\(\int \frac {\sqrt {2 x^2+\sqrt {3+4 x^4}}}{(c+d x) \sqrt {3+4 x^4}} \, dx\) [255]

Optimal result

Integrand size = 40, antiderivative size = 169 \[ \int \frac {\sqrt {2 x^2+\sqrt {3+4 x^4}}}{(c+d x) \sqrt {3+4 x^4}} \, dx=\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \arctan \left (\frac {\sqrt {3} d+2 i c x}{\sqrt {2 i c^2-\sqrt {3} d^2} \sqrt {\sqrt {3}-2 i x^2}}\right )}{\sqrt {2 i c^2-\sqrt {3} d^2}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \text {arctanh}\left (\frac {\sqrt {3} d-2 i c x}{\sqrt {2 i c^2+\sqrt {3} d^2} \sqrt {\sqrt {3}+2 i x^2}}\right )}{\sqrt {2 i c^2+\sqrt {3} d^2}} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 1.41 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.15 \[ \int \frac {\sqrt {2 x^2+\sqrt {3+4 x^4}}}{(c+d x) \sqrt {3+4 x^4}} \, dx=-\frac {\sqrt {-2 c^2-\sqrt {4 c^4+3 d^4}} \arctan \left (\frac {d \sqrt {2 x^2+\sqrt {3+4 x^4}}}{\sqrt {-2 c^2-\sqrt {4 c^4+3 d^4}}}\right )}{\sqrt {4 c^4+3 d^4}}+\frac {\sqrt {-2 c^2+\sqrt {4 c^4+3 d^4}} \arctan \left (\frac {d \sqrt {2 x^2+\sqrt {3+4 x^4}}}{\sqrt {-2 c^2+\sqrt {4 c^4+3 d^4}}}\right )}{\sqrt {4 c^4+3 d^4}}+c \text {RootSum}\left [9 d^2-24 c^2 \text {$\#$1}-6 d^2 \text {$\#$1}^2-8 c^2 \text {$\#$1}^3+d^2 \text {$\#$1}^4\&,\frac {-3 \log \left (2 x^2+\sqrt {3+4 x^4}+2 x \sqrt {2 x^2+\sqrt {3+4 x^4}}-\text {$\#$1}\right )-\log \left (2 x^2+\sqrt {3+4 x^4}+2 x \sqrt {2 x^2+\sqrt {3+4 x^4}}-\text {$\#$1}\right ) \text {$\#$1}^2}{-6 c^2-3 d^2 \text {$\#$1}-6 c^2 \text {$\#$1}^2+d^2 \text {$\#$1}^3}\&\right ] \] Input:

Integrate[Sqrt[2*x^2 + Sqrt[3 + 4*x^4]]/((c + d*x)*Sqrt[3 + 4*x^4]),x]
 

Output:

-((Sqrt[-2*c^2 - Sqrt[4*c^4 + 3*d^4]]*ArcTan[(d*Sqrt[2*x^2 + Sqrt[3 + 4*x^ 
4]])/Sqrt[-2*c^2 - Sqrt[4*c^4 + 3*d^4]]])/Sqrt[4*c^4 + 3*d^4]) + (Sqrt[-2* 
c^2 + Sqrt[4*c^4 + 3*d^4]]*ArcTan[(d*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]])/Sqrt[- 
2*c^2 + Sqrt[4*c^4 + 3*d^4]]])/Sqrt[4*c^4 + 3*d^4] + c*RootSum[9*d^2 - 24* 
c^2*#1 - 6*d^2*#1^2 - 8*c^2*#1^3 + d^2*#1^4 & , (-3*Log[2*x^2 + Sqrt[3 + 4 
*x^4] + 2*x*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] - #1] - Log[2*x^2 + Sqrt[3 + 4*x 
^4] + 2*x*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] - #1]*#1^2)/(-6*c^2 - 3*d^2*#1 - 6 
*c^2*#1^2 + d^2*#1^3) & ]
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2558, 488, 217, 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.

\(\displaystyle \int \frac {\sqrt {\sqrt {4 x^4+3}+2 x^2}}{\sqrt {4 x^4+3} (c+d x)} \, dx\)

\(\Big \downarrow \) 2558

\(\displaystyle \left (\frac {1}{2}-\frac {i}{2}\right ) \int \frac {1}{(c+d x) \sqrt {\sqrt {3}-2 i x^2}}dx+\left (\frac {1}{2}+\frac {i}{2}\right ) \int \frac {1}{(c+d x) \sqrt {2 i x^2+\sqrt {3}}}dx\)

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \arctan \left (\frac {\sqrt {3} d+2 i c x}{\sqrt {\sqrt {3}-2 i x^2} \sqrt {-\sqrt {3} d^2+2 i c^2}}\right )}{\sqrt {-\sqrt {3} d^2+2 i c^2}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \text {arctanh}\left (\frac {\sqrt {3} d-2 i c x}{\sqrt {\sqrt {3}+2 i x^2} \sqrt {\sqrt {3} d^2+2 i c^2}}\right )}{\sqrt {\sqrt {3} d^2+2 i c^2}}\)

Input:

Int[Sqrt[2*x^2 + Sqrt[3 + 4*x^4]]/((c + d*x)*Sqrt[3 + 4*x^4]),x]
 

Output:

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

Defintions of rubi rules used

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 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 488

Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ 
Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ 
[{a, b, c, d}, x]
 

rule 2558

Int[(((c_.) + (d_.)*(x_))^(m_.)*Sqrt[(b_.)*(x_)^2 + Sqrt[(a_) + (e_.)*(x_)^ 
4]])/Sqrt[(a_) + (e_.)*(x_)^4], x_Symbol] :> Simp[(1 - I)/2   Int[(c + d*x) 
^m/Sqrt[Sqrt[a] - I*b*x^2], x], x] + Simp[(1 + I)/2   Int[(c + d*x)^m/Sqrt[ 
Sqrt[a] + I*b*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[e, b^2] && G 
tQ[a, 0]
 

Maple [F]

\[\int \frac {\sqrt {2 x^{2}+\sqrt {4 x^{4}+3}}}{\left (d x +c \right ) \sqrt {4 x^{4}+3}}d x\]

Input:

int((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)/(4*x^4+3)^(1/2),x)
 

Output:

int((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)/(4*x^4+3)^(1/2),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt {2 x^2+\sqrt {3+4 x^4}}}{(c+d x) \sqrt {3+4 x^4}} \, dx=\text {Timed out} \] Input:

integrate((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)/(4*x^4+3)^(1/2),x, algorit 
hm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\sqrt {2 x^2+\sqrt {3+4 x^4}}}{(c+d x) \sqrt {3+4 x^4}} \, dx=\int \frac {\sqrt {2 x^{2} + \sqrt {4 x^{4} + 3}}}{\left (c + d x\right ) \sqrt {4 x^{4} + 3}}\, dx \] Input:

integrate((2*x**2+(4*x**4+3)**(1/2))**(1/2)/(d*x+c)/(4*x**4+3)**(1/2),x)
 

Output:

Integral(sqrt(2*x**2 + sqrt(4*x**4 + 3))/((c + d*x)*sqrt(4*x**4 + 3)), x)
 

Maxima [F]

\[ \int \frac {\sqrt {2 x^2+\sqrt {3+4 x^4}}}{(c+d x) \sqrt {3+4 x^4}} \, dx=\int { \frac {\sqrt {2 \, x^{2} + \sqrt {4 \, x^{4} + 3}}}{\sqrt {4 \, x^{4} + 3} {\left (d x + c\right )}} \,d x } \] Input:

integrate((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)/(4*x^4+3)^(1/2),x, algorit 
hm="maxima")
 

Output:

integrate(sqrt(2*x^2 + sqrt(4*x^4 + 3))/(sqrt(4*x^4 + 3)*(d*x + c)), x)
 

Giac [F]

\[ \int \frac {\sqrt {2 x^2+\sqrt {3+4 x^4}}}{(c+d x) \sqrt {3+4 x^4}} \, dx=\int { \frac {\sqrt {2 \, x^{2} + \sqrt {4 \, x^{4} + 3}}}{\sqrt {4 \, x^{4} + 3} {\left (d x + c\right )}} \,d x } \] Input:

integrate((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)/(4*x^4+3)^(1/2),x, algorit 
hm="giac")
 

Output:

integrate(sqrt(2*x^2 + sqrt(4*x^4 + 3))/(sqrt(4*x^4 + 3)*(d*x + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {2 x^2+\sqrt {3+4 x^4}}}{(c+d x) \sqrt {3+4 x^4}} \, dx=\int \frac {\sqrt {2\,x^2+\sqrt {4\,x^4+3}}}{\sqrt {4\,x^4+3}\,\left (c+d\,x\right )} \,d x \] Input:

int((2*x^2 + (4*x^4 + 3)^(1/2))^(1/2)/((4*x^4 + 3)^(1/2)*(c + d*x)),x)
 

Output:

int((2*x^2 + (4*x^4 + 3)^(1/2))^(1/2)/((4*x^4 + 3)^(1/2)*(c + d*x)), x)
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 1008, normalized size of antiderivative = 5.96 \[ \int \frac {\sqrt {2 x^2+\sqrt {3+4 x^4}}}{(c+d x) \sqrt {3+4 x^4}} \, dx =\text {Too large to display} \] Input:

int((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)/(4*x^4+3)^(1/2),x)
 

Output:

(sqrt(4*c**4 + 3*d**4)*(sqrt(sqrt(4*c**4 + 3*d**4) - 2*c**2)*atan((2*sqrt( 
4*x**4 + 3)*sqrt(4*c**4 + 3*d**4)*sqrt(sqrt(4*x**4 + 3) + 2*x**2)*sqrt(sqr 
t(4*c**4 + 3*d**4) - 2*c**2)*c*x - 4*sqrt(4*c**4 + 3*d**4)*sqrt(sqrt(4*x** 
4 + 3) + 2*x**2)*sqrt(sqrt(4*c**4 + 3*d**4) - 2*c**2)*c*x**3 - 3*sqrt(4*c* 
*4 + 3*d**4)*sqrt(sqrt(4*x**4 + 3) + 2*x**2)*sqrt(sqrt(4*c**4 + 3*d**4) - 
2*c**2)*d + 4*sqrt(4*x**4 + 3)*sqrt(sqrt(4*x**4 + 3) + 2*x**2)*sqrt(sqrt(4 
*c**4 + 3*d**4) - 2*c**2)*c**3*x + 3*sqrt(4*x**4 + 3)*sqrt(sqrt(4*x**4 + 3 
) + 2*x**2)*sqrt(sqrt(4*c**4 + 3*d**4) - 2*c**2)*d**3 - 8*sqrt(sqrt(4*x**4 
 + 3) + 2*x**2)*sqrt(sqrt(4*c**4 + 3*d**4) - 2*c**2)*c**3*x**3 - 6*sqrt(sq 
rt(4*x**4 + 3) + 2*x**2)*sqrt(sqrt(4*c**4 + 3*d**4) - 2*c**2)*c**2*d + 6*s 
qrt(sqrt(4*x**4 + 3) + 2*x**2)*sqrt(sqrt(4*c**4 + 3*d**4) - 2*c**2)*c*d**2 
*x - 6*sqrt(sqrt(4*x**4 + 3) + 2*x**2)*sqrt(sqrt(4*c**4 + 3*d**4) - 2*c**2 
)*d**3*x**2)/(24*c**2*d**2*x**2 - 18*d**4)) - sqrt(sqrt(4*c**4 + 3*d**4) + 
 2*c**2)*log(c**2 + 2*c*d*x + d**2*x**2) + sqrt(sqrt(4*c**4 + 3*d**4) + 2* 
c**2)*log( - 2*sqrt(4*x**4 + 3)*sqrt(4*c**4 + 3*d**4)*sqrt(sqrt(4*x**4 + 3 
) + 2*x**2)*sqrt(sqrt(4*c**4 + 3*d**4) + 2*c**2)*c*x + 4*sqrt(4*c**4 + 3*d 
**4)*sqrt(sqrt(4*x**4 + 3) + 2*x**2)*sqrt(sqrt(4*c**4 + 3*d**4) + 2*c**2)* 
c*x**3 - 3*sqrt(4*c**4 + 3*d**4)*sqrt(sqrt(4*x**4 + 3) + 2*x**2)*sqrt(sqrt 
(4*c**4 + 3*d**4) + 2*c**2)*d + 4*sqrt(4*x**4 + 3)*sqrt(sqrt(4*x**4 + 3) + 
 2*x**2)*sqrt(sqrt(4*c**4 + 3*d**4) + 2*c**2)*c**3*x - 3*sqrt(4*x**4 + ...