3.414 \(\int \frac{\cot ^6(e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx\)

Optimal. Leaf size=172 \[ -\frac{\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 f (a+b)^3}-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{\sqrt{a} f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{5 f (a+b)}+\frac{(5 a+9 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 f (a+b)^2} \]

[Out]

-(ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(Sqrt[a]*f)) - ((15*a^2 + 40*a*b + 33*b^2)*Cot
[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(15*(a + b)^3*f) + ((5*a + 9*b)*Cot[e + f*x]^3*Sqrt[a + b + b*Tan[e
+ f*x]^2])/(15*(a + b)^2*f) - (Cot[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/(5*(a + b)*f)

________________________________________________________________________________________

Rubi [A]  time = 0.344836, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {4141, 1975, 480, 583, 12, 377, 203} \[ -\frac{\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 f (a+b)^3}-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{\sqrt{a} f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{5 f (a+b)}+\frac{(5 a+9 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^6/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

-(ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(Sqrt[a]*f)) - ((15*a^2 + 40*a*b + 33*b^2)*Cot
[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(15*(a + b)^3*f) + ((5*a + 9*b)*Cot[e + f*x]^3*Sqrt[a + b + b*Tan[e
+ f*x]^2])/(15*(a + b)^2*f) - (Cot[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/(5*(a + b)*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 480

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*e*(m + 1)), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[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 \frac{\cot ^6(e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \sqrt{a+b \left (1+x^2\right )}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{-5 a-9 b-4 b x^2}{x^4 \left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f}\\ &=\frac{(5 a+9 b) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 (a+b) f}-\frac{\operatorname{Subst}\left (\int \frac{-15 a^2-40 a b-33 b^2-2 b (5 a+9 b) x^2}{x^2 \left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f}\\ &=-\frac{\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}+\frac{(5 a+9 b) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 (a+b) f}+\frac{\operatorname{Subst}\left (\int -\frac{15 (a+b)^3}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^3 f}\\ &=-\frac{\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}+\frac{(5 a+9 b) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 (a+b) 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 )}{f}\\ &=-\frac{\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}+\frac{(5 a+9 b) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 (a+b) 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 )}{f}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{\sqrt{a} f}-\frac{\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}+\frac{(5 a+9 b) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 (a+b) f}\\ \end{align*}

Mathematica [A]  time = 4.5927, size = 199, normalized size = 1.16 \[ -\frac{\csc (e+f x) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-\left (11 a^2+26 a b+15 b^2\right ) \csc ^2(e+f x)+23 a^2+3 (a+b)^2 \csc ^4(e+f x)+60 a b+45 b^2\right )}{30 f (a+b)^3 \sqrt{a+b \sec ^2(e+f x)}}-\frac{\sec (e+f x) \sqrt{a \cos (2 e+2 f x)+a+2 b} \tan ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{-a \sin ^2(e+f x)+a+b}}\right )}{\sqrt{2} \sqrt{a} f \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^6/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

-((ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]*Sec[e + f*
x])/(Sqrt[2]*Sqrt[a]*f*Sqrt[a + b*Sec[e + f*x]^2])) - ((a + 2*b + a*Cos[2*(e + f*x)])*Csc[e + f*x]*(23*a^2 + 6
0*a*b + 45*b^2 - (11*a^2 + 26*a*b + 15*b^2)*Csc[e + f*x]^2 + 3*(a + b)^2*Csc[e + f*x]^4)*Sec[e + f*x])/(30*(a
+ b)^3*f*Sqrt[a + b*Sec[e + f*x]^2])

________________________________________________________________________________________

Maple [C]  time = 0.641, size = 11267, normalized size = 65.5 \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)^6/(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

result too large to display

________________________________________________________________________________________

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)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [B]  time = 6.61052, size = 2334, normalized size = 13.57 \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)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/120*(15*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*(a^3 + 3*a^2*b
+ 3*a*b^2 + b^3)*cos(f*x + e)^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(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*((23*a^3 + 60*a^2*b + 45*a*b^2)*cos(f*x + e)^5 - (
35*a^3 + 94*a^2*b + 75*a*b^2)*cos(f*x + e)^3 + (15*a^3 + 40*a^2*b + 33*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x +
e)^2 + b)/cos(f*x + e)^2))/(((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^4 - 2*(a^4 + 3*a^3*b + 3*a^2*b
^2 + a*b^3)*f*cos(f*x + e)^2 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f)*sin(f*x + e)), 1/60*(15*((a^3 + 3*a^2*b
+ 3*a*b^2 + b^3)*cos(f*x + e)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*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*((23*a^3 + 60*a^2*b + 45*a*b^2)*cos(f*x + e)^5 - (35*a^3 +
94*a^2*b + 75*a*b^2)*cos(f*x + e)^3 + (15*a^3 + 40*a^2*b + 33*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)
/cos(f*x + e)^2))/(((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^4 - 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^
3)*f*cos(f*x + e)^2 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f)*sin(f*x + e))]

________________________________________________________________________________________

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)**6/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{6}}{\sqrt{b \sec \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(cot(f*x + e)^6/sqrt(b*sec(f*x + e)^2 + a), x)