + 1/2*e)^2 - 12*b*tan(1/2*f*x + 1/2*e)^2 - a)/tan(1/2*f*x + 1/2*e)^3)/f