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

Optimal. Leaf size=146 \[ \frac {3 \sqrt {d} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 b^{5/2} e^{3/2}}-\frac {3 (b c-a d)}{2 b^2 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {c+d x^2}{2 b e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 146, 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 = {1960, 290, 325, 208} \[ \frac {3 \sqrt {d} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 b^{5/2} e^{3/2}}-\frac {3 (b c-a d)}{2 b^2 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {c+d x^2}{2 b e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 290

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

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Mathematica [C]  time = 0.07, size = 86, normalized size = 0.59 \[ -\frac {\left (a+b x^2\right ) \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};\frac {d \left (b x^2+a\right )}{a d-b c}\right )}{b \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{3/2} \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b*x^2)*Hypergeometric2F1[-3/2, -1/2, 1/2, (d*(a + b*x^2))/(-(b*c) + a*d)])/(b*((e*(a + b*x^2))/(c + d*
x^2))^(3/2)*((b*(c + d*x^2))/(b*c - a*d))^(3/2)))

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fricas [A]  time = 0.86, size = 443, normalized size = 3.03 \[ \left [-\frac {3 \, {\left ({\left (b^{2} c - a b d\right )} e x^{2} + {\left (a b c - a^{2} d\right )} e\right )} \sqrt {\frac {d}{b e}} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} - 4 \, {\left (2 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + a b c d + {\left (3 \, b^{2} c d + a b d^{2}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {d}{b e}}\right ) - 4 \, {\left (b d^{2} x^{4} - 2 \, b c^{2} + 3 \, a c d - {\left (b c d - 3 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{8 \, {\left (b^{3} e^{2} x^{2} + a b^{2} e^{2}\right )}}, -\frac {3 \, {\left ({\left (b^{2} c - a b d\right )} e x^{2} + {\left (a b c - a^{2} d\right )} e\right )} \sqrt {-\frac {d}{b e}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {d}{b e}}}{2 \, {\left (b d x^{2} + a d\right )}}\right ) - 2 \, {\left (b d^{2} x^{4} - 2 \, b c^{2} + 3 \, a c d - {\left (b c d - 3 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, {\left (b^{3} e^{2} x^{2} + a b^{2} e^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep*d+c)]Unable to divide, perhaps due to rounding error%%%{%%%{2,[1,2,2]%%%},[2,1,2,0]%%%}+%%%{%%%{-4,
[2,1,2]%%%},[2,1,1,1]%%%}+%%%{%%%{2,[3,0,2]%%%},[2,1,0,2]%%%}+%%%{%%{[%%%{-4,[0,2,2]%%%},0]:[1,0,%%%{-1,[1,1,1
]%%%}]%%},[1,1,3,0]%%%}+%%%{%%{[%%%{8,[1,1,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,1,2,1]%%%}+%%%{%%{[%%%{-4,
[2,0,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,1,1,2]%%%}+%%%{%%%{2,[0,3,3]%%%},[0,1,4,0]%%%}+%%%{%%%{-4,[1,2,3
]%%%},[0,1,3,1]%%%}+%%%{%%%{2,[2,1,3]%%%},[0,1,2,2]%%%} / %%%{%%%{1,[2,0,2]%%%},[2,0,0,0]%%%}+%%%{%%{[%%%{-2,[
1,0,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,0,1,0]%%%}+%%%{%%%{1,[1,1,3]%%%},[0,0,2,0]%%%} Error: Bad Argumen
t Value

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maple [B]  time = 0.05, size = 432, normalized size = 2.96 \[ \frac {\left (-3 a b \,d^{2} x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 b^{2} c d \,x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-3 a^{2} d^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 a b c d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a d +4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, a d -4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, b c \right ) \left (b \,x^{2}+a \right )}{4 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (d \,x^{2}+c \right ) \left (\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}\right )^{\frac {3}{2}} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.74, size = 199, normalized size = 1.36 \[ \frac {1}{4} \, e {\left (\frac {2 \, {\left (2 \, {\left (b^{2} c - a b d\right )} e - \frac {3 \, {\left (b c d - a d^{2}\right )} {\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )}}{b^{2} d \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {3}{2}} e^{2} - b^{3} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} e^{3}} - \frac {3 \, {\left (b c d - a d^{2}\right )} \log \left (\frac {d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {b d e}}{d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {b d e}}\right )}{\sqrt {b d e} b^{2} e^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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