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

Optimal. Leaf size=165 \[ -\frac {3 \sqrt {a} e^{3/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{2 c^{5/2}}+\frac {3 e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 c^2}+\frac {(b c-a d) \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 c \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 165, 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, 288, 321, 208} \[ -\frac {3 \sqrt {a} e^{3/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{2 c^{5/2}}+\frac {3 e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 c^2}+\frac {(b c-a d) \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 c \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(2*c^2) + ((b*c - a*d)*((e*(a + b*x^2))/(c + d*x^2))^(3/2)
)/(2*c*(a - (c*(a + b*x^2))/(c + d*x^2))) - (3*Sqrt[a]*(b*c - a*d)*e^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2
))/(c + d*x^2)])/(Sqrt[a]*Sqrt[e])])/(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 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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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 {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx &=((b c-a d) e) \operatorname {Subst}\left (\int \frac {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 )\\ &=\frac {(b c-a d) \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 c \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}+\frac {(3 (b c-a d) e) \operatorname {Subst}\left (\int \frac {x^2}{-a e+c x^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{2 c}\\ &=\frac {3 (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 c^2}+\frac {(b c-a d) \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 c \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}+\frac {\left (3 a (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 )}{2 c^2}\\ &=\frac {3 (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 c^2}+\frac {(b c-a d) \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 c \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {3 \sqrt {a} (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 )}{2 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 146, normalized size = 0.88 \[ \frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt {c} \sqrt {a+b x^2} \left (2 b c x^2-a \left (c+3 d x^2\right )\right )-3 \sqrt {a} x^2 \sqrt {c+d x^2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )\right )}{2 c^{5/2} x^2 \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.36, size = 350, normalized size = 2.12 \[ \left [-\frac {3 \, {\left (b c - a d\right )} \sqrt {\frac {a e}{c}} e x^{2} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e x^{4} + 8 \, a^{2} c^{2} e + 8 \, {\left (a b c^{2} + a^{2} c d\right )} e x^{2} + 4 \, {\left ({\left (b c^{2} d + a c d^{2}\right )} x^{4} + 2 \, a c^{3} + {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2}\right )} \sqrt {\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{x^{4}}\right ) - 4 \, {\left ({\left (2 \, b c - 3 \, a d\right )} e x^{2} - a c e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{8 \, c^{2} x^{2}}, \frac {3 \, {\left (b c - a d\right )} \sqrt {-\frac {a e}{c}} e x^{2} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {-\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (a b e x^{2} + a^{2} e\right )}}\right ) + 2 \, {\left ({\left (2 \, b c - 3 \, a d\right )} e x^{2} - a c e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, c^{2} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(3*(b*c - a*d)*sqrt(a*e/c)*e*x^2*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*((b*c^2*d + a*c*d^2)*x^4 + 2*a*c^3 + (b*c^3 + 3*a*c^2*d)*x^2)*sqrt(a*e/c)*sqrt((b*e*x^2 +
a*e)/(d*x^2 + c)))/x^4) - 4*((2*b*c - 3*a*d)*e*x^2 - a*c*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(c^2*x^2), 1/4*
(3*(b*c - a*d)*sqrt(-a*e/c)*e*x^2*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(-a*e/c)*sqrt((b*e*x^2 + a*e)/(d*x^
2 + c))/(a*b*e*x^2 + a^2*e)) + 2*((2*b*c - 3*a*d)*e*x^2 - a*c*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(c^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((e*(b*x^2+a)/(d*x^2+c))^(3/2)/x^3,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: 0.55Unable to divide, perhaps due to rounding error%%%{%%%{2,[0,1,0]%%%},[6,0,0]%%
%}+%%%{%%{[-4,0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[5,1,0]%%%}+%%%{%%%{2,[1,0,1]%%%},[4,2,0]%%%}+%%%{%%%{-4,[0,1,1]%
%%},[4,1,1]%%%}+%%%{%%{[%%%{8,[0,0,1]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[3,2,1]%%%}+%%%{%%%{-4,[1,0,2]%%%},[2
,3,1]%%%}+%%%{%%%{2,[0,1,2]%%%},[2,2,2]%%%}+%%%{%%{[%%%{-4,[0,0,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,3,2]%
%%}+%%%{%%%{2,[1,0,3]%%%},[0,4,2]%%%} / %%%{%%%{1,[0,2,0]%%%},[6,0,0]%%%}+%%%{%%{[%%%{-2,[0,1,0]%%%},0]:[1,0,%
%%{-1,[1,1,1]%%%}]%%},[5,1,0]%%%}+%%%{%%%{1,[1,1,1]%%%},[4,2,0]%%%}+%%%{%%%{-2,[0,2,1]%%%},[4,1,1]%%%}+%%%{%%{
[%%%{4,[0,1,1]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[3,2,1]%%%}+%%%{%%%{-2,[1,1,2]%%%},[2,3,1]%%%}+%%%{%%%{1,[0,
2,2]%%%},[2,2,2]%%%}+%%%{%%{[%%%{-2,[0,1,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,3,2]%%%}+%%%{%%%{1,[1,1,3]%%
%},[0,4,2]%%%} Error: Bad Argument Value

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maple [B]  time = 0.08, size = 641, normalized size = 3.88 \[ -\frac {\left (-3 a^{2} c \,d^{2} x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+3 a b \,c^{2} d \,x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b \,d^{2} x^{6}-3 a^{2} c^{2} d \,x^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+3 a b \,c^{3} x^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a \,d^{2} x^{4}-4 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b c d \,x^{4}+4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, a c d \,x^{2}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a c d \,x^{2}-4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, b \,c^{2} x^{2}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b \,c^{2} x^{2}+2 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, d \,x^{2}+2 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, c \right ) \left (d \,x^{2}+c \right ) \left (\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}\right )^{\frac {3}{2}}}{4 \sqrt {a c}\, \left (b \,x^{2}+a \right ) \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, c^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 2.26, size = 189, normalized size = 1.15 \[ \frac {1}{4} \, {\left (\frac {2 \, {\left (a b c - a^{2} d\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} e}{a c^{2} e - \frac {{\left (b x^{2} + a\right )} c^{3} e}{d x^{2} + c}} + \frac {3 \, {\left (a b c - a^{2} d\right )} e \log \left (\frac {c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {a c e}}{c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {a c e}}\right )}{\sqrt {a c e} c^{2}} + \frac {4 \, {\left (b c - a d\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}{c^{2}}\right )} e \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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