\(\int \frac {\sqrt {2 a b x^2+b^2 x^4}}{c+d x^2} \, dx\) [272]

Optimal result

Integrand size = 29, antiderivative size = 111 \[ \int \frac {\sqrt {2 a b x^2+b^2 x^4}}{c+d x^2} \, dx=\frac {\sqrt {b x^2 \left (2 a+b x^2\right )}}{d x}-\frac {\sqrt {b c-2 a d} \sqrt {b x^2 \left (2 a+b x^2\right )} \arctan \left (\frac {\sqrt {d} \sqrt {2 a+b x^2}}{\sqrt {b c-2 a d}}\right )}{d^{3/2} x \sqrt {2 a+b x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {2 a b x^2+b^2 x^4}}{c+d x^2} \, dx=\frac {b x \sqrt {2 a+b x^2} \left (\sqrt {d} \sqrt {2 a+b x^2}-\sqrt {b c-2 a d} \arctan \left (\frac {\sqrt {d} \sqrt {2 a+b x^2}}{\sqrt {b c-2 a d}}\right )\right )}{d^{3/2} \sqrt {b x^2 \left (2 a+b x^2\right )}} \] Input:

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

Output:

(b*x*Sqrt[2*a + b*x^2]*(Sqrt[d]*Sqrt[2*a + b*x^2] - Sqrt[b*c - 2*a*d]*ArcT 
an[(Sqrt[d]*Sqrt[2*a + b*x^2])/Sqrt[b*c - 2*a*d]]))/(d^(3/2)*Sqrt[b*x^2*(2 
*a + b*x^2)])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1466, 353, 60, 73, 218}

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.

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

\(\Big \downarrow \) 1466

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

\(\Big \downarrow \) 353

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 218

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

rule 353

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ 
{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
 

rule 1466

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbo 
l] :> Simp[(b*x^2 + c*x^4)^FracPart[p]/(x^(2*FracPart[p])*(b + c*x^2)^FracP 
art[p])   Int[x^(2*p)*(d + e*x^2)^q*(b + c*x^2)^p, x], x] /; FreeQ[{b, c, d 
, e, p, q}, x] &&  !IntegerQ[p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(363\) vs. \(2(95)=190\).

Time = 0.50 (sec) , antiderivative size = 364, normalized size of antiderivative = 3.28

method	result	size
default	\(\frac {\sqrt {b^{2} x^{4}+2 a b \,x^{2}}\, \left (-2 \ln \left (-\frac {2 \left (\sqrt {-c d}\, b^{2} x +\sqrt {b \left (b \,x^{2}+2 a \right )}\, \sqrt {\frac {b \left (2 a d -b c \right )}{d}}\, d +2 a b d \right )}{-d x +\sqrt {-c d}}\right ) a b d +\ln \left (-\frac {2 \left (\sqrt {-c d}\, b^{2} x +\sqrt {b \left (b \,x^{2}+2 a \right )}\, \sqrt {\frac {b \left (2 a d -b c \right )}{d}}\, d +2 a b d \right )}{-d x +\sqrt {-c d}}\right ) b^{2} c -2 \ln \left (-\frac {2 \left (\sqrt {-c d}\, b^{2} x -\sqrt {b \left (b \,x^{2}+2 a \right )}\, \sqrt {\frac {b \left (2 a d -b c \right )}{d}}\, d -2 a b d \right )}{d x +\sqrt {-c d}}\right ) a b d +\ln \left (-\frac {2 \left (\sqrt {-c d}\, b^{2} x -\sqrt {b \left (b \,x^{2}+2 a \right )}\, \sqrt {\frac {b \left (2 a d -b c \right )}{d}}\, d -2 a b d \right )}{d x +\sqrt {-c d}}\right ) b^{2} c +2 \sqrt {b \left (b \,x^{2}+2 a \right )}\, \sqrt {\frac {b \left (2 a d -b c \right )}{d}}\, d \right )}{2 x \sqrt {b \left (b \,x^{2}+2 a \right )}\, d^{2} \sqrt {\frac {b \left (2 a d -b c \right )}{d}}}\)	\(364\)
pseudoelliptic	\(\frac {b \left (2 \ln \left (\frac {\sqrt {b^{2} x^{2}+2 a b}\, \sqrt {\frac {2 a b d -c \,b^{2}}{d}}\, d -\sqrt {-b^{2} c d}\, b x +2 a b d}{b d x +\sqrt {-b^{2} c d}}\right ) a b d -2 \ln \left (\frac {\sqrt {b^{2} x^{2}+2 a b}\, \sqrt {\frac {2 a b d -c \,b^{2}}{d}}\, d +\sqrt {-b^{2} c d}\, b x +2 a b d}{b d x -\sqrt {-b^{2} c d}}\right ) a b d -\ln \left (\frac {\sqrt {b^{2} x^{2}+2 a b}\, \sqrt {\frac {2 a b d -c \,b^{2}}{d}}\, d -\sqrt {-b^{2} c d}\, b x +2 a b d}{b d x +\sqrt {-b^{2} c d}}\right ) b^{2} c +\ln \left (\frac {\sqrt {b^{2} x^{2}+2 a b}\, \sqrt {\frac {2 a b d -c \,b^{2}}{d}}\, d +\sqrt {-b^{2} c d}\, b x +2 a b d}{b d x -\sqrt {-b^{2} c d}}\right ) b^{2} c +2 \sqrt {-b^{2} c d}\, \ln \left (b x +\sqrt {b^{2} x^{2}+2 a b}\right ) \sqrt {\frac {2 a b d -c \,b^{2}}{d}}\right )}{2 \sqrt {-b^{2} c d}\, \sqrt {\frac {2 a b d -c \,b^{2}}{d}}\, d}\)	\(390\)
risch	\(\frac {\sqrt {b \,x^{2} \left (b \,x^{2}+2 a \right )}}{d x}+\frac {\left (2 a d -b c \right ) \left (-\frac {\ln \left (\frac {\frac {2 b \left (2 a d -b c \right )}{d}+\frac {2 b^{2} \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {b \left (2 a d -b c \right )}{d}}\, \sqrt {b^{2} \left (x -\frac {\sqrt {-c d}}{d}\right )^{2}+\frac {2 b^{2} \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {b \left (2 a d -b c \right )}{d}}}{x -\frac {\sqrt {-c d}}{d}}\right )}{2 d \sqrt {\frac {b \left (2 a d -b c \right )}{d}}}-\frac {\ln \left (\frac {\frac {2 b \left (2 a d -b c \right )}{d}-\frac {2 b^{2} \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {b \left (2 a d -b c \right )}{d}}\, \sqrt {b^{2} \left (x +\frac {\sqrt {-c d}}{d}\right )^{2}-\frac {2 b^{2} \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {b \left (2 a d -b c \right )}{d}}}{x +\frac {\sqrt {-c d}}{d}}\right )}{2 d \sqrt {\frac {b \left (2 a d -b c \right )}{d}}}\right ) \sqrt {b \,x^{2} \left (b \,x^{2}+2 a \right )}\, \sqrt {b \left (b \,x^{2}+2 a \right )}}{d x \left (b \,x^{2}+2 a \right )}\)	\(402\)

Input:

int((b^2*x^4+2*a*b*x^2)^(1/2)/(d*x^2+c),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.99 \[ \int \frac {\sqrt {2 a b x^2+b^2 x^4}}{c+d x^2} \, dx=\left [\frac {x \sqrt {-\frac {b^{2} c - 2 \, a b d}{d}} \log \left (\frac {b^{2} d x^{3} - {\left (b^{2} c - 4 \, a b d\right )} x - 2 \, \sqrt {b^{2} x^{4} + 2 \, a b x^{2}} d \sqrt {-\frac {b^{2} c - 2 \, a b d}{d}}}{d x^{3} + c x}\right ) + 2 \, \sqrt {b^{2} x^{4} + 2 \, a b x^{2}}}{2 \, d x}, \frac {x \sqrt {\frac {b^{2} c - 2 \, a b d}{d}} \arctan \left (-\frac {\sqrt {b^{2} x^{4} + 2 \, a b x^{2}} d \sqrt {\frac {b^{2} c - 2 \, a b d}{d}}}{{\left (b^{2} c - 2 \, a b d\right )} x}\right ) + \sqrt {b^{2} x^{4} + 2 \, a b x^{2}}}{d x}\right ] \] Input:

integrate((b^2*x^4+2*a*b*x^2)^(1/2)/(d*x^2+c),x, algorithm="fricas")
 

Output:

[1/2*(x*sqrt(-(b^2*c - 2*a*b*d)/d)*log((b^2*d*x^3 - (b^2*c - 4*a*b*d)*x - 
2*sqrt(b^2*x^4 + 2*a*b*x^2)*d*sqrt(-(b^2*c - 2*a*b*d)/d))/(d*x^3 + c*x)) + 
 2*sqrt(b^2*x^4 + 2*a*b*x^2))/(d*x), (x*sqrt((b^2*c - 2*a*b*d)/d)*arctan(- 
sqrt(b^2*x^4 + 2*a*b*x^2)*d*sqrt((b^2*c - 2*a*b*d)/d)/((b^2*c - 2*a*b*d)*x 
)) + sqrt(b^2*x^4 + 2*a*b*x^2))/(d*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

\[ \int \frac {\sqrt {2 a b x^2+b^2 x^4}}{c+d x^2} \, dx=\int { \frac {\sqrt {b^{2} x^{4} + 2 \, a b x^{2}}}{d x^{2} + c} \,d x } \] Input:

integrate((b^2*x^4+2*a*b*x^2)^(1/2)/(d*x^2+c),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (95) = 190\).

Time = 0.14 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.83 \[ \int \frac {\sqrt {2 a b x^2+b^2 x^4}}{c+d x^2} \, dx=-\frac {{\left (b^{2} c \mathrm {sgn}\left (x\right ) - 2 \, a b d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {b^{2} x^{2} + 2 \, a b} d}{\sqrt {b^{2} c d - 2 \, a b d^{2}}}\right )}{\sqrt {b^{2} c d - 2 \, a b d^{2}} d} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b} \mathrm {sgn}\left (x\right )}{d} + \frac {{\left (b^{2} c \arctan \left (\frac {\sqrt {2} \sqrt {a b} d}{\sqrt {b^{2} c d - 2 \, a b d^{2}}}\right ) - 2 \, a b d \arctan \left (\frac {\sqrt {2} \sqrt {a b} d}{\sqrt {b^{2} c d - 2 \, a b d^{2}}}\right ) - \sqrt {2} \sqrt {b^{2} c d - 2 \, a b d^{2}} \sqrt {a b}\right )} \mathrm {sgn}\left (x\right )}{\sqrt {b^{2} c d - 2 \, a b d^{2}} d} \] Input:

integrate((b^2*x^4+2*a*b*x^2)^(1/2)/(d*x^2+c),x, algorithm="giac")
 

Output:

-(b^2*c*sgn(x) - 2*a*b*d*sgn(x))*arctan(sqrt(b^2*x^2 + 2*a*b)*d/sqrt(b^2*c 
*d - 2*a*b*d^2))/(sqrt(b^2*c*d - 2*a*b*d^2)*d) + sqrt(b^2*x^2 + 2*a*b)*sgn 
(x)/d + (b^2*c*arctan(sqrt(2)*sqrt(a*b)*d/sqrt(b^2*c*d - 2*a*b*d^2)) - 2*a 
*b*d*arctan(sqrt(2)*sqrt(a*b)*d/sqrt(b^2*c*d - 2*a*b*d^2)) - sqrt(2)*sqrt( 
b^2*c*d - 2*a*b*d^2)*sqrt(a*b))*sgn(x)/(sqrt(b^2*c*d - 2*a*b*d^2)*d)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {2 a b x^2+b^2 x^4}}{c+d x^2} \, dx=\int \frac {\sqrt {b^2\,x^4+2\,a\,b\,x^2}}{d\,x^2+c} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {2 a b x^2+b^2 x^4}}{c+d x^2} \, dx=\frac {\sqrt {b}\, \left (-\sqrt {d}\, \sqrt {-2 a d +b c}\, \mathit {atan} \left (\frac {\sqrt {b}\, \sqrt {b \,x^{2}+2 a}\, d x +2 a d +b d \,x^{2}}{\sqrt {d}\, \sqrt {b \,x^{2}+2 a}\, \sqrt {-2 a d +b c}+\sqrt {d}\, \sqrt {b}\, \sqrt {-2 a d +b c}\, x}\right )+\sqrt {b \,x^{2}+2 a}\, d \right )}{d^{2}} \] Input:

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

Output:

(sqrt(b)*( - sqrt(d)*sqrt( - 2*a*d + b*c)*atan((sqrt(b)*sqrt(2*a + b*x**2) 
*d*x + 2*a*d + b*d*x**2)/(sqrt(d)*sqrt(2*a + b*x**2)*sqrt( - 2*a*d + b*c) 
+ sqrt(d)*sqrt(b)*sqrt( - 2*a*d + b*c)*x)) + sqrt(2*a + b*x**2)*d))/d**2