\(\int \sqrt {b \tan (c+d x)} \, dx\) [12]

Optimal result

Integrand size = 12, antiderivative size = 137 \[ \int \sqrt {b \tan (c+d x)} \, dx=-\frac {\sqrt {b} \arctan \left (1-\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} d}+\frac {\sqrt {b} \arctan \left (1+\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} d}-\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}+\sqrt {b} \tan (c+d x)}\right )}{\sqrt {2} d} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.52 \[ \int \sqrt {b \tan (c+d x)} \, dx=\frac {\left (\arctan \left (\sqrt [4]{-\tan ^2(c+d x)}\right )-\text {arctanh}\left (\sqrt [4]{-\tan ^2(c+d x)}\right )\right ) \sqrt [4]{-\tan (c+d x)} \sqrt {b \tan (c+d x)}}{d \tan ^{\frac {3}{4}}(c+d x)} \] Input:

Integrate[Sqrt[b*Tan[c + d*x]],x]
 

Output:

((ArcTan[(-Tan[c + d*x]^2)^(1/4)] - ArcTanh[(-Tan[c + d*x]^2)^(1/4)])*(-Ta 
n[c + d*x])^(1/4)*Sqrt[b*Tan[c + d*x]])/(d*Tan[c + d*x]^(3/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.34 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.27, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {3042, 3957, 266, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 b \int \frac {b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {2 b \left (\frac {1}{2} \int \frac {b^2 \tan ^2(c+d x)+b}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}-\frac {1}{2} \int \frac {b-b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}\right )}{d}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{b^2 \tan ^2(c+d x)-\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}+\frac {1}{2} \int \frac {1}{b^2 \tan ^2(c+d x)+\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}\right )-\frac {1}{2} \int \frac {b-b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}\right )}{d}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (\frac {\int \frac {1}{-b^2 \tan ^2(c+d x)-1}d\left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}-\frac {\int \frac {1}{-b^2 \tan ^2(c+d x)-1}d\left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}\right )-\frac {1}{2} \int \frac {b-b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}\right )}{d}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )-\frac {1}{2} \int \frac {b-b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}\right )}{d}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2} \sqrt {b}-2 \sqrt {b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {b}+\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{b^2 \tan ^2(c+d x)+\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {b}-2 \sqrt {b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {b}+\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{b^2 \tan ^2(c+d x)+\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {b}-2 \sqrt {b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}-\frac {\int \frac {\sqrt {b}+\sqrt {2} \sqrt {b \tan (c+d x)}}{b^2 \tan ^2(c+d x)+\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {b}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )\right )}{d}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} b^{3/2} \tan (c+d x)+b^2 \tan ^2(c+d x)+b\right )}{2 \sqrt {2} \sqrt {b}}-\frac {\log \left (\sqrt {2} b^{3/2} \tan (c+d x)+b^2 \tan ^2(c+d x)+b\right )}{2 \sqrt {2} \sqrt {b}}\right )\right )}{d}\)

Input:

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

Output:

(2*b*((-(ArcTan[1 - Sqrt[2]*Sqrt[b]*Tan[c + d*x]]/(Sqrt[2]*Sqrt[b])) + Arc 
Tan[1 + Sqrt[2]*Sqrt[b]*Tan[c + d*x]]/(Sqrt[2]*Sqrt[b]))/2 + (Log[b - Sqrt 
[2]*b^(3/2)*Tan[c + d*x] + b^2*Tan[c + d*x]^2]/(2*Sqrt[2]*Sqrt[b]) - Log[b 
 + Sqrt[2]*b^(3/2)*Tan[c + d*x] + b^2*Tan[c + d*x]^2]/(2*Sqrt[2]*Sqrt[b])) 
/2))/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 217

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

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 3042

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

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {b \sqrt {2}\, \left (\ln \left (\frac {b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d \left (b^{2}\right )^{\frac {1}{4}}}\)	\(136\)
default	\(\frac {b \sqrt {2}\, \left (\ln \left (\frac {b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d \left (b^{2}\right )^{\frac {1}{4}}}\)	\(136\)

Input:

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

Output:

1/4/d*b/(b^2)^(1/4)*2^(1/2)*(ln((b*tan(d*x+c)-(b^2)^(1/4)*(b*tan(d*x+c))^( 
1/2)*2^(1/2)+(b^2)^(1/2))/(b*tan(d*x+c)+(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)*2 
^(1/2)+(b^2)^(1/2)))+2*arctan(2^(1/2)/(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)+1)- 
2*arctan(-2^(1/2)/(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)+1))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.05 \[ \int \sqrt {b \tan (c+d x)} \, dx=\frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b}{b}\right ) + 2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} - b}{b}\right ) - \sqrt {2} \sqrt {b} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right ) + \sqrt {2} \sqrt {b} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right )}{4 \, d} \] Input:

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

Output:

1/4*(2*sqrt(2)*sqrt(b)*arctan((sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(b) + b)/b 
) + 2*sqrt(2)*sqrt(b)*arctan((sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(b) - b)/b) 
 - sqrt(2)*sqrt(b)*log(b*tan(d*x + c) + sqrt(2)*sqrt(b*tan(d*x + c))*sqrt( 
b) + b) + sqrt(2)*sqrt(b)*log(b*tan(d*x + c) - sqrt(2)*sqrt(b*tan(d*x + c) 
)*sqrt(b) + b))/d
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12 \[ \int \sqrt {b \tan (c+d x)} \, dx=\frac {b {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right )}{\sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right )}{\sqrt {b}} - \frac {\sqrt {2} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right )}{\sqrt {b}} + \frac {\sqrt {2} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right )}{\sqrt {b}}\right )}}{4 \, d} \] Input:

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

Output:

1/4*b*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(b) + 2*sqrt(b*tan(d*x + 
c)))/sqrt(b))/sqrt(b) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(b) - 2 
*sqrt(b*tan(d*x + c)))/sqrt(b))/sqrt(b) - sqrt(2)*log(b*tan(d*x + c) + sqr 
t(2)*sqrt(b*tan(d*x + c))*sqrt(b) + b)/sqrt(b) + sqrt(2)*log(b*tan(d*x + c 
) - sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(b) + b)/sqrt(b))/d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.28 \[ \int \sqrt {b \tan (c+d x)} \, dx=\frac {\frac {2 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{d} + \frac {2 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{d} - \frac {\sqrt {2} {\left | b \right |}^{\frac {3}{2}} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{d} + \frac {\sqrt {2} {\left | b \right |}^{\frac {3}{2}} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{d}}{4 \, b} \] Input:

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

Output:

1/4*(2*sqrt(2)*abs(b)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) + 2*s 
qrt(b*tan(d*x + c)))/sqrt(abs(b)))/d + 2*sqrt(2)*abs(b)^(3/2)*arctan(-1/2* 
sqrt(2)*(sqrt(2)*sqrt(abs(b)) - 2*sqrt(b*tan(d*x + c)))/sqrt(abs(b)))/d - 
sqrt(2)*abs(b)^(3/2)*log(b*tan(d*x + c) + sqrt(2)*sqrt(b*tan(d*x + c))*sqr 
t(abs(b)) + abs(b))/d + sqrt(2)*abs(b)^(3/2)*log(b*tan(d*x + c) - sqrt(2)* 
sqrt(b*tan(d*x + c))*sqrt(abs(b)) + abs(b))/d)/b
 

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.36 \[ \int \sqrt {b \tan (c+d x)} \, dx=\frac {{\left (-1\right )}^{1/4}\,\sqrt {b}\,\left (\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {b}}\right )-\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {b}}\right )\right )}{d} \] Input:

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

Output:

((-1)^(1/4)*b^(1/2)*(atan(((-1)^(1/4)*(b*tan(c + d*x))^(1/2))/b^(1/2)) - a 
tanh(((-1)^(1/4)*(b*tan(c + d*x))^(1/2))/b^(1/2))))/d
 

Reduce [F]

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

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

Output:

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