3.3.98 \(\int \frac {x}{\sqrt {\frac {e (a+b x^2)}{c+d x^2}}} \, dx\) [298]

Optimal. Leaf size=106 \[ \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{2 b e}+\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 b^{3/2} \sqrt {d} \sqrt {e}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1981, 1979, 205, 214} \begin {gather*} \frac {(b c-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 )}{2 b^{3/2} \sqrt {d} \sqrt {e}}+\frac {\left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1979

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

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}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx &=((b c-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 )\\ &=\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{2 b e}+\frac {(b c-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 )}{2 b}\\ &=\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{2 b e}+\frac {(b c-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 )}{2 b^{3/2} \sqrt {d} \sqrt {e}}\\ \end {align*}

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Mathematica [A]
time = 0.40, size = 136, normalized size = 1.28 \begin {gather*} \frac {\sqrt {b} \sqrt {d} \left (a+b x^2\right ) \left (c+d x^2\right )+(b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 b^{3/2} \sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(199\) vs. \(2(86)=172\).
time = 0.06, size = 200, normalized size = 1.89

method	result	size
default	\(\frac {\left (b \,x^{2}+a \right ) \left (-d \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a +c \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\right )}{4 \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, b \sqrt {b d}}\)	\(200\)
risch	\(\frac {b \,x^{2}+a}{2 b \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}+\frac {\left (-\frac {\ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a d}{4 b \sqrt {d e b}}+\frac {\ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) c}{4 \sqrt {d e b}}\right ) \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{\sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 138, normalized size = 1.30 \begin {gather*} \frac {1}{4} \, {\left (\frac {2 \, {\left (b c - a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{b^{2} - \frac {{\left (b x^{2} + a\right )} b d}{d x^{2} + c}} - \frac {{\left (b c - a d\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}\right )} e^{\left (-\frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 288, normalized size = 2.72 \begin {gather*} \left [-\frac {{\left ({\left (b c - a d\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 (b d^{2} x^{2} + b c d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {1}{2}\right )}}{8 \, b^{2} d}, -\frac {{\left ({\left (b c - a d\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 (b d^{2} x^{2} + b c d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {1}{2}\right )}}{4 \, b^{2} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*((b*c - a*d)*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*(b*d^2*x^2
+ b*c*d)*sqrt((b*x^2 + a)/(d*x^2 + c)))*e^(-1/2)/(b^2*d), -1/4*((b*c - a*d)*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*(b*d^2*x^2 + b*c*d)*sqrt((b*x^2
+ a)/(d*x^2 + c)))*e^(-1/2)/(b^2*d)]

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

[Out]

Timed out

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Giac [A]
time = 6.34, size = 126, normalized size = 1.19 \begin {gather*} \frac {{\left (\frac {2 \, \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c}}{b} - \frac {{\left (b c - a d\right )} \sqrt {b d} \log \left ({\left | -2 \, {\left (\sqrt {b d} x^{2} - \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c}\right )} b d - \sqrt {b d} b c - \sqrt {b d} a d \right |}\right )}{b^{2} d}\right )} e^{\left (-\frac {1}{2}\right )}}{4 \, \mathrm {sgn}\left (d x^{2} + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*sqrt(b*d*x^4 + b*c*x^2 + a*d*x^2 + a*c)/b - (b*c - a*d)*sqrt(b*d)*log(abs(-2*(sqrt(b*d)*x^2 - sqrt(b*d*
x^4 + b*c*x^2 + a*d*x^2 + a*c))*b*d - sqrt(b*d)*b*c - sqrt(b*d)*a*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}{\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/((e*(a + b*x^2))/(c + d*x^2))^(1/2),x)

[Out]

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

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