3.4.12 \(\int \frac {1}{x^5 (\frac {e (a+b x^2)}{c+d x^2})^{3/2}} \, dx\) [312]

Optimal. Leaf size=255 \[ \frac {b (b c-a d)}{a^3 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}-\frac {(b c-a d)^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 a^2 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )^2}-\frac {(7 b c-3 a d) (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 a^3 \left (a e^2-\frac {c e^2 \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {3 (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{8 a^{7/2} \sqrt {c} e^{3/2}} \]

[Out]

-3/8*(-a*d+b*c)*(-a*d+5*b*c)*arctanh(c^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/a^(1/2)/e^(1/2))/a^(7/2)/e^(3/2)/c^
(1/2)+b*(-a*d+b*c)/a^3/e/(e*(b*x^2+a)/(d*x^2+c))^(1/2)-1/4*(-a*d+b*c)^2*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/a^2/(a*e
-c*e*(b*x^2+a)/(d*x^2+c))^2-1/8*(-3*a*d+7*b*c)*(-a*d+b*c)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/a^3/(a*e^2-c*e^2*(b*x^
2+a)/(d*x^2+c))

________________________________________________________________________________________

Rubi [A]
time = 0.25, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1981, 1980, 467, 464, 214} \begin {gather*} -\frac {3 (5 b c-a d) (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^{7/2} \sqrt {c} e^{3/2}}-\frac {(7 b c-3 a d) (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 a^3 \left (a e^2-\frac {c e^2 \left (a+b x^2\right )}{c+d x^2}\right )}+\frac {b (b c-a d)}{a^3 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}-\frac {(b c-a d)^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 a^2 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*((e*(a + b*x^2))/(c + d*x^2))^(3/2)),x]

[Out]

(b*(b*c - a*d))/(a^3*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) - ((b*c - a*d)^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/
(4*a^2*(a*e - (c*e*(a + b*x^2))/(c + d*x^2))^2) - ((7*b*c - 3*a*d)*(b*c - a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2
)])/(8*a^3*(a*e^2 - (c*e^2*(a + b*x^2))/(c + d*x^2))) - (3*(b*c - a*d)*(5*b*c - a*d)*ArcTanh[(Sqrt[c]*Sqrt[(e*
(a + b*x^2))/(c + d*x^2)])/(Sqrt[a]*Sqrt[e])])/(8*a^(7/2)*Sqrt[c]*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 464

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

Rule 467

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[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

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*} \int \frac {1}{x^5 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx &=((b c-a d) e) \text {Subst}\left (\int \frac {b e-d x^2}{x^2 \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 a^2 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )^2}-\frac {1}{4} ((b c-a d) e) \text {Subst}\left (\int \frac {\frac {4 b}{a}+\frac {3 (b c-a d) x^2}{a^2 e}}{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)^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 a^2 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )^2}-\frac {(7 b c-3 a d) (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 a^3 \left (a e^2-\frac {c e^2 \left (a+b x^2\right )}{c+d x^2}\right )}+\frac {1}{8} ((b c-a d) e) \text {Subst}\left (\int \frac {\frac {8 b}{a^2 e}+\frac {(7 b c-3 a d) x^2}{a^3 e^2}}{x^2 \left (-a e+c x^2\right )} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac {b (b c-a d)}{a^3 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}-\frac {(b c-a d)^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 a^2 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )^2}-\frac {(7 b c-3 a d) (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 a^3 \left (a e^2-\frac {c e^2 \left (a+b x^2\right )}{c+d x^2}\right )}+\frac {(3 (b c-a d) (5 b c-a d)) \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 )}{8 a^3 e}\\ &=\frac {b (b c-a d)}{a^3 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}-\frac {(b c-a d)^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 a^2 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )^2}-\frac {(7 b c-3 a d) (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 a^3 \left (a e^2-\frac {c e^2 \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {3 (b c-a d) (5 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^{7/2} \sqrt {c} e^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 4.43, size = 189, normalized size = 0.74 \begin {gather*} \frac {\sqrt {a} \sqrt {c} \sqrt {c+d x^2} \left (15 b^2 c x^4+a b x^2 \left (5 c-13 d x^2\right )-a^2 \left (2 c+5 d x^2\right )\right )-3 \left (5 b^2 c^2-6 a b c d+a^2 d^2\right ) x^4 \sqrt {a+b x^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 a^{7/2} \sqrt {c} e x^4 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*((e*(a + b*x^2))/(c + d*x^2))^(3/2)),x]

[Out]

(Sqrt[a]*Sqrt[c]*Sqrt[c + d*x^2]*(15*b^2*c*x^4 + a*b*x^2*(5*c - 13*d*x^2) - a^2*(2*c + 5*d*x^2)) - 3*(5*b^2*c^
2 - 6*a*b*c*d + a^2*d^2)*x^4*Sqrt[a + b*x^2]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(8*
a^(7/2)*Sqrt[c]*e*x^4*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*Sqrt[c + d*x^2])

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1041\) vs. \(2(229)=458\).
time = 0.14, size = 1042, normalized size = 4.09

method	result	size
risch	\(-\frac {\left (b \,x^{2}+a \right ) \left (5 a d \,x^{2}-7 b c \,x^{2}+2 a c \right )}{8 a^{3} x^{4} e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}+\frac {\left (-\frac {3 \ln \left (\frac {2 a c e +\left (a d e +b c e \right ) x^{2}+2 \sqrt {a c e}\, \sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}}{x^{2}}\right ) d^{2}}{16 a \sqrt {a c e}}+\frac {9 \ln \left (\frac {2 a c e +\left (a d e +b c e \right ) x^{2}+2 \sqrt {a c e}\, \sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}}{x^{2}}\right ) b c d}{8 a^{2} \sqrt {a c e}}-\frac {15 \ln \left (\frac {2 a c e +\left (a d e +b c e \right ) x^{2}+2 \sqrt {a c e}\, \sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}}{x^{2}}\right ) b^{2} c^{2}}{16 a^{3} \sqrt {a c e}}-\frac {b \,x^{2} d^{3}}{a \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {2 b^{2} x^{2} c \,d^{2}}{a^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {b^{3} x^{2} c^{2} d}{a^{3} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {b c \,d^{2}}{a \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {2 b^{2} c^{2} d}{a^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {b^{3} c^{3}}{a^{3} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}\right ) \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(636\)
default	\(-\frac {\left (-6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a \,b^{2} d^{2} x^{8}+18 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b^{3} c d \,x^{8}+3 \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^{3} b c \,d^{2} x^{6}-18 \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^{2} b^{2} c^{2} d \,x^{6}+15 \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^{3} c^{3} x^{6}-12 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a^{2} b \,d^{2} x^{6}+26 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a \,b^{2} c d \,x^{6}+18 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b^{3} c^{2} x^{6}+3 \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^{4} c \,d^{2} x^{4}-18 \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^{3} b \,c^{2} d \,x^{4}+15 \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^{2} b^{2} c^{3} x^{4}+6 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, a b d \,x^{4}-18 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, b^{2} c \,x^{4}-6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a^{3} d^{2} x^{4}+8 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a^{2} b c d \,x^{4}+18 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a \,b^{2} c^{2} x^{4}+16 \sqrt {a c}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a^{2} b c d \,x^{4}-16 \sqrt {a c}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a \,b^{2} c^{2} x^{4}+6 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, a^{2} d \,x^{2}-14 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, a 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^{2} c \right ) \left (b \,x^{2}+a \right )}{16 c \sqrt {a c}\, x^{4} a^{4} \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}}}\)	\(1042\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/16*(-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*a*b^2*d^2*x^8+18*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)
*(a*c)^(1/2)*b^3*c*d*x^8+3*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
^3*b*c*d^2*x^6-18*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^2*b^2*c^
2*d*x^6+15*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^3*c^3*x^6-12*
(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*a^2*b*d^2*x^6+26*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/
2)*a*b^2*c*d*x^6+18*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*b^3*c^2*x^6+3*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^4*c*d^2*x^4-18*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^3*b*c^2*d*x^4+15*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^2*b^2*c^3*x^4+6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*a*b*d*x^4-
18*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*b^2*c*x^4-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)
*a^3*d^2*x^4+8*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*a^2*b*c*d*x^4+18*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^
(1/2)*(a*c)^(1/2)*a*b^2*c^2*x^4+16*(a*c)^(1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*a^2*b*c*d*x^4-16*(a*c)^(1/2)*((d*x^
2+c)*(b*x^2+a))^(1/2)*a*b^2*c^2*x^4+6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*a^2*d*x^2-14*(b*d*x^4+a*
d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*a*b*c*x^2+4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*a^2*c)/c*(b*x
^2+a)/(a*c)^(1/2)/x^4/a^4/((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)

________________________________________________________________________________________

Maxima [A]
time = 0.54, size = 281, normalized size = 1.10 \begin {gather*} \frac {1}{16} \, {\left (\frac {2 \, {\left (8 \, a^{2} b^{2} c - 8 \, a^{3} b d + \frac {3 \, {\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2}\right )} {\left (b x^{2} + a\right )}^{2}}{{\left (d x^{2} + c\right )}^{2}} - \frac {5 \, {\left (5 \, a b^{2} c^{2} - 6 \, a^{2} b c d + a^{3} d^{2}\right )} {\left (b x^{2} + a\right )}}{d x^{2} + c}\right )}}{a^{3} c^{2} \left (\frac {b x^{2} + a}{d x^{2} + c}\right )^{\frac {5}{2}} - 2 \, a^{4} c \left (\frac {b x^{2} + a}{d x^{2} + c}\right )^{\frac {3}{2}} + a^{5} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}} + \frac {3 \, {\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} \log \left (\frac {c \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} - \sqrt {a c}}{c \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} + \sqrt {a c}}\right )}{\sqrt {a c} a^{3}}\right )} e^{\left (-\frac {3}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="maxima")

[Out]

1/16*(2*(8*a^2*b^2*c - 8*a^3*b*d + 3*(5*b^2*c^3 - 6*a*b*c^2*d + a^2*c*d^2)*(b*x^2 + a)^2/(d*x^2 + c)^2 - 5*(5*
a*b^2*c^2 - 6*a^2*b*c*d + a^3*d^2)*(b*x^2 + a)/(d*x^2 + c))/(a^3*c^2*((b*x^2 + a)/(d*x^2 + c))^(5/2) - 2*a^4*c
*((b*x^2 + a)/(d*x^2 + c))^(3/2) + a^5*sqrt((b*x^2 + a)/(d*x^2 + c))) + 3*(5*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*lo
g((c*sqrt((b*x^2 + a)/(d*x^2 + c)) - sqrt(a*c))/(c*sqrt((b*x^2 + a)/(d*x^2 + c)) + sqrt(a*c)))/(sqrt(a*c)*a^3)
)*e^(-3/2)

________________________________________________________________________________________

Fricas [A]
time = 1.86, size = 584, normalized size = 2.29 \begin {gather*} \left [\frac {{\left (3 \, {\left ({\left (5 \, b^{3} c^{2} - 6 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{6} + {\left (5 \, a b^{2} c^{2} - 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{4}\right )} \sqrt {a c} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} 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} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{x^{4}}\right ) + 4 \, {\left ({\left (15 \, a b^{2} c^{2} d - 13 \, a^{2} b c d^{2}\right )} x^{6} - 2 \, a^{3} c^{3} + {\left (15 \, a b^{2} c^{3} - 8 \, a^{2} b c^{2} d - 5 \, a^{3} c d^{2}\right )} x^{4} + {\left (5 \, a^{2} b c^{3} - 7 \, a^{3} c^{2} d\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {3}{2}\right )}}{32 \, {\left (a^{4} b c x^{6} + a^{5} c x^{4}\right )}}, \frac {{\left (3 \, {\left ({\left (5 \, b^{3} c^{2} - 6 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{6} + {\left (5 \, a b^{2} c^{2} - 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{4}\right )} \sqrt {-a c} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {-a c} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{2 \, {\left (a b c x^{2} + a^{2} c\right )}}\right ) + 2 \, {\left ({\left (15 \, a b^{2} c^{2} d - 13 \, a^{2} b c d^{2}\right )} x^{6} - 2 \, a^{3} c^{3} + {\left (15 \, a b^{2} c^{3} - 8 \, a^{2} b c^{2} d - 5 \, a^{3} c d^{2}\right )} x^{4} + {\left (5 \, a^{2} b c^{3} - 7 \, a^{3} c^{2} d\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {3}{2}\right )}}{16 \, {\left (a^{4} b c x^{6} + a^{5} c x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="fricas")

[Out]

[1/32*(3*((5*b^3*c^2 - 6*a*b^2*c*d + a^2*b*d^2)*x^6 + (5*a*b^2*c^2 - 6*a^2*b*c*d + a^3*d^2)*x^4)*sqrt(a*c)*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*((b*c*d + a*d^2)*x^4 + 2*a*c
^2 + (b*c^2 + 3*a*c*d)*x^2)*sqrt(a*c)*sqrt((b*x^2 + a)/(d*x^2 + c)))/x^4) + 4*((15*a*b^2*c^2*d - 13*a^2*b*c*d^
2)*x^6 - 2*a^3*c^3 + (15*a*b^2*c^3 - 8*a^2*b*c^2*d - 5*a^3*c*d^2)*x^4 + (5*a^2*b*c^3 - 7*a^3*c^2*d)*x^2)*sqrt(
(b*x^2 + a)/(d*x^2 + c)))*e^(-3/2)/(a^4*b*c*x^6 + a^5*c*x^4), 1/16*(3*((5*b^3*c^2 - 6*a*b^2*c*d + a^2*b*d^2)*x
^6 + (5*a*b^2*c^2 - 6*a^2*b*c*d + a^3*d^2)*x^4)*sqrt(-a*c)*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(-a*c)*sqr
t((b*x^2 + a)/(d*x^2 + c))/(a*b*c*x^2 + a^2*c)) + 2*((15*a*b^2*c^2*d - 13*a^2*b*c*d^2)*x^6 - 2*a^3*c^3 + (15*a
*b^2*c^3 - 8*a^2*b*c^2*d - 5*a^3*c*d^2)*x^4 + (5*a^2*b*c^3 - 7*a^3*c^2*d)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c)))*
e^(-3/2)/(a^4*b*c*x^6 + a^5*c*x^4)]

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(e*(b*x**2+a)/(d*x**2+c))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(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:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^5\,{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*((e*(a + b*x^2))/(c + d*x^2))^(3/2)),x)

[Out]

int(1/(x^5*((e*(a + b*x^2))/(c + d*x^2))^(3/2)), x)

________________________________________________________________________________________