\(\int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) [612]

Optimal result

Integrand size = 25, antiderivative size = 139 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {\sqrt {i a-b} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {i a+b} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \] 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 
))/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))/d-2*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {\sqrt [4]{-1} \sqrt {-a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\sqrt [4]{-1} \sqrt {a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\frac {2 \sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}}}{d} \] Input:

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

Output:

-(((-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]]] + (-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]]] + (2*Sq 
rt[a + b*Tan[c + d*x]])/Sqrt[Tan[c + d*x]])/d)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4051, 27, 3042, 4099, 3042, 4098, 104, 216, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4051

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4099

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4098

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-((Sqrt[I*a - b]*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[ 
c + d*x]]])/d) + (Sqrt[I*a + b]*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]]) 
/Sqrt[a + b*Tan[c + d*x]]])/d - (2*Sqrt[a + b*Tan[c + d*x]])/(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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 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 3042

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

rule 4051

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 + 1)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(a^2 + b^2 
))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c 
*(m + 1) - b*d*n - (b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(m + n + 1)*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] && LtQ[m, -1] && GtQ[n, 0] && Int 
egerQ[2*m]
 

rule 4098

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

rule 4099

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

Maple [B] (warning: unable to verify)

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

Time = 0.58 (sec) , antiderivative size = 1089777, normalized size of antiderivative = 7840.12

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

Input:

int((a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(3/2),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. 2696 vs. \(2 (111) = 222\).

Time = 0.31 (sec) , antiderivative size = 2696, normalized size of antiderivative = 19.40 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/8*(d*sqrt((d^2*sqrt(-a^2/d^4) + b)/d^2)*log((2*(2*a^4*b + 4*a^2*b^3 + ( 
a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x 
 + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c) 
 + a)*sqrt(tan(d*x + c)) + (a^5*d - (a^5 + 6*a^3*b^2 + 8*a*b^4)*d*tan(d*x 
+ c)^2 - 4*(a^4*b + 2*a^2*b^3)*d*tan(d*x + c) - ((a^3*b + 4*a*b^3)*d^3*tan 
(d*x + c)^2 - 2*(a^4 + 3*a^2*b^2 + 4*b^4)*d^3*tan(d*x + c) - (3*a^3*b + 4* 
a*b^3)*d^3)*sqrt(-a^2/d^4))*sqrt((d^2*sqrt(-a^2/d^4) + b)/d^2))/(tan(d*x + 
 c)^2 + 1))*tan(d*x + c) + d*sqrt((d^2*sqrt(-a^2/d^4) + b)/d^2)*log(-(2*(2 
*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x + c) + (2*(a^3*b 
+ 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4 
))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + (a^5*d - (a^5 + 6*a^3*b^2 
 + 8*a*b^4)*d*tan(d*x + c)^2 - 4*(a^4*b + 2*a^2*b^3)*d*tan(d*x + c) - ((a^ 
3*b + 4*a*b^3)*d^3*tan(d*x + c)^2 - 2*(a^4 + 3*a^2*b^2 + 4*b^4)*d^3*tan(d* 
x + c) - (3*a^3*b + 4*a*b^3)*d^3)*sqrt(-a^2/d^4))*sqrt((d^2*sqrt(-a^2/d^4) 
 + b)/d^2))/(tan(d*x + c)^2 + 1))*tan(d*x + c) - d*sqrt((d^2*sqrt(-a^2/d^4 
) + b)/d^2)*log((2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan( 
d*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^ 
4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) - (a^5 
*d - (a^5 + 6*a^3*b^2 + 8*a*b^4)*d*tan(d*x + c)^2 - 4*(a^4*b + 2*a^2*b^3)* 
d*tan(d*x + c) - ((a^3*b + 4*a*b^3)*d^3*tan(d*x + c)^2 - 2*(a^4 + 3*a^2...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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