3.268 \(\int \frac {\sqrt {\frac {e (a+b x^2)}{c+d x^2}}}{x^5} \, dx\)

Optimal. Leaf size=208 \[ \frac {\sqrt {e} (3 a d+b c) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{8 a^{3/2} c^{5/2}}-\frac {(b c-a d)^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 c^2 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )^2}+\frac {(b c-5 a d) (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 a c^2 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )} \]

[Out]

1/8*(-a*d+b*c)*(3*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)/a^(3/2)/c^(5
/2)-1/4*(-a*d+b*c)^2*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/c^2/(a-c*(b*x^2+a)/(d*x^2+c))^2+1/8*(-5*a*d+b*c)*(-a*d+b*c)
*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/a/c^2/(a-c*(b*x^2+a)/(d*x^2+c))

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Rubi [A]  time = 0.17, antiderivative size = 208, 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 = {1960, 455, 385, 208} \[ \frac {\sqrt {e} (3 a d+b c) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{8 a^{3/2} c^{5/2}}-\frac {(b c-a d)^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 c^2 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )^2}+\frac {(b c-5 a d) (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 a c^2 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(e*(a + b*x^2))/(c + d*x^2)]/x^5,x]

[Out]

-((b*c - a*d)^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(4*c^2*(a - (c*(a + b*x^2))/(c + d*x^2))^2) + ((b*c - 5*a*d
)*(b*c - a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(8*a*c^2*(a - (c*(a + b*x^2))/(c + d*x^2))) + ((b*c - a*d)*(b
*c + 3*a*d)*Sqrt[e]*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[a]*Sqrt[e])])/(8*a^(3/2)*c^(5/2)
)

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

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^5} \, dx &=((b c-a d) e) \operatorname {Subst}\left (\int \frac {x^2 \left (b e-d x^2\right )}{\left (-a e+c x^2\right )^3} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=-\frac {(b c-a d)^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 c^2 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )^2}-\frac {((b c-a d) e) \operatorname {Subst}\left (\int \frac {-(b c-a d) e+4 c d 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 )}{4 c^2}\\ &=-\frac {(b c-a d)^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 c^2 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )^2}+\frac {(b c-5 a d) (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 a c^2 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {((b c-a d) (b c+3 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 )}{8 a c^2}\\ &=-\frac {(b c-a d)^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 c^2 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )^2}+\frac {(b c-5 a d) (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 a c^2 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}+\frac {(b c-a d) (b c+3 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 )}{8 a^{3/2} c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 174, normalized size = 0.84 \[ \frac {\sqrt {c+d x^2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (x^4 \left (-3 a^2 d^2+2 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )+\sqrt {a} \sqrt {c} \sqrt {a+b x^2} \sqrt {c+d x^2} \left (-2 a c+3 a d x^2-b c x^2\right )\right )}{8 a^{3/2} c^{5/2} x^4 \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(e*(a + b*x^2))/(c + d*x^2)]/x^5,x]

[Out]

(Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*Sqrt[c + d*x^2]*(Sqrt[a]*Sqrt[c]*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(-2*a*c -
b*c*x^2 + 3*a*d*x^2) + (b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*x^4*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c
+ d*x^2])]))/(8*a^(3/2)*c^(5/2)*x^4*Sqrt[a + b*x^2])

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fricas [A]  time = 1.19, size = 427, normalized size = 2.05 \[ \left [-\frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} x^{4} \sqrt {\frac {e}{a c}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e x^{4} + 8 \, a^{2} c^{2} e + 8 \, {\left (a b c^{2} + a^{2} c d\right )} e x^{2} - 4 \, {\left (2 \, a^{2} c^{3} + {\left (a b c^{2} d + a^{2} c d^{2}\right )} x^{4} + {\left (a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {e}{a c}}}{x^{4}}\right ) + 4 \, {\left ({\left (b c d - 3 \, a d^{2}\right )} x^{4} + 2 \, a c^{2} + {\left (b c^{2} - a c d\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{32 \, a c^{2} x^{4}}, -\frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} x^{4} \sqrt {-\frac {e}{a c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {e}{a c}}}{2 \, {\left (b e x^{2} + a e\right )}}\right ) + 2 \, {\left ({\left (b c d - 3 \, a d^{2}\right )} x^{4} + 2 \, a c^{2} + {\left (b c^{2} - a c d\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{16 \, a c^{2} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*(b*x^2+a)/(d*x^2+c))^(1/2)/x^5,x, algorithm="fricas")

[Out]

[-1/32*((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*x^4*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*((b*c*d - 3*a*d^2)*x^4 + 2*a*c^2 + (b*c^2 - a*c*d
)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(a*c^2*x^4), -1/16*((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*x^4*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*((b*c*d - 3*a*d^2)*x^4 + 2*a*c^2 + (b*c^2 - a*c*d)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(a*c^2*x^4)]

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giac [B]  time = 0.82, size = 598, normalized size = 2.88 \[ -\frac {1}{8} \, {\left (\frac {{\left (b^{2} c^{2} e + 2 \, a b c d e - 3 \, a^{2} d^{2} e\right )} \arctan \left (-\frac {\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}}{\sqrt {-a c e}}\right )}{\sqrt {-a c e} a c^{2}} - \frac {{\left (\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )} a b^{2} c^{3} e^{2} + 10 \, {\left (\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )} a^{2} b c^{2} d e^{2} + 5 \, {\left (\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )} a^{3} c d^{2} e^{2} + {\left (\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )}^{3} b^{2} c^{2} e + 2 \, {\left (\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )}^{3} a b c d e - 3 \, {\left (\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )}^{3} a^{2} d^{2} e + 8 \, \sqrt {b d} a^{3} c^{2} d e^{\frac {5}{2}} + 8 \, {\left (\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )}^{2} \sqrt {b d} a b c^{2} e^{\frac {3}{2}}}{{\left (a c e - {\left (\sqrt {b d} x^{2} e^{\frac {1}{2}} - \sqrt {b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )}^{2}\right )}^{2} a c^{2}}\right )} \mathrm {sgn}\left (d x^{2} + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*(b*x^2+a)/(d*x^2+c))^(1/2)/x^5,x, algorithm="giac")

[Out]

-1/8*((b^2*c^2*e + 2*a*b*c*d*e - 3*a^2*d^2*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)*a*c^2) - ((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*b^2*c^3*e^2 + 10*(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^2*b*c^2*d*e^2 + 5*(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^3*c*d^2*e^2
+ (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))^3*b^2*c^2*e + 2*(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))^3*a*b*c*d*e - 3*(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))^3*a^2*d^2*e + 8*sqrt(b*d)*a^3*c^2*d*e^(5/2) + 8*(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*sqrt(b*d)*a*b*c^2*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)^2*a*c^2))*sgn(d*x^2 + c)

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maple [B]  time = 0.07, size = 558, normalized size = 2.68 \[ \frac {\sqrt {\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-3 a^{3} c \,d^{2} x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+2 a^{2} b \,c^{2} d \,x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+a \,b^{2} c^{3} x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )-10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a b \,d^{2} x^{6}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b^{2} c d \,x^{6}-10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a^{2} d^{2} x^{4}-8 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a b c d \,x^{4}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b^{2} c^{2} x^{4}+10 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, a d \,x^{2}+2 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, b c \,x^{2}-4 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, a c \right )}{16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, a^{2} c^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x^2+a)/(d*x^2+c)*e)^(1/2)/x^5,x)

[Out]

1/16*((b*x^2+a)/(d*x^2+c)*e)^(1/2)*(d*x^2+c)*(-10*b*d^2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^6*a*(a*c)^(1/2)-
2*b^2*d*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^6*c*(a*c)^(1/2)-3*a^3*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)*d^2*c*x^4+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)*d*b*a^2*c^2*x^4+c^3*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^2*a*x^4-10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*d^2*a^2*x^4*(a*c)^(1/2)-8*(b*d*x^4+a*d*x^2+b*
c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^4*a*b*c*d-2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*b^2*c^2*x^4*(a*c)^(1/2)+10*(b*d
*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*d*a*x^2*(a*c)^(1/2)+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*b*c*x^2*(a*c)^(1/2)-
4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*a*c*(a*c)^(1/2))/((d*x^2+c)*(b*x^2+a))^(1/2)/c^3/a^2/x^4/(a*c)^(1/2)

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maxima [A]  time = 2.17, size = 265, normalized size = 1.27 \[ -\frac {1}{16} \, e {\left (\frac {2 \, {\left ({\left (b^{2} c^{3} - 6 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {3}{2}} + {\left (a b^{2} c^{2} + 2 \, a^{2} b c d - 3 \, a^{3} d^{2}\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} e\right )}}{a^{3} c^{2} e^{2} - \frac {2 \, {\left (b x^{2} + a\right )} a^{2} c^{3} e^{2}}{d x^{2} + c} + \frac {{\left (b x^{2} + a\right )}^{2} a c^{4} e^{2}}{{\left (d x^{2} + c\right )}^{2}}} + \frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \log \left (\frac {c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {a c e}}{c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {a c e}}\right )}{\sqrt {a c e} a c^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*(b*x^2+a)/(d*x^2+c))^(1/2)/x^5,x, algorithm="maxima")

[Out]

-1/16*e*(2*((b^2*c^3 - 6*a*b*c^2*d + 5*a^2*c*d^2)*((b*x^2 + a)*e/(d*x^2 + c))^(3/2) + (a*b^2*c^2 + 2*a^2*b*c*d
- 3*a^3*d^2)*sqrt((b*x^2 + a)*e/(d*x^2 + c))*e)/(a^3*c^2*e^2 - 2*(b*x^2 + a)*a^2*c^3*e^2/(d*x^2 + c) + (b*x^2
+ a)^2*a*c^4*e^2/(d*x^2 + c)^2) + (b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*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)*a*c^2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}}}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*(a + b*x^2))/(c + d*x^2))^(1/2)/x^5,x)

[Out]

int(((e*(a + b*x^2))/(c + d*x^2))^(1/2)/x^5, 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**5,x)

[Out]

Timed out

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