\(\int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx\) [1315]

Optimal result

Integrand size = 29, antiderivative size = 276 \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {i (a-i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c-i d)^{5/2} f}+\frac {i (a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c+i d)^{5/2} f}+\frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {4 \left (b c^2-3 a c d-2 b d^2\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \] Output:

-I*(a-I*b)^(3/2)*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2 
)/(c+d*tan(f*x+e))^(1/2))/(c-I*d)^(5/2)/f+I*(a+I*b)^(3/2)*arctanh((c+I*d)^ 
(1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))/(c+I*d) 
^(5/2)/f+2/3*(-a*d+b*c)*(a+b*tan(f*x+e))^(1/2)/(c^2+d^2)/f/(c+d*tan(f*x+e) 
)^(3/2)+4/3*(-3*a*c*d+b*c^2-2*b*d^2)*(a+b*tan(f*x+e))^(1/2)/(c^2+d^2)^2/f/ 
(c+d*tan(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 2.12 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {\frac {3 i (-a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(-c+i d)^{5/2}}+\frac {3 i (a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c+i d)^{5/2}}-\frac {2 \sqrt {a+b \tan (e+f x)} \left (-3 b c^3+7 a c^2 d+3 b c d^2+a d^3+2 d \left (-b c^2+3 a c d+2 b d^2\right ) \tan (e+f x)\right )}{\left (c^2+d^2\right )^2 (c+d \tan (e+f x))^{3/2}}}{3 f} \] Input:

Integrate[(a + b*Tan[e + f*x])^(3/2)/(c + d*Tan[e + f*x])^(5/2),x]
 

Output:

(((3*I)*(-a + I*b)^(3/2)*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e + f*x]]) 
/(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(-c + I*d)^(5/2) + ((3*I)*(a 
+ I*b)^(3/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I* 
b]*Sqrt[c + d*Tan[e + f*x]])])/(c + I*d)^(5/2) - (2*Sqrt[a + b*Tan[e + f*x 
]]*(-3*b*c^3 + 7*a*c^2*d + 3*b*c*d^2 + a*d^3 + 2*d*(-(b*c^2) + 3*a*c*d + 2 
*b*d^2)*Tan[e + f*x]))/((c^2 + d^2)^2*(c + d*Tan[e + f*x])^(3/2)))/(3*f)
 

Rubi [A] (verified)

Time = 1.88 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.28, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 4050, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 4050

\(\displaystyle \frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac {2 \int -\frac {3 c a^2+4 b d a+2 b (b c-a d) \tan ^2(e+f x)-b^2 c+3 \left (-d a^2+2 b c a+b^2 d\right ) \tan (e+f x)}{2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}dx}{3 \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {3 c a^2+4 b d a+2 b (b c-a d) \tan ^2(e+f x)-b^2 c+3 \left (-d a^2+2 b c a+b^2 d\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}dx}{3 \left (c^2+d^2\right )}+\frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {3 c a^2+4 b d a+2 b (b c-a d) \tan (e+f x)^2-b^2 c+3 \left (-d a^2+2 b c a+b^2 d\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}dx}{3 \left (c^2+d^2\right )}+\frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4132

\(\displaystyle \frac {\frac {2 \int -\frac {3 \left ((b c-a d) (b (c-d)-a (c+d)) (a (c-d)+b (c+d))-2 (b c-a d)^2 (a c+b d) \tan (e+f x)\right )}{2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}+\frac {4 \left (-3 a c d+b c^2-2 b d^2\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 \left (c^2+d^2\right )}+\frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {4 \left (-3 a c d+b c^2-2 b d^2\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {3 \int \frac {(b c-a d) (b (c-d)-a (c+d)) (a (c-d)+b (c+d))-2 (b c-a d)^2 (a c+b d) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}}{3 \left (c^2+d^2\right )}+\frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {4 \left (-3 a c d+b c^2-2 b d^2\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {3 \int \frac {(b c-a d) (b (c-d)-a (c+d)) (a (c-d)+b (c+d))-2 (b c-a d)^2 (a c+b d) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}}{3 \left (c^2+d^2\right )}+\frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4099

\(\displaystyle \frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 \left (-3 a c d+b c^2-2 b d^2\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {3 \left (\frac {1}{2} (b+i a)^2 (c+i d)^2 (b c-a d) \int \frac {i \tan (e+f x)+1}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx-\frac {1}{2} (a+i b)^2 (c-i d)^2 (b c-a d) \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx\right )}{\left (c^2+d^2\right ) (b c-a d)}}{3 \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 \left (-3 a c d+b c^2-2 b d^2\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {3 \left (\frac {1}{2} (b+i a)^2 (c+i d)^2 (b c-a d) \int \frac {i \tan (e+f x)+1}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx-\frac {1}{2} (a+i b)^2 (c-i d)^2 (b c-a d) \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx\right )}{\left (c^2+d^2\right ) (b c-a d)}}{3 \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 4098

\(\displaystyle \frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 \left (-3 a c d+b c^2-2 b d^2\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {3 \left (\frac {(b+i a)^2 (c+i d)^2 (b c-a d) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{2 f}-\frac {(a+i b)^2 (c-i d)^2 (b c-a d) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{2 f}\right )}{\left (c^2+d^2\right ) (b c-a d)}}{3 \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 \left (-3 a c d+b c^2-2 b d^2\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {3 \left (\frac {(b+i a)^2 (c+i d)^2 (b c-a d) \int \frac {1}{i a+b-\frac {(i c+d) (a+b \tan (e+f x))}{c+d \tan (e+f x)}}d\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{f}-\frac {(a+i b)^2 (c-i d)^2 (b c-a d) \int \frac {1}{-i a+b+\frac {(i c-d) (a+b \tan (e+f x))}{c+d \tan (e+f x)}}d\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{f}\right )}{\left (c^2+d^2\right ) (b c-a d)}}{3 \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 \left (-3 a c d+b c^2-2 b d^2\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {3 \left (-\frac {i (a+i b)^{3/2} (c-i d)^2 (b c-a d) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {c+i d}}-\frac {i (b+i a)^2 (c+i d)^2 (b c-a d) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a-i b} \sqrt {c-i d}}\right )}{\left (c^2+d^2\right ) (b c-a d)}}{3 \left (c^2+d^2\right )}\)

Input:

Int[(a + b*Tan[e + f*x])^(3/2)/(c + d*Tan[e + f*x])^(5/2),x]
 

Output:

(2*(b*c - a*d)*Sqrt[a + b*Tan[e + f*x]])/(3*(c^2 + d^2)*f*(c + d*Tan[e + f 
*x])^(3/2)) + ((-3*(((-I)*(I*a + b)^2*(c + I*d)^2*(b*c - a*d)*ArcTanh[(Sqr 
t[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x 
]])])/(Sqrt[a - I*b]*Sqrt[c - I*d]*f) - (I*(a + I*b)^(3/2)*(c - I*d)^2*(b* 
c - a*d)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*S 
qrt[c + d*Tan[e + f*x]])])/(Sqrt[c + I*d]*f)))/((b*c - a*d)*(c^2 + d^2)) + 
 (4*(b*c^2 - 3*a*c*d - 2*b*d^2)*Sqrt[a + b*Tan[e + f*x]])/((c^2 + d^2)*f*S 
qrt[c + d*Tan[e + f*x]]))/(3*(c^2 + d^2))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 221

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4050

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

rule 4098

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[A^2/f   Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f* 
x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && 
 NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
 

rule 4099

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(A + I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T 
an[e + f*x]), x], x] + Simp[(A - I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d 
*Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A 
, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^2, 
0]
 

rule 4132

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ 
[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

Maple [F(-1)]

Timed out.

\[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{\left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}d x\]

Input:

int((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x)
 

Output:

int((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 22531 vs. \(2 (220) = 440\).

Time = 51.51 (sec) , antiderivative size = 22531, normalized size of antiderivative = 81.63 \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="fric 
as")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((a+b*tan(f*x+e))**(3/2)/(c+d*tan(f*x+e))**(5/2),x)
 

Output:

Integral((a + b*tan(e + f*x))**(3/2)/(c + d*tan(e + f*x))**(5/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxi 
ma")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(((2*b*d+2*a*c)^2>0)', see `assum 
e?` for mo
 

Giac [F]

\[ \int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac 
")
 

Output:

integrate((b*tan(f*x + e) + a)^(3/2)/(d*tan(f*x + e) + c)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:

int((a + b*tan(e + f*x))^(3/2)/(c + d*tan(e + f*x))^(5/2),x)
 

Output:

int((a + b*tan(e + f*x))^(3/2)/(c + d*tan(e + f*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {too large to display} \] Input:

int((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x)
 

Output:

(4*sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x)*b**3*d - 
 2*sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*a*b**2*d + 6*sqrt(tan 
(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*b**3*c + 6*int((sqrt(tan(e + f*x 
)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(tan(e + f*x)**4*b*d**3 + 
tan(e + f*x)**3*a*d**3 + 3*tan(e + f*x)**3*b*c*d**2 + 3*tan(e + f*x)**2*a* 
c*d**2 + 3*tan(e + f*x)**2*b*c**2*d + 3*tan(e + f*x)*a*c**2*d + tan(e + f* 
x)*b*c**3 + a*c**3),x)*tan(e + f*x)**2*a**3*b*d**4*f - 12*int((sqrt(tan(e 
+ f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(tan(e + f*x)**4*b*d* 
*3 + tan(e + f*x)**3*a*d**3 + 3*tan(e + f*x)**3*b*c*d**2 + 3*tan(e + f*x)* 
*2*a*c*d**2 + 3*tan(e + f*x)**2*b*c**2*d + 3*tan(e + f*x)*a*c**2*d + tan(e 
 + f*x)*b*c**3 + a*c**3),x)*tan(e + f*x)**2*a**2*b**2*c*d**3*f + 6*int((sq 
rt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(tan(e + f*x 
)**4*b*d**3 + tan(e + f*x)**3*a*d**3 + 3*tan(e + f*x)**3*b*c*d**2 + 3*tan( 
e + f*x)**2*a*c*d**2 + 3*tan(e + f*x)**2*b*c**2*d + 3*tan(e + f*x)*a*c**2* 
d + tan(e + f*x)*b*c**3 + a*c**3),x)*tan(e + f*x)**2*a*b**3*c**2*d**2*f + 
12*int((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(t 
an(e + f*x)**4*b*d**3 + tan(e + f*x)**3*a*d**3 + 3*tan(e + f*x)**3*b*c*d** 
2 + 3*tan(e + f*x)**2*a*c*d**2 + 3*tan(e + f*x)**2*b*c**2*d + 3*tan(e + f* 
x)*a*c**2*d + tan(e + f*x)*b*c**3 + a*c**3),x)*tan(e + f*x)*a**3*b*c*d**3* 
f - 24*int((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f...