3.438 \(\int \frac{\cot ^2(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=174 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{a^{5/2} f}-\frac{(a-3 b) (3 a+b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{3 a^2 f (a+b)^3}-\frac{b (7 a+3 b) \cot (e+f x)}{3 a^2 f (a+b)^2 \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{b \cot (e+f x)}{3 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

[Out]

-(ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(a^(5/2)*f)) - (b*Cot[e + f*x])/(3*a*(a + b)*f
*(a + b + b*Tan[e + f*x]^2)^(3/2)) - (b*(7*a + 3*b)*Cot[e + f*x])/(3*a^2*(a + b)^2*f*Sqrt[a + b + b*Tan[e + f*
x]^2]) - ((a - 3*b)*(3*a + b)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(3*a^2*(a + b)^3*f)

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Rubi [A]  time = 0.377997, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {4141, 1975, 472, 579, 583, 12, 377, 203} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{a^{5/2} f}-\frac{(a-3 b) (3 a+b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{3 a^2 f (a+b)^3}-\frac{b (7 a+3 b) \cot (e+f x)}{3 a^2 f (a+b)^2 \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{b \cot (e+f x)}{3 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(a^(5/2)*f)) - (b*Cot[e + f*x])/(3*a*(a + b)*f
*(a + b + b*Tan[e + f*x]^2)^(3/2)) - (b*(7*a + 3*b)*Cot[e + f*x])/(3*a^2*(a + b)^2*f*Sqrt[a + b + b*Tan[e + f*
x]^2]) - ((a - 3*b)*(3*a + b)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(3*a^2*(a + b)^3*f)

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 472

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

Rule 579

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

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 \frac{\cot ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x^2\right ) \left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \cot (e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{3 a-b-4 b x^2}{x^2 \left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 a (a+b) f}\\ &=-\frac{b \cot (e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (7 a+3 b) \cot (e+f x)}{3 a^2 (a+b)^2 f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{(a-3 b) (3 a+b)-2 b (7 a+3 b) x^2}{x^2 \left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a^2 (a+b)^2 f}\\ &=-\frac{b \cot (e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (7 a+3 b) \cot (e+f x)}{3 a^2 (a+b)^2 f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{(a-3 b) (3 a+b) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{3 a^2 (a+b)^3 f}-\frac{\operatorname{Subst}\left (\int \frac{3 (a+b)^3}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a^2 (a+b)^3 f}\\ &=-\frac{b \cot (e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (7 a+3 b) \cot (e+f x)}{3 a^2 (a+b)^2 f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{(a-3 b) (3 a+b) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{3 a^2 (a+b)^3 f}-\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a^2 f}\\ &=-\frac{b \cot (e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (7 a+3 b) \cot (e+f x)}{3 a^2 (a+b)^2 f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{(a-3 b) (3 a+b) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{3 a^2 (a+b)^3 f}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{a^2 f}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{a^{5/2} f}-\frac{b \cot (e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (7 a+3 b) \cot (e+f x)}{3 a^2 (a+b)^2 f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{(a-3 b) (3 a+b) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{3 a^2 (a+b)^3 f}\\ \end{align*}

Mathematica [A]  time = 7.82932, size = 247, normalized size = 1.42 \[ -\frac{\csc (e+f x) \sec ^5(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (4 \left (6 a^3 b+3 a^4+8 a b^3+3 b^4\right ) \cos (2 (e+f x))+a \left (3 a^3+9 a b^2+4 b^3\right ) \cos (4 (e+f x))+3 \left (5 a^2 b^2+8 a^3 b+3 a^4-12 a b^3-4 b^4\right )\right )}{48 a^2 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^{5/2}}-\frac{\sec ^5(e+f x) (a \cos (2 e+2 f x)+a+2 b)^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{-a \sin ^2(e+f x)+a+b}}\right )}{4 \sqrt{2} a^{5/2} f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]]*(a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f
*x]^5)/(4*Sqrt[2]*a^(5/2)*f*(a + b*Sec[e + f*x]^2)^(5/2)) - ((a + 2*b + a*Cos[2*(e + f*x)])*(3*(3*a^4 + 8*a^3*
b + 5*a^2*b^2 - 12*a*b^3 - 4*b^4) + 4*(3*a^4 + 6*a^3*b + 8*a*b^3 + 3*b^4)*Cos[2*(e + f*x)] + a*(3*a^3 + 9*a*b^
2 + 4*b^3)*Cos[4*(e + f*x)])*Csc[e + f*x]*Sec[e + f*x]^5)/(48*a^2*(a + b)^3*f*(a + b*Sec[e + f*x]^2)^(5/2))

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Maple [C]  time = 0.595, size = 7586, normalized size = 43.6 \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)^2/(a+b*sec(f*x+e)^2)^(5/2),x)

[Out]

result too large to display

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Maxima [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)^2/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 7.65613, size = 2473, normalized size = 14.21 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(3*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5 + (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cos(f*x + e)^4 + 2*(a^4
*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(f*x + e)^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*co
s(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4
- 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x
+ e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-
a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))*sin(f*x + e) + 8*((3*a^5 + 9*a^3*b^2 + 4*a^2*b^3)
*cos(f*x + e)^5 + (6*a^4*b - 9*a^3*b^2 + 4*a^2*b^3 + 3*a*b^4)*cos(f*x + e)^3 + (3*a^3*b^2 - 8*a^2*b^3 - 3*a*b^
4)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*f*cos(f*
x + e)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*f*cos(f*x + e)^2 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a
^3*b^5)*f)*sin(f*x + e)), 1/12*(3*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5 + (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)
*cos(f*x + e)^4 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(f*x + e)^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x
+ e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/
cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e)))*sin(f*
x + e) - 4*((3*a^5 + 9*a^3*b^2 + 4*a^2*b^3)*cos(f*x + e)^5 + (6*a^4*b - 9*a^3*b^2 + 4*a^2*b^3 + 3*a*b^4)*cos(f
*x + e)^3 + (3*a^3*b^2 - 8*a^2*b^3 - 3*a*b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a^
8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*f*cos(f*x + e)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*f*cos(f*x +
e)^2 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*f)*sin(f*x + e))]

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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)**2/(a+b*sec(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{2}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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