3.298 \(\int \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=163 \[ -\frac {\left (8 a^2-4 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{3/2} f}+\frac {\sqrt {a-b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}-\frac {\cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}+\frac {(4 a-b) \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 a f} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 163, 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 = {3670, 446, 99, 151, 156, 63, 208} \[ -\frac {\left (8 a^2-4 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{3/2} f}+\frac {\sqrt {a-b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}-\frac {\cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}+\frac {(4 a-b) \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

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 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 \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{x^5 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^3 (1+x)} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac {\cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} (-4 a+b)-\frac {3 b x}{2}}{x^2 (1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{4 f}\\ &=\frac {(4 a-b) \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 a f}-\frac {\cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{4} \left (-8 a^2+4 a b+b^2\right )-\frac {1}{4} (4 a-b) b x}{x (1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{4 a f}\\ &=\frac {(4 a-b) \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 a f}-\frac {\cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}-\frac {(a-b) \operatorname {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}+\frac {\left (8 a^2-4 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{16 a f}\\ &=\frac {(4 a-b) \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 a f}-\frac {\cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}-\frac {(a-b) \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 )}{b f}+\frac {\left (8 a^2-4 a b-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 b f}\\ &=-\frac {\left (8 a^2-4 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{3/2} f}+\frac {\sqrt {a-b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}+\frac {(4 a-b) \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 a f}-\frac {\cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}\\ \end {align*}

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Mathematica [A]  time = 1.32, size = 138, normalized size = 0.85 \[ \frac {\left (-8 a^2+4 a b+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )+\sqrt {a} \left (8 a \sqrt {a-b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )-\cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)} \left (2 a \cot ^2(e+f x)-4 a+b\right )\right )}{8 a^{3/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 729, normalized size = 4.47 \[ \left [\frac {8 \, \sqrt {a - b} a^{2} \log \left (\frac {b \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 2 \, a - b}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{4} - {\left (8 \, a^{2} - 4 \, a b - b^{2}\right )} \sqrt {a} \log \left (\frac {b \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a} + 2 \, a}{\tan \left (f x + e\right )^{2}}\right ) \tan \left (f x + e\right )^{4} + 2 \, {\left ({\left (4 \, a^{2} - a b\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{16 \, a^{2} f \tan \left (f x + e\right )^{4}}, \frac {16 \, a^{2} \sqrt {-a + b} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{a - b}\right ) \tan \left (f x + e\right )^{4} - {\left (8 \, a^{2} - 4 \, a b - b^{2}\right )} \sqrt {a} \log \left (\frac {b \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a} + 2 \, a}{\tan \left (f x + e\right )^{2}}\right ) \tan \left (f x + e\right )^{4} + 2 \, {\left ({\left (4 \, a^{2} - a b\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{16 \, a^{2} f \tan \left (f x + e\right )^{4}}, \frac {4 \, \sqrt {a - b} a^{2} \log \left (\frac {b \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 2 \, a - b}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{4} + {\left (8 \, a^{2} - 4 \, a b - b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) \tan \left (f x + e\right )^{4} + {\left ({\left (4 \, a^{2} - a b\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{8 \, a^{2} f \tan \left (f x + e\right )^{4}}, \frac {8 \, a^{2} \sqrt {-a + b} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{a - b}\right ) \tan \left (f x + e\right )^{4} + {\left (8 \, a^{2} - 4 \, a b - b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) \tan \left (f x + e\right )^{4} + {\left ({\left (4 \, a^{2} - a b\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{8 \, a^{2} f \tan \left (f x + e\right )^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(8*sqrt(a - b)*a^2*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))*tan(f*x + e)^4 - (8*a^2 - 4*a*b - b^2)*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)*tan(f*x + e)^4 + 2*((4*a^2 - a*b)*tan(f*x + e)^2 - 2*a^2)*sqrt(b*tan(f*x + e
)^2 + a))/(a^2*f*tan(f*x + e)^4), 1/16*(16*a^2*sqrt(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(a
- b))*tan(f*x + e)^4 - (8*a^2 - 4*a*b - b^2)*sqrt(a)*log((b*tan(f*x + e)^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*sqr
t(a) + 2*a)/tan(f*x + e)^2)*tan(f*x + e)^4 + 2*((4*a^2 - a*b)*tan(f*x + e)^2 - 2*a^2)*sqrt(b*tan(f*x + e)^2 +
a))/(a^2*f*tan(f*x + e)^4), 1/8*(4*sqrt(a - b)*a^2*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))*tan(f*x + e)^4 + (8*a^2 - 4*a*b - b^2)*sqrt(-a)*arctan(sqrt(b*tan(f*x +
e)^2 + a)*sqrt(-a)/a)*tan(f*x + e)^4 + ((4*a^2 - a*b)*tan(f*x + e)^2 - 2*a^2)*sqrt(b*tan(f*x + e)^2 + a))/(a^
2*f*tan(f*x + e)^4), 1/8*(8*a^2*sqrt(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(a - b))*tan(f*x
+ e)^4 + (8*a^2 - 4*a*b - b^2)*sqrt(-a)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a)*tan(f*x + e)^4 + ((4*a^2
- a*b)*tan(f*x + e)^2 - 2*a^2)*sqrt(b*tan(f*x + e)^2 + a))/(a^2*f*tan(f*x + e)^4)]

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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)^(1/2),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
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)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
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)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
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)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
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)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
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)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
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)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
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)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
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t
_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-
2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep
/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_
nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2
*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check si
gn: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to ch
eck sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable
to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)
Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nos
tep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi
/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>
(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nost
ep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/
t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign:
(2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check
sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to
check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unab
le to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/
2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_n
ostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*
pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2
)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_no
step/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*p
i/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign
: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to chec
k sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable t
o check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Un
able to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_noste
p/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, integration of abs or sign assumes cons
tant sign by intervals (correct if the argument is real):Check [abs(t_nostep^2-1)]Discontinuities at zeroes of
t_nostep^2-1 were not checkedUnable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign:
(2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check
sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable 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*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Un
able to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to
check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Warning, integration of abs or sign assumes constant sign by i
ntervals (correct if the argument is real):Check [abs(t_nostep^2-1)]Evaluation time: 2.25Error: Bad Argument T
ype

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maple [B]  time = 1.28, size = 5676, normalized size = 34.82 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^5*(a+b*tan(f*x+e)^2)^(1/2),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 12.02, size = 542, normalized size = 3.33 \[ \frac {\mathrm {atanh}\left (\frac {3\,b^7\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{64\,\sqrt {a^3}\,\left (\frac {a\,b^5}{4}-\frac {11\,b^6}{32}+\frac {3\,b^7}{64\,a}+\frac {11\,b^8}{256\,a^2}+\frac {b^9}{256\,a^3}\right )}-\frac {11\,b^6\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{32\,\sqrt {a^3}\,\left (\frac {b^5}{4}-\frac {11\,b^6}{32\,a}+\frac {3\,b^7}{64\,a^2}+\frac {11\,b^8}{256\,a^3}+\frac {b^9}{256\,a^4}\right )}+\frac {11\,b^8\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{256\,\sqrt {a^3}\,\left (\frac {3\,b^7}{64}-\frac {11\,a\,b^6}{32}+\frac {a^2\,b^5}{4}+\frac {11\,b^8}{256\,a}+\frac {b^9}{256\,a^2}\right )}+\frac {b^9\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{256\,\sqrt {a^3}\,\left (\frac {3\,a\,b^7}{64}+\frac {11\,b^8}{256}-\frac {11\,a^2\,b^6}{32}+\frac {a^3\,b^5}{4}+\frac {b^9}{256\,a}\right )}+\frac {a\,b^5\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{4\,\sqrt {a^3}\,\left (\frac {b^5}{4}-\frac {11\,b^6}{32\,a}+\frac {3\,b^7}{64\,a^2}+\frac {11\,b^8}{256\,a^3}+\frac {b^9}{256\,a^4}\right )}\right )\,\left (-8\,a^2+4\,a\,b+b^2\right )}{8\,f\,\sqrt {a^3}}-\frac {\mathrm {atanh}\left (\frac {b^5\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a-b}}{4\,\left (\frac {7\,b^6}{32}-\frac {a\,b^5}{4}+\frac {b^7}{32\,a}\right )}+\frac {b^6\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a-b}}{32\,\left (-\frac {a^2\,b^5}{4}+\frac {7\,a\,b^6}{32}+\frac {b^7}{32}\right )}\right )\,\sqrt {a-b}}{f}-\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\left (\frac {b^2}{8}+\frac {a\,b}{2}\right )-\frac {b\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}\,\left (4\,a-b\right )}{8\,a}}{f\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^2+a^2\,f-2\,a\,f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^5*(a + b*tan(e + f*x)^2)^(1/2),x)

[Out]

(atanh((3*b^7*(a + b*tan(e + f*x)^2)^(1/2))/(64*(a^3)^(1/2)*((a*b^5)/4 - (11*b^6)/32 + (3*b^7)/(64*a) + (11*b^
8)/(256*a^2) + b^9/(256*a^3))) - (11*b^6*(a + b*tan(e + f*x)^2)^(1/2))/(32*(a^3)^(1/2)*(b^5/4 - (11*b^6)/(32*a
) + (3*b^7)/(64*a^2) + (11*b^8)/(256*a^3) + b^9/(256*a^4))) + (11*b^8*(a + b*tan(e + f*x)^2)^(1/2))/(256*(a^3)
^(1/2)*((3*b^7)/64 - (11*a*b^6)/32 + (a^2*b^5)/4 + (11*b^8)/(256*a) + b^9/(256*a^2))) + (b^9*(a + b*tan(e + f*
x)^2)^(1/2))/(256*(a^3)^(1/2)*((3*a*b^7)/64 + (11*b^8)/256 - (11*a^2*b^6)/32 + (a^3*b^5)/4 + b^9/(256*a))) + (
a*b^5*(a + b*tan(e + f*x)^2)^(1/2))/(4*(a^3)^(1/2)*(b^5/4 - (11*b^6)/(32*a) + (3*b^7)/(64*a^2) + (11*b^8)/(256
*a^3) + b^9/(256*a^4))))*(4*a*b - 8*a^2 + b^2))/(8*f*(a^3)^(1/2)) - (atanh((b^5*(a + b*tan(e + f*x)^2)^(1/2)*(
a - b)^(1/2))/(4*((7*b^6)/32 - (a*b^5)/4 + b^7/(32*a))) + (b^6*(a + b*tan(e + f*x)^2)^(1/2)*(a - b)^(1/2))/(32
*((7*a*b^6)/32 + b^7/32 - (a^2*b^5)/4)))*(a - b)^(1/2))/f - ((a + b*tan(e + f*x)^2)^(1/2)*((a*b)/2 + b^2/8) -
(b*(a + b*tan(e + f*x)^2)^(3/2)*(4*a - b))/(8*a))/(f*(a + b*tan(e + f*x)^2)^2 + a^2*f - 2*a*f*(a + b*tan(e + f
*x)^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**5*(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

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

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