\(\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x)) \, dx\) [557]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [F(-2)]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 21, antiderivative size = 125 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x)) \, dx=-\frac {(a-b) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a-b) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {2 b \sqrt {\tan (c+d x)}}{d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.63 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x)) \, dx=\frac {\sqrt [4]{-1} (i a+b) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-(-1)^{3/4} (a+i b) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 b \sqrt {\tan (c+d x)}}{d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.30, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 4011, 3042, 4017, 25, 1482, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(2*(((a - b)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 
 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 - ((a + b)*(-1/2*Log[1 - Sqrt[2 
]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Tan[c 
+ d*x]] + Tan[c + d*x]]/(2*Sqrt[2])))/2))/d + (2*b*Sqrt[Tan[c + d*x]])/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 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 1482
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a*c, 2]}, Simp[(d*q + a*e)/(2*a*c)   Int[(q + c*x^2)/(a + c*x^4), x], x] + 
Simp[(d*q - a*e)/(2*a*c)   Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a 
, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- 
a)*c]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4011
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int 
[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 
 0] && GtQ[m, 0]
 

rule 4017
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2/f   Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq 
rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & 
& NeQ[c^2 + d^2, 0]
 
Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.51

method result size
derivativedivides \(\frac {2 b \sqrt {\tan \left (d x +c \right )}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {a \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{d}\) \(189\)
default \(\frac {2 b \sqrt {\tan \left (d x +c \right )}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {a \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{d}\) \(189\)
parts \(\frac {a \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4 d}+\frac {b \left (2 \sqrt {\tan \left (d x +c \right )}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}\) \(192\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (105) = 210\).

Time = 0.09 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.98 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x)) \, dx=-\frac {2 \, \sqrt {\frac {1}{2}} d \sqrt {\frac {a^{2} - 2 \, a b + b^{2}}{d^{2}}} \arctan \left (-\frac {2 \, \sqrt {\frac {1}{2}} {\left (a + b\right )} d \sqrt {\frac {a^{2} - 2 \, a b + b^{2}}{d^{2}}} \sqrt {\tan \left (d x + c\right )} + d^{2} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{d^{2}}} \sqrt {\frac {a^{2} - 2 \, a b + b^{2}}{d^{2}}}}{a^{2} - b^{2}}\right ) + 2 \, \sqrt {\frac {1}{2}} d \sqrt {\frac {a^{2} - 2 \, a b + b^{2}}{d^{2}}} \arctan \left (-\frac {2 \, \sqrt {\frac {1}{2}} {\left (a + b\right )} d \sqrt {\frac {a^{2} - 2 \, a b + b^{2}}{d^{2}}} \sqrt {\tan \left (d x + c\right )} - d^{2} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{d^{2}}} \sqrt {\frac {a^{2} - 2 \, a b + b^{2}}{d^{2}}}}{a^{2} - b^{2}}\right ) + \sqrt {\frac {1}{2}} d \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{d^{2}}} \log \left (2 \, \sqrt {\frac {1}{2}} d \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{d^{2}}} \sqrt {\tan \left (d x + c\right )} + {\left (a + b\right )} \tan \left (d x + c\right ) + a + b\right ) - \sqrt {\frac {1}{2}} d \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{d^{2}}} \log \left (-2 \, \sqrt {\frac {1}{2}} d \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{d^{2}}} \sqrt {\tan \left (d x + c\right )} + {\left (a + b\right )} \tan \left (d x + c\right ) + a + b\right ) - 4 \, b \sqrt {\tan \left (d x + c\right )}}{2 \, d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.08 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x)) \, dx=\frac {2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (a + b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (a + b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 8 \, b \sqrt {\tan \left (d x + c\right )}}{4 \, d} \] Input:

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

Output:

1/4*(2*sqrt(2)*(a - b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c))) 
) + 2*sqrt(2)*(a - b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c))) 
) - sqrt(2)*(a + b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + s 
qrt(2)*(a + b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 8*b*s 
qrt(tan(d*x + c)))/d
 

Giac [F(-2)]

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

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c)),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 [B] (verification not implemented)

Time = 1.80 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.22 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x)) \, dx=\frac {2\,b\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {\sqrt {2}\,a\,\left (\ln \left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}-\mathrm {tan}\left (c+d\,x\right )-1\right )-\ln \left (\mathrm {tan}\left (c+d\,x\right )+\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}+1\right )\right )}{4\,d}+\frac {\sqrt {2}\,a\,\left (\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}-1\right )+\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}+1\right )\right )}{2\,d}+\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )\,1{}\mathrm {i}}{d}+\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )\,1{}\mathrm {i}}{d} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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