3.4.9 \(\int \frac {x}{(\frac {e (a+b x^2)}{c+d x^2})^{3/2}} \, dx\) [309]

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

[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 = 0.12, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1981, 1979, 296, 331, 214} \begin {gather*} \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}}} \end {gather*}

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 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 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

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 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}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx &=((b c-a d) e) \text {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)) \text {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)) \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^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 [A]
time = 1.41, size = 138, normalized size = 0.95 \begin {gather*} \frac {\sqrt {b} \sqrt {c+d x^2} \left (-2 b c+3 a d+b d x^2\right )+3 \sqrt {d} (b c-a d) \sqrt {a+b x^2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 b^{5/2} e \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/((e*(a + b*x^2))/(c + d*x^2))^(3/2),x]

[Out]

(Sqrt[b]*Sqrt[c + d*x^2]*(-2*b*c + 3*a*d + b*d*x^2) + 3*Sqrt[d]*(b*c - a*d)*Sqrt[a + b*x^2]*ArcTanh[(Sqrt[d]*S
qrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])])/(2*b^(5/2)*e*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. \(431\) vs. \(2(122)=244\).
time = 0.10, size = 432, normalized size = 2.96

method	result	size
default	\(-\frac {\left (3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b \,d^{2} x^{2}-3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c d \,x^{2}-2 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b d \,x^{2}+3 d^{2} \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2}-3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a c b d -2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a d -4 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a d +4 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, b c \right ) \left (b \,x^{2}+a \right )}{4 b^{2} \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (d \,x^{2}+c \right ) \left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )^{\frac {3}{2}}}\)	\(432\)
risch	\(\frac {\left (b \,x^{2}+a \right ) d}{2 b^{2} e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}+\frac {\left (-\frac {3 d^{2} \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a}{4 b^{2} \sqrt {d e b}}+\frac {3 d \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) c}{4 b \sqrt {d e b}}+\frac {x^{2} a^{2} d^{3}}{b^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {2 x^{2} a c \,d^{2}}{b \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {x^{2} c^{2} d}{\left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {a^{2} c \,d^{2}}{b^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {2 a \,c^{2} d}{b \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {c^{3}}{\left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}\right ) \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(513\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 180, normalized size = 1.23 \begin {gather*} \frac {1}{4} \, {\left (\frac {2 \, {\left (2 \, b^{2} c - 2 \, a b d - \frac {3 \, {\left (b c d - a d^{2}\right )} {\left (b x^{2} + a\right )}}{d x^{2} + c}\right )}}{b^{2} d \left (\frac {b x^{2} + a}{d x^{2} + c}\right )^{\frac {3}{2}} - b^{3} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}} - \frac {3 \, {\left (b c d - a 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}}\right )} e^{\left (-\frac {3}{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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.52, size = 403, normalized size = 2.76 \begin {gather*} \left [-\frac {{\left (3 \, {\left (a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x^{2}\right )} \sqrt {\frac {d}{b}} \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 x^{2} + a}{d x^{2} + c}} \sqrt {\frac {d}{b}}\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 x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {3}{2}\right )}}{8 \, {\left (b^{3} x^{2} + a b^{2}\right )}}, -\frac {{\left (3 \, {\left (a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x^{2}\right )} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {-\frac {d}{b}}}{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 x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {3}{2}\right )}}{4 \, {\left (b^{3} x^{2} + a b^{2}\right )}}\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))^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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

[Out]

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

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