3.2.31 \(\int \frac {\sqrt {c+d \tan (e+f x)} (A+B \tan (e+f x)+C \tan ^2(e+f x))}{\sqrt {a+b \tan (e+f x)}} \, dx\) [131]

3.2.31.1 Optimal result

Integrand size = 49, antiderivative size = 287 \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx=-\frac {(i A+B-i C) \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a-i b} f}-\frac {(B-i (A-C)) \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} f}+\frac {(b c C+2 b B d-a C d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{3/2} \sqrt {d} f}+\frac {C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f} \]

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))*(c-I*d)^(1/2)/f/(a-I*b)^(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))*( 
c+I*d)^(1/2)/f/(a+I*b)^(1/2)+(2*B*b*d-C*a*d+C*b*c)*arctanh(d^(1/2)*(a+b*ta 
n(f*x+e))^(1/2)/b^(1/2)/(c+d*tan(f*x+e))^(1/2))/b^(3/2)/f/d^(1/2)+C*(a+b*t 
an(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/b/f
 

3.2.31.2 Mathematica [A] (verified)

Time = 4.70 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.54 \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx=\frac {\frac {b \left (b B c+b (A-C) d+\sqrt {-b^2} (A c-c C-B d)\right ) \text {arctanh}\left (\frac {\sqrt {-c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+\sqrt {-b^2}} \sqrt {-c+\frac {\sqrt {-b^2} d}{b}}}+\frac {b \left (\sqrt {-b^2} (A c-c C-B d)-b (B c+(A-C) d)\right ) \text {arctanh}\left (\frac {\sqrt {c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+\frac {\sqrt {-b^2} d}{b}}}+b C \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}+\frac {\sqrt {b} \sqrt {c-\frac {a d}{b}} (b c C+2 b B d-a C d) \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right ) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}{\sqrt {d} \sqrt {c+d \tan (e+f x)}}}{b^2 f} \]

input

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

output

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

3.2.31.3 Rubi [A] (verified)

Time = 1.26 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4130, 27, 3042, 4138, 2348, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4138

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

\(\Big \downarrow \) 2348

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.2.31.3.1 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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2348

Int[(Px_)*((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_.) + (b_. 
)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(c + d*x)^m*(e + f*x)^ 
n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[P 
x, x] && (IntegerQ[p] || (IntegerQ[2*p] && IntegerQ[m] && ILtQ[n, 0])) && 
!(IGtQ[m, 0] && IGtQ[n, 0])
 

rule 3042

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

rule 4130

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[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ 
e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a 
+ b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

rule 4138

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] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 
2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f 
, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + 
 d^2, 0]
 

3.2.31.4 Maple [F(-1)]

Timed out.

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

input

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

output

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

3.2.31.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39018 vs. \(2 (225) = 450\).

Time = 105.64 (sec) , antiderivative size = 78051, normalized size of antiderivative = 271.95 \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx=\text {Too large to display} \]

input

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

output

Too large to include
 

3.2.31.6 Sympy [F]

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

input

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

output

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

3.2.31.7 Maxima [F]

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

input

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

output

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

3.2.31.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.2.31.9 Mupad [F(-1)]

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

input

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

output

\text{Hanged}