3.3.97 \(\int \frac {x^3}{\sqrt {\frac {e (a+b x^2)}{c+d x^2}}} \, dx\) [297]

Optimal. Leaf size=169 \[ -\frac {(b c+3 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 b^2 d e}+\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 b d e}-\frac {(b c-a d) (b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 b^{5/2} d^{3/2} \sqrt {e}} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 169, 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, 393, 205, 214} \begin {gather*} -\frac {(b c-a d) (3 a d+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^{5/2} d^{3/2} \sqrt {e}}-\frac {\left (c+d x^2\right ) (3 a d+b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 b^2 d e}+\frac {\left (c+d x^2\right )^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((b*c + 3*a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(b^2*d*e) + (Sqrt[(e*(a + b*x^2))/(c + d*x^
2)]*(c + d*x^2)^2)/(4*b*d*e) - ((b*c - a*d)*(b*c + 3*a*d)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/
(Sqrt[b]*Sqrt[e])])/(8*b^(5/2)*d^(3/2)*Sqrt[e])

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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

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Mathematica [A]
time = 0.92, size = 161, normalized size = 0.95 \begin {gather*} \frac {\sqrt {b} \sqrt {d} \left (a+b x^2\right ) \sqrt {c+d x^2} \left (-3 a d+b \left (c+2 d x^2\right )\right )-\left (b^2 c^2+2 a b c d-3 a^2 d^2\right ) \sqrt {a+b x^2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 b^{5/2} d^{3/2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d 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[b]*Sqrt[d]*(a + b*x^2)*Sqrt[c + d*x^2]*(-3*a*d + b*(c + 2*d*x^2)) - (b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*Sq
rt[a + b*x^2]*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])])/(8*b^(5/2)*d^(3/2)*Sqrt[(e*(a + b*
x^2))/(c + d*x^2)]*Sqrt[c + d*x^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(340\) vs. \(2(145)=290\).
time = 0.07, size = 341, normalized size = 2.02

method	result	size
risch	\(-\frac {\left (-2 b d \,x^{2}+3 a d -b c \right ) \left (b \,x^{2}+a \right )}{8 b^{2} d \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}+\frac {\left (\frac {3 d \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a^{2}}{16 b^{2} \sqrt {d e b}}-\frac {\ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a c}{8 b \sqrt {d e b}}-\frac {\ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) c^{2}}{16 d \sqrt {d e b}}\right ) \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{\sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(306\)
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (-4 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b d \,x^{2}-3 d^{2} \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2}+2 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a c b d +\ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2}+6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a d -2 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b c \right )}{16 \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, b^{2} d \sqrt {b d}}\)	\(341\)

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

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Maxima [A]
time = 0.52, size = 249, normalized size = 1.47 \begin {gather*} \frac {1}{16} \, {\left (\frac {2 \, {\left ({\left (b^{2} c^{2} d + 2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} \left (\frac {b x^{2} + a}{d x^{2} + c}\right )^{\frac {3}{2}} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )}}{b^{4} d - \frac {2 \, {\left (b x^{2} + a\right )} b^{3} d^{2}}{d x^{2} + c} + \frac {{\left (b x^{2} + a\right )}^{2} b^{2} d^{3}}{{\left (d x^{2} + c\right )}^{2}}} + \frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, 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^{2} d}\right )} e^{\left (-\frac {1}{2}\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="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 388, normalized size = 2.30 \begin {gather*} \left [-\frac {{\left ({\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {b d} \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} + b^{2} c^{2} d - 3 \, a b c d^{2} + 3 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {1}{2}\right )}}{32 \, b^{3} d^{2}}, \frac {{\left ({\left (b^{2} c^{2} + 2 \, a b c d - 3 \, 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 ) + 2 \, {\left (2 \, b^{2} d^{3} x^{4} + b^{2} c^{2} d - 3 \, a b c d^{2} + 3 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {1}{2}\right )}}{16 \, b^{3} d^{2}}\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*((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*sqrt(b*d)*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 + b^2*c^2*d - 3*a*b*c*d^2 + 3*(b^2*c*d^2 - a*b*d^3)*x^2)*sqrt((b*x^2 + a)/(d*x^2 +
c)))*e^(-1/2)/(b^3*d^2), 1/16*((b^2*c^2 + 2*a*b*c*d - 3*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)) + 2*(2*b^2*d^3*x^4 + b^2*c^2*d - 3*a*b*c*d^2 +
3*(b^2*c*d^2 - a*b*d^3)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c)))*e^(-1/2)/(b^3*d^2)]

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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 = 5.55, size = 153, normalized size = 0.91 \begin {gather*} \frac {{\left (2 \, \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c} {\left (\frac {2 \, x^{2}}{b} + \frac {b c - 3 \, a d}{b^{2} d}\right )} + \frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, 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^{2} d}\right )} e^{\left (-\frac {1}{2}\right )}}{16 \, \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/b + (b*c - 3*a*d)/(b^2*d)) + (b^2*c^2 + 2*a*b*c*d - 3*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^2*d))*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 \frac {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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