3.3.64 \(\int x^3 \sqrt {\frac {e (a+b x^2)}{c+d x^2}} \, dx\) [264]

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

[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

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Rubi [A]
time = 0.17, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1981, 1980, 466, 393, 214} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((5*b*c - a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(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 214

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 393

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 466

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 1980

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_Symbol] :> With[{q = Denominator[p]}
, Dist[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 1981

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Dist[1/n, Sub
st[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]]

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) \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 0.89, size = 161, normalized size = 1.00 \begin {gather*} \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} \tanh ^{-1}\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}} \end {gather*}

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)]*(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])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(341\) vs. \(2(137)=274\).
time = 0.08, size = 342, normalized size = 2.12

method	result	size
risch	\(\frac {\left (2 b d \,x^{2}+a d -3 b c \right ) \left (d \,x^{2}+c \right ) \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}{8 b \,d^{2}}+\frac {\left (-\frac {\ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a^{2}}{16 b \sqrt {d e b}}-\frac {\ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a c}{8 d \sqrt {d e b}}+\frac {3 b \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) c^{2}}{16 d^{2} \sqrt {d e b}}\right ) \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{b \,x^{2}+a}\)	\(305\)
default	\(\frac {\sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (4 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b d \,x^{2}-d^{2} \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2}-2 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a c b d +3 b^{2} \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) c^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a d -6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b c \right )}{16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, d^{2} b \sqrt {b d}}\)	\(342\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 253, normalized size = 1.57 \begin {gather*} \frac {1}{16} \, {\left (\frac {2 \, {\left ({\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\right )} \left (\frac {b x^{2} + a}{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 {b x^{2} + a}{d x^{2} + c}}\right )}}{b^{3} d^{2} - \frac {2 \, {\left (b x^{2} + a\right )} b^{2} d^{3}}{d x^{2} + c} + \frac {{\left (b x^{2} + a\right )}^{2} b d^{4}}{{\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 {b x^{2} + a}{d x^{2} + c}} - \sqrt {b d}}{d \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} + \sqrt {b d}}\right )}{\sqrt {b d} b d^{2}}\right )} e^{\frac {1}{2}} \end {gather*}

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

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Fricas [A]
time = 0.39, size = 394, normalized size = 2.45 \begin {gather*} \left [-\frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {b d} e^{\frac {1}{2}} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} - 4 \, {\left (2 \, b d^{2} x^{4} + b c^{2} + a c d + {\left (3 \, b c d + a d^{2}\right )} x^{2}\right )} \sqrt {b d} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right ) - 4 \, {\left (2 \, b^{2} d^{3} x^{4} - 3 \, b^{2} c^{2} d + a b c d^{2} - {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} e^{\frac {1}{2}}}{32 \, b^{2} d^{3}}, -\frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-b d} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{2 \, {\left (b^{2} d x^{2} + a b d\right )}}\right ) e^{\frac {1}{2}} - 2 \, {\left (2 \, b^{2} d^{3} x^{4} - 3 \, b^{2} c^{2} d + a b c d^{2} - {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} e^{\frac {1}{2}}}{16 \, b^{2} d^{3}}\right ] \end {gather*}

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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Giac [A]
time = 7.74, size = 155, normalized size = 0.96 \begin {gather*} \frac {1}{16} \, {\left (2 \, \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c} {\left (\frac {2 \, x^{2}}{d} - \frac {3 \, b c - a d}{b d^{2}}\right )} - \frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \log \left ({\left | -b c - a d - 2 \, {\left (\sqrt {b d} x^{2} - \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c}\right )} \sqrt {b d} \right |}\right )}{\sqrt {b d} b d^{2}}\right )} e^{\frac {1}{2}} \mathrm {sgn}\left (d x^{2} + c\right ) \end {gather*}

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

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

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)

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