3.4.58 \(\int \frac {\cot ^6(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [358]

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

[Out]

-arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(5/2)/f-1/3*(11*a-8*b)*b*cot(f*x+e)^5/a^2/(a-b)
^2/f/(a+b*tan(f*x+e)^2)^(1/2)-1/15*(15*a^4+10*a^3*b+8*a^2*b^2-176*a*b^3+128*b^4)*cot(f*x+e)*(a+b*tan(f*x+e)^2)
^(1/2)/a^5/(a-b)^2/f+1/15*(5*a^3+4*a^2*b-88*a*b^2+64*b^3)*cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2)/a^4/(a-b)^2/f-
1/5*(a^2-22*a*b+16*b^2)*cot(f*x+e)^5*(a+b*tan(f*x+e)^2)^(1/2)/a^3/(a-b)^2/f-1/3*b*cot(f*x+e)^5/a/(a-b)/f/(a+b*
tan(f*x+e)^2)^(3/2)

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Rubi [A]
time = 0.31, antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3751, 483, 593, 597, 12, 385, 209} \begin {gather*} -\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 {\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 (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 f (a-b)^2}-\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 f (a-b)^2}-\frac {\text {ArcTan}\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}} \end {gather*}

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 385

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 483

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 593

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 597

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 3751

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]))

Rubi steps

\begin {align*} \int \frac {\cot ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\text {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 {\text {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 {\text {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 {\text {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 {\text {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 {\text {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 {\text {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 {\text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 14.49, size = 441, normalized size = 1.35 \begin {gather*} \frac {\sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (-\frac {15 a^5 b \left (\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}\right )^{3/2} \left (2 (a-b) F\left (\left .\text {ArcSin}\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 \Pi \left (-\frac {b}{a-b};\left .\text {ArcSin}\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 ) \sin ^2(e+f x) \sin (2 (e+f x))}{2 \sqrt {2}}-(a-b) \left ((a-b)^2 \left (23 a^2+54 a b+73 b^2\right ) (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 ^2(e+f x)+3 a^2 (a-b)^2 (a+b+(a-b) \cos (2 (e+f x)))^2 \cot (e+f x) \csc ^4(e+f x)+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))\right )\right )}{15 \sqrt {2} a^5 (a-b)^3 f (a+b+(a-b) \cos (2 (e+f x)))^2} \end {gather*}

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)

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Maple [F]
time = 0.36, size = 0, normalized size = 0.00 \[\int \frac {\cot ^{6}\left (f x +e \right )}{\left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {5}{2}}}\, dx\]

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)

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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Fricas [A]
time = 2.65, size = 1055, normalized size = 3.23 \begin {gather*} \left [-\frac {15 \, {\left (a^{5} b^{2} \tan \left (f x + e\right )^{9} + 2 \, a^{6} b \tan \left (f x + e\right )^{7} + a^{7} \tan \left (f x + e\right )^{5}\right )} \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 ) + 4 \, {\left ({\left (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}\right )} \tan \left (f x + e\right )^{8} + 3 \, a^{7} - 9 \, a^{6} b + 9 \, a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, {\left (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}\right )} \tan \left (f x + e\right )^{6} + 3 \, {\left (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}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{7} - 7 \, a^{6} b - 9 \, a^{5} b^{2} + 19 \, a^{4} b^{3} - 8 \, a^{3} b^{4}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{60 \, {\left ({\left (a^{8} b^{2} - 3 \, a^{7} b^{3} + 3 \, a^{6} b^{4} - a^{5} b^{5}\right )} f \tan \left (f x + e\right )^{9} + 2 \, {\left (a^{9} b - 3 \, a^{8} b^{2} + 3 \, a^{7} b^{3} - a^{6} b^{4}\right )} f \tan \left (f x + e\right )^{7} + {\left (a^{10} - 3 \, a^{9} b + 3 \, a^{8} b^{2} - a^{7} b^{3}\right )} f \tan \left (f x + e\right )^{5}\right )}}, -\frac {15 \, {\left (a^{5} b^{2} \tan \left (f x + e\right )^{9} + 2 \, a^{6} b \tan \left (f x + e\right )^{7} + a^{7} \tan \left (f x + e\right )^{5}\right )} \sqrt {a - b} \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 ) + 2 \, {\left ({\left (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}\right )} \tan \left (f x + e\right )^{8} + 3 \, a^{7} - 9 \, a^{6} b + 9 \, a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, {\left (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}\right )} \tan \left (f x + e\right )^{6} + 3 \, {\left (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}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{7} - 7 \, a^{6} b - 9 \, a^{5} b^{2} + 19 \, a^{4} b^{3} - 8 \, a^{3} b^{4}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{30 \, {\left ({\left (a^{8} b^{2} - 3 \, a^{7} b^{3} + 3 \, a^{6} b^{4} - a^{5} b^{5}\right )} f \tan \left (f x + e\right )^{9} + 2 \, {\left (a^{9} b - 3 \, a^{8} b^{2} + 3 \, a^{7} b^{3} - a^{6} b^{4}\right )} f \tan \left (f x + e\right )^{7} + {\left (a^{10} - 3 \, a^{9} b + 3 \, a^{8} b^{2} - a^{7} b^{3}\right )} f \tan \left (f x + e\right )^{5}\right )}}\right ] \end {gather*}

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)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{6}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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