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

Optimal. Leaf size=161 \[ \frac {\sqrt {e} (b c-a d) (a d+3 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 b^{3/2} d^{5/2}}+\frac {\left (c+d x^2\right )^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^2}-\frac {\left (c+d x^2\right ) (5 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 b d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1960, 455, 385, 208} \[ \frac {\sqrt {e} (b c-a d) (a d+3 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 b^{3/2} d^{5/2}}+\frac {\left (c+d x^2\right )^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^2}-\frac {\left (c+d x^2\right ) (5 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((5*b*c - a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(8*b*d^2) + (Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*
(c + d*x^2)^2)/(4*d^2) + ((b*c - a*d)*(3*b*c + a*d)*Sqrt[e]*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]
)/(Sqrt[b]*Sqrt[e])])/(8*b^(3/2)*d^(5/2))

Rule 208

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

Rule 385

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

Rule 455

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] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[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 1960

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

Rubi steps

\begin {align*} \int x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx &=((b c-a d) e) \operatorname {Subst}\left (\int \frac {x^2 \left (-a e+c x^2\right )}{\left (b e-d x^2\right )^3} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 d^2}-\frac {((b c-a d) e) \operatorname {Subst}\left (\int \frac {(b c-a d) e+4 c d x^2}{\left (b e-d x^2\right )^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{4 d^2}\\ &=-\frac {(5 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 b d^2}+\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 d^2}+\frac {((b c-a d) (3 b c+a d) e) \operatorname {Subst}\left (\int \frac {1}{b e-d x^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{8 b d^2}\\ &=-\frac {(5 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 b d^2}+\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 d^2}+\frac {(b c-a d) (3 b c+a d) \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 b^{3/2} d^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.38, size = 149, normalized size = 0.93 \[ \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (b \sqrt {d} \left (c+d x^2\right ) \left (a d-3 b c+2 b d x^2\right )+\frac {(a d+3 b c) (b c-a d)^{3/2} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )}{\sqrt {a+b x^2}}\right )}{8 b^2 d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 407, normalized size = 2.53 \[ \left [-\frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {\frac {e}{b d}} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e - 4 \, {\left (2 \, b^{2} d^{3} x^{4} + b^{2} c^{2} d + a b c d^{2} + {\left (3 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {e}{b d}}\right ) - 4 \, {\left (2 \, b d^{2} x^{4} - 3 \, b c^{2} + a c d - {\left (b c d - a d^{2}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{32 \, b d^{2}}, -\frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {-\frac {e}{b d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {e}{b d}}}{2 \, {\left (b e x^{2} + a e\right )}}\right ) - 2 \, {\left (2 \, b d^{2} x^{4} - 3 \, b c^{2} + a c d - {\left (b c d - a d^{2}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{16 \, b d^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-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)]

________________________________________________________________________________________

giac [A]  time = 0.82, size = 185, normalized size = 1.15 \[ \frac {1}{16} \, {\left (2 \, \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e} {\left (\frac {2 \, x^{2}}{d} - \frac {3 \, b c - a d}{b d^{2}}\right )} - \frac {{\left (3 \, b^{2} c^{2} e - 2 \, a b c d e - a^{2} d^{2} e\right )} e^{\left (-\frac {1}{2}\right )} \log \left ({\left | -b c e - a d e - 2 \, {\left (\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )} \sqrt {b d} e^{\frac {1}{2}} \right |}\right )}{\sqrt {b d} b d^{2}}\right )} \mathrm {sgn}\left (d x^{2} + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.04, size = 342, normalized size = 2.12 \[ \frac {\sqrt {\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-a^{2} d^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-2 a b c d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 b^{2} c^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+4 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a d -6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b c \right )}{16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, b \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((b*x^2+a)/(d*x^2+c)*e)^(1/2)*(d*x^2+c)*(4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^2*b*d-d^2*ln
(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^2-2*ln(1/2*(2*b*d*x^
2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a*c*b*d+3*b^2*ln(1/2*(2*b*d*x^2+a*d+
b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*c^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*
(b*d)^(1/2)*a*d-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b*c)/((d*x^2+c)*(b*x^2+a))^(1/2)/d^2/b/(b*d)
^(1/2)

________________________________________________________________________________________

maxima [A]  time = 2.25, size = 269, normalized size = 1.67 \[ \frac {1}{16} \, e {\left (\frac {2 \, {\left ({\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\right )} \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {3}{2}} - {\left (3 \, b^{3} c^{2} - 2 \, a b^{2} c d - a^{2} b d^{2}\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} e\right )}}{b^{3} d^{2} e^{2} - \frac {2 \, {\left (b x^{2} + a\right )} b^{2} d^{3} e^{2}}{d x^{2} + c} + \frac {{\left (b x^{2} + a\right )}^{2} b d^{4} e^{2}}{{\left (d x^{2} + c\right )}^{2}}} - \frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \log \left (\frac {d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {b d e}}{d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {b d e}}\right )}{\sqrt {b d e} b d^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*e*(2*((5*b^2*c^2*d - 6*a*b*c*d^2 + a^2*d^3)*((b*x^2 + a)*e/(d*x^2 + c))^(3/2) - (3*b^3*c^2 - 2*a*b^2*c*d
- a^2*b*d^2)*sqrt((b*x^2 + a)*e/(d*x^2 + c))*e)/(b^3*d^2*e^2 - 2*(b*x^2 + a)*b^2*d^3*e^2/(d*x^2 + c) + (b*x^2
+ a)^2*b*d^4*e^2/(d*x^2 + c)^2) - (3*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*log((d*sqrt((b*x^2 + a)*e/(d*x^2 + c)) - s
qrt(b*d*e))/(d*sqrt((b*x^2 + a)*e/(d*x^2 + c)) + sqrt(b*d*e)))/(sqrt(b*d*e)*b*d^2))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\sqrt {\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________