3.345 \(\int \frac{\cot ^6(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=252 \[ \frac{\left (5 a^2+4 a b-24 b^2\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 f (a-b)}-\frac{\left (10 a^2 b+15 a^3+8 a b^2-48 b^3\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^4 f (a-b)}-\frac{(a-6 b) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^2 f (a-b)}-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{3/2}}-\frac{b \cot ^5(e+f x)}{a f (a-b) \sqrt{a+b \tan ^2(e+f x)}} \]
[Out]
-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(3/2)*f)) - (b*Cot[e + f*x]^5)/(a*(a
- b)*f*Sqrt[a + b*Tan[e + f*x]^2]) - ((15*a^3 + 10*a^2*b + 8*a*b^2 - 48*b^3)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f
*x]^2])/(15*a^4*(a - b)*f) + ((5*a^2 + 4*a*b - 24*b^2)*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^3*(a -
b)*f) - ((a - 6*b)*Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(5*a^2*(a - b)*f)
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Rubi [A] time = 0.368314, antiderivative size = 252, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used =
{3670, 472, 583, 12, 377, 203} \[ \frac{\left (5 a^2+4 a b-24 b^2\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 f (a-b)}-\frac{\left (10 a^2 b+15 a^3+8 a b^2-48 b^3\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^4 f (a-b)}-\frac{(a-6 b) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^2 f (a-b)}-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{3/2}}-\frac{b \cot ^5(e+f x)}{a f (a-b) \sqrt{a+b \tan ^2(e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[Cot[e + f*x]^6/(a + b*Tan[e + f*x]^2)^(3/2),x]
[Out]
-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(3/2)*f)) - (b*Cot[e + f*x]^5)/(a*(a
- b)*f*Sqrt[a + b*Tan[e + f*x]^2]) - ((15*a^3 + 10*a^2*b + 8*a*b^2 - 48*b^3)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f
*x]^2])/(15*a^4*(a - b)*f) + ((5*a^2 + 4*a*b - 24*b^2)*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^3*(a -
b)*f) - ((a - 6*b)*Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(5*a^2*(a - b)*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 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 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)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \cot ^5(e+f x)}{a (a-b) f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{a-6 b-6 b x^2}{x^6 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{a (a-b) f}\\ &=-\frac{b \cot ^5(e+f x)}{a (a-b) f \sqrt{a+b \tan ^2(e+f x)}}-\frac{(a-6 b) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^2 (a-b) f}-\frac{\operatorname{Subst}\left (\int \frac{5 a^2+4 a b-24 b^2+4 (a-6 b) b x^2}{x^4 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 a^2 (a-b) f}\\ &=-\frac{b \cot ^5(e+f x)}{a (a-b) f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\left (5 a^2+4 a b-24 b^2\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 (a-b) f}-\frac{(a-6 b) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^2 (a-b) f}+\frac{\operatorname{Subst}\left (\int \frac{15 a^3+10 a^2 b+8 a b^2-48 b^3+2 b \left (5 a^2+4 a b-24 b^2\right ) x^2}{x^2 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a^3 (a-b) f}\\ &=-\frac{b \cot ^5(e+f x)}{a (a-b) f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (15 a^3+10 a^2 b+8 a b^2-48 b^3\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^4 (a-b) f}+\frac{\left (5 a^2+4 a b-24 b^2\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 (a-b) f}-\frac{(a-6 b) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^2 (a-b) f}-\frac{\operatorname{Subst}\left (\int \frac{15 a^4}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a^4 (a-b) f}\\ &=-\frac{b \cot ^5(e+f x)}{a (a-b) f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (15 a^3+10 a^2 b+8 a b^2-48 b^3\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^4 (a-b) f}+\frac{\left (5 a^2+4 a b-24 b^2\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 (a-b) f}-\frac{(a-6 b) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^2 (a-b) 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 )}{(a-b) f}\\ &=-\frac{b \cot ^5(e+f x)}{a (a-b) f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (15 a^3+10 a^2 b+8 a b^2-48 b^3\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^4 (a-b) f}+\frac{\left (5 a^2+4 a b-24 b^2\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 (a-b) f}-\frac{(a-6 b) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^2 (a-b) 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 )}{(a-b) f}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2} f}-\frac{b \cot ^5(e+f x)}{a (a-b) f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (15 a^3+10 a^2 b+8 a b^2-48 b^3\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^4 (a-b) f}+\frac{\left (5 a^2+4 a b-24 b^2\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 (a-b) f}-\frac{(a-6 b) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^2 (a-b) f}\\ \end{align*}
Mathematica [C] time = 16.4838, size = 850, normalized size = 3.37 \[ \frac{\sqrt{\frac{\cos (2 (e+f x)) a+a+b-b \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \left (\frac{\sin (2 (e+f x)) b^4}{a^4 (a-b) (\cos (2 (e+f x)) a+a+b-b \cos (2 (e+f x)))}-\frac{\cot (e+f x) \csc ^4(e+f x)}{5 a^2}+\frac{(11 a \cos (e+f x)+9 b \cos (e+f x)) \csc ^3(e+f x)}{15 a^3}+\frac{\left (-23 \cos (e+f x) a^2-34 b \cos (e+f x) a-33 b^2 \cos (e+f x)\right ) \csc (e+f x)}{15 a^4}\right )}{f}-\frac{-\frac{b \sqrt{\frac{a+b+(a-b) \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \sqrt{-\frac{a \cot ^2(e+f x)}{b}} \sqrt{-\frac{a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right ),1\right ) \sin ^4(e+f x)}{a (a+b+(a-b) \cos (2 (e+f x)))}-\frac{4 b \sqrt{\cos (2 (e+f x))+1} \sqrt{\frac{a+b+(a-b) \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \left (\frac{\sqrt{-\frac{a \cot ^2(e+f x)}{b}} \sqrt{-\frac{a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right ),1\right ) \sin ^4(e+f x)}{4 a \sqrt{\cos (2 (e+f x))+1} \sqrt{a+b+(a-b) \cos (2 (e+f x))}}-\frac{\sqrt{-\frac{a \cot ^2(e+f x)}{b}} \sqrt{-\frac{a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \Pi \left (-\frac{b}{a-b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right )\right |1\right ) \sin ^4(e+f x)}{2 (a-b) \sqrt{\cos (2 (e+f x))+1} \sqrt{a+b+(a-b) \cos (2 (e+f x))}}\right )}{\sqrt{a+b+(a-b) \cos (2 (e+f x))}}}{(a-b) f} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cot[e + f*x]^6/(a + b*Tan[e + f*x]^2)^(3/2),x]
[Out]
-((-((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)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e +
f*x]^4)/(a*(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)]]) - (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)]*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)]])))/Sqrt[a + b + (a - b)*Cos[2*(e
+ f*x)]])/((a - b)*f)) + (Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(((-2
3*a^2*Cos[e + f*x] - 34*a*b*Cos[e + f*x] - 33*b^2*Cos[e + f*x])*Csc[e + f*x])/(15*a^4) + ((11*a*Cos[e + f*x] +
9*b*Cos[e + f*x])*Csc[e + f*x]^3)/(15*a^3) - (Cot[e + f*x]*Csc[e + f*x]^4)/(5*a^2) + (b^4*Sin[2*(e + f*x)])/(
a^4*(a - b)*(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)]))))/f
________________________________________________________________________________________
Maple [C] time = 0.355, size = 3925, normalized size = 15.6 \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)^6/(a+b*tan(f*x+e)^2)^(3/2),x)
[Out]
-1/15/f/a^4/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/(a-b)/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*(-30*cos(f*x+e
)^5*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)*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))*a^4+15*cos(f*x+e)^5*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^4-30*
cos(f*x+e)^4*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*co
s(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))*a^4+15*cos(f*x+e)^4*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^4+60*cos(f*x+e)^3*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)*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))*a^4-30*cos(f*x+e)^3*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^4+60*cos(f*x+e)^2*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)*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))*a^4-30*cos(f*x+e)^2*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^4-30*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)*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))*a^4+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^4+10*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b^2+8*((2*I*b^(1/2)*(a-b)^(1/2)+
a-2*b)/a)^(1/2)*a*b^3-48*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^4+15*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((c
os(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^4-30*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)*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))*a^4-12*cos(f*x+e)^6*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2
)*a^3*b+34*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^3*b-40*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1
/2)+a-2*b)/a)^(1/2)*a^3*b-26*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b^2-48*cos(f*x+e)^2*((
2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b^3-12*cos(f*x+e)^6*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b^
2-32*cos(f*x+e)^6*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b^3+28*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-
2*b)/a)^(1/2)*a^2*b^2+72*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b^3+23*cos(f*x+e)^6*((2*I*b^
(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^4-144*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^4+48*cos(f*
x+e)^6*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^4-35*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)
*a^4+15*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^4+15*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2
)*a^3*b+144*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^4)*cos(f*x+e)^3*((a*cos(f*x+e)^2-cos(f*x+
e)^2*b+b)/cos(f*x+e)^2)^(3/2)/sin(f*x+e)^5
________________________________________________________________________________________
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)^(3/2),x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
Fricas [A] time = 2.85181, size = 1569, normalized size = 6.23 \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*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")
[Out]
[1/60*(15*(a^4*b*tan(f*x + e)^7 + a^5*tan(f*x + e)^5)*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)) - 4*((15*a^4*b - 5*a^3*b^2 - 2*a^2*b^3 - 56*a*b^4
+ 48*b^5)*tan(f*x + e)^6 + 3*a^5 - 6*a^4*b + 3*a^3*b^2 + (15*a^5 - 10*a^4*b - a^3*b^2 - 28*a^2*b^3 + 24*a*b^4
)*tan(f*x + e)^4 - (5*a^5 - 4*a^4*b - 7*a^3*b^2 + 6*a^2*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^6
*b - 2*a^5*b^2 + a^4*b^3)*f*tan(f*x + e)^7 + (a^7 - 2*a^6*b + a^5*b^2)*f*tan(f*x + e)^5), -1/30*(15*(a^4*b*tan
(f*x + e)^7 + a^5*tan(f*x + e)^5)*sqrt(a - b)*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)) + 2*((15*a^4*b - 5*a^3*b^2 - 2*a^2*b^3 - 56*a*b^4 + 48*b^5)*tan(f*x + e)^6 + 3*a
^5 - 6*a^4*b + 3*a^3*b^2 + (15*a^5 - 10*a^4*b - a^3*b^2 - 28*a^2*b^3 + 24*a*b^4)*tan(f*x + e)^4 - (5*a^5 - 4*a
^4*b - 7*a^3*b^2 + 6*a^2*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^6*b - 2*a^5*b^2 + a^4*b^3)*f*tan
(f*x + e)^7 + (a^7 - 2*a^6*b + a^5*b^2)*f*tan(f*x + e)^5)]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{6}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, 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)**(3/2),x)
[Out]
Integral(cot(e + f*x)**6/(a + b*tan(e + f*x)**2)**(3/2), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{6}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="giac")
[Out]
integrate(cot(f*x + e)^6/(b*tan(f*x + e)^2 + a)^(3/2), x)