3.3.77 \(\int x^3 (\frac {e (a+b x^2)}{c+d x^2})^{3/2} \, dx\) [277]

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

[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 = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1981, 1980, 466, 1171, 396, 214} \begin {gather*} \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} \end {gather*}

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 396

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 466

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 1171

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] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1980

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_Symbol] :> With[{q = Denominator[p]}
, Dist[q*e*(b*c - a*d), Subst[Int[x^(q*(p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a
+ b*x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]

Rule 1981

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Dist[1/n, Sub
st[Int[x^(Simplify[(m + 1)/n] - 1)*(e*((a + b*x)/(c + d*x)))^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n,
p}, x] && 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) \text {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) \text {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) \text {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 ) \text {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 = 2.25, size = 177, normalized size = 0.89 \begin {gather*} \frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt {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 )+3 \left (5 b^2 c^2-6 a b c d+a^2 d^2\right ) \sqrt {c+d x^2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )\right )}{8 \sqrt {b} d^{7/2} \sqrt {a+b x^2}} \end {gather*}

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)]*(Sqrt[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*(5*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*Sqrt[c + d*x^2]*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(
Sqrt[b]*Sqrt[c + d*x^2])]))/(8*Sqrt[b]*d^(7/2)*Sqrt[a + b*x^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(678\) vs. \(2(173)=346\).
time = 0.12, size = 679, normalized size = 3.41

method	result	size
risch	\(\frac {\left (2 b d \,x^{2}+5 a d -7 b c \right ) \left (d \,x^{2}+c \right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}{8 d^{3}}+\frac {\left (\frac {3 \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a^{2}}{16 d \sqrt {d e b}}-\frac {9 \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a b c}{8 d^{2} \sqrt {d e b}}+\frac {15 \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) b^{2} c^{2}}{16 d^{3} \sqrt {d e b}}+\frac {c \,x^{2} a^{2} b}{d \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {2 c^{2} x^{2} a \,b^{2}}{d^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {c^{3} x^{2} b^{3}}{d^{3} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {c \,a^{3}}{d \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {2 c^{2} a^{2} b}{d^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {c^{3} a \,b^{2}}{d^{3} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}\right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{b \,x^{2}+a}\)	\(607\)
default	\(\frac {\left (4 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b \,d^{2} x^{4}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} d^{3} x^{2}-18 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c \,d^{2} x^{2}+15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2} d \,x^{2}+10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a \,d^{2} x^{2}-10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b c d \,x^{2}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} c \,d^{2}-18 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b \,c^{2} d +15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{3}+10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a c d -14 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b \,c^{2}+16 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a c d -16 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, b \,c^{2}\right ) \left (d \,x^{2}+c \right ) \left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )^{\frac {3}{2}}}{16 d^{3} \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (b \,x^{2}+a \right )}\)	\(679\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b*d^2*x^4+3*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x
^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*d^3*x^2-18*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*
c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b*c*d^2*x^2+15*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^
(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^2*c^2*d*x^2+10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*d^2
*x^2-10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b*c*d*x^2+3*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x
^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*c*d^2-18*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)
^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b*c^2*d+15*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*
(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^2*c^3+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+16*(b*d)^(1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*a*c*d-16*(b*d)^(1/2)*
((d*x^2+c)*(b*x^2+a))^(1/2)*b*c^2)/d^3*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(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 = 0.52, size = 282, normalized size = 1.42 \begin {gather*} \frac {1}{16} \, {\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 {b x^{2} + a}{d x^{2} + c}\right )^{\frac {3}{2}} - {\left (7 \, b^{3} c^{2} - 10 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )}}{b^{2} d^{3} - \frac {2 \, {\left (b x^{2} + a\right )} b d^{4}}{d x^{2} + c} + \frac {{\left (b x^{2} + a\right )}^{2} d^{5}}{{\left (d x^{2} + c\right )}^{2}}} - \frac {16 \, {\left (b c^{2} - a c d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{d^{3}} - \frac {3 \, {\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} \log \left (\frac {d \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} - \sqrt {b d}}{d \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} + \sqrt {b d}}\right )}{\sqrt {b d} d^{3}}\right )} e^{\frac {3}{2}} \end {gather*}

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

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Fricas [A]
time = 0.53, size = 396, normalized size = 1.99 \begin {gather*} \left [\frac {3 \, {\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} \sqrt {b d} e^{\frac {3}{2}} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 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} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right ) + 4 \, {\left (2 \, b^{2} d^{3} x^{4} - 15 \, b^{2} c^{2} d + 13 \, a b c d^{2} - 5 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} e^{\frac {3}{2}}}{32 \, b d^{4}}, -\frac {3 \, {\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-b d} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{2 \, {\left (b^{2} d x^{2} + a b d\right )}}\right ) e^{\frac {3}{2}} - 2 \, {\left (2 \, b^{2} d^{3} x^{4} - 15 \, b^{2} c^{2} d + 13 \, a b c d^{2} - 5 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} e^{\frac {3}{2}}}{16 \, b d^{4}}\right ] \end {gather*}

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)*sqrt(b*d)*e^(3/2)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2
+ 8*(b^2*c*d + a*b*d^2)*x^2 + 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)*sqrt((b*x^2 +
a)/(d*x^2 + c))) + 4*(2*b^2*d^3*x^4 - 15*b^2*c^2*d + 13*a*b*c*d^2 - 5*(b^2*c*d^2 - a*b*d^3)*x^2)*sqrt((b*x^2
+ a)/(d*x^2 + c))*e^(3/2))/(b*d^4), -1/16*(3*(5*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*sqrt(-b*d)*arctan(1/2*(2*b*d*x^
2 + b*c + a*d)*sqrt(-b*d)*sqrt((b*x^2 + a)/(d*x^2 + c))/(b^2*d*x^2 + a*b*d))*e^(3/2) - 2*(2*b^2*d^3*x^4 - 15*b
^2*c^2*d + 13*a*b*c*d^2 - 5*(b^2*c*d^2 - a*b*d^3)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c))*e^(3/2))/(b*d^4)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Giac [A]
time = 5.64, size = 321, normalized size = 1.61 \begin {gather*} \frac {1}{16} \, {\left (2 \, \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c} {\left (\frac {2 \, b x^{2} \mathrm {sgn}\left (d x^{2} + c\right )}{d^{2}} - \frac {7 \, b^{2} c d^{5} \mathrm {sgn}\left (d x^{2} + c\right ) - 5 \, a b d^{6} \mathrm {sgn}\left (d x^{2} + c\right )}{b d^{8}}\right )} - \frac {16 \, {\left (b^{2} c^{3} \mathrm {sgn}\left (d x^{2} + c\right ) - 2 \, a b c^{2} d \mathrm {sgn}\left (d x^{2} + c\right ) + a^{2} c d^{2} \mathrm {sgn}\left (d x^{2} + c\right )\right )}}{{\left ({\left (\sqrt {b d} x^{2} - \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c}\right )} d + \sqrt {b d} c\right )} d^{3}} - \frac {3 \, {\left (5 \, \sqrt {b d} b^{2} c^{2} \mathrm {sgn}\left (d x^{2} + c\right ) - 6 \, \sqrt {b d} a b c d \mathrm {sgn}\left (d x^{2} + c\right ) + \sqrt {b d} a^{2} d^{2} \mathrm {sgn}\left (d x^{2} + c\right )\right )} \log \left ({\left | -2 \, {\left (\sqrt {b d} x^{2} - \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c}\right )} b d - \sqrt {b d} b c - \sqrt {b d} a d \right |}\right )}{b d^{4}}\right )} e^{\frac {3}{2}} \end {gather*}

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]

1/16*(2*sqrt(b*d*x^4 + b*c*x^2 + a*d*x^2 + a*c)*(2*b*x^2*sgn(d*x^2 + c)/d^2 - (7*b^2*c*d^5*sgn(d*x^2 + c) - 5*
a*b*d^6*sgn(d*x^2 + c))/(b*d^8)) - 16*(b^2*c^3*sgn(d*x^2 + c) - 2*a*b*c^2*d*sgn(d*x^2 + c) + a^2*c*d^2*sgn(d*x
^2 + c))/(((sqrt(b*d)*x^2 - sqrt(b*d*x^4 + b*c*x^2 + a*d*x^2 + a*c))*d + sqrt(b*d)*c)*d^3) - 3*(5*sqrt(b*d)*b^
2*c^2*sgn(d*x^2 + c) - 6*sqrt(b*d)*a*b*c*d*sgn(d*x^2 + c) + sqrt(b*d)*a^2*d^2*sgn(d*x^2 + c))*log(abs(-2*(sqrt
(b*d)*x^2 - sqrt(b*d*x^4 + b*c*x^2 + a*d*x^2 + a*c))*b*d - sqrt(b*d)*b*c - sqrt(b*d)*a*d))/(b*d^4))*e^(3/2)

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

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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