3.2.51 \(\int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx\) [151]

3.2.51.1 Optimal result

Integrand size = 49, antiderivative size = 251 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {(i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{3/2} \sqrt {c-i d} f}-\frac {(B-i (A-C)) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2} \sqrt {c+i d} f}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)}} \]

output

-(I*A+B-I*C)*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))/(a-I*b)^(3/2)/f/(c-I*d)^(1/2)-(B-I*(A-C))*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))/( 
a+I*b)^(3/2)/f/(c+I*d)^(1/2)-2*(A*b^2-a*(B*b-C*a))*(c+d*tan(f*x+e))^(1/2)/ 
(a^2+b^2)/(-a*d+b*c)/f/(a+b*tan(f*x+e))^(1/2)
 

3.2.51.2 Mathematica [A] (verified)

Time = 2.87 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.05 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=\frac {\frac {(a+i b) (i A+B-i C) \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+i b} \sqrt {-c+i d}}+\frac {(i a+b) (A+i B-C) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} \sqrt {c+i d}}+\frac {2 \left (A b^2+a (-b B+a C)\right ) \sqrt {c+d \tan (e+f x)}}{(-b c+a d) \sqrt {a+b \tan (e+f x)}}}{\left (a^2+b^2\right ) f} \]

input

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

output

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

3.2.51.3 Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.21, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4099

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4098

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

input

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

output

(((-I)*(a + I*b)*(A - I*B - C)*(b*c - a*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[a - I*b 
]*Sqrt[c - I*d]*f) + (I*(a - I*b)*(A + I*B - C)*(b*c - a*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[a + I*b]*Sqrt[c + I*d]*f))/((a^2 + b^2)*(b*c - a*d)) - (2*(A*b^2 
 - a*(b*B - a*C))*Sqrt[c + d*Tan[e + f*x]])/((a^2 + b^2)*(b*c - a*d)*f*Sqr 
t[a + b*Tan[e + f*x]])
 

3.2.51.3.1 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 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])))
 

3.2.51.4 Maple [F(-1)]

Timed out.

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

input

int((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e) 
)^(3/2),x)
 

output

int((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e) 
)^(3/2),x)
 

3.2.51.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 83974 vs. \(2 (200) = 400\).

Time = 274.81 (sec) , antiderivative size = 83974, normalized size of antiderivative = 334.56 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]

input

integrate((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1/2)/(a+b*tan( 
f*x+e))^(3/2),x, algorithm="fricas")
 

output

Too large to include
 

3.2.51.6 Sympy [F]

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

input

integrate((A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(c+d*tan(f*x+e))**(1/2)/(a+b*ta 
n(f*x+e))**(3/2),x)
 

output

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

3.2.51.7 Maxima [F]

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

input

integrate((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1/2)/(a+b*tan( 
f*x+e))^(3/2),x, algorithm="maxima")
 

output

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

3.2.51.8 Giac [F(-1)]

Timed out. \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=\text {Timed out} \]

input

integrate((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1/2)/(a+b*tan( 
f*x+e))^(3/2),x, algorithm="giac")
 

output

Timed out
 

3.2.51.9 Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=\text {Hanged} \]

input

int((A + B*tan(e + f*x) + C*tan(e + f*x)^2)/((a + b*tan(e + f*x))^(3/2)*(c 
 + d*tan(e + f*x))^(1/2)),x)
 

output

\text{Hanged}