∫ cot5 (e+fx)___ (a+b tan2(e+fx))5∕2 dx

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

Optimal. Leaf size=272 \[ \frac {(4 a+7 b) \cot ^2(e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\left (8 a^2+20 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{9/2} f}+\frac {b \left (12 a^2+15 a b-35 b^2\right )}{24 a^3 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {b \left (4 a^3+3 a^2 b-50 a b^2+35 b^3\right )}{8 a^4 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f (a-b)^{5/2}}-\frac {\cot ^4(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

[Out]

-1/8*(8*a^2+20*a*b+35*b^2)*arctanh((a+b*tan(f*x+e)^2)^(1/2)/a^(1/2))/a^(9/2)/f+arctanh((a+b*tan(f*x+e)^2)^(1/2

)/(a-b)^(1/2))/(a-b)^(5/2)/f+1/8*b*(4*a^3+3*a^2*b-50*a*b^2+35*b^3)/a^4/(a-b)^2/f/(a+b*tan(f*x+e)^2)^(1/2)+1/24

*b*(12*a^2+15*a*b-35*b^2)/a^3/(a-b)/f/(a+b*tan(f*x+e)^2)^(3/2)+1/8*(4*a+7*b)*cot(f*x+e)^2/a^2/f/(a+b*tan(f*x+e

)^2)^(3/2)-1/4*cot(f*x+e)^4/a/f/(a+b*tan(f*x+e)^2)^(3/2)

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Rubi [A] time = 0.44, antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3670, 446, 103, 151, 152, 156, 63, 208} \[ \frac {b \left (3 a^2 b+4 a^3-50 a b^2+35 b^3\right )}{8 a^4 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}}+\frac {b \left (12 a^2+15 a b-35 b^2\right )}{24 a^3 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\left (8 a^2+20 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{9/2} f}+\frac {(4 a+7 b) \cot ^2(e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f (a-b)^{5/2}}-\frac {\cot ^4(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((8*a^2 + 20*a*b + 35*b^2)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]])/(8*a^(9/2)*f) + ArcTanh[Sqrt[a + b*Ta

n[e + f*x]^2]/Sqrt[a - b]]/((a - b)^(5/2)*f) + (b*(12*a^2 + 15*a*b - 35*b^2))/(24*a^3*(a - b)*f*(a + b*Tan[e +

f*x]^2)^(3/2)) + ((4*a + 7*b)*Cot[e + f*x]^2)/(8*a^2*f*(a + b*Tan[e + f*x]^2)^(3/2)) - Cot[e + f*x]^4/(4*a*f*

(a + b*Tan[e + f*x]^2)^(3/2)) + (b*(4*a^3 + 3*a^2*b - 50*a*b^2 + 35*b^3))/(8*a^4*(a - b)^2*f*Sqrt[a + b*Tan[e

+ f*x]^2])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub

st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &

& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,

b, c, d, m, n, x]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +

b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*

c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +

c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&

IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb

ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*

f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*

g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p

+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb

ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*

f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*

g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p

+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>

Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)

^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,

b}, x] && NegQ[a/b]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int

[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&

NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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]))

Rubi steps

\begin {align*} \int \frac {\cot ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^5 \left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^3 (1+x) (a+b x)^{5/2}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac {\cot ^4(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} (4 a+7 b)+\frac {7 b x}{2}}{x^2 (1+x) (a+b x)^{5/2}} \, dx,x,\tan ^2(e+f x)\right )}{4 a f}\\ &=\frac {(4 a+7 b) \cot ^2(e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^4(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{4} \left (8 a^2+20 a b+35 b^2\right )+\frac {5}{4} b (4 a+7 b) x}{x (1+x) (a+b x)^{5/2}} \, dx,x,\tan ^2(e+f x)\right )}{4 a^2 f}\\ &=\frac {b \left (12 a^2+15 a b-35 b^2\right )}{24 a^3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(4 a+7 b) \cot ^2(e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^4(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\operatorname {Subst}\left (\int \frac {-\frac {3}{8} (a-b) \left (8 a^2+20 a b+35 b^2\right )-\frac {3}{8} b \left (12 a^2+15 a b-35 b^2\right ) x}{x (1+x) (a+b x)^{3/2}} \, dx,x,\tan ^2(e+f x)\right )}{6 a^3 (a-b) f}\\ &=\frac {b \left (12 a^2+15 a b-35 b^2\right )}{24 a^3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(4 a+7 b) \cot ^2(e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^4(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {b \left (4 a^3+3 a^2 b-50 a b^2+35 b^3\right )}{8 a^4 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {\frac {3}{16} (a-b)^2 \left (8 a^2+20 a b+35 b^2\right )+\frac {3}{16} b \left (4 a^3+3 a^2 b-50 a b^2+35 b^3\right ) x}{x (1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{3 a^4 (a-b)^2 f}\\ &=\frac {b \left (12 a^2+15 a b-35 b^2\right )}{24 a^3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(4 a+7 b) \cot ^2(e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^4(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {b \left (4 a^3+3 a^2 b-50 a b^2+35 b^3\right )}{8 a^4 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 (a-b)^2 f}+\frac {\left (8 a^2+20 a b+35 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{16 a^4 f}\\ &=\frac {b \left (12 a^2+15 a b-35 b^2\right )}{24 a^3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(4 a+7 b) \cot ^2(e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^4(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {b \left (4 a^3+3 a^2 b-50 a b^2+35 b^3\right )}{8 a^4 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {1}{1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \tan ^2(e+f x)}\right )}{(a-b)^2 b f}+\frac {\left (8 a^2+20 a b+35 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \tan ^2(e+f x)}\right )}{8 a^4 b f}\\ &=-\frac {\left (8 a^2+20 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{9/2} f}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2} f}+\frac {b \left (12 a^2+15 a b-35 b^2\right )}{24 a^3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(4 a+7 b) \cot ^2(e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^4(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {b \left (4 a^3+3 a^2 b-50 a b^2+35 b^3\right )}{8 a^4 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}\\ \end {align*}

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Mathematica [C] time = 1.99, size = 165, normalized size = 0.61 \[ \frac {\cot ^2(e+f x) \left (8 a^3 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b \tan ^2(e+f x)+a}{a-b}\right )+(a-b) \left (3 a \cot ^2(e+f x) \left (2 a \cot ^2(e+f x)-4 a-7 b\right )-\left (8 a^2+20 a b+35 b^2\right ) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b \tan ^2(e+f x)}{a}+1\right )\right )\right )}{24 a^3 f (b-a) \sqrt {a+b \tan ^2(e+f x)} \left (a \cot ^2(e+f x)+b\right )} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Cot[e + f*x]^2*(8*a^3*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[e + f*x]^2)/(a - b)] + (a - b)*(3*a*Cot[e +

f*x]^2*(-4*a - 7*b + 2*a*Cot[e + f*x]^2) - (8*a^2 + 20*a*b + 35*b^2)*Hypergeometric2F1[-3/2, 1, -1/2, 1 + (b*

Tan[e + f*x]^2)/a])))/(24*a^3*(-a + b)*f*(b + a*Cot[e + f*x]^2)*Sqrt[a + b*Tan[e + f*x]^2])

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fricas [B] time = 0.57, size = 2433, normalized size = 8.94 \[ \text {result too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/48*(24*(a^5*b^2*tan(f*x + e)^8 + 2*a^6*b*tan(f*x + e)^6 + a^7*tan(f*x + e)^4)*sqrt(a - b)*log((b*tan(f*x +

e)^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) + 2*a - b)/(tan(f*x + e)^2 + 1)) + 3*((8*a^5*b^2 - 4*a^4*b^3 -

a^3*b^4 - 53*a^2*b^5 + 85*a*b^6 - 35*b^7)*tan(f*x + e)^8 + 2*(8*a^6*b - 4*a^5*b^2 - a^4*b^3 - 53*a^3*b^4 + 85

*a^2*b^5 - 35*a*b^6)*tan(f*x + e)^6 + (8*a^7 - 4*a^6*b - a^5*b^2 - 53*a^4*b^3 + 85*a^3*b^4 - 35*a^2*b^5)*tan(f

*x + e)^4)*sqrt(a)*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a) + 2*a)/tan(f*x + e)^2) - 2*(6*

a^7 - 18*a^6*b + 18*a^5*b^2 - 6*a^4*b^3 - 3*(4*a^5*b^2 - a^4*b^3 - 53*a^3*b^4 + 85*a^2*b^5 - 35*a*b^6)*tan(f*x

+ e)^6 - 4*(6*a^6*b - 3*a^5*b^2 - 53*a^4*b^3 + 85*a^3*b^4 - 35*a^2*b^5)*tan(f*x + e)^4 - 3*(4*a^7 - 5*a^6*b -

9*a^5*b^2 + 17*a^4*b^3 - 7*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)^8 + 2*(a^9*b - 3*a^8*b^2 + 3*a^7*b^3 - a^6*b^4)*f*tan(f*x + e)^6 + (a^10 - 3*a^

9*b + 3*a^8*b^2 - a^7*b^3)*f*tan(f*x + e)^4), 1/48*(48*(a^5*b^2*tan(f*x + e)^8 + 2*a^6*b*tan(f*x + e)^6 + a^7*

tan(f*x + e)^4)*sqrt(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(a - b)) + 3*((8*a^5*b^2 - 4*a^4*

b^3 - a^3*b^4 - 53*a^2*b^5 + 85*a*b^6 - 35*b^7)*tan(f*x + e)^8 + 2*(8*a^6*b - 4*a^5*b^2 - a^4*b^3 - 53*a^3*b^4

+ 85*a^2*b^5 - 35*a*b^6)*tan(f*x + e)^6 + (8*a^7 - 4*a^6*b - a^5*b^2 - 53*a^4*b^3 + 85*a^3*b^4 - 35*a^2*b^5)*

tan(f*x + e)^4)*sqrt(a)*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a) + 2*a)/tan(f*x + e)^2) -

2*(6*a^7 - 18*a^6*b + 18*a^5*b^2 - 6*a^4*b^3 - 3*(4*a^5*b^2 - a^4*b^3 - 53*a^3*b^4 + 85*a^2*b^5 - 35*a*b^6)*ta

n(f*x + e)^6 - 4*(6*a^6*b - 3*a^5*b^2 - 53*a^4*b^3 + 85*a^3*b^4 - 35*a^2*b^5)*tan(f*x + e)^4 - 3*(4*a^7 - 5*a^

6*b - 9*a^5*b^2 + 17*a^4*b^3 - 7*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)^8 + 2*(a^9*b - 3*a^8*b^2 + 3*a^7*b^3 - a^6*b^4)*f*tan(f*x + e)^6 + (a^10 -

3*a^9*b + 3*a^8*b^2 - a^7*b^3)*f*tan(f*x + e)^4), 1/24*(3*((8*a^5*b^2 - 4*a^4*b^3 - a^3*b^4 - 53*a^2*b^5 + 85

*a*b^6 - 35*b^7)*tan(f*x + e)^8 + 2*(8*a^6*b - 4*a^5*b^2 - a^4*b^3 - 53*a^3*b^4 + 85*a^2*b^5 - 35*a*b^6)*tan(f

*x + e)^6 + (8*a^7 - 4*a^6*b - a^5*b^2 - 53*a^4*b^3 + 85*a^3*b^4 - 35*a^2*b^5)*tan(f*x + e)^4)*sqrt(-a)*arctan

(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a) + 12*(a^5*b^2*tan(f*x + e)^8 + 2*a^6*b*tan(f*x + e)^6 + a^7*tan(f*x +

e)^4)*sqrt(a - b)*log((b*tan(f*x + e)^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) + 2*a - b)/(tan(f*x + e)^2

+ 1)) - (6*a^7 - 18*a^6*b + 18*a^5*b^2 - 6*a^4*b^3 - 3*(4*a^5*b^2 - a^4*b^3 - 53*a^3*b^4 + 85*a^2*b^5 - 35*a*b

^6)*tan(f*x + e)^6 - 4*(6*a^6*b - 3*a^5*b^2 - 53*a^4*b^3 + 85*a^3*b^4 - 35*a^2*b^5)*tan(f*x + e)^4 - 3*(4*a^7

- 5*a^6*b - 9*a^5*b^2 + 17*a^4*b^3 - 7*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)^8 + 2*(a^9*b - 3*a^8*b^2 + 3*a^7*b^3 - a^6*b^4)*f*tan(f*x + e)^6 + (

a^10 - 3*a^9*b + 3*a^8*b^2 - a^7*b^3)*f*tan(f*x + e)^4), 1/24*(3*((8*a^5*b^2 - 4*a^4*b^3 - a^3*b^4 - 53*a^2*b^

5 + 85*a*b^6 - 35*b^7)*tan(f*x + e)^8 + 2*(8*a^6*b - 4*a^5*b^2 - a^4*b^3 - 53*a^3*b^4 + 85*a^2*b^5 - 35*a*b^6)

*tan(f*x + e)^6 + (8*a^7 - 4*a^6*b - a^5*b^2 - 53*a^4*b^3 + 85*a^3*b^4 - 35*a^2*b^5)*tan(f*x + e)^4)*sqrt(-a)*

arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a) + 24*(a^5*b^2*tan(f*x + e)^8 + 2*a^6*b*tan(f*x + e)^6 + a^7*tan(

f*x + e)^4)*sqrt(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(a - b)) - (6*a^7 - 18*a^6*b + 18*a^5

*b^2 - 6*a^4*b^3 - 3*(4*a^5*b^2 - a^4*b^3 - 53*a^3*b^4 + 85*a^2*b^5 - 35*a*b^6)*tan(f*x + e)^6 - 4*(6*a^6*b -

3*a^5*b^2 - 53*a^4*b^3 + 85*a^3*b^4 - 35*a^2*b^5)*tan(f*x + e)^4 - 3*(4*a^7 - 5*a^6*b - 9*a^5*b^2 + 17*a^4*b^3

- 7*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)^8 + 2*(a^9*b - 3*a^8*b^2 + 3*a^7*b^3 - a^6*b^4)*f*tan(f*x + e)^6 + (a^10 - 3*a^9*b + 3*a^8*b^2 - a^7*b

^3)*f*tan(f*x + e)^4)]

________________________________________________________________________________________

giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab

le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p