\(\int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\) [637]

Optimal result

Integrand size = 25, antiderivative size = 147 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=-\frac {i \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {\frac {\sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}-\frac {\sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}-\frac {2 \sqrt {a+b \tan (c+d x)}}{a \sqrt {\tan (c+d x)}}}{d} \] Input:

Integrate[1/(Tan[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4052, 27, 3042, 4058, 613, 104, 218, 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.

\(\displaystyle \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx\)

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4058

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

\(\Big \downarrow \) 613

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(Tan[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]]),x]
 

Output:

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

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

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 613

Int[Sqrt[(e_.)*(x_)]/(Sqrt[(c_) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^2)), x_Sym 
bol] :> Simp[e/(2*b)   Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] + x)), x 
], x] - Simp[e/(2*b)   Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] - x)), x 
], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 3042

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

rule 4052

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

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]
 

Maple [B] (warning: unable to verify)

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

Time = 0.68 (sec) , antiderivative size = 943928, normalized size of antiderivative = 6421.28

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

Input:

int(1/tan(d*x+c)^(3/2)/(a+b*tan(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. 4031 vs. \(2 (115) = 230\).

Time = 0.49 (sec) , antiderivative size = 4031, normalized size of antiderivative = 27.42 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {1}{\sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \] Input:

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

Output:

Integral(1/(sqrt(a + b*tan(c + d*x))*tan(c + d*x)**(3/2)), x)
 

Maxima [F]

\[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {1}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{3} b +\tan \left (d x +c \right )^{2} a}d x \] Input:

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

Output:

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