3.332 \(\int \frac{\cot ^6(e+f x)}{\sqrt{a+b \tan ^2(e+f x)}} \, dx\)
Optimal. Leaf size=170 \[ -\frac{\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 f}+\frac{(5 a+4 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f \sqrt{a-b}}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a f} \]
[Out]
-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/(Sqrt[a - b]*f)) - ((15*a^2 + 10*a*b + 8*b^2)*
Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^3*f) + ((5*a + 4*b)*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/
(15*a^2*f) - (Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(5*a*f)
________________________________________________________________________________________
Rubi [A] time = 0.245944, antiderivative size = 170, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used =
{3670, 480, 583, 12, 377, 203} \[ -\frac{\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 f}+\frac{(5 a+4 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f \sqrt{a-b}}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a f} \]
Antiderivative was successfully verified.
[In]
Int[Cot[e + f*x]^6/Sqrt[a + b*Tan[e + f*x]^2],x]
[Out]
-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/(Sqrt[a - b]*f)) - ((15*a^2 + 10*a*b + 8*b^2)*
Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^3*f) + ((5*a + 4*b)*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/
(15*a^2*f) - (Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(5*a*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 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 \tan ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a f}+\frac{\operatorname{Subst}\left (\int \frac{-5 a-4 b-4 b x^2}{x^4 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=\frac{(5 a+4 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a f}-\frac{\operatorname{Subst}\left (\int \frac{-15 a^2-10 a b-8 b^2-2 b (5 a+4 b) x^2}{x^2 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a^2 f}\\ &=-\frac{\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 f}+\frac{(5 a+4 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a f}+\frac{\operatorname{Subst}\left (\int -\frac{15 a^3}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a^3 f}\\ &=-\frac{\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 f}+\frac{(5 a+4 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a f}-\frac{\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{\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 f}+\frac{(5 a+4 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a f}-\frac{\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{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{\sqrt{a-b} f}-\frac{\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 f}+\frac{(5 a+4 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a f}\\ \end{align*}
Mathematica [C] time = 16.286, size = 794, normalized size = 4.67 \[ \frac{b \sin ^4(e+f x) \csc (2 (e+f x)) \sqrt{\frac{(a-b) \cos (2 (e+f x))+a+b}{\cos (2 (e+f x))+1}} \sqrt{-\frac{a \cot ^2(e+f x)}{b}} \sqrt{-\frac{a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}{\sqrt{2}}\right ),1\right )}{a f ((a-b) \cos (2 (e+f x))+a+b)}+\frac{4 b \sqrt{\cos (2 (e+f x))+1} \sqrt{\frac{(a-b) \cos (2 (e+f x))+a+b}{\cos (2 (e+f x))+1}} \left (\frac{\sin ^4(e+f x) \csc (2 (e+f x)) \sqrt{-\frac{a \cot ^2(e+f x)}{b}} \sqrt{-\frac{a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}{\sqrt{2}}\right ),1\right )}{4 a \sqrt{\cos (2 (e+f x))+1} \sqrt{(a-b) \cos (2 (e+f x))+a+b}}-\frac{\sin ^4(e+f x) \csc (2 (e+f x)) \sqrt{-\frac{a \cot ^2(e+f x)}{b}} \sqrt{-\frac{a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \Pi \left (-\frac{b}{a-b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right )\right |1\right )}{2 (a-b) \sqrt{\cos (2 (e+f x))+1} \sqrt{(a-b) \cos (2 (e+f x))+a+b}}\right )}{f \sqrt{(a-b) \cos (2 (e+f x))+a+b}}+\frac{\sqrt{\frac{a \cos (2 (e+f x))+a-b \cos (2 (e+f x))+b}{\cos (2 (e+f x))+1}} \left (\frac{\csc (e+f x) \left (-23 a^2 \cos (e+f x)-14 a b \cos (e+f x)-8 b^2 \cos (e+f x)\right )}{15 a^3}+\frac{\csc ^3(e+f x) (11 a \cos (e+f x)+4 b \cos (e+f x))}{15 a^2}-\frac{\cot (e+f x) \csc ^4(e+f x)}{5 a}\right )}{f} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[e + f*x]^6/Sqrt[a + b*Tan[e + f*x]^2],x]
[Out]
(Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(((-23*a^2*Cos[e + f*x] - 14*a
*b*Cos[e + f*x] - 8*b^2*Cos[e + f*x])*Csc[e + f*x])/(15*a^3) + ((11*a*Cos[e + f*x] + 4*b*Cos[e + f*x])*Csc[e +
f*x]^3)/(15*a^2) - (Cot[e + f*x]*Csc[e + f*x]^4)/(5*a)))/f + (b*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 +
Cos[2*(e + f*x)])]*Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a
+ b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*C
os[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(a*f*(a + b + (a - b)*Cos[2*(e + f*x)])) + (4
*b*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*((Sqrt[-((a*Cot[
e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*
Csc[e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/
b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(4*a*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]) - (Sqr
t[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(
e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e
+ f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(2*(a - b)*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + (
a - b)*Cos[2*(e + f*x)]])))/(f*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])
________________________________________________________________________________________
Maple [C] time = 0.346, size = 3741, normalized size = 22. \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*tan(f*x+e)^2)^(1/2),x)
[Out]
-1/15/f/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/a^3*(23*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)
^6*a^3-30*a^3*2^(1/2)*(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b
)/(cos(f*x+e)+1))^(1/2)*(-2/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e
)-b)/(cos(f*x+e)+1))^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(
2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a
)^(1/2))*sin(f*x+e)+15*2^(1/2)*(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos
(f*x+e)+b)/(cos(f*x+e)+1))^(1/2)*(-2/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*
cos(f*x+e)-b)/(cos(f*x+e)+1))^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x
+e),((8*I*b^(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*a^3*sin(f*x+e)+8*((2*I*b^
(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^3+15*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^2*a^3-8*((2*I*b^
(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^6*b^3+24*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^4*b
^3-24*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^2*b^3-35*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*
cos(f*x+e)^4*a^3+15*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b+10*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/
2)*a*b^2-9*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^6*a^2*b-6*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^
(1/2)*cos(f*x+e)^6*a*b^2+34*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^4*a^2*b+22*((2*I*b^(1/2)*(a-b
)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^4*a*b^2-40*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^2*a^2*b-26*
((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^2*a*b^2+60*2^(1/2)*(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)
-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1/2)*(-2/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1
/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b
^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-
a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*cos(f*x+e)^2*sin(f*x+e)*a^3+15*cos(f*x+e)*sin(f*x+e
)*2^(1/2)*(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e
)+1))^(1/2)*(-2/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*
x+e)+1))^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-
b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*a^3-30*cos(f*x+e)*sin(f*x+e)*2^(1/2)*(1/a*(I*c
os(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1/2)*(-2/a*(
I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))^(1/2)*Elli
pticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*
a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*a^3+15*2^(1/2)*(1/a*(I
*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1/2)*(-2/a
*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))^(1/2)*El
lipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/2)-4*I*b^(1
/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*cos(f*x+e)^5*sin(f*x+e)*a^3-30*2^(1/2)*(1/a*(I*cos(f*x+e)*b^(1/
2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1/2)*(-2/a*(I*cos(f*x+e)*b^
(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))^(1/2)*EllipticPi((cos(f*x
+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2
)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*cos(f*x+e)^5*sin(f*x+e)*a^3+15*2^(1/2
)*(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1
/2)*(-2/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))
^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/2)
-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*cos(f*x+e)^4*sin(f*x+e)*a^3-30*2^(1/2)*(1/a*(I*cos(f*x
+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1/2)*(-2/a*(I*cos(
f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))^(1/2)*EllipticPi
((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2
*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*cos(f*x+e)^4*sin(f*x+e)*a^3-
30*2^(1/2)*(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+
e)+1))^(1/2)*(-2/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f
*x+e)+1))^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a
-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*cos(f*x+e)^3*sin(f*x+e)*a^3+60*2^(1/2)*(1/a*(
I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1/2)*(-2/
a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))^(1/2)*E
llipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*
b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*cos(f*x+e)^3*sin(f*
x+e)*a^3-30*2^(1/2)*(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/
(cos(f*x+e)+1))^(1/2)*(-2/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-
b)/(cos(f*x+e)+1))^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b
^(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*cos(f*x+e)^2*sin(f*x+e)*a^3)/cos(f*x
+e)/sin(f*x+e)^5/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)
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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)^6/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [A] time = 2.85725, size = 1022, normalized size = 6.01 \begin{align*} \left [-\frac{15 \, a^{3} \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 )^{5} + 4 \,{\left ({\left (15 \, a^{3} - 5 \, a^{2} b - 2 \, a b^{2} - 8 \, b^{3}\right )} \tan \left (f x + e\right )^{4} + 3 \, a^{3} - 3 \, a^{2} b -{\left (5 \, a^{3} - a^{2} b - 4 \, a b^{2}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{60 \,{\left (a^{4} - a^{3} b\right )} f \tan \left (f x + e\right )^{5}}, -\frac{15 \, \sqrt{a - b} a^{3} \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 )^{5} + 2 \,{\left ({\left (15 \, a^{3} - 5 \, a^{2} b - 2 \, a b^{2} - 8 \, b^{3}\right )} \tan \left (f x + e\right )^{4} + 3 \, a^{3} - 3 \, a^{2} b -{\left (5 \, a^{3} - a^{2} b - 4 \, a b^{2}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{30 \,{\left (a^{4} - a^{3} b\right )} f \tan \left (f x + e\right )^{5}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
[Out]
[-1/60*(15*a^3*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)^5 + 4*((15*a^3 - 5*a^2*b - 2*a*b^2 - 8*b^3)*tan(f*x + e)^4 + 3*a^3 - 3*a^2*
b - (5*a^3 - a^2*b - 4*a*b^2)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^4 - a^3*b)*f*tan(f*x + e)^5), -1
/30*(15*sqrt(a - b)*a^3*arctan(-2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b)*tan(f*x + e)/((a - 2*b)*tan(f*x + e)^
2 - a))*tan(f*x + e)^5 + 2*((15*a^3 - 5*a^2*b - 2*a*b^2 - 8*b^3)*tan(f*x + e)^4 + 3*a^3 - 3*a^2*b - (5*a^3 - a
^2*b - 4*a*b^2)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^4 - a^3*b)*f*tan(f*x + e)^5)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{6}{\left (e + f x \right )}}{\sqrt{a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)**6/(a+b*tan(f*x+e)**2)**(1/2),x)
[Out]
Integral(cot(e + f*x)**6/sqrt(a + b*tan(e + f*x)**2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{6}}{\sqrt{b \tan \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*tan(f*x+e)^2)^(1/2),x, algorithm="giac")
[Out]
integrate(cot(f*x + e)^6/sqrt(b*tan(f*x + e)^2 + a), x)