3.317 \(\int \cot ^4(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=115 \[ \frac{(a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}-\frac{a \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 f}+\frac{(3 a-4 b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 f} \]

[Out]

((a - b)^(3/2)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/f + ((3*a - 4*b)*Cot[e + f*x]*Sq
rt[a + b*Tan[e + f*x]^2])/(3*f) - (a*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(3*f)

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Rubi [A]  time = 0.173331, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3670, 474, 583, 12, 377, 203} \[ \frac{(a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}-\frac{a \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 f}+\frac{(3 a-4 b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^4*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

((a - b)^(3/2)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/f + ((3*a - 4*b)*Cot[e + f*x]*Sq
rt[a + b*Tan[e + f*x]^2])/(3*f) - (a*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(3*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 474

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(c*(e*x)^
(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)
*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*(c*b - a*d)*(m + 1) + c*n*(b*c*(p + 1) + a*d*(q - 1)) + d*((c*b - a*
d)*(m + 1) + c*b*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0]
&& GtQ[q, 1] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 12

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

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.289824, size = 78, normalized size = 0.68 \[ -\frac{\cot (e+f x) \sqrt{a+b \tan ^2(e+f x)} \left (a \cot ^2(e+f x)+b\right ) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-\frac{(a-b) \tan ^2(e+f x)}{a+b \tan ^2(e+f x)}\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^4*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

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

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Maple [C]  time = 0.319, size = 6591, normalized size = 57.3 \begin{align*} \text{output too large to display} \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)^(3/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \cot \left (f x + e\right )^{4}\,{d x} \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)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2)*cot(f*x + e)^4, x)

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Fricas [A]  time = 2.36605, size = 760, normalized size = 6.61 \begin{align*} \left [-\frac{3 \,{\left (a - b\right )} \sqrt{-a + b} \log \left (-\frac{{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \,{\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} - 4 \,{\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{3} - 4 \,{\left ({\left (3 \, a - 4 \, b\right )} \tan \left (f x + e\right )^{2} - a\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{12 \, f \tan \left (f x + e\right )^{3}}, \frac{3 \,{\left (a - b\right )}^{\frac{3}{2}} \arctan \left (-\frac{2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) \tan \left (f x + e\right )^{3} + 2 \,{\left ({\left (3 \, a - 4 \, b\right )} \tan \left (f x + e\right )^{2} - a\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{6 \, f \tan \left (f x + e\right )^{3}}\right ] \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)^(3/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(a - b)*sqrt(-a + b)*log(-((a^2 - 8*a*b + 8*b^2)*tan(f*x + e)^4 - 2*(3*a^2 - 4*a*b)*tan(f*x + e)^2 +
a^2 - 4*((a - 2*b)*tan(f*x + e)^3 - a*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b))/(tan(f*x + e)^4
+ 2*tan(f*x + e)^2 + 1))*tan(f*x + e)^3 - 4*((3*a - 4*b)*tan(f*x + e)^2 - a)*sqrt(b*tan(f*x + e)^2 + a))/(f*ta
n(f*x + e)^3), 1/6*(3*(a - b)^(3/2)*arctan(-2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b)*tan(f*x + e)/((a - 2*b)*t
an(f*x + e)^2 - a))*tan(f*x + e)^3 + 2*((3*a - 4*b)*tan(f*x + e)^2 - a)*sqrt(b*tan(f*x + e)^2 + a))/(f*tan(f*x
+ e)^3)]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \cot \left (f x + e\right )^{4}\,{d x} \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)^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2)*cot(f*x + e)^4, x)