\(\int \frac {1}{x (\frac {e (a+b x^2)}{c+d x^2})^{3/2}} \, dx\) [310]

Optimal result

Integrand size = 26, antiderivative size = 152 \[ \int \frac {1}{x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\frac {b c-a d}{a b e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}-\frac {c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{a^{3/2} e^{3/2}}+\frac {d^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{b^{3/2} e^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1981, 1980, 491, 536, 214} \[ \int \frac {1}{x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=-\frac {c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{a^{3/2} e^{3/2}}+\frac {d^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{b^{3/2} e^{3/2}}+\frac {b c-a d}{a b e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \]

[In]

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

[Out]

(b*c - a*d)/(a*b*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) - (c^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c + d*
x^2)])/(Sqrt[a]*Sqrt[e])])/(a^(3/2)*e^(3/2)) + (d^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(S
qrt[b]*Sqrt[e])])/(b^(3/2)*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 491

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*e*(m + 1))), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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

Mathematica [A] (verified)

Time = 2.09 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.24 \[ \int \frac {1}{x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\frac {-b^{3/2} c^{3/2} \sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )+\sqrt {a} \left (\sqrt {b} (b c-a d) \sqrt {c+d x^2}+a d^{3/2} \sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )\right )}{a^{3/2} b^{3/2} e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(400\) vs. \(2(122)=244\).

Time = 0.14 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.64

method	result	size
default	\(-\frac {\left (\sqrt {b d}\, \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) b^{2} c^{2} x^{2}-\ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a b \,d^{2} x^{2}+\sqrt {b d}\, \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a b \,c^{2}-\ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a^{2} d^{2}+2 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, a d -2 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, b c \right ) \left (b \,x^{2}+a \right )}{2 a b \sqrt {a c}\, \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (d \,x^{2}+c \right ) {\left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )}^{\frac {3}{2}}}\)	\(401\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (122) = 244\).

Time = 0.91 (sec) , antiderivative size = 1293, normalized size of antiderivative = 8.51 \[ \int \frac {1}{x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/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:Unable to divide, perhaps due to rounding error%%%{%%%{2,[1,2,2]%%%},[2,1,3,0]%%%}+%%%{%%%{-4,[2,1,2]%%%},[
2,1,2,1]%%%

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\int \frac {1}{x\,{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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