\(\int \sqrt {a+b \tan (c+d x)} \, dx\) [507]

Optimal result
Mathematica [C] (verified)
Rubi [A] (warning: unable to verify)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F(-2)]
Giac [F(-1)]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 14, antiderivative size = 268 \[ \int \sqrt {a+b \tan (c+d x)} \, dx=-\frac {b \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} \sqrt {a+b \tan (c+d x)}}{a+\sqrt {a^2+b^2}+b \tan (c+d x)}\right )}{\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} d}+\frac {b \text {arctanh}\left (\frac {\sqrt {a+\sqrt {a^2+b^2}}-\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} \sqrt {a-\sqrt {a^2+b^2}} d}-\frac {b \text {arctanh}\left (\frac {\sqrt {a+\sqrt {a^2+b^2}}+\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} \sqrt {a-\sqrt {a^2+b^2}} d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.32 \[ \int \sqrt {a+b \tan (c+d x)} \, dx=-\frac {i \left (\sqrt {a-i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )-\sqrt {a+i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )\right )}{d} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.63 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.50, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 3966, 483, 1449, 1142, 25, 27, 1083, 219, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3966

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

\(\Big \downarrow \) 483

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

\(\Big \downarrow \) 1449

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(2*b*((-((Sqrt[a + Sqrt[a^2 + b^2]]*ArcTanh[(-(Sqrt[2]*Sqrt[a + Sqrt[a^2 + 
 b^2]]) + 2*Sqrt[a + b*Tan[c + d*x]])/(Sqrt[2]*Sqrt[a - Sqrt[a^2 + b^2]])] 
)/Sqrt[a - Sqrt[a^2 + b^2]]) + Log[Sqrt[a^2 + b^2] + b^2*Tan[c + d*x]^2 - 
Sqrt[2]*Sqrt[a + Sqrt[a^2 + b^2]]*Sqrt[a + b*Tan[c + d*x]]]/2)/(2*Sqrt[2]* 
Sqrt[a + Sqrt[a^2 + b^2]]) - ((Sqrt[a + Sqrt[a^2 + b^2]]*ArcTanh[(Sqrt[2]* 
Sqrt[a + Sqrt[a^2 + b^2]] + 2*Sqrt[a + b*Tan[c + d*x]])/(Sqrt[2]*Sqrt[a - 
Sqrt[a^2 + b^2]])])/Sqrt[a - Sqrt[a^2 + b^2]] + Log[Sqrt[a^2 + b^2] + b^2* 
Tan[c + d*x]^2 + Sqrt[2]*Sqrt[a + Sqrt[a^2 + b^2]]*Sqrt[a + b*Tan[c + d*x] 
]]/2)/(2*Sqrt[2]*Sqrt[a + Sqrt[a^2 + 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 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 483
Int[Sqrt[(c_) + (d_.)*(x_)]/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[2*d 
Subst[Int[x^2/(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4), x], x, Sqrt[c + d*x]], x 
] /; FreeQ[{a, b, c, d}, x]
 

rule 1083
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{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 1142
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1449
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = 
Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}, Simp[1/(2*c*r)   Int[x^(m - 1)/(q 
 - r*x + x^2), x], x] - Simp[1/(2*c*r)   Int[x^(m - 1)/(q + r*x + x^2), x], 
 x]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 1] && LtQ[m, 
3] && NegQ[b^2 - 4*a*c]
 

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

rule 3966
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Su 
bst[Int[(a + x)^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c 
, d, n}, x] && NeQ[a^2 + b^2, 0]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(649\) vs. \(2(217)=434\).

Time = 4.27 (sec) , antiderivative size = 650, normalized size of antiderivative = 2.43

method result size
derivativedivides \(-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {\left (a^{2}+b^{2}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d b \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}-\frac {a^{2} \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d b \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {\left (a^{2}+b^{2}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d b \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}-\frac {a^{2} \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d b \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\) \(650\)
default \(-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {\left (a^{2}+b^{2}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d b \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}-\frac {a^{2} \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d b \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {\left (a^{2}+b^{2}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d b \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}-\frac {a^{2} \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d b \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\) \(650\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.19 \[ \int \sqrt {a+b \tan (c+d x)} \, dx=-\frac {1}{2} \, \sqrt {-\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} + a}{d^{2}}} \log \left (d^{3} \sqrt {-\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} + a}{d^{2}}} \sqrt {-\frac {b^{2}}{d^{4}}} + \sqrt {b \tan \left (d x + c\right ) + a} b\right ) + \frac {1}{2} \, \sqrt {-\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} + a}{d^{2}}} \log \left (-d^{3} \sqrt {-\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} + a}{d^{2}}} \sqrt {-\frac {b^{2}}{d^{4}}} + \sqrt {b \tan \left (d x + c\right ) + a} b\right ) + \frac {1}{2} \, \sqrt {\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} - a}{d^{2}}} \log \left (d^{3} \sqrt {\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} - a}{d^{2}}} \sqrt {-\frac {b^{2}}{d^{4}}} + \sqrt {b \tan \left (d x + c\right ) + a} b\right ) - \frac {1}{2} \, \sqrt {\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} - a}{d^{2}}} \log \left (-d^{3} \sqrt {\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} - a}{d^{2}}} \sqrt {-\frac {b^{2}}{d^{4}}} + \sqrt {b \tan \left (d x + c\right ) + a} b\right ) \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more 
details)Is
 

Giac [F(-1)]

Timed out. \[ \int \sqrt {a+b \tan (c+d x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 1.39 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.79 \[ \int \sqrt {a+b \tan (c+d x)} \, dx=-\mathrm {atanh}\left (\frac {d^3\,\sqrt {-\frac {a-b\,1{}\mathrm {i}}{d^2}}\,\left (\frac {16\,\left (b^4-a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}+\frac {16\,a\,b^2\,\left (a-b\,1{}\mathrm {i}\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}\right )}{16\,\left (a^2\,b^3+b^5\right )}\right )\,\sqrt {-\frac {a-b\,1{}\mathrm {i}}{d^2}}-\mathrm {atanh}\left (\frac {d^3\,\sqrt {-\frac {a+b\,1{}\mathrm {i}}{d^2}}\,\left (\frac {16\,\left (b^4-a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}+\frac {16\,a\,b^2\,\left (a+b\,1{}\mathrm {i}\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}\right )}{16\,\left (a^2\,b^3+b^5\right )}\right )\,\sqrt {-\frac {a+b\,1{}\mathrm {i}}{d^2}} \] Input:

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

Output:

- atanh((d^3*(-(a - b*1i)/d^2)^(1/2)*((16*(b^4 - a^2*b^2)*(a + b*tan(c + d 
*x))^(1/2))/d^2 + (16*a*b^2*(a - b*1i)*(a + b*tan(c + d*x))^(1/2))/d^2))/( 
16*(b^5 + a^2*b^3)))*(-(a - b*1i)/d^2)^(1/2) - atanh((d^3*(-(a + b*1i)/d^2 
)^(1/2)*((16*(b^4 - a^2*b^2)*(a + b*tan(c + d*x))^(1/2))/d^2 + (16*a*b^2*( 
a + b*1i)*(a + b*tan(c + d*x))^(1/2))/d^2))/(16*(b^5 + a^2*b^3)))*(-(a + b 
*1i)/d^2)^(1/2)
 

Reduce [F]

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

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

Output:

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