3.311 \(\int \frac {1}{x^3 (\frac {e (a+b x^2)}{c+d x^2})^{3/2}} \, dx\)

Optimal. Leaf size=170 \[ \frac {3 \sqrt {c} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{2 a^{5/2} e^{3/2}}-\frac {3 (b c-a d)}{2 a^2 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {b c-a d}{2 a \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 170, 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, 290, 325, 208} \[ \frac {3 \sqrt {c} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{2 a^{5/2} e^{3/2}}-\frac {3 (b c-a d)}{2 a^2 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {b c-a d}{2 a \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b*c - a*d))/(2*a^2*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) + (b*c - a*d)/(2*a*Sqrt[(e*(a + b*x^2))/(c + d*x^
2)]*(a*e - (c*e*(a + b*x^2))/(c + d*x^2))) + (3*Sqrt[c]*(b*c - a*d)*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c +
d*x^2)])/(Sqrt[a]*Sqrt[e])])/(2*a^(5/2)*e^(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 290

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 {1}{x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx &=((b c-a d) e) \operatorname {Subst}\left (\int \frac {1}{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 a \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {(3 (b c-a d)) \operatorname {Subst}\left (\int \frac {1}{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 )}{2 a}\\ &=-\frac {3 (b c-a d)}{2 a^2 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {b c-a d}{2 a \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {(3 c (b c-a d)) \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 a^2 e}\\ &=-\frac {3 (b c-a d)}{2 a^2 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {b c-a d}{2 a \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (a e-\frac {c e \left (a+b x^2\right )}{c+d x^2}\right )}+\frac {3 \sqrt {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 )}{2 a^{5/2} e^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 148, normalized size = 0.87 \[ \frac {3 \sqrt {c} x^2 \sqrt {a+b x^2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )-\sqrt {a} \sqrt {c+d x^2} \left (a \left (c-2 d x^2\right )+3 b c x^2\right )}{2 a^{5/2} e x^2 \sqrt {c+d x^2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 2.18, size = 469, normalized size = 2.76 \[ \left [-\frac {3 \, {\left ({\left (b^{2} c - a b d\right )} e x^{4} + {\left (a b c - a^{2} d\right )} e x^{2}\right )} \sqrt {\frac {c}{a e}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (a b c d + a^{2} d^{2}\right )} x^{4} + 2 \, a^{2} c^{2} + {\left (a b c^{2} + 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {c}{a e}}}{x^{4}}\right ) + 4 \, {\left ({\left (3 \, b c d - 2 \, a d^{2}\right )} x^{4} + a c^{2} + {\left (3 \, b c^{2} - a c d\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{8 \, {\left (a^{2} b e^{2} x^{4} + a^{3} e^{2} x^{2}\right )}}, -\frac {3 \, {\left ({\left (b^{2} c - a b d\right )} e x^{4} + {\left (a b c - a^{2} d\right )} e x^{2}\right )} \sqrt {-\frac {c}{a e}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {c}{a e}}}{2 \, {\left (b c x^{2} + a c\right )}}\right ) + 2 \, {\left ({\left (3 \, b c d - 2 \, a d^{2}\right )} x^{4} + a c^{2} + {\left (3 \, b c^{2} - a c d\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, {\left (a^{2} b e^{2} x^{4} + a^{3} e^{2} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(1/x^3/(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,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep*d+c)]Unable to divide, perhaps due to rounding error%%%{%%%{2,[1,0,0]%%%},[6,1,0,0]%%%}+%%%{%%{[-4,
0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[5,1,1,0]%%%}+%%%{%%%{2,[0,1,1]%%%},[4,1,2,0]%%%}+%%%{%%%{-4,[1,0,1]%%%},[4,1,1
,1]%%%}+%%%{%%{[%%%{8,[0,0,1]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[3,1,2,1]%%%}+%%%{%%%{-4,[0,1,2]%%%},[2,1,3,1
]%%%}+%%%{%%%{2,[1,0,2]%%%},[2,1,2,2]%%%}+%%%{%%{[%%%{-4,[0,0,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,1,3,2]%
%%}+%%%{%%%{2,[0,1,3]%%%},[0,1,4,2]%%%} / %%%{%%%{1,[2,0,0]%%%},[6,0,0,0]%%%}+%%%{%%{[%%%{-2,[1,0,0]%%%},0]:[1
,0,%%%{-1,[1,1,1]%%%}]%%},[5,0,1,0]%%%}+%%%{%%%{1,[1,1,1]%%%},[4,0,2,0]%%%}+%%%{%%%{-2,[2,0,1]%%%},[4,0,1,1]%%
%}+%%%{%%{[%%%{4,[1,0,1]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[3,0,2,1]%%%}+%%%{%%%{-2,[1,1,2]%%%},[2,0,3,1]%%%}
+%%%{%%%{1,[2,0,2]%%%},[2,0,2,2]%%%}+%%%{%%{[%%%{-2,[1,0,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,0,3,2]%%%}+%
%%{%%%{1,[1,1,3]%%%},[0,0,4,2]%%%} Error: Bad Argument Value

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maple [B]  time = 0.07, size = 641, normalized size = 3.77 \[ \frac {\left (-3 a^{2} b c d \,x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+3 a \,b^{2} c^{2} x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b^{2} d \,x^{6}-3 a^{3} c d \,x^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+3 a^{2} b \,c^{2} x^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+4 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a b d \,x^{4}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b^{2} c \,x^{4}+4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, a^{2} d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a^{2} d \,x^{2}-4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, a b c \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a b c \,x^{2}-2 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, b \,x^{2}-2 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, a \right ) \left (b \,x^{2}+a \right )}{4 \sqrt {a c}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (d \,x^{2}+c \right ) \left (\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}\right )^{\frac {3}{2}} a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.97, size = 197, normalized size = 1.16 \[ \frac {1}{4} \, e {\left (\frac {2 \, {\left (2 \, {\left (a b c - a^{2} d\right )} e - \frac {3 \, {\left (b c^{2} - a c d\right )} {\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )}}{a^{2} c \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {3}{2}} e^{2} - a^{3} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} e^{3}} - \frac {3 \, {\left (b c - a d\right )} c \log \left (\frac {c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {a c e}}{c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {a c e}}\right )}{\sqrt {a c e} a^{2} e^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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