3.3.100 \(\int \frac {1}{x^3 \sqrt {\frac {e (a+b x^2)}{c+d x^2}}} \, dx\) [300]

Optimal. Leaf size=130 \[ \frac {(b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 a \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )}+\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{2 a^{3/2} \sqrt {c} \sqrt {e}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1981, 1980, 205, 214} \begin {gather*} \frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{2 a^{3/2} \sqrt {c} \sqrt {e}}+\frac {(b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 a \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(2*a*(a*e - (c*e*(a + b*x^2))/(c + d*x^2))) + ((b*c - a*d)*Arc
Tanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[a]*Sqrt[e])])/(2*a^(3/2)*Sqrt[c]*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 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 {1}{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 {1}{\left (-a e+c x^2\right )^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac {(b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 a \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {(b c-a d) \text {Subst}\left (\int \frac {1}{-a e+c x^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{2 a}\\ &=\frac {(b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 a \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )}+\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{2 a^{3/2} \sqrt {c} \sqrt {e}}\\ \end {align*}

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Mathematica [A]
time = 0.66, size = 143, normalized size = 1.10 \begin {gather*} \frac {-\sqrt {a} \sqrt {c} \left (a+b x^2\right ) \left (c+d x^2\right )+(b c-a d) x^2 \sqrt {a+b x^2} \sqrt {c+d x^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{3/2} \sqrt {c} x^2 \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[1/(x^3*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]),x]

[Out]

(-(Sqrt[a]*Sqrt[c]*(a + b*x^2)*(c + d*x^2)) + (b*c - a*d)*x^2*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*ArcTanh[(Sqrt[c]
*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(3/2)*Sqrt[c]*x^2*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. \(325\) vs. \(2(110)=220\).
time = 0.08, size = 326, normalized size = 2.51

method	result	size
risch	\(-\frac {b \,x^{2}+a}{2 a \,x^{2} \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}+\frac {\left (-\frac {\ln \left (\frac {2 a c e +\left (a d e +b c e \right ) x^{2}+2 \sqrt {a c e}\, \sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}}{x^{2}}\right ) d}{4 \sqrt {a c e}}+\frac {\ln \left (\frac {2 a c e +\left (a d e +b c e \right ) x^{2}+2 \sqrt {a c e}\, \sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}}{x^{2}}\right ) b c}{4 a \sqrt {a c e}}\right ) \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{\sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(226\)
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (-2 b d \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, x^{4} \sqrt {a c}+a^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) d c \,x^{2}-c^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) b a \,x^{2}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, d a \,x^{2} \sqrt {a c}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b c \,x^{2} \sqrt {a c}+2 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\right )}{4 \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a^{2} c \,x^{2} \sqrt {a c}}\)	\(326\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 138, normalized size = 1.06 \begin {gather*} \frac {1}{4} \, {\left (\frac {2 \, {\left (b c - a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{a^{2} - \frac {{\left (b x^{2} + a\right )} a c}{d x^{2} + c}} - \frac {{\left (b c - a d\right )} \log \left (\frac {c \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} - \sqrt {a c}}{c \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} + \sqrt {a c}}\right )}{\sqrt {a c} a}\right )} e^{\left (-\frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]
time = 0.44, size = 310, normalized size = 2.38 \begin {gather*} \left [-\frac {{\left (\sqrt {a c} {\left (b c - a d\right )} x^{2} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b c d + a d^{2}\right )} x^{4} + 2 \, a c^{2} + {\left (b c^{2} + 3 \, a c d\right )} x^{2}\right )} \sqrt {a c} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{x^{4}}\right ) + 4 \, {\left (a c d x^{2} + a c^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {1}{2}\right )}}{8 \, a^{2} c x^{2}}, -\frac {{\left (\sqrt {-a c} {\left (b c - a d\right )} x^{2} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {-a c} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{2 \, {\left (a b c x^{2} + a^{2} c\right )}}\right ) + 2 \, {\left (a c d x^{2} + a c^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {1}{2}\right )}}{4 \, a^{2} c x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]
time = 6.01, size = 216, normalized size = 1.66 \begin {gather*} -\frac {{\left (\frac {{\left (b c - a d\right )} \arctan \left (-\frac {\sqrt {b d} x^{2} - \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c}}{\sqrt {-a c}}\right )}{\sqrt {-a c} a} - \frac {{\left (\sqrt {b d} x^{2} - \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c}\right )} b c + {\left (\sqrt {b d} x^{2} - \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c}\right )} a d + 2 \, \sqrt {b d} a c}{{\left ({\left (\sqrt {b d} x^{2} - \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c}\right )}^{2} - a c\right )} a}\right )} e^{\left (-\frac {1}{2}\right )}}{2 \, \mathrm {sgn}\left (d x^{2} + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="giac")

[Out]

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

[Out]

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

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