\(\int \frac {\sqrt {a+b \tan (c+d x)}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\) [845]

Optimal result

Integrand size = 25, antiderivative size = 244 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\frac {i \sqrt {i a-b} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {b} d}+\frac {i \sqrt {i a+b} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {\sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.35 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-(-1)^{3/4} \sqrt {-a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+(-1)^{3/4} \sqrt {a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {a^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}\right )}{d} \] Input:

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

Output:

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-((-1)^(3/4)*Sqrt[-a + I*b]*ArcTan 
[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]) 
 + (-1)^(3/4)*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[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d 
*x]] + (a^(3/2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b* 
Tan[c + d*x])/a])/(Sqrt[b]*Sqrt[a + b*Tan[c + d*x]])))/d
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.83, 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, 4053, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}dx\)

\(\Big \downarrow \) 4053

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4138

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-(((-I)*Sqrt[I*a - b]*ArcTan[(Sqrt[ 
I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - (a*ArcTanh[(Sqrt[ 
b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[b] - I*Sqrt[I*a + b 
]*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d) 
 + (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 4053

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a + b*Tan[e + f*x])^(m - 1)*((c + 
d*Tan[e + f*x])^n/(f*(m + n - 1))), x] + Simp[1/(m + n - 1)   Int[(a + b*Ta 
n[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a^2*c*(m + n - 1) - b 
*(b*c*(m - 1) + a*d*n) + (2*a*b*c + a^2*d - b^2*d)*(m + n - 1)*Tan[e + f*x] 
 + b*(b*c*n + a*d*(2*m + n - 2))*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] && GtQ[m, 1] && GtQ[n, 0] && IntegerQ[2*n]
 

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] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1286\) vs. \(2(198)=396\).

Time = 3.27 (sec) , antiderivative size = 1287, normalized size of antiderivative = 5.27

method	result	size
default	\(\text {Expression too large to display}\)	\(1287\)

Input:

int((a+b*tan(d*x+c))^(1/2)/cot(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4/d*2^(1/2)/b^(1/2)/(-b+(a^2+b^2)^(1/2))^(1/2)/a*((b+(a^2+b^2)^(1/2))^( 
1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*b^(3/2)*ln((a*cot(d*x+c)*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*a*cot(d*x+c)+2*(a^2+b^2)^(1/2)*cos(d*x+c) 
+2*cos(d*x+c)*b-sin(d*x+c)*a+a*csc(d*x+c)-2*(a^2+b^2)^(1/2)-2*b)/(-1+cos(d 
*x+c)))*cos(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*b^ 
(3/2)*ln(-(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)-a*cot(d*x+c)*cos(d*x+c)+2* 
a*cot(d*x+c)-2*(a^2+b^2)^(1/2)*cos(d*x+c)-2*cos(d*x+c)*b+sin(d*x+c)*a-a*cs 
c(d*x+c)+2*(a^2+b^2)^(1/2)+2*b)/(-1+cos(d*x+c)))*cos(d*x+c)-(b+(a^2+b^2)^( 
1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*b^(1/2)*ln((a*cot(d*x+c)*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*a*cot(d*x+c)+2*(a^2+b^2)^(1/2)*cos 
(d*x+c)+2*cos(d*x+c)*b-sin(d*x+c)*a+a*csc(d*x+c)-2*(a^2+b^2)^(1/2)-2*b)/(- 
1+cos(d*x+c)))*cos(d*x+c)*(a^2+b^2)^(1/2)+(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a 
^2+b^2)^(1/2))^(1/2)*b^(1/2)*ln(-(2*2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*s 
in(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*sin(d*x+c)-a*c 
ot(d*x+c)*cos(d*x+c)+2*a*cot(d*x+c)-2*(a^2+b^2)^(1/2)*cos(d*x+c)-2*cos(d*x 
+c)*b+sin(d*x+c)*a-a*csc(d*x+c)+2*(a^2+b^2)^(1/2)+2*b)/(-1+cos(d*x+c)))*co 
s(d*x+c)*(a^2+b^2)^(1/2)+2*(-b+(a^2+b^2)^(1/2))^(1/2)*arctanh(1/b^(1/2)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2897 vs. \(2 (192) = 384\).

Time = 0.66 (sec) , antiderivative size = 5827, normalized size of antiderivative = 23.88 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(d*x+c))^(1/2)/cot(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

integrate((a+b*tan(d*x+c))**(1/2)/cot(d*x+c)**(3/2),x)
 

Output:

Integral(sqrt(a + b*tan(c + d*x))/cot(c + d*x)**(3/2), x)
 

Maxima [F]

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

integrate((a+b*tan(d*x+c))^(1/2)/cot(d*x+c)^(3/2),x, algorithm="maxima")
 

Output:

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

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+b*tan(d*x+c))^(1/2)/cot(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

int((a + b*tan(c + d*x))^(1/2)/cot(c + d*x)^(3/2),x)
 

Output:

int((a + b*tan(c + d*x))^(1/2)/cot(c + d*x)^(3/2), x)
 

Reduce [F]

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

int((a+b*tan(d*x+c))^(1/2)/cot(d*x+c)^(3/2),x)
 

Output:

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