3.281 \(\int \frac {(\frac {e (a+b x^2)}{c+d x^2})^{3/2}}{x^5} \, dx\)

Optimal. Leaf size=256 \[ -\frac {3 e^{3/2} (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{8 \sqrt {a} c^{7/2}}-\frac {a e^3 (b c-a d)^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 c^3 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )^2}+\frac {e^2 (5 b c-9 a d) (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 c^3 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {d e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c^3} \]

[Out]

-3/8*(-5*a*d+b*c)*(-a*d+b*c)*e^(3/2)*arctanh(c^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/a^(1/2)/e^(1/2))/c^(7/2)/a^
(1/2)-d*(-a*d+b*c)*e*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/c^3-1/4*a*(-a*d+b*c)^2*e^3*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/c^
3/(a*e-c*e*(b*x^2+a)/(d*x^2+c))^2+1/8*(-9*a*d+5*b*c)*(-a*d+b*c)*e^2*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/c^3/(a*e-c*e
*(b*x^2+a)/(d*x^2+c))

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Rubi [A]  time = 0.22, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1960, 455, 1157, 388, 208} \[ -\frac {a e^3 (b c-a d)^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 c^3 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )^2}+\frac {e^2 (5 b c-9 a d) (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 c^3 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {3 e^{3/2} (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{8 \sqrt {a} c^{7/2}}-\frac {d e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*(b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/c^3) - (a*(b*c - a*d)^2*e^3*Sqrt[(e*(a + b*x^2))/(c + d*
x^2)])/(4*c^3*(a*e - (c*e*(a + b*x^2))/(c + d*x^2))^2) + ((5*b*c - 9*a*d)*(b*c - a*d)*e^2*Sqrt[(e*(a + b*x^2))
/(c + d*x^2)])/(8*c^3*(a*e - (c*e*(a + b*x^2))/(c + d*x^2))) - (3*(b*c - 5*a*d)*(b*c - a*d)*e^(3/2)*ArcTanh[(S
qrt[c]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[a]*Sqrt[e])])/(8*Sqrt[a]*c^(7/2))

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -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 {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^5} \, dx &=((b c-a d) e) \operatorname {Subst}\left (\int \frac {x^4 \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 {a (b c-a d)^2 e^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 c^3 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )^2}-\frac {((b c-a d) e) \operatorname {Subst}\left (\int \frac {-a (b c-a d) e^2-4 c (b c-a d) e x^2+4 c^2 d x^4}{\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^3}\\ &=-\frac {a (b c-a d)^2 e^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 c^3 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )^2}+\frac {(5 b c-9 a d) (b c-a d) e^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 c^3 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {(b c-a d) \operatorname {Subst}\left (\int \frac {-a (3 b c-7 a d) e^2+8 a c d e x^2}{-a e+c x^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{8 a c^3}\\ &=-\frac {d (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c^3}-\frac {a (b c-a d)^2 e^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 c^3 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )^2}+\frac {(5 b c-9 a d) (b c-a d) e^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 c^3 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )}+\frac {\left (3 (b c-5 a d) (b c-a d) e^2\right ) \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 c^3}\\ &=-\frac {d (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c^3}-\frac {a (b c-a d)^2 e^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 c^3 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )^2}+\frac {(5 b c-9 a d) (b c-a d) e^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 c^3 \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {3 (b c-5 a d) (b c-a d) e^{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 )}{8 \sqrt {a} c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 186, normalized size = 0.73 \[ -\frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (3 x^4 \sqrt {c+d x^2} \left (5 a^2 d^2-6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )+\sqrt {a} \sqrt {c} \sqrt {a+b x^2} \left (a \left (2 c^2-5 c d x^2-15 d^2 x^4\right )+b c x^2 \left (5 c+13 d x^2\right )\right )\right )}{8 \sqrt {a} c^{7/2} x^4 \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(Sqrt[a]*Sqrt[c]*Sqrt[a + b*x^2]*(b*c*x^2*(5*c + 13*d*x^2) + a*(2*c^
2 - 5*c*d*x^2 - 15*d^2*x^4)) + 3*(b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*x^4*Sqrt[c + d*x^2]*ArcTanh[(Sqrt[c]*Sqrt[a
+ b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])]))/(Sqrt[a]*c^(7/2)*x^4*Sqrt[a + b*x^2])

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fricas [A]  time = 3.45, size = 435, normalized size = 1.70 \[ \left [\frac {3 \, {\left (b^{2} c^{2} - 6 \, a b c d + 5 \, a^{2} d^{2}\right )} e x^{4} \sqrt {\frac {e}{a c}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e x^{4} + 8 \, a^{2} c^{2} e + 8 \, {\left (a b c^{2} + a^{2} c d\right )} e x^{2} - 4 \, {\left (2 \, a^{2} c^{3} + {\left (a b c^{2} d + a^{2} c d^{2}\right )} x^{4} + {\left (a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {e}{a c}}}{x^{4}}\right ) - 4 \, {\left ({\left (13 \, b c d - 15 \, a d^{2}\right )} e x^{4} + 2 \, a c^{2} e + 5 \, {\left (b c^{2} - a c d\right )} e x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{32 \, c^{3} x^{4}}, \frac {3 \, {\left (b^{2} c^{2} - 6 \, a b c d + 5 \, a^{2} d^{2}\right )} e x^{4} \sqrt {-\frac {e}{a c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {e}{a c}}}{2 \, {\left (b e x^{2} + a e\right )}}\right ) - 2 \, {\left ({\left (13 \, b c d - 15 \, a d^{2}\right )} e x^{4} + 2 \, a c^{2} e + 5 \, {\left (b c^{2} - a c d\right )} e x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{16 \, c^{3} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))^(3/2)/x^5,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)]Evaluation time: 0.56Unable to divide, perhaps due to rounding error%%%{%%%{2,[4,1,4]%%%},[2,7,0]%%
%}+%%%{%%%{-8,[3,2,4]%%%},[2,6,1]%%%}+%%%{%%%{12,[2,3,4]%%%},[2,5,2]%%%}+%%%{%%%{-8,[1,4,4]%%%},[2,4,3]%%%}+%%
%{%%%{2,[0,5,4]%%%},[2,3,4]%%%}+%%%{%%{[%%%{-4,[4,0,4]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,8,0]%%%}+%%%{%%{[
%%%{16,[3,1,4]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,7,1]%%%}+%%%{%%{[%%%{-24,[2,2,4]%%%},0]:[1,0,%%%{-1,[1,1,
1]%%%}]%%},[1,6,2]%%%}+%%%{%%{[%%%{16,[1,3,4]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,5,3]%%%}+%%%{%%{[%%%{-4,[0
,4,4]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,4,4]%%%}+%%%{%%%{2,[5,0,5]%%%},[0,9,0]%%%}+%%%{%%%{-8,[4,1,5]%%%},
[0,8,1]%%%}+%%%{%%%{12,[3,2,5]%%%},[0,7,2]%%%}+%%%{%%%{-8,[2,3,5]%%%},[0,6,3]%%%}+%%%{%%%{2,[1,4,5]%%%},[0,5,4
]%%%} / %%%{%%%{1,[0,2,0]%%%},[2,0,0]%%%}+%%%{%%{[%%%{-2,[0,1,0]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,1,0]%%%
}+%%%{%%%{1,[1,1,1]%%%},[0,2,0]%%%} Error: Bad Argument Value

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maple [B]  time = 0.07, size = 1042, normalized size = 4.07 \[ \frac {\left (-15 a^{3} c \,d^{3} x^{6} \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 )+18 a^{2} b \,c^{2} d^{2} x^{6} \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 )-3 a \,b^{2} c^{3} d \,x^{6} \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 )-18 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a b \,d^{3} x^{8}+6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b^{2} c \,d^{2} x^{8}-15 a^{3} c^{2} 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 )+18 a^{2} b \,c^{3} 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 )-3 a \,b^{2} c^{4} 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 )-18 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a^{2} d^{3} x^{6}-26 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a b c \,d^{2} x^{6}+12 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b^{2} c^{2} d \,x^{6}+16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, a^{2} c \,d^{2} x^{4}-18 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a^{2} c \,d^{2} x^{4}-16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, a b \,c^{2} d \,x^{4}-8 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a b \,c^{2} d \,x^{4}+6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b^{2} c^{3} 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}\, a \,d^{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}\, b c d \,x^{4}+14 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, a c d \,x^{2}-6 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, b \,c^{2} 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^{2}\right ) \left (d \,x^{2}+c \right ) \left (\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}\right )^{\frac {3}{2}}}{16 \sqrt {a c}\, \left (b \,x^{2}+a \right ) \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a \,c^{4} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.48, size = 303, normalized size = 1.18 \[ -\frac {1}{16} \, e {\left (\frac {2 \, {\left ({\left (5 \, b^{2} c^{3} - 14 \, a b c^{2} d + 9 \, a^{2} c d^{2}\right )} \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {3}{2}} e - {\left (3 \, a b^{2} c^{2} - 10 \, a^{2} b c d + 7 \, a^{3} d^{2}\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} e^{2}\right )}}{a^{2} c^{3} e^{2} - \frac {2 \, {\left (b x^{2} + a\right )} a c^{4} e^{2}}{d x^{2} + c} + \frac {{\left (b x^{2} + a\right )}^{2} c^{5} e^{2}}{{\left (d x^{2} + c\right )}^{2}}} - \frac {3 \, {\left (b^{2} c^{2} - 6 \, a b c d + 5 \, a^{2} d^{2}\right )} e \log \left (\frac {c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {a c e}}{c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {a c e}}\right )}{\sqrt {a c e} c^{3}} + \frac {16 \, {\left (b c d - a d^{2}\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}{c^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*e*(2*((5*b^2*c^3 - 14*a*b*c^2*d + 9*a^2*c*d^2)*((b*x^2 + a)*e/(d*x^2 + c))^(3/2)*e - (3*a*b^2*c^2 - 10*a
^2*b*c*d + 7*a^3*d^2)*sqrt((b*x^2 + a)*e/(d*x^2 + c))*e^2)/(a^2*c^3*e^2 - 2*(b*x^2 + a)*a*c^4*e^2/(d*x^2 + c)
+ (b*x^2 + a)^2*c^5*e^2/(d*x^2 + c)^2) - 3*(b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*e*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^3) + 16*(b*c*d - a*d^
2)*sqrt((b*x^2 + a)*e/(d*x^2 + c))/c^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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