3.3.76 \(\int x^5 (\frac {e (a+b x^2)}{c+d x^2})^{3/2} \, dx\) [276]

Optimal. Leaf size=282 \[ \frac {c^2 (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d^4}+\frac {\left (79 b^2 c^2-50 a b c d-5 a^2 d^2\right ) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{48 b d^4}-\frac {(11 b c+a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 d^4}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac {(b c-a d) \left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{16 b^{3/2} d^{9/2}} \]

[Out]

1/6*(e*(b*x^2+a)/(d*x^2+c))^(5/2)*(d*x^2+c)^3/b/d^2/e-1/16*(-a*d+b*c)*(-a^2*d^2-10*a*b*c*d+35*b^2*c^2)*e^(3/2)
*arctanh(d^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^(1/2)/e^(1/2))/b^(3/2)/d^(9/2)+c^2*(-a*d+b*c)*e*(e*(b*x^2+a)/
(d*x^2+c))^(1/2)/d^4+1/48*(-5*a^2*d^2-50*a*b*c*d+79*b^2*c^2)*e*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b/d^4-1
/24*(a*d+11*b*c)*e*(d*x^2+c)^2*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/d^4

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Rubi [A]
time = 0.35, antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1981, 1980, 474, 466, 1171, 396, 214} \begin {gather*} \frac {e \left (c+d x^2\right ) \left (-5 a^2 d^2-50 a b c d+79 b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{48 b d^4}-\frac {e^{3/2} (b c-a d) \left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{16 b^{3/2} d^{9/2}}+\frac {c^2 e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d^4}-\frac {e \left (c+d x^2\right )^2 (a d+11 b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{24 d^4}+\frac {\left (c+d x^2\right )^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2}}{6 b d^2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*((e*(a + b*x^2))/(c + d*x^2))^(3/2),x]

[Out]

(c^2*(b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/d^4 + ((79*b^2*c^2 - 50*a*b*c*d - 5*a^2*d^2)*e*Sqrt[(e*(
a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(48*b*d^4) - ((11*b*c + a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*
x^2)^2)/(24*d^4) + (((e*(a + b*x^2))/(c + d*x^2))^(5/2)*(c + d*x^2)^3)/(6*b*d^2*e) - ((b*c - a*d)*(35*b^2*c^2
- 10*a*b*c*d - a^2*d^2)*e^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(16*b^
(3/2)*d^(9/2))

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 474

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(-(b*c - a*
d)^2)*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b^2*e*n*(p + 1))), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a +
b*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a
, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

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^5 \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 )^2}{\left (b e-d x^2\right )^4} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac {(b c-a d) \text {Subst}\left (\int \frac {x^4 \left (-6 a^2 d^2 e^2+5 (b c e-a d e)^2+6 b c^2 d e 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 )}{6 b d^2}\\ &=-\frac {(11 b c+a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 d^4}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac {(b c-a d) \text {Subst}\left (\int \frac {-b d (b c-a d) (11 b c+a d) e^3-4 d^2 (b c-a d) (11 b c+a d) e^2 x^2-24 b c^2 d^3 e 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 )}{24 b d^5}\\ &=\frac {\left (79 b^2 c^2-50 a b c d-5 a^2 d^2\right ) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{48 b d^4}-\frac {(11 b c+a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 d^4}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}+\frac {(b c-a d) \text {Subst}\left (\int \frac {-3 b d \left (19 b^2 c^2-10 a b c d-a^2 d^2\right ) e^3-48 b^2 c^2 d^2 e^2 x^2}{b e-d x^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{48 b^2 d^5 e}\\ &=\frac {c^2 (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d^4}+\frac {\left (79 b^2 c^2-50 a b c d-5 a^2 d^2\right ) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{48 b d^4}-\frac {(11 b c+a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 d^4}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac {\left ((b c-a d) \left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) 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 )}{16 b d^4}\\ &=\frac {c^2 (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d^4}+\frac {\left (79 b^2 c^2-50 a b c d-5 a^2 d^2\right ) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{48 b d^4}-\frac {(11 b c+a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 d^4}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac {(b c-a d) \left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) 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 )}{16 b^{3/2} d^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 4.64, size = 294, normalized size = 1.04 \begin {gather*} \frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (b \sqrt {d} \sqrt {b c-a d} \left (3 a^3 d^2 \left (c+d x^2\right )+a^2 b d \left (-100 c^2-35 c d x^2+17 d^2 x^4\right )+b^3 x^2 \left (105 c^3+35 c^2 d x^2-14 c d^2 x^4+8 d^3 x^6\right )+a b^2 \left (105 c^3-65 c^2 d x^2-52 c d^2 x^4+22 d^3 x^6\right )\right )-3 (b c-a d)^2 \left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) \sqrt {a+b x^2} \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 )\right )}{48 b^2 d^{9/2} \sqrt {b c-a d} \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*((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[b*c - a*d]*(3*a^3*d^2*(c + d*x^2) + a^2*b*d*(-100*c^2 - 3
5*c*d*x^2 + 17*d^2*x^4) + b^3*x^2*(105*c^3 + 35*c^2*d*x^2 - 14*c*d^2*x^4 + 8*d^3*x^6) + a*b^2*(105*c^3 - 65*c^
2*d*x^2 - 52*c*d^2*x^4 + 22*d^3*x^6)) - 3*(b*c - a*d)^2*(35*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*Sqrt[a + b*x^2]*Sq
rt[(b*(c + d*x^2))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]]))/(48*b^2*d^(9/2)*Sqrt[b*c
- a*d]*(a + b*x^2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1026\) vs. \(2(252)=504\).
time = 0.15, size = 1027, normalized size = 3.64

method	result	size
risch	\(\frac {\left (8 b^{2} d^{2} x^{4}+14 a b \,d^{2} x^{2}-22 b^{2} c d \,x^{2}+3 a^{2} d^{2}-52 a b c d +57 b^{2} c^{2}\right ) \left (d \,x^{2}+c \right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}{48 b \,d^{4}}+\frac {\left (-\frac {\ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a^{3}}{32 d b \sqrt {d e b}}-\frac {9 \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a^{2} c}{32 d^{2} \sqrt {d e b}}+\frac {45 b \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) c^{2} a}{32 d^{3} \sqrt {d e b}}-\frac {35 b^{2} \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) c^{3}}{32 d^{4} \sqrt {d e b}}-\frac {b \,c^{2} x^{2} a^{2}}{d^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {2 b^{2} c^{3} x^{2} a}{d^{3} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {b^{3} c^{4} x^{2}}{d^{4} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {c^{2} a^{3}}{d^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {2 b \,c^{3} a^{2}}{d^{3} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {b^{2} c^{4} a}{d^{4} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}\right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{b \,x^{2}+a}\)	\(730\)
default	\(\frac {\left (12 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b \,d^{3} x^{4}-60 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} c \,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^{3} d^{4} x^{2}-27 \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} b c \,d^{3} x^{2}+135 \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^{2} c^{2} d^{2} x^{2}-105 \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^{3} c^{3} d \,x^{2}+16 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {b d}\, b \,d^{2} x^{2}+6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{2} d^{3} x^{2}-108 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b c \,d^{2} x^{2}+54 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} c^{2} 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^{3} c \,d^{3}-27 \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} b \,c^{2} d^{2}+135 \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^{2} c^{3} d -105 \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^{3} c^{4}+16 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {b d}\, b c d +6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{2} c \,d^{2}-120 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b \,c^{2} d +114 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} c^{3}-96 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a b \,c^{2} d +96 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, b^{2} c^{3}\right ) \left (d \,x^{2}+c \right ) \left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )^{\frac {3}{2}}}{96 d^{4} b \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (b \,x^{2}+a \right )}\)	\(1027\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(e*(b*x^2+a)/(d*x^2+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/96*(12*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*b*d^3*x^4-60*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b
*d)^(1/2)*b^2*c*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^3*d^4*x^2-27*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*b*c*d^3*x^2+135*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^2*c^2*d^2*x^2-105*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^3*c^3*d*x^2+16*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(b*d)^(1/2)*b*d^2*x^2+6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^
(1/2)*(b*d)^(1/2)*a^2*d^3*x^2-108*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*b*c*d^2*x^2+54*(b*d*x^4+a*
d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b^2*c^2*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^3*c*d^3-27*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*b*c^2*d^2+135*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^2*c^3*d-105*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^3*c^4+16*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(b*d)^(1/2)*b*c*d+6*(b*d*x^4+a*d*x^2+b*c*x^
2+a*c)^(1/2)*(b*d)^(1/2)*a^2*c*d^2-120*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*b*c^2*d+114*(b*d*x^4+
a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b^2*c^3-96*(b*d)^(1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*a*b*c^2*d+96*(b*d)^(
1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*b^2*c^3)/d^4/b*(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 = 426, normalized size = 1.51 \begin {gather*} \frac {1}{96} \, {\left (\frac {2 \, {\left (3 \, {\left (29 \, b^{3} c^{3} d^{2} - 51 \, a b^{2} c^{2} d^{3} + 23 \, a^{2} b c d^{4} - a^{3} d^{5}\right )} \left (\frac {b x^{2} + a}{d x^{2} + c}\right )^{\frac {5}{2}} - 8 \, {\left (17 \, b^{4} c^{3} d - 27 \, a b^{3} c^{2} d^{2} + 9 \, a^{2} b^{2} c d^{3} + a^{3} b d^{4}\right )} \left (\frac {b x^{2} + a}{d x^{2} + c}\right )^{\frac {3}{2}} + 3 \, {\left (19 \, b^{5} c^{3} - 29 \, a b^{4} c^{2} d + 9 \, a^{2} b^{3} c d^{2} + a^{3} b^{2} d^{3}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )}}{b^{4} d^{4} - \frac {3 \, {\left (b x^{2} + a\right )} b^{3} d^{5}}{d x^{2} + c} + \frac {3 \, {\left (b x^{2} + a\right )}^{2} b^{2} d^{6}}{{\left (d x^{2} + c\right )}^{2}} - \frac {{\left (b x^{2} + a\right )}^{3} b d^{7}}{{\left (d x^{2} + c\right )}^{3}}} + \frac {96 \, {\left (b c^{3} - a c^{2} d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{d^{4}} + \frac {3 \, {\left (35 \, b^{3} c^{3} - 45 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\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} b d^{4}}\right )} e^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="maxima")

[Out]

1/96*(2*(3*(29*b^3*c^3*d^2 - 51*a*b^2*c^2*d^3 + 23*a^2*b*c*d^4 - a^3*d^5)*((b*x^2 + a)/(d*x^2 + c))^(5/2) - 8*
(17*b^4*c^3*d - 27*a*b^3*c^2*d^2 + 9*a^2*b^2*c*d^3 + a^3*b*d^4)*((b*x^2 + a)/(d*x^2 + c))^(3/2) + 3*(19*b^5*c^
3 - 29*a*b^4*c^2*d + 9*a^2*b^3*c*d^2 + a^3*b^2*d^3)*sqrt((b*x^2 + a)/(d*x^2 + c)))/(b^4*d^4 - 3*(b*x^2 + a)*b^
3*d^5/(d*x^2 + c) + 3*(b*x^2 + a)^2*b^2*d^6/(d*x^2 + c)^2 - (b*x^2 + a)^3*b*d^7/(d*x^2 + c)^3) + 96*(b*c^3 - a
*c^2*d)*sqrt((b*x^2 + a)/(d*x^2 + c))/d^4 + 3*(35*b^3*c^3 - 45*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*log((d*s
qrt((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)*b*d^4))*e^
(3/2)

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Fricas [A]
time = 0.58, size = 526, normalized size = 1.87 \begin {gather*} \left [\frac {3 \, {\left (35 \, b^{3} c^{3} - 45 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\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 (8 \, b^{3} d^{4} x^{6} + 105 \, b^{3} c^{3} d - 100 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - 14 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{4} + {\left (35 \, b^{3} c^{2} d^{2} - 38 \, a b^{2} c d^{3} + 3 \, a^{2} b d^{4}\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} e^{\frac {3}{2}}}{192 \, b^{2} d^{5}}, \frac {3 \, {\left (35 \, b^{3} c^{3} - 45 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\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 (8 \, b^{3} d^{4} x^{6} + 105 \, b^{3} c^{3} d - 100 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - 14 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{4} + {\left (35 \, b^{3} c^{2} d^{2} - 38 \, a b^{2} c d^{3} + 3 \, a^{2} b d^{4}\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} e^{\frac {3}{2}}}{96 \, b^{2} d^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="fricas")

[Out]

[1/192*(3*(35*b^3*c^3 - 45*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*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*(8*b^3*d^4*x^6 + 105*b^3*c^3*d - 100*a*b^2*c^2*d^2 + 3*a^2*b*c*d^
3 - 14*(b^3*c*d^3 - a*b^2*d^4)*x^4 + (35*b^3*c^2*d^2 - 38*a*b^2*c*d^3 + 3*a^2*b*d^4)*x^2)*sqrt((b*x^2 + a)/(d*
x^2 + c))*e^(3/2))/(b^2*d^5), 1/96*(3*(35*b^3*c^3 - 45*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*sqrt(-b*d)*arcta
n(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*(8*b^3
*d^4*x^6 + 105*b^3*c^3*d - 100*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - 14*(b^3*c*d^3 - a*b^2*d^4)*x^4 + (35*b^3*c^2*d^
2 - 38*a*b^2*c*d^3 + 3*a^2*b*d^4)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c))*e^(3/2))/(b^2*d^5)]

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

[Out]

Timed out

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Giac [A]
time = 4.81, size = 420, normalized size = 1.49 \begin {gather*} \frac {1}{96} \, {\left (2 \, \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c} {\left (2 \, {\left (\frac {4 \, b x^{2} \mathrm {sgn}\left (d x^{2} + c\right )}{d^{2}} - \frac {11 \, b^{3} c d^{10} \mathrm {sgn}\left (d x^{2} + c\right ) - 7 \, a b^{2} d^{11} \mathrm {sgn}\left (d x^{2} + c\right )}{b^{2} d^{13}}\right )} x^{2} + \frac {57 \, b^{3} c^{2} d^{9} \mathrm {sgn}\left (d x^{2} + c\right ) - 52 \, a b^{2} c d^{10} \mathrm {sgn}\left (d x^{2} + c\right ) + 3 \, a^{2} b d^{11} \mathrm {sgn}\left (d x^{2} + c\right )}{b^{2} d^{13}}\right )} + \frac {96 \, {\left (b^{2} c^{4} \mathrm {sgn}\left (d x^{2} + c\right ) - 2 \, a b c^{3} d \mathrm {sgn}\left (d x^{2} + c\right ) + a^{2} c^{2} 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^{4}} + \frac {3 \, {\left (35 \, \sqrt {b d} b^{3} c^{3} \mathrm {sgn}\left (d x^{2} + c\right ) - 45 \, \sqrt {b d} a b^{2} c^{2} d \mathrm {sgn}\left (d x^{2} + c\right ) + 9 \, \sqrt {b d} a^{2} b c d^{2} \mathrm {sgn}\left (d x^{2} + c\right ) + \sqrt {b d} a^{3} d^{3} \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^{2} d^{5}}\right )} e^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="giac")

[Out]

1/96*(2*sqrt(b*d*x^4 + b*c*x^2 + a*d*x^2 + a*c)*(2*(4*b*x^2*sgn(d*x^2 + c)/d^2 - (11*b^3*c*d^10*sgn(d*x^2 + c)
- 7*a*b^2*d^11*sgn(d*x^2 + c))/(b^2*d^13))*x^2 + (57*b^3*c^2*d^9*sgn(d*x^2 + c) - 52*a*b^2*c*d^10*sgn(d*x^2 +
c) + 3*a^2*b*d^11*sgn(d*x^2 + c))/(b^2*d^13)) + 96*(b^2*c^4*sgn(d*x^2 + c) - 2*a*b*c^3*d*sgn(d*x^2 + c) + a^2
*c^2*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^4) + 3
*(35*sqrt(b*d)*b^3*c^3*sgn(d*x^2 + c) - 45*sqrt(b*d)*a*b^2*c^2*d*sgn(d*x^2 + c) + 9*sqrt(b*d)*a^2*b*c*d^2*sgn(
d*x^2 + c) + sqrt(b*d)*a^3*d^3*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^2*d^5))*e^(3/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^5\,{\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^5*((e*(a + b*x^2))/(c + d*x^2))^(3/2),x)

[Out]

int(x^5*((e*(a + b*x^2))/(c + d*x^2))^(3/2), x)

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