\(\int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx\) [621]

Optimal result

Integrand size = 35, antiderivative size = 261 \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\frac {\sqrt {i a-b} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(2 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {b} d}-\frac {\sqrt {i a+b} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \] Output:

(I*a-b)^(1/2)*(A+I*B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c 
))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d+(2*A*b+B*a)*arctanh(b^(1/2)* 
tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2) 
/b^(1/2)/d-(I*a+b)^(1/2)*(A-I*B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a 
+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d+B*(a+b*tan(d*x+c 
))^(1/2)/d/cot(d*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 2.39 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\sqrt [4]{-1} \sqrt {-a+i b} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+\sqrt [4]{-1} \sqrt {a+i b} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))+\frac {(2 A b+a B) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) (a+b \tan (c+d x))}{\sqrt {a} \sqrt {b} \sqrt {1+\frac {b \tan (c+d x)}{a}}}\right )}{d \sqrt {a+b \tan (c+d x)}} \] Input:

Integrate[(Sqrt[a + b*Tan[c + d*x]]*(A + B*Tan[c + d*x]))/Sqrt[Cot[c + d*x 
]],x]
 

Output:

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-1)^(1/4)*Sqrt[-a + I*b]*(A - I*B 
)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + 
 d*x]]]*Sqrt[a + b*Tan[c + d*x]] + (-1)^(1/4)*Sqrt[a + I*b]*(A + I*B)*ArcT 
an[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] 
*Sqrt[a + b*Tan[c + d*x]] + B*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x]) + (( 
2*A*b + a*B)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*(a + b*Tan[c + 
d*x]))/(Sqrt[a]*Sqrt[b]*Sqrt[1 + (b*Tan[c + d*x])/a])))/(d*Sqrt[a + b*Tan[ 
c + d*x]])
 

Rubi [A] (verified)

Time = 1.89 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.84, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4729, 3042, 4093, 27, 3042, 4138, 2035, 2257, 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.

\(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}}dx\)

\(\Big \downarrow \) 4729

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4093

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4138

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

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

Input:

Int[(Sqrt[a + b*Tan[c + d*x]]*(A + B*Tan[c + d*x]))/Sqrt[Cot[c + d*x]],x]
 

Output:

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

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 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

rule 3042

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

rule 4093

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[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp 
[1/(m + n)   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n - 1)* 
Simp[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (A*b*c + a*B*c + a*A*d - b*B*d)*(m 
 + n)*Tan[e + f*x] + (A*b*d*(m + n) + B*(a*d*m + b*c*n))*Tan[e + f*x]^2, x] 
, x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[ 
a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[0, m, 1] && LtQ[0, n, 1]
 

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]
 

rule 4729

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a 
+ b*x])^m*(c*Tan[a + b*x])^m   Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ[u, 
x]
 

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 1.01 (sec) , antiderivative size = 2180708, normalized size of antiderivative = 8355.20

\[\text {output too large to display}\]

Input:

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

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8069 vs. \(2 (209) = 418\).

Time = 3.20 (sec) , antiderivative size = 16171, normalized size of antiderivative = 61.96 \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algo 
rithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {a + b \tan {\left (c + d x \right )}}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx \] Input:

integrate((a+b*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c))/cot(d*x+c)**(1/2),x)
 

Output:

Integral((A + B*tan(c + d*x))*sqrt(a + b*tan(c + d*x))/sqrt(cot(c + d*x)), 
 x)
 

Maxima [F]

\[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a}}{\sqrt {\cot \left (d x + c\right )}} \,d x } \] Input:

integrate((a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algo 
rithm="maxima")
 

Output:

integrate((B*tan(d*x + c) + A)*sqrt(b*tan(d*x + c) + a)/sqrt(cot(d*x + c)) 
, x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algo 
rithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\left (\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \sqrt {\cot \left (d x +c \right )}\, \tan \left (d x +c \right )}{\cot \left (d x +c \right )}d x \right ) b +\left (\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \sqrt {\cot \left (d x +c \right )}}{\cot \left (d x +c \right )}d x \right ) a \] Input:

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

Output:

int((sqrt(tan(c + d*x)*b + a)*sqrt(cot(c + d*x))*tan(c + d*x))/cot(c + d*x 
),x)*b + int((sqrt(tan(c + d*x)*b + a)*sqrt(cot(c + d*x)))/cot(c + d*x),x) 
*a