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

Optimal. Leaf size=112 \[ \frac {\sqrt {b} \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{\sqrt {d}}-\frac {\sqrt {a} \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {c}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1960, 481, 208} \[ \frac {\sqrt {b} \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{\sqrt {d}}-\frac {\sqrt {a} \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 481

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

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

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Mathematica [A]  time = 0.18, size = 173, normalized size = 1.54 \[ \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt {c} \sqrt {b c-a d} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )-\sqrt {a} \sqrt {d} \sqrt {c+d x^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )\right )}{\sqrt {c} \sqrt {d} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.70, size = 865, normalized size = 7.72 \[ \left [\frac {1}{4} \, \sqrt {\frac {b e}{d}} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \, {\left (2 \, b d^{3} x^{4} + b c^{2} d + a c d^{2} + {\left (3 \, b c d^{2} + a d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e}{d}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}\right ) + \frac {1}{4} \, \sqrt {\frac {a e}{c}} \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^{2} d + a c d^{2}\right )} x^{4} + 2 \, a c^{3} + {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2}\right )} \sqrt {\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{x^{4}}\right ), -\frac {1}{2} \, \sqrt {-\frac {b e}{d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-\frac {b e}{d}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (b^{2} e x^{2} + a b e\right )}}\right ) + \frac {1}{4} \, \sqrt {\frac {a e}{c}} \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^{2} d + a c d^{2}\right )} x^{4} + 2 \, a c^{3} + {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2}\right )} \sqrt {\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{x^{4}}\right ), \frac {1}{2} \, \sqrt {-\frac {a e}{c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {-\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (a b e x^{2} + a^{2} e\right )}}\right ) + \frac {1}{4} \, \sqrt {\frac {b e}{d}} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \, {\left (2 \, b d^{3} x^{4} + b c^{2} d + a c d^{2} + {\left (3 \, b c d^{2} + a d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e}{d}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}\right ), \frac {1}{2} \, \sqrt {-\frac {a e}{c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {-\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (a b e x^{2} + a^{2} e\right )}}\right ) - \frac {1}{2} \, \sqrt {-\frac {b e}{d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-\frac {b e}{d}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (b^{2} e x^{2} + a b e\right )}}\right )\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((e*(b*x^2+a)/(d*x^2+c))^(1/2)/x,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(x^2*d+c)]Error: Bad Argument Type

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maple [B]  time = 0.04, size = 179, normalized size = 1.60 \[ \frac {\sqrt {\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-\sqrt {b d}\, a \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+\sqrt {a c}\, b \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )\right )}{2 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, \sqrt {a c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.24, size = 149, normalized size = 1.33 \[ \frac {1}{2} \, {\left (\frac {a \log \left (\frac {c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {a c e}}{c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {a c e}}\right )}{\sqrt {a c e}} - \frac {b \log \left (\frac {d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {b d e}}{d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {b d e}}\right )}{\sqrt {b d e}}\right )} e \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*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))
)/sqrt(a*c*e) - b*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)) + s
qrt(b*d*e)))/sqrt(b*d*e))*e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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