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

Optimal. Leaf size=199 \[ \frac {3 e^{3/2} (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 \sqrt {b} d^{7/2}}+\frac {b e \left (c+d x^2\right )^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^3}-\frac {e \left (c+d x^2\right ) (9 b c-5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 d^3}-\frac {c e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d^3} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 199, 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} \[ \frac {3 e^{3/2} (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 \sqrt {b} d^{7/2}}+\frac {b e \left (c+d x^2\right )^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^3}-\frac {e \left (c+d x^2\right ) (9 b c-5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 d^3}-\frac {c e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*(b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/d^3) - ((9*b*c - 5*a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^
2)]*(c + d*x^2))/(8*d^3) + (b*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^2)/(4*d^3) + (3*(b*c - a*d)*(5*b
*c - a*d)*e^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(8*Sqrt[b]*d^(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 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 {x^4 \left (-a e+c x^2\right )}{\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 {b e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 d^3}+\frac {((b c-a d) e) \operatorname {Subst}\left (\int \frac {-b (b c-a d) e^2-4 d (b c-a d) e x^2-4 c d^2 x^4}{\left (b e-d x^2\right )^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{4 d^3}\\ &=-\frac {(9 b c-5 a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 d^3}+\frac {b e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 d^3}-\frac {(b c-a d) \operatorname {Subst}\left (\int \frac {-b (7 b c-3 a d) e^2-8 b c d e x^2}{b e-d x^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{8 b d^3}\\ &=-\frac {c (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d^3}-\frac {(9 b c-5 a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 d^3}+\frac {b e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 d^3}+\frac {\left (3 (b c-a d) (5 b c-a d) e^2\right ) \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 d^3}\\ &=-\frac {c (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d^3}-\frac {(9 b c-5 a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 d^3}+\frac {b e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 d^3}+\frac {3 (b c-a d) (5 b c-a d) e^{3/2} \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 \sqrt {b} d^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 191, normalized size = 0.96 \[ \frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (3 \sqrt {b c-a d} \left (a^2 d^2-6 a b c d+5 b^2 c^2\right ) \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )+b \sqrt {d} \sqrt {a+b x^2} \left (a d \left (13 c+5 d x^2\right )+b \left (-15 c^2-5 c d x^2+2 d^2 x^4\right )\right )\right )}{8 b d^{7/2} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.98, size = 417, normalized size = 2.10 \[ \left [\frac {3 \, {\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} e \sqrt {\frac {e}{b d}} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \, {\left (2 \, b^{2} d^{3} x^{4} + b^{2} c^{2} d + a b c d^{2} + {\left (3 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {e}{b d}}\right ) + 4 \, {\left (2 \, b d^{2} e x^{4} - 5 \, {\left (b c d - a d^{2}\right )} e x^{2} - {\left (15 \, b c^{2} - 13 \, a c d\right )} e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{32 \, d^{3}}, -\frac {3 \, {\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} e \sqrt {-\frac {e}{b d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {e}{b d}}}{2 \, {\left (b e x^{2} + a e\right )}}\right ) - 2 \, {\left (2 \, b d^{2} e x^{4} - 5 \, {\left (b c d - a d^{2}\right )} e x^{2} - {\left (15 \, b c^{2} - 13 \, a c d\right )} e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{16 \, d^{3}}\right ] \]

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

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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(x^2*d+c)]Evaluation time: 0.76Unable to divide, perhaps due to rounding error%%%{%%%{2,[0,4,0]%%%},[2,0,0,0]
%%%}+%%%{%%{[%%%{-4,[0,3,0]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,0,0,1]%%%}+%%%{%%%{2,[1,3,1]%%%},[0,0,0,2]%%
%} / %%%{%%%{1,[0,2,2]%%%},[2,0,0,0]%%%}+%%%{%%{[%%%{-2,[0,1,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,0,0,1]%%
%}+%%%{%%%{1,[1,1,3]%%%},[0,0,0,2]%%%} Error: Bad Argument Value

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maple [B]  time = 0.06, size = 679, normalized size = 3.41 \[ \frac {\left (3 a^{2} d^{3} x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-18 a b c \,d^{2} x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+15 b^{2} c^{2} d \,x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+4 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b \,d^{2} x^{4}+3 a^{2} c \,d^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-18 a b \,c^{2} d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+15 b^{2} c^{3} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a \,d^{2} x^{2}-10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b c d \,x^{2}+16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, a c d +10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a c d -16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, b \,c^{2}-14 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b \,c^{2}\right ) \left (d \,x^{2}+c \right ) \left (\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}\right )^{\frac {3}{2}}}{16 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (b \,x^{2}+a \right ) d^{3}} \]

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*d^2+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*a^2*d^3-18*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*c*d^2+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*b^2*c^2*d+10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^2*a
*d^2-10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^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^2*c*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*b*c^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))*b^2*c^3+16*((d*x^2+c)*(b*x^2+a))^(1/2)*(b*d)^(1/2)*a*c*d-16*((d*x^2+c)*(b*x^
2+a))^(1/2)*(b*d)^(1/2)*b*c^2+10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*c*d-14*(b*d*x^4+a*d*x^2+b*c
*x^2+a*c)^(1/2)*(b*d)^(1/2)*b*c^2)/d^3*(d*x^2+c)*((b*x^2+a)/(d*x^2+c)*e)^(3/2)/(b*d)^(1/2)/((d*x^2+c)*(b*x^2+a
))^(1/2)/(b*x^2+a)

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maxima [A]  time = 2.29, size = 303, normalized size = 1.52 \[ \frac {1}{16} \, e {\left (\frac {2 \, {\left ({\left (9 \, b^{2} c^{2} d - 14 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {3}{2}} e - {\left (7 \, b^{3} c^{2} - 10 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} e^{2}\right )}}{b^{2} d^{3} e^{2} - \frac {2 \, {\left (b x^{2} + a\right )} b d^{4} e^{2}}{d x^{2} + c} + \frac {{\left (b x^{2} + a\right )}^{2} d^{5} e^{2}}{{\left (d x^{2} + c\right )}^{2}}} - \frac {3 \, {\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} e \log \left (\frac {d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {b d e}}{d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {b d e}}\right )}{\sqrt {b d e} d^{3}} - \frac {16 \, {\left (b c^{2} - a c d\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}{d^{3}}\right )} \]

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

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

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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