\(\int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1307]

Optimal result

Integrand size = 29, antiderivative size = 213 \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i (a-i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}+\frac {i (a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c+i d)^{3/2} f}+\frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{\left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}} \] 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)^(3/2)/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) 
^(3/2)/f+2*(-a*d+b*c)*(a+b*tan(f*x+e))^(1/2)/(c^2+d^2)/f/(c+d*tan(f*x+e))^ 
(1/2)
 

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {-\frac {i (-a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(-c+i d)^{3/2}}+\frac {\frac {(a+i b)^{3/2} (i c+d) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c+i d)^{3/2}}+\frac {2 (b c-a d) \sqrt {a+b \tan (e+f x)}}{(c+i d) \sqrt {c+d \tan (e+f x)}}}{c-i d}}{f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.16 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 4050, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4050

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4099

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4098

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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]
 

Maple [F(-1)]

Timed out.

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15120 vs. \(2 (165) = 330\).

Time = 20.38 (sec) , antiderivative size = 15120, normalized size of antiderivative = 70.99 \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 2*sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*b**2 + 2*int((sqrt 
(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(tan(e + f*x)* 
*3*b*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 2*tan(e + f 
*x)*a*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*tan(e + f*x)*a**2*b*d**2*f - 
2*int((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(ta 
n(e + f*x)**3*b*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 
2*tan(e + f*x)*a*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*tan(e + f*x)*a*b** 
2*c*d*f + 2*int((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + 
 f*x))/(tan(e + f*x)**3*b*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)** 
2*b*c*d + 2*tan(e + f*x)*a*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*a**2*b*c 
*d*f - 2*int((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f* 
x))/(tan(e + f*x)**3*b*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b 
*c*d + 2*tan(e + f*x)*a*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*a*b**2*c**2 
*f + int((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a))/(tan(e + f*x) 
**3*b*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 2*tan(e + 
f*x)*a*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*tan(e + f*x)*a**3*d**2*f - i 
nt((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a))/(tan(e + f*x)**3*b* 
d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 2*tan(e + f*x)*a 
*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*tan(e + f*x)*a**2*b*c*d*f - int((s 
qrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a))/(tan(e + f*x)**3*b*d*...