\(\int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx\) [844]

Optimal result

Integrand size = 25, antiderivative size = 211 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx=\frac {\sqrt {i a-b} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {\sqrt {i a+b} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d} \] Output:

(I*a-b)^(1/2)*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*b^(1/2)*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)/d-(I*a+b) 
^(1/2)*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
 

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\sqrt [4]{-1} \left (\sqrt {-a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\sqrt {a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )+\frac {2 \sqrt {a} \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \] Input:

Integrate[Sqrt[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]*ArcTan[ 
((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 
 Sqrt[a + I*b]*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a 
 + b*Tan[c + d*x]]]) + (2*Sqrt[a]*Sqrt[b]*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d* 
x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/Sqrt[a + b*Tan[c + d*x]]))/d
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.82, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4729, 3042, 4058, 609, 65, 219, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4058

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

\(\Big \downarrow \) 609

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-2 \left (\frac {1}{2} \sqrt {b+i a} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {1}{2} \sqrt {-b+i a} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )\right )}{d}\)

Input:

Int[Sqrt[a + b*Tan[c + d*x]]/Sqrt[Cot[c + d*x]],x]
 

Output:

((2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] 
 - 2*(-1/2*(Sqrt[I*a - b]*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a 
 + b*Tan[c + d*x]]]) + (Sqrt[I*a + b]*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + 
d*x]])/Sqrt[a + b*Tan[c + d*x]]])/2))*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x] 
])/d
 

Defintions of rubi rules used

rule 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 609

Int[(((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_))/((a_) + (b_.)*(x_)^2), x_S 
ymbol] :> Simp[d*(e/b)   Int[(e*x)^(m - 1)*(c + d*x)^(n - 1), x], x] - Simp 
[e/b   Int[(e*x)^(m - 1)*(c + d*x)^(n - 1)*((a*d - b*c*x)/(a + b*x^2)), x], 
 x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[0, n, 1] && LtQ[0, m, 1] &&  !Integ 
erQ[m] &&  !IntegerQ[n]
 

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 4058

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), 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/(1 + ff^2*x^2)), x], x, 
 Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, 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)

Leaf count of result is larger than twice the leaf count of optimal. \(971\) vs. \(2(171)=342\).

Time = 5.22 (sec) , antiderivative size = 972, normalized size of antiderivative = 4.61

method	result	size
default	\(\text {Expression too large to display}\)	\(972\)

Input:

int((a+b*tan(d*x+c))^(1/2)/cot(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/8/d*2^(1/2)/(-b+(a^2+b^2)^(1/2))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1) 
*(a+b*tan(d*x+c))^(1/2)*(-4*b^(1/2)*2^(1/2)*arctanh(1/b^(1/2)*((a*cos(d*x+ 
c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)/(1-cos(d*x+c))*sin(d*x 
+c))*(-b+(a^2+b^2)^(1/2))^(1/2)+(b+(a^2+b^2)^(1/2))^(1/2)*ln(1/(1-cos(d*x+ 
c))*(-a*(1-cos(d*x+c))^2*csc(d*x+c)+2*(a^2+b^2)^(1/2)*(1-cos(d*x+c))+2*2^( 
1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(b+(a 
^2+b^2)^(1/2))^(1/2)*sin(d*x+c)+2*b*(1-cos(d*x+c))+sin(d*x+c)*a))*(-b+(a^2 
+b^2)^(1/2))^(1/2)-(b+(a^2+b^2)^(1/2))^(1/2)*ln(-1/(1-cos(d*x+c))*(-a*(1-c 
os(d*x+c))^2*csc(d*x+c)+2*(a^2+b^2)^(1/2)*(1-cos(d*x+c))-2*2^(1/2)*((a*cos 
(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(b+(a^2+b^2)^(1/2 
))^(1/2)*sin(d*x+c)+2*b*(1-cos(d*x+c))+sin(d*x+c)*a))*(-b+(a^2+b^2)^(1/2)) 
^(1/2)+2*arctan(1/(-b+(a^2+b^2)^(1/2))^(1/2)*((b+(a^2+b^2)^(1/2))^(1/2)*(- 
cot(d*x+c)+csc(d*x+c))+2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(co 
s(d*x+c)+1)^2)^(1/2))/(1-cos(d*x+c))*sin(d*x+c))*(a^2+b^2)^(1/2)-2*arctan( 
1/(-b+(a^2+b^2)^(1/2))^(1/2)*((b+(a^2+b^2)^(1/2))^(1/2)*(-cot(d*x+c)+csc(d 
*x+c))-2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^( 
1/2))/(1-cos(d*x+c))*sin(d*x+c))*(a^2+b^2)^(1/2)-2*arctan(1/(-b+(a^2+b^2)^ 
(1/2))^(1/2)*((b+(a^2+b^2)^(1/2))^(1/2)*(-cot(d*x+c)+csc(d*x+c))+2^(1/2)*( 
(a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2))/(1-cos(d*x 
+c))*sin(d*x+c))*b+2*arctan(1/(-b+(a^2+b^2)^(1/2))^(1/2)*((b+(a^2+b^2)^...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2838 vs. \(2 (167) = 334\).

Time = 0.49 (sec) , antiderivative size = 5709, normalized size of antiderivative = 27.06 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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