\(\int \cot ^3(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\) [310]

Optimal result

Integrand size = 25, antiderivative size = 116 \[ \int \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {a} (2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{2 f}-\frac {(a-b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}-\frac {a \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3751, 457, 100, 162, 65, 214} \[ \int \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {a} (2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{2 f}-\frac {(a-b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}-\frac {a \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f} \]

[In]

Int[Cot[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

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

Rule 65

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 100

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

Rule 162

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 214

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

Rule 457

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.94 \[ \int \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {a} (2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )-2 (a-b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )-a \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f} \]

[In]

Integrate[Cot[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1990\) vs. \(2(98)=196\).

Time = 1.07 (sec) , antiderivative size = 1991, normalized size of antiderivative = 17.16

method	result	size
default	\(\text {Expression too large to display}\)	\(1991\)

[In]

int(cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/4/f/(a-b)^(1/2)*(-2*cos(f*x+e)*ln(2/a^(1/2)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+
e)+1)^2)^(1/2)+((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+b)
/(cos(f*x+e)+1))*(a-b)^(1/2)*a^(3/2)+2*cos(f*x+e)*ln(-4*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)
/(cos(f*x+e)+1)^2)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*
a^(1/2)+b)/(cos(f*x+e)-1))*(a-b)^(1/2)*a^(3/2)+2*a^(3/2)*ln(2/a^(1/2)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-b*c
os(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(
f*x+e)*a+b*cos(f*x+e)+b)/(cos(f*x+e)+1))*(a-b)^(1/2)+3*cos(f*x+e)*ln(2/a^(1/2)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x
+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(
1/2)-cos(f*x+e)*a+b*cos(f*x+e)+b)/(cos(f*x+e)+1))*(a-b)^(1/2)*a^(1/2)*b-2*a^(3/2)*ln(-4*(cos(f*x+e)*a^(1/2)*((
a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-b*cos(f*x+
e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*(a-b)^(1/2)-3*cos(f*x+e)*ln(-4*(cos(f*x+e)*a^(1/2)*
((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-b*cos(f*
x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*(a-b)^(1/2)*a^(1/2)*b+2*cos(f*x+e)*((a*cos(f*x+e)
^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*(a-b)^(1/2)*a-3*a^(1/2)*b*ln(2/a^(1/2)*(cos(f*x+e)*a^(1/2)*((a*co
s(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2
)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+b)/(cos(f*x+e)+1))*(a-b)^(1/2)+3*a^(1/2)*ln(-4*(cos(f*x+e)*a^(1/2)*((a*cos
(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-b*cos(f*x+e)^2+
b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*b*(a-b)^(1/2)-4*cos(f*x+e)*ln(4*cos(f*x+e)*(a-b)^(1/2)*(
(a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f*x+e)*a-4*b*cos(f*x+e)+4*(a-b)^(1/2)*((a*cos(
f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2))*a^2+8*cos(f*x+e)*ln(4*cos(f*x+e)*(a-b)^(1/2)*((a*cos(f*x+e
)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f*x+e)*a-4*b*cos(f*x+e)+4*(a-b)^(1/2)*((a*cos(f*x+e)^2-b*c
os(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2))*a*b-4*cos(f*x+e)*ln(4*cos(f*x+e)*(a-b)^(1/2)*((a*cos(f*x+e)^2-b*cos(f*
x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f*x+e)*a-4*b*cos(f*x+e)+4*(a-b)^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+
b)/(cos(f*x+e)+1)^2)^(1/2))*b^2+4*ln(4*cos(f*x+e)*(a-b)^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1
)^2)^(1/2)+4*cos(f*x+e)*a-4*b*cos(f*x+e)+4*(a-b)^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1
/2))*a^2-8*ln(4*cos(f*x+e)*(a-b)^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f*x+e)
*a-4*b*cos(f*x+e)+4*(a-b)^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2))*a*b+4*ln(4*cos(f*x
+e)*(a-b)^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f*x+e)*a-4*b*cos(f*x+e)+4*(a-
b)^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2))*b^2)*(a+b*tan(f*x+e)^2)^(3/2)*cos(f*x+e)^
3/(cos(f*x+e)-1)/(cos(f*x+e)+1)/(a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+
e)+1)^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 584, normalized size of antiderivative = 5.03 \[ \int \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {2 \, {\left (a - b\right )}^{\frac {3}{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 )^{2} + {\left (2 \, a - 3 \, b\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 )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} a}{4 \, f \tan \left (f x + e\right )^{2}}, -\frac {4 \, {\left (a - b\right )} \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 )^{2} + {\left (2 \, a - 3 \, b\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 )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} a}{4 \, f \tan \left (f x + e\right )^{2}}, -\frac {\sqrt {-a} {\left (2 \, a - 3 \, b\right )} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) \tan \left (f x + e\right )^{2} + {\left (a - b\right )}^{\frac {3}{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 )^{2} + \sqrt {b \tan \left (f x + e\right )^{2} + a} a}{2 \, f \tan \left (f x + e\right )^{2}}, -\frac {\sqrt {-a} {\left (2 \, a - 3 \, b\right )} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) \tan \left (f x + e\right )^{2} + 2 \, {\left (a - b\right )} \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 )^{2} + \sqrt {b \tan \left (f x + e\right )^{2} + a} a}{2 \, f \tan \left (f x + e\right )^{2}}\right ] \]

[In]

integrate(cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(2*(a - b)^(3/2)*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)^2 + (2*a - 3*b)*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)^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*a)/(f*tan(f*x + e)^2), -1/4*(4*(a - b)*sqr
t(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(a - b))*tan(f*x + e)^2 + (2*a - 3*b)*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)^2 + 2*sqrt(b*tan(f*
x + e)^2 + a)*a)/(f*tan(f*x + e)^2), -1/2*(sqrt(-a)*(2*a - 3*b)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a)*
tan(f*x + e)^2 + (a - b)^(3/2)*log((b*tan(f*x + e)^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) + 2*a - b)/(ta
n(f*x + e)^2 + 1))*tan(f*x + e)^2 + sqrt(b*tan(f*x + e)^2 + a)*a)/(f*tan(f*x + e)^2), -1/2*(sqrt(-a)*(2*a - 3*
b)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a)*tan(f*x + e)^2 + 2*(a - b)*sqrt(-a + b)*arctan(-sqrt(b*tan(f*
x + e)^2 + a)*sqrt(-a + b)/(a - b))*tan(f*x + e)^2 + sqrt(b*tan(f*x + e)^2 + a)*a)/(f*tan(f*x + e)^2)]

Sympy [F]

\[ \int \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \cot ^{3}{\left (e + f x \right )}\, dx \]

[In]

integrate(cot(f*x+e)**3*(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Integral((a + b*tan(e + f*x)**2)**(3/2)*cot(e + f*x)**3, x)

Maxima [F]

\[ \int \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cot \left (f x + e\right )^{3} \,d x } \]

[In]

integrate(cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2)*cot(f*x + e)^3, x)

Giac [F(-1)]

Timed out. \[ \int \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 10.98 (sec) , antiderivative size = 447, normalized size of antiderivative = 3.85 \[ \int \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {\mathrm {atanh}\left (\frac {3\,a^2\,b^4\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3-3\,a^2\,b+3\,a\,b^2-b^3}}{2\,\left (-\frac {3\,a^4\,b^4}{2}+5\,a^3\,b^5-\frac {11\,a^2\,b^6}{2}+2\,a\,b^7\right )}-\frac {2\,a\,b^5\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3-3\,a^2\,b+3\,a\,b^2-b^3}}{-\frac {3\,a^4\,b^4}{2}+5\,a^3\,b^5-\frac {11\,a^2\,b^6}{2}+2\,a\,b^7}\right )\,\sqrt {{\left (a-b\right )}^3}}{f}+\frac {\sqrt {a}\,\mathrm {atanh}\left (\frac {3\,\sqrt {a}\,b^7\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{-\frac {3\,a^4\,b^4}{2}+\frac {23\,a^3\,b^5}{4}-\frac {29\,a^2\,b^6}{4}+3\,a\,b^7}-\frac {29\,a^{3/2}\,b^6\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{4\,\left (-\frac {3\,a^4\,b^4}{2}+\frac {23\,a^3\,b^5}{4}-\frac {29\,a^2\,b^6}{4}+3\,a\,b^7\right )}+\frac {23\,a^{5/2}\,b^5\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{4\,\left (-\frac {3\,a^4\,b^4}{2}+\frac {23\,a^3\,b^5}{4}-\frac {29\,a^2\,b^6}{4}+3\,a\,b^7\right )}-\frac {3\,a^{7/2}\,b^4\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{2\,\left (-\frac {3\,a^4\,b^4}{2}+\frac {23\,a^3\,b^5}{4}-\frac {29\,a^2\,b^6}{4}+3\,a\,b^7\right )}\right )\,\left (2\,a-3\,b\right )}{2\,f}-\frac {a\,b\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{2\,\left (f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )-a\,f\right )} \]

[In]

int(cot(e + f*x)^3*(a + b*tan(e + f*x)^2)^(3/2),x)

[Out]

(atanh((3*a^2*b^4*(a + b*tan(e + f*x)^2)^(1/2)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)^(1/2))/(2*(2*a*b^7 - (11*a^2*b^
6)/2 + 5*a^3*b^5 - (3*a^4*b^4)/2)) - (2*a*b^5*(a + b*tan(e + f*x)^2)^(1/2)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)^(1/
2))/(2*a*b^7 - (11*a^2*b^6)/2 + 5*a^3*b^5 - (3*a^4*b^4)/2))*((a - b)^3)^(1/2))/f + (a^(1/2)*atanh((3*a^(1/2)*b
^7*(a + b*tan(e + f*x)^2)^(1/2))/(3*a*b^7 - (29*a^2*b^6)/4 + (23*a^3*b^5)/4 - (3*a^4*b^4)/2) - (29*a^(3/2)*b^6
*(a + b*tan(e + f*x)^2)^(1/2))/(4*(3*a*b^7 - (29*a^2*b^6)/4 + (23*a^3*b^5)/4 - (3*a^4*b^4)/2)) + (23*a^(5/2)*b
^5*(a + b*tan(e + f*x)^2)^(1/2))/(4*(3*a*b^7 - (29*a^2*b^6)/4 + (23*a^3*b^5)/4 - (3*a^4*b^4)/2)) - (3*a^(7/2)*
b^4*(a + b*tan(e + f*x)^2)^(1/2))/(2*(3*a*b^7 - (29*a^2*b^6)/4 + (23*a^3*b^5)/4 - (3*a^4*b^4)/2)))*(2*a - 3*b)
)/(2*f) - (a*b*(a + b*tan(e + f*x)^2)^(1/2))/(2*(f*(a + b*tan(e + f*x)^2) - a*f))