\(\int x^3 \sqrt {\frac {e (a+b x^2)}{c+d x^2}} \, dx\) [43]

Optimal result

Integrand size = 26, antiderivative size = 197 \[ \int x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx=-\frac {(5 b c-a d) \left (c+d x^2\right ) \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{8 b d^2}+\frac {\left (c+d x^2\right )^2 \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{4 d^2}+\frac {(b c-a d) (3 b c+a d) \sqrt {e} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{\sqrt {b} \sqrt {e}}\right )}{8 b^{3/2} d^{5/2}} \] Output:

-1/8*(-a*d+5*b*c)*(d*x^2+c)*(b*e/d-(-a*d+b*c)*e/d/(d*x^2+c))^(1/2)/b/d^2+1 
/4*(d*x^2+c)^2*(b*e/d-(-a*d+b*c)*e/d/(d*x^2+c))^(1/2)/d^2+1/8*(-a*d+b*c)*( 
a*d+3*b*c)*e^(1/2)*arctanh(d^(1/2)*(b*e/d-(-a*d+b*c)*e/d/(d*x^2+c))^(1/2)/ 
b^(1/2)/e^(1/2))/b^(3/2)/d^(5/2)
 

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.82 \[ \int x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx=\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt {b} \sqrt {d} \sqrt {a+b x^2} \left (c+d x^2\right ) \left (-3 b c+a d+2 b d x^2\right )+\left (3 b^2 c^2-2 a b c d-a^2 d^2\right ) \sqrt {c+d x^2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )\right )}{8 b^{3/2} d^{5/2} \sqrt {a+b x^2}} \] Input:

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

Output:

(Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(Sqrt[b]*Sqrt[d]*Sqrt[a + b*x^2]*(c + d 
*x^2)*(-3*b*c + a*d + 2*b*d*x^2) + (3*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*Sqrt[ 
c + d*x^2]*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])]))/ 
(8*b^(3/2)*d^(5/2)*Sqrt[a + b*x^2])
 

Rubi [A] (warning: unable to verify)

Time = 0.59 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {2053, 2052, 25, 360, 25, 298, 221}

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.

\(\displaystyle \int x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx\)

\(\Big \downarrow \) 2053

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

\(\Big \downarrow \) 2052

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 360

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 298

\(\displaystyle e (b c-a d) \left (\frac {e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^2 \left (b e-d x^4\right )^2}-\frac {\frac {(5 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b \left (b e-d x^4\right )}-\frac {(a d+3 b c) \int \frac {1}{b e-d x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{2 b}}{4 d^2}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle e (b c-a d) \left (\frac {e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^2 \left (b e-d x^4\right )^2}-\frac {\frac {(5 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b \left (b e-d x^4\right )}-\frac {(a d+3 b c) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 b^{3/2} \sqrt {d} \sqrt {e}}}{4 d^2}\right )\)

Input:

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

Output:

(b*c - a*d)*e*(((b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(4*d^2*(b 
*e - d*x^4)^2) - (((5*b*c - a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(2*b*( 
b*e - d*x^4)) - ((3*b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + 
d*x^2)])/(Sqrt[b]*Sqrt[e])])/(2*b^(3/2)*Sqrt[d]*Sqrt[e]))/(4*d^2))
 

Defintions of rubi rules used

rule 25

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

rule 221

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

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 360

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : 
> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p 
+ 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1))   Int[(a + b*x^2)^(p + 1)*Expan 
dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 
- 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; 
FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & 
& (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

rule 2052

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_S 
ymbol] :> With[{q = Denominator[p]}, Simp[q*e*(b*c - a*d)   Subst[Int[x^(q* 
(p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a + b* 
x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] 
&& IntegerQ[m]
 

rule 2053

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.) 
))^(p_), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(e*( 
(a + b*x)/(c + d*x)))^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p}, 
 x] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {\left (2 b d \,x^{2}+a d -3 b c \right ) \left (d \,x^{2}+c \right ) \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}{8 b \,d^{2}}-\frac {\left (a^{2} d^{2}+2 a b c d -3 b^{2} c^{2}\right ) \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +b d \,x^{2} e}{\sqrt {b d e}}+\sqrt {b d e \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right ) e}}{16 b \,d^{2} \sqrt {b d e}\, \left (b \,x^{2}+a \right )}\)	\(189\)
default	\(\frac {\sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (4 \sqrt {b d}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b d \,x^{2}-\ln \left (\frac {2 b d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} d^{2}-2 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c d +3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2}+2 \sqrt {b d}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, a d -6 \sqrt {b d}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b c \right )}{16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, d^{2} b \sqrt {b d}}\)	\(342\)

Input:

int(x^3*(e*(b*x^2+a)/(d*x^2+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/8*(2*b*d*x^2+a*d-3*b*c)*(d*x^2+c)/b/d^2*(e*(b*x^2+a)/(d*x^2+c))^(1/2)-1/ 
16*(a^2*d^2+2*a*b*c*d-3*b^2*c^2)/b/d^2*ln((1/2*a*d*e+1/2*b*c*e+b*d*x^2*e)/ 
(b*d*e)^(1/2)+(b*d*e*x^4+(a*d*e+b*c*e)*x^2+a*c*e)^(1/2))/(b*d*e)^(1/2)*(e* 
(b*x^2+a)/(d*x^2+c))^(1/2)*((d*x^2+c)*(b*x^2+a)*e)^(1/2)/(b*x^2+a)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 407, normalized size of antiderivative = 2.07 \[ \int x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx=\left [-\frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {\frac {e}{b d}} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e - 4 \, {\left (2 \, b^{2} d^{3} x^{4} + b^{2} c^{2} d + a b c d^{2} + {\left (3 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {e}{b d}}\right ) - 4 \, {\left (2 \, b d^{2} x^{4} - 3 \, b c^{2} + a c d - {\left (b c d - a d^{2}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{32 \, b d^{2}}, -\frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {-\frac {e}{b d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {e}{b d}}}{2 \, {\left (b e x^{2} + a e\right )}}\right ) - 2 \, {\left (2 \, b d^{2} x^{4} - 3 \, b c^{2} + a c d - {\left (b c d - a d^{2}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{16 \, b d^{2}}\right ] \] Input:

integrate(x^3*(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="fricas")
 

Output:

[-1/32*((3*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*sqrt(e/(b*d))*log(8*b^2*d^2*e*x^ 
4 + 8*(b^2*c*d + a*b*d^2)*e*x^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*e - 4*(2 
*b^2*d^3*x^4 + b^2*c^2*d + a*b*c*d^2 + (3*b^2*c*d^2 + a*b*d^3)*x^2)*sqrt(( 
b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(e/(b*d))) - 4*(2*b*d^2*x^4 - 3*b*c^2 + a* 
c*d - (b*c*d - a*d^2)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b*d^2), -1/ 
16*((3*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*sqrt(-e/(b*d))*arctan(1/2*(2*b*d*x^2 
 + b*c + a*d)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(-e/(b*d))/(b*e*x^2 + 
a*e)) - 2*(2*b*d^2*x^4 - 3*b*c^2 + a*c*d - (b*c*d - a*d^2)*x^2)*sqrt((b*e* 
x^2 + a*e)/(d*x^2 + c)))/(b*d^2)]
 

Sympy [F(-1)]

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

integrate(x**3*(e*(b*x**2+a)/(d*x**2+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^3*(e*(b*x^2+a)/(d*x^2+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(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.86 \[ \int x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx=\frac {1}{16} \, {\left (2 \, \sqrt {b d e x^{4} + b c e x^{2} + a d e x^{2} + a c e} {\left (\frac {2 \, x^{2}}{d} - \frac {3 \, b c - a d}{b d^{2}}\right )} - \frac {{\left (3 \, b^{2} c^{2} e - 2 \, a b c d e - a^{2} d^{2} e\right )} \log \left ({\left | -b c e - a d e - 2 \, \sqrt {b d e} {\left (\sqrt {b d e} x^{2} - \sqrt {b d e x^{4} + b c e x^{2} + a d e x^{2} + a c e}\right )} \right |}\right )}{\sqrt {b d e} b d^{2}}\right )} \mathrm {sgn}\left (d x^{2} + c\right ) \] Input:

integrate(x^3*(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="giac")
 

Output:

1/16*(2*sqrt(b*d*e*x^4 + b*c*e*x^2 + a*d*e*x^2 + a*c*e)*(2*x^2/d - (3*b*c 
- a*d)/(b*d^2)) - (3*b^2*c^2*e - 2*a*b*c*d*e - a^2*d^2*e)*log(abs(-b*c*e - 
 a*d*e - 2*sqrt(b*d*e)*(sqrt(b*d*e)*x^2 - sqrt(b*d*e*x^4 + b*c*e*x^2 + a*d 
*e*x^2 + a*c*e))))/(sqrt(b*d*e)*b*d^2))*sgn(d*x^2 + c)
 

Mupad [F(-1)]

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

int(x^3*((e*(a + b*x^2))/(c + d*x^2))^(1/2),x)
 

Output:

int(x^3*((e*(a + b*x^2))/(c + d*x^2))^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.00 \[ \int x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx=\frac {\sqrt {e}\, \left (\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a b \,d^{2}-3 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{2} c d +2 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{2} d^{2} x^{2}+\sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d +\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a^{2} d^{2}+2 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d +\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a b c d -3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d +\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) b^{2} c^{2}\right )}{8 b^{2} d^{3}} \] Input:

int(x^3*(e*(b*x^2+a)/(d*x^2+c))^(1/2),x)
 

Output:

(sqrt(e)*(sqrt(c + d*x**2)*sqrt(a + b*x**2)*a*b*d**2 - 3*sqrt(c + d*x**2)* 
sqrt(a + b*x**2)*b**2*c*d + 2*sqrt(c + d*x**2)*sqrt(a + b*x**2)*b**2*d**2* 
x**2 + sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2)*d + sqrt(d)*sqrt(c 
+ d*x**2)*b)*a**2*d**2 + 2*sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2) 
*d + sqrt(d)*sqrt(c + d*x**2)*b)*a*b*c*d - 3*sqrt(d)*sqrt(b)*log( - sqrt(b 
)*sqrt(a + b*x**2)*d + sqrt(d)*sqrt(c + d*x**2)*b)*b**2*c**2))/(8*b**2*d** 
3)