3.358 \(\int \frac{\cot ^6(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=327 \[ -\frac{\left (a^2-22 a b+16 b^2\right ) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^3 f (a-b)^2}+\frac{\left (4 a^2 b+5 a^3-88 a b^2+64 b^3\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^4 f (a-b)^2}-\frac{\left (8 a^2 b^2+10 a^3 b+15 a^4-176 a b^3+128 b^4\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^5 f (a-b)^2}-\frac{b (11 a-8 b) \cot ^5(e+f x)}{3 a^2 f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}-\frac{b \cot ^5(e+f x)}{3 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

[Out]

-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(5/2)*f)) - (b*Cot[e + f*x]^5)/(3*a*(
a - b)*f*(a + b*Tan[e + f*x]^2)^(3/2)) - ((11*a - 8*b)*b*Cot[e + f*x]^5)/(3*a^2*(a - b)^2*f*Sqrt[a + b*Tan[e +
f*x]^2]) - ((15*a^4 + 10*a^3*b + 8*a^2*b^2 - 176*a*b^3 + 128*b^4)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(1
5*a^5*(a - b)^2*f) + ((5*a^3 + 4*a^2*b - 88*a*b^2 + 64*b^3)*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^4
*(a - b)^2*f) - ((a^2 - 22*a*b + 16*b^2)*Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(5*a^3*(a - b)^2*f)

________________________________________________________________________________________

Rubi [A]  time = 0.49624, antiderivative size = 327, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3670, 472, 579, 583, 12, 377, 203} \[ -\frac{\left (a^2-22 a b+16 b^2\right ) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^3 f (a-b)^2}+\frac{\left (4 a^2 b+5 a^3-88 a b^2+64 b^3\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^4 f (a-b)^2}-\frac{\left (8 a^2 b^2+10 a^3 b+15 a^4-176 a b^3+128 b^4\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^5 f (a-b)^2}-\frac{b (11 a-8 b) \cot ^5(e+f x)}{3 a^2 f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}-\frac{b \cot ^5(e+f x)}{3 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^6/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(5/2)*f)) - (b*Cot[e + f*x]^5)/(3*a*(
a - b)*f*(a + b*Tan[e + f*x]^2)^(3/2)) - ((11*a - 8*b)*b*Cot[e + f*x]^5)/(3*a^2*(a - b)^2*f*Sqrt[a + b*Tan[e +
f*x]^2]) - ((15*a^4 + 10*a^3*b + 8*a^2*b^2 - 176*a*b^3 + 128*b^4)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(1
5*a^5*(a - b)^2*f) + ((5*a^3 + 4*a^2*b - 88*a*b^2 + 64*b^3)*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^4
*(a - b)^2*f) - ((a^2 - 22*a*b + 16*b^2)*Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(5*a^3*(a - b)^2*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 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 ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \cot ^5(e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{3 a-8 b-8 b x^2}{x^6 \left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 a (a-b) f}\\ &=-\frac{b \cot ^5(e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(11 a-8 b) b \cot ^5(e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{3 \left (a^2-22 a b+16 b^2\right )-6 (11 a-8 b) b x^2}{x^6 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a^2 (a-b)^2 f}\\ &=-\frac{b \cot ^5(e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(11 a-8 b) b \cot ^5(e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (a^2-22 a b+16 b^2\right ) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^3 (a-b)^2 f}-\frac{\operatorname{Subst}\left (\int \frac{3 \left (5 a^3+4 a^2 b-88 a b^2+64 b^3\right )+12 b \left (a^2-22 a b+16 b^2\right ) x^2}{x^4 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a^3 (a-b)^2 f}\\ &=-\frac{b \cot ^5(e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(11 a-8 b) b \cot ^5(e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\left (5 a^3+4 a^2 b-88 a b^2+64 b^3\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^4 (a-b)^2 f}-\frac{\left (a^2-22 a b+16 b^2\right ) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^3 (a-b)^2 f}+\frac{\operatorname{Subst}\left (\int \frac{3 \left (15 a^4+10 a^3 b+8 a^2 b^2-176 a b^3+128 b^4\right )+6 b \left (5 a^3+4 a^2 b-88 a b^2+64 b^3\right ) x^2}{x^2 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{45 a^4 (a-b)^2 f}\\ &=-\frac{b \cot ^5(e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(11 a-8 b) b \cot ^5(e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (15 a^4+10 a^3 b+8 a^2 b^2-176 a b^3+128 b^4\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^5 (a-b)^2 f}+\frac{\left (5 a^3+4 a^2 b-88 a b^2+64 b^3\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^4 (a-b)^2 f}-\frac{\left (a^2-22 a b+16 b^2\right ) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^3 (a-b)^2 f}-\frac{\operatorname{Subst}\left (\int \frac{45 a^5}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{45 a^5 (a-b)^2 f}\\ &=-\frac{b \cot ^5(e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(11 a-8 b) b \cot ^5(e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (15 a^4+10 a^3 b+8 a^2 b^2-176 a b^3+128 b^4\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^5 (a-b)^2 f}+\frac{\left (5 a^3+4 a^2 b-88 a b^2+64 b^3\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^4 (a-b)^2 f}-\frac{\left (a^2-22 a b+16 b^2\right ) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^3 (a-b)^2 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)^2 f}\\ &=-\frac{b \cot ^5(e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(11 a-8 b) b \cot ^5(e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (15 a^4+10 a^3 b+8 a^2 b^2-176 a b^3+128 b^4\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^5 (a-b)^2 f}+\frac{\left (5 a^3+4 a^2 b-88 a b^2+64 b^3\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^4 (a-b)^2 f}-\frac{\left (a^2-22 a b+16 b^2\right ) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^3 (a-b)^2 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)^2 f}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{(a-b)^{5/2} f}-\frac{b \cot ^5(e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(11 a-8 b) b \cot ^5(e+f x)}{3 a^2 (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (15 a^4+10 a^3 b+8 a^2 b^2-176 a b^3+128 b^4\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^5 (a-b)^2 f}+\frac{\left (5 a^3+4 a^2 b-88 a b^2+64 b^3\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^4 (a-b)^2 f}-\frac{\left (a^2-22 a b+16 b^2\right ) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a^3 (a-b)^2 f}\\ \end{align*}

Mathematica [C]  time = 16.3189, size = 441, normalized size = 1.35 \[ \frac{\sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (-\frac{15 a^5 b \sin ^2(e+f x) \sin (2 (e+f x)) \left (\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}\right )^{3/2} \left (2 (a-b) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}{\sqrt{2}}\right ),1\right )-2 a \Pi \left (-\frac{b}{a-b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right )\right |1\right )\right )}{2 \sqrt{2}}-(a-b) \left ((a-b)^2 \left (23 a^2+54 a b+73 b^2\right ) \cot (e+f x) ((a-b) \cos (2 (e+f x))+a+b)^2+3 a^2 (a-b)^2 \cot (e+f x) \csc ^4(e+f x) ((a-b) \cos (2 (e+f x))+a+b)^2+10 a b^5 \sin (2 (e+f x))-5 b^4 (15 a-11 b) \sin (2 (e+f x)) ((a-b) \cos (2 (e+f x))+a+b)-a (a-b)^2 (11 a+14 b) \cot (e+f x) \csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)^2\right )\right )}{15 \sqrt{2} a^5 f (a-b)^3 ((a-b) \cos (2 (e+f x))+a+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^6/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*((-15*a^5*b*(((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[
e + f*x]^2)/b)^(3/2)*(2*(a - b)*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/S
qrt[2]], 1] - 2*a*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]^2*Sin[2*(e + f*x)])/(2*Sqrt[2]) - (a - b)*((a - b)^2*(23*a^2 + 54*a*b + 73*b^2)*(a
+ b + (a - b)*Cos[2*(e + f*x)])^2*Cot[e + f*x] - a*(a - b)^2*(11*a + 14*b)*(a + b + (a - b)*Cos[2*(e + f*x)])^
2*Cot[e + f*x]*Csc[e + f*x]^2 + 3*a^2*(a - b)^2*(a + b + (a - b)*Cos[2*(e + f*x)])^2*Cot[e + f*x]*Csc[e + f*x]
^4 + 10*a*b^5*Sin[2*(e + f*x)] - 5*(15*a - 11*b)*b^4*(a + b + (a - b)*Cos[2*(e + f*x)])*Sin[2*(e + f*x)])))/(1
5*Sqrt[2]*a^5*(a - b)^3*f*(a + b + (a - b)*Cos[2*(e + f*x)])^2)

________________________________________________________________________________________

Maple [F]  time = 0.255, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cot \left ( fx+e \right ) \right ) ^{6} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{5}{2}}}}\, dx \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)^(5/2),x)

[Out]

int(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(5/2),x)

________________________________________________________________________________________

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)^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 3.35924, size = 2286, normalized size = 6.99 \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)^(5/2),x, algorithm="fricas")

[Out]

[-1/60*(15*(a^5*b^2*tan(f*x + e)^9 + 2*a^6*b*tan(f*x + e)^7 + a^7*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^5*b^2 - 5*
a^4*b^3 - 2*a^3*b^4 - 184*a^2*b^5 + 304*a*b^6 - 128*b^7)*tan(f*x + e)^8 + 3*a^7 - 9*a^6*b + 9*a^5*b^2 - 3*a^4*
b^3 + 3*(10*a^6*b - 5*a^5*b^2 - a^4*b^3 - 92*a^3*b^4 + 152*a^2*b^5 - 64*a*b^6)*tan(f*x + e)^6 + 3*(5*a^7 - 5*a
^6*b + a^5*b^2 - 23*a^4*b^3 + 38*a^3*b^4 - 16*a^2*b^5)*tan(f*x + e)^4 - (5*a^7 - 7*a^6*b - 9*a^5*b^2 + 19*a^4*
b^3 - 8*a^3*b^4)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^8*b^2 - 3*a^7*b^3 + 3*a^6*b^4 - a^5*b^5)*f*ta
n(f*x + e)^9 + 2*(a^9*b - 3*a^8*b^2 + 3*a^7*b^3 - a^6*b^4)*f*tan(f*x + e)^7 + (a^10 - 3*a^9*b + 3*a^8*b^2 - a^
7*b^3)*f*tan(f*x + e)^5), -1/30*(15*(a^5*b^2*tan(f*x + e)^9 + 2*a^6*b*tan(f*x + e)^7 + a^7*tan(f*x + e)^5)*sqr
t(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*((1
5*a^5*b^2 - 5*a^4*b^3 - 2*a^3*b^4 - 184*a^2*b^5 + 304*a*b^6 - 128*b^7)*tan(f*x + e)^8 + 3*a^7 - 9*a^6*b + 9*a^
5*b^2 - 3*a^4*b^3 + 3*(10*a^6*b - 5*a^5*b^2 - a^4*b^3 - 92*a^3*b^4 + 152*a^2*b^5 - 64*a*b^6)*tan(f*x + e)^6 +
3*(5*a^7 - 5*a^6*b + a^5*b^2 - 23*a^4*b^3 + 38*a^3*b^4 - 16*a^2*b^5)*tan(f*x + e)^4 - (5*a^7 - 7*a^6*b - 9*a^5
*b^2 + 19*a^4*b^3 - 8*a^3*b^4)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^8*b^2 - 3*a^7*b^3 + 3*a^6*b^4 -
a^5*b^5)*f*tan(f*x + e)^9 + 2*(a^9*b - 3*a^8*b^2 + 3*a^7*b^3 - a^6*b^4)*f*tan(f*x + e)^7 + (a^10 - 3*a^9*b +
3*a^8*b^2 - a^7*b^3)*f*tan(f*x + e)^5)]

________________________________________________________________________________________

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*tan(f*x+e)**2)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{6}}{{\left (b \tan \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)^6/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(cot(f*x + e)^6/(b*tan(f*x + e)^2 + a)^(5/2), x)