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

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

[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 = 0.08, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1960, 199, 208} \[ \frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{2 a^{3/2} \sqrt {c} \sqrt {e}}+\frac {(b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 a \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )} \]

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 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 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 {1}{x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx &=((b c-a d) e) \operatorname {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) \operatorname {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 = 0.12, size = 133, normalized size = 1.02 \[ \frac {\sqrt {a+b x^2} \left (-\frac {(a d-b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} \sqrt {c}}-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{a x^2}\right )}{2 \sqrt {c+d x^2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 333, normalized size = 2.56 \[ \left [-\frac {\sqrt {a c e} {\left (b c - a d\right )} x^{2} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e x^{4} + 8 \, a^{2} c^{2} e + 8 \, {\left (a b c^{2} + a^{2} c d\right )} e x^{2} - 4 \, {\left ({\left (b c d + a d^{2}\right )} x^{4} + 2 \, a c^{2} + {\left (b c^{2} + 3 \, a c d\right )} x^{2}\right )} \sqrt {a c e} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{x^{4}}\right ) + 4 \, {\left (a c d x^{2} + a c^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{8 \, a^{2} c e x^{2}}, -\frac {\sqrt {-a c e} {\left (b c - a d\right )} x^{2} \arctan \left (\frac {\sqrt {-a c e} {\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (a b c e x^{2} + a^{2} c e\right )}}\right ) + 2 \, {\left (a c d x^{2} + a c^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, a^{2} c e x^{2}}\right ] \]

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*e)*(b*c - a*d)*x^2*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*e*x^4 + 8*a^2*c^2*e + 8*(a*b*c^2 + a^2
*c*d)*e*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*e)*sqrt((b*e*x^2 + a*e)/(d*x^
2 + c)))/x^4) + 4*(a*c*d*x^2 + a*c^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(a^2*c*e*x^2), -1/4*(sqrt(-a*c*e)*(b*
c - a*d)*x^2*arctan(1/2*sqrt(-a*c*e)*((b*c + a*d)*x^2 + 2*a*c)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/(a*b*c*e*x^2
+ a^2*c*e)) + 2*(a*c*d*x^2 + a*c^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(a^2*c*e*x^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(1/x^3/(e*(b*x^2+a)/(d*x^2+c))^(1/2),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%%%{%%%{1,[0,1,0]%%%},[6,0,0]%%%}+%%%{%%{[-2,0]
:[1,0,%%%{-1,[1,1,1]%%%}]%%},[5,1,0]%%%}+%%%{%%%{1,[1,0,1]%%%},[4,2,0]%%%}+%%%{%%%{-2,[0,1,1]%%%},[4,1,1]%%%}+
%%%{%%{[%%%{4,[0,0,1]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[3,2,1]%%%}+%%%{%%%{-2,[1,0,2]%%%},[2,3,1]%%%}+%%%{%%
%{1,[0,1,2]%%%},[2,2,2]%%%}+%%%{%%{[%%%{-2,[0,0,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,3,2]%%%}+%%%{%%%{1,[1
,0,3]%%%},[0,4,2]%%%} / %%%{%%%{1,[0,2,0]%%%},[6,0,0]%%%}+%%%{%%{[%%%{-2,[0,1,0]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%
}]%%},[5,1,0]%%%}+%%%{%%%{1,[1,1,1]%%%},[4,2,0]%%%}+%%%{%%%{-2,[0,2,1]%%%},[4,1,1]%%%}+%%%{%%{[%%%{4,[0,1,1]%%
%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[3,2,1]%%%}+%%%{%%%{-2,[1,1,2]%%%},[2,3,1]%%%}+%%%{%%%{1,[0,2,2]%%%},[2,2,2]
%%%}+%%%{%%{[%%%{-2,[0,1,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,3,2]%%%}+%%%{%%%{1,[1,1,3]%%%},[0,4,2]%%%} E
rror: Bad Argument Value

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maple [B]  time = 0.05, size = 326, normalized size = 2.51 \[ -\frac {\left (b \,x^{2}+a \right ) \left (a^{2} c d \,x^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )-a b \,c^{2} x^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b d \,x^{4}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a d \,x^{2}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b c \,x^{2}+2 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\right )}{4 \sqrt {\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, a^{2} c \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*c*d*x^2*ln((a*d*x^2+b*c*x^2+2*a
*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2))/x^2)-a*b*c^2*x^2*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)
*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2))/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)*(a*c)^(1/2)*b*c*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2))/((b*x^2+a)
/(d*x^2+c)*e)^(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 = 1.52, size = 153, normalized size = 1.18 \[ \frac {1}{4} \, e {\left (\frac {2 \, {\left (b c - a d\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}{a^{2} e^{2} - \frac {{\left (b x^{2} + a\right )} a c e^{2}}{d x^{2} + c}} - \frac {{\left (b c - a d\right )} \log \left (\frac {c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {a c e}}{c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {a c e}}\right )}{\sqrt {a c e} a e}\right )} \]

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*e*(2*(b*c - a*d)*sqrt((b*x^2 + a)*e/(d*x^2 + c))/(a^2*e^2 - (b*x^2 + a)*a*c*e^2/(d*x^2 + c)) - (b*c - a*d)
*log((c*sqrt((b*x^2 + a)*e/(d*x^2 + c)) - sqrt(a*c*e))/(c*sqrt((b*x^2 + a)*e/(d*x^2 + c)) + sqrt(a*c*e)))/(sqr
t(a*c*e)*a*e))

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

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

[Out]

Timed out

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