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

Optimal. Leaf size=202 \[ \frac {3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 b^{7/2} \sqrt {d} e^{3/2}}+\frac {\left (c+d x^2\right ) (3 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 b^3 e^2}+\frac {a (b c-a d)}{b^3 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {\left (c+d x^2\right )^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b^2 e^2} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1960, 456, 453, 208} \[ \frac {\left (c+d x^2\right )^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b^2 e^2}+\frac {\left (c+d x^2\right ) (3 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 b^3 e^2}+\frac {3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 b^{7/2} \sqrt {d} e^{3/2}}+\frac {a (b c-a d)}{b^3 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(b*c - a*d))/(b^3*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) + ((3*b*c - 7*a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]
*(c + d*x^2))/(8*b^3*e^2) + (Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^2)/(4*b^2*e^2) + (3*(b*c - 5*a*d)*(
b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(8*b^(7/2)*Sqrt[d]*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 453

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 456

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

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Mathematica [A]  time = 0.32, size = 190, normalized size = 0.94 \[ \frac {\sqrt {d} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \left (-15 a^2 d+a b \left (13 c-5 d x^2\right )+b^2 x^2 \left (5 c+2 d x^2\right )\right )+3 \sqrt {a+b x^2} (b c-5 a d) \sqrt {b c-a d} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )}{8 b^3 \sqrt {d} e \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.09, size = 585, normalized size = 2.90 \[ \left [\frac {3 \, {\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {b d e} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \, {\left (2 \, b d^{2} x^{4} + b c^{2} + a c d + {\left (3 \, b c d + a d^{2}\right )} x^{2}\right )} \sqrt {b d e} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}\right ) + 4 \, {\left (2 \, b^{3} d^{3} x^{6} + 13 \, a b^{2} c^{2} d - 15 \, a^{2} b c d^{2} + {\left (7 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x^{4} + {\left (5 \, b^{3} c^{2} d + 8 \, a b^{2} c d^{2} - 15 \, a^{2} b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{32 \, {\left (b^{5} d e^{2} x^{2} + a b^{4} d e^{2}\right )}}, -\frac {3 \, {\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {-b d e} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-b d e} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (b^{2} d e x^{2} + a b d e\right )}}\right ) - 2 \, {\left (2 \, b^{3} d^{3} x^{6} + 13 \, a b^{2} c^{2} d - 15 \, a^{2} b c d^{2} + {\left (7 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x^{4} + {\left (5 \, b^{3} c^{2} d + 8 \, a b^{2} c d^{2} - 15 \, a^{2} b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{16 \, {\left (b^{5} d e^{2} x^{2} + a b^{4} d e^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(3*(a*b^2*c^2 - 6*a^2*b*c*d + 5*a^3*d^2 + (b^3*c^2 - 6*a*b^2*c*d + 5*a^2*b*d^2)*x^2)*sqrt(b*d*e)*log(8*b
^2*d^2*e*x^4 + 8*(b^2*c*d + a*b*d^2)*e*x^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*e + 4*(2*b*d^2*x^4 + b*c^2 + a*c*
d + (3*b*c*d + a*d^2)*x^2)*sqrt(b*d*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))) + 4*(2*b^3*d^3*x^6 + 13*a*b^2*c^2*d
- 15*a^2*b*c*d^2 + (7*b^3*c*d^2 - 5*a*b^2*d^3)*x^4 + (5*b^3*c^2*d + 8*a*b^2*c*d^2 - 15*a^2*b*d^3)*x^2)*sqrt((b
*e*x^2 + a*e)/(d*x^2 + c)))/(b^5*d*e^2*x^2 + a*b^4*d*e^2), -1/16*(3*(a*b^2*c^2 - 6*a^2*b*c*d + 5*a^3*d^2 + (b^
3*c^2 - 6*a*b^2*c*d + 5*a^2*b*d^2)*x^2)*sqrt(-b*d*e)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(-b*d*e)*sqrt((b*e
*x^2 + a*e)/(d*x^2 + c))/(b^2*d*e*x^2 + a*b*d*e)) - 2*(2*b^3*d^3*x^6 + 13*a*b^2*c^2*d - 15*a^2*b*c*d^2 + (7*b^
3*c*d^2 - 5*a*b^2*d^3)*x^4 + (5*b^3*c^2*d + 8*a*b^2*c*d^2 - 15*a^2*b*d^3)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c
)))/(b^5*d*e^2*x^2 + a*b^4*d*e^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(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,[1,2,2]%%%},[2,1,2,0]%%%}+%%%{%%%{-4,
[2,1,2]%%%},[2,1,1,1]%%%}+%%%{%%%{2,[3,0,2]%%%},[2,1,0,2]%%%}+%%%{%%{[%%%{-4,[0,2,2]%%%},0]:[1,0,%%%{-1,[1,1,1
]%%%}]%%},[1,1,3,0]%%%}+%%%{%%{[%%%{8,[1,1,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,1,2,1]%%%}+%%%{%%{[%%%{-4,
[2,0,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,1,1,2]%%%}+%%%{%%%{2,[0,3,3]%%%},[0,1,4,0]%%%}+%%%{%%%{-4,[1,2,3
]%%%},[0,1,3,1]%%%}+%%%{%%%{2,[2,1,3]%%%},[0,1,2,2]%%%} / %%%{%%%{1,[2,0,2]%%%},[2,0,0,0]%%%}+%%%{%%{[%%%{-2,[
1,0,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,0,1,0]%%%}+%%%{%%%{1,[1,1,3]%%%},[0,0,2,0]%%%} Error: Bad Argumen
t Value

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maple [B]  time = 0.06, size = 679, normalized size = 3.36 \[ -\frac {\left (-15 a^{2} b \,d^{2} x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+18 a \,b^{2} c d \,x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-3 b^{3} c^{2} x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-4 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} d \,x^{4}-15 a^{3} d^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+18 a^{2} b c d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-3 a \,b^{2} c^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b d \,x^{2}-10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} c \,x^{2}+16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, a^{2} d +14 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{2} d -16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, a b c -10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b c \right ) \left (b \,x^{2}+a \right )}{16 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (d \,x^{2}+c \right ) \left (\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}\right )^{\frac {3}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(-4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^4*b^2*d-15*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*
d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*a^2*b*d^2+18*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x
^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*a*b^2*c*d-3*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b
*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*b^3*c^2+10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^2
*a*b*d-10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^2*b^2*c-15*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*
d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^3*d^2+18*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c
*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^2*b*c*d-3*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c
)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a*b^2*c^2+16*((d*x^2+c)*(b*x^2+a))^(1/2)*(b*d)^(1/2)*a^2*d-16*((d*x^2+c)*(b*
x^2+a))^(1/2)*(b*d)^(1/2)*a*b*c+14*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a^2*d-10*(b*d*x^4+a*d*x^2+b
*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*b*c)/b^3*(b*x^2+a)/(b*d)^(1/2)/((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 = 1.81, size = 311, normalized size = 1.54 \[ \frac {1}{16} \, e {\left (\frac {2 \, {\left (8 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} e^{2} - \frac {3 \, {\left (b^{2} c^{2} d - 6 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} {\left (b x^{2} + a\right )}^{2} e^{2}}{{\left (d x^{2} + c\right )}^{2}} + \frac {5 \, {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} {\left (b x^{2} + a\right )} e^{2}}{d x^{2} + c}\right )}}{b^{3} d^{2} \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {5}{2}} e^{2} - 2 \, b^{4} d \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {3}{2}} e^{3} + b^{5} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} e^{4}} - \frac {3 \, {\left (b^{2} c^{2} - 6 \, a b c d + 5 \, a^{2} d^{2}\right )} \log \left (\frac {d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {b d e}}{d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {b d e}}\right )}{\sqrt {b d e} b^{3} e^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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