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

Optimal. Leaf size=127 \[ \frac {(b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 c \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {\sqrt {e} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{2 \sqrt {a} c^{3/2}} \]

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

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Rubi [A]  time = 0.09, antiderivative size = 127, 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, 288, 208} \[ \frac {(b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 c \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {\sqrt {e} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{2 \sqrt {a} c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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

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Mathematica [A]  time = 0.10, size = 133, normalized size = 1.05 \[ \frac {\sqrt {c+d x^2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (-\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} c^{3/2}}-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{c x^2}\right )}{2 \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.64, size = 333, normalized size = 2.62 \[ \left [-\frac {{\left (b c - a d\right )} x^{2} \sqrt {\frac {e}{a c}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e x^{4} + 8 \, a^{2} c^{2} e + 8 \, {\left (a b c^{2} + a^{2} c d\right )} e x^{2} + 4 \, {\left (2 \, a^{2} c^{3} + {\left (a b c^{2} d + a^{2} c d^{2}\right )} x^{4} + {\left (a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {e}{a c}}}{x^{4}}\right ) + 4 \, {\left (d x^{2} + c\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{8 \, c x^{2}}, \frac {{\left (b c - a d\right )} x^{2} \sqrt {-\frac {e}{a c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {e}{a c}}}{2 \, {\left (b e x^{2} + a e\right )}}\right ) - 2 \, {\left (d x^{2} + c\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, c x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.68, size = 269, normalized size = 2.12 \[ \frac {1}{2} \, {\left (\frac {{\left (b c e - a d e\right )} \arctan \left (-\frac {\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}}{\sqrt {-a c e}}\right )}{\sqrt {-a c e} c} - \frac {{\left (\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )} b c e + {\left (\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )} a d e + 2 \, \sqrt {b d} a c e^{\frac {3}{2}}}{{\left (a c e - {\left (\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )}^{2}\right )} c}\right )} \mathrm {sgn}\left (d x^{2} + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 326, normalized size = 2.57 \[ -\frac {\sqrt {\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-a^{2} c d \,x^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+a b \,c^{2} x^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b d \,x^{4}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a d \,x^{2}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b c \,x^{2}+2 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\right )}{4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, a \,c^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.23, size = 145, normalized size = 1.14 \[ \frac {1}{4} \, e {\left (\frac {2 \, {\left (b c - a d\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}{a c e - \frac {{\left (b x^{2} + a\right )} c^{2} e}{d x^{2} + c}} + \frac {{\left (b c - a d\right )} \log \left (\frac {c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {a c e}}{c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {a c e}}\right )}{\sqrt {a c e} c}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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