\(\int \frac {\sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx\) [643]

Optimal result

Integrand size = 25, antiderivative size = 149 \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}-\frac {2 b \sqrt {\tan (c+d x)}}{\left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \] Output:

-arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a-b)^(3/ 
2)/d+arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a+b 
)^(3/2)/d-2*b*tan(d*x+c)^(1/2)/(a^2+b^2)/d/(a+b*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {\frac {\sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{3/2}}+\frac {\sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{3/2}}-\frac {2 b \sqrt {\tan (c+d x)}}{\left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{d} \] Input:

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

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.18, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4051

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4099

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4098

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(((I*a + b)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d 
*x]]])/(Sqrt[I*a - b]*d) - ((I*a - b)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + 
d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d))/(a^2 + b^2) - (2*b*Sq 
rt[Tan[c + d*x]])/((a^2 + b^2)*d*Sqrt[a + b*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 = 1.06 (sec) , antiderivative size = 798609, normalized size of antiderivative = 5359.79

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

Input:

int(tan(d*x+c)^(1/2)/(a+b*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. 7445 vs. \(2 (121) = 242\).

Time = 1.22 (sec) , antiderivative size = 7445, normalized size of antiderivative = 49.97 \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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