3.3.96 \(\int \frac {x^5}{\sqrt {\frac {e (a+b x^2)}{c+d x^2}}} \, dx\) [296]

Optimal. Leaf size=281 \[ \frac {\left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{16 b^3 d^2 e}-\frac {(3 b c+5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 b^2 d^2 e}-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^3 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}{6 b d (b c-a d) e}+\frac {(b c-a d) \left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{16 b^{7/2} d^{5/2} \sqrt {e}} \]

[Out]

1/16*(-a*d+b*c)*(5*a^2*d^2+2*a*b*c*d+b^2*c^2)*arctanh(d^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^(1/2)/e^(1/2))/b
^(7/2)/d^(5/2)/e^(1/2)+1/16*(5*a^2*d^2+2*a*b*c*d+b^2*c^2)*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^3/d^2/e-1/
24*(5*a*d+3*b*c)*(d*x^2+c)^2*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^2/d^2/e-1/6*(d*x^2+c)^3*(a-c*(b*x^2+a)/(d*x^2+c))
*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b/d/(-a*d+b*c)/e

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Rubi [A]
time = 0.29, antiderivative size = 281, 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, 424, 393, 205, 214} \begin {gather*} \frac {(b c-a d) \left (5 a^2 d^2+2 a b c d+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^{7/2} d^{5/2} \sqrt {e}}+\frac {\left (c+d x^2\right ) \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{16 b^3 d^2 e}-\frac {\left (c+d x^2\right )^2 (5 a d+3 b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{24 b^2 d^2 e}-\frac {\left (c+d x^2\right )^3 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{6 b d e (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(16*b^3*d^2*e) - ((3*b*c + 5
*a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^2)/(24*b^2*d^2*e) - (Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c
+ d*x^2)^3*(a - (c*(a + b*x^2))/(c + d*x^2)))/(6*b*d*(b*c - a*d)*e) + ((b*c - a*d)*(b^2*c^2 + 2*a*b*c*d + 5*a^
2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(16*b^(7/2)*d^(5/2)*Sqrt[e])

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 393

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

Rule 424

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

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 \frac {x^5}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx &=((b c-a d) e) \text {Subst}\left (\int \frac {\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 {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^3 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}{6 b d (b c-a d) e}-\frac {(b c-a d) \text {Subst}\left (\int \frac {-a (b c+5 a d) e^2+3 c (b c+a d) e 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 )}{6 b d}\\ &=-\frac {(3 b c+5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 b^2 d^2 e}-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^3 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}{6 b d (b c-a d) e}+\frac {\left ((b c-a d) \left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) e\right ) \text {Subst}\left (\int \frac {1}{\left (b e-d x^2\right )^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{8 b^2 d^2}\\ &=\frac {\left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{16 b^3 d^2 e}-\frac {(3 b c+5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 b^2 d^2 e}-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^3 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}{6 b d (b c-a d) e}+\frac {\left ((b c-a d) \left (b^2 c^2+2 a b c d+5 a^2 d^2\right )\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^3 d^2}\\ &=\frac {\left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{16 b^3 d^2 e}-\frac {(3 b c+5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 b^2 d^2 e}-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^3 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}{6 b d (b c-a d) e}+\frac {(b c-a d) \left (b^2 c^2+2 a b c d+5 a^2 d^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^{7/2} d^{5/2} \sqrt {e}}\\ \end {align*}

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Mathematica [A]
time = 1.54, size = 224, normalized size = 0.80 \begin {gather*} \frac {\sqrt {a+b x^2} \left (\sqrt {d} \sqrt {a+b x^2} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \left (15 a^2 d^2-2 a b d \left (2 c+5 d x^2\right )+b^2 \left (-3 c^2+2 c d x^2+8 d^2 x^4\right )\right )+3 \sqrt {b c-a d} \left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )\right )}{48 b^3 d^{5/2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(Sqrt[d]*Sqrt[a + b*x^2]*Sqrt[(b*(c + d*x^2))/(b*c - a*d)]*(15*a^2*d^2 - 2*a*b*d*(2*c + 5*d*x
^2) + b^2*(-3*c^2 + 2*c*d*x^2 + 8*d^2*x^4)) + 3*Sqrt[b*c - a*d]*(b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*ArcSinh[(Sqr
t[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]]))/(48*b^3*d^(5/2)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*Sqrt[(b*(c + d*x^2)
)/(b*c - a*d)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(526\) vs. \(2(253)=506\).
time = 0.09, size = 527, normalized size = 1.88

method	result	size
risch	\(\frac {\left (8 b^{2} d^{2} x^{4}-10 a b \,d^{2} x^{2}+2 b^{2} c d \,x^{2}+15 a^{2} d^{2}-4 a b c d -3 b^{2} c^{2}\right ) \left (b \,x^{2}+a \right )}{48 b^{3} d^{2} \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}+\frac {\left (-\frac {5 d \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a^{3}}{32 b^{3} \sqrt {d e b}}+\frac {3 \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a^{2} c}{32 b^{2} \sqrt {d e b}}+\frac {\ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) c^{2} a}{32 b d \sqrt {d e b}}+\frac {\ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) c^{3}}{32 d^{2} \sqrt {d e b}}\right ) \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{\sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(418\)
default	\(\frac {\left (b \,x^{2}+a \right ) \left (-36 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, x^{2} a b \,d^{2} \sqrt {b d}-12 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, x^{2} c \,b^{2} d \sqrt {b d}-15 d^{3} \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3}+9 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} c b \,d^{2}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,c^{2} b^{2} d +3 b^{3} \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) c^{3}+16 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} b d \sqrt {b d}+30 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, a^{2} d^{2} \sqrt {b d}-24 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, a c b d \sqrt {b d}-6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, c^{2} b^{2} \sqrt {b d}\right )}{96 \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, b^{3} d^{2} \sqrt {b d}}\)	\(527\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/96*(b*x^2+a)*(-36*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*a*b*d^2*(b*d)^(1/2)-12*(b*d*x^4+a*d*x^2+b*c*x^2+a*
c)^(1/2)*x^2*c*b^2*d*(b*d)^(1/2)-15*d^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+9*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*b*d^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*c^2*b^2*d+3*b^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))*c^
3+16*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*b*d*(b*d)^(1/2)+30*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*a^2*d^2*(b*d)^
(1/2)-24*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*a*c*b*d*(b*d)^(1/2)-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*c^2*b^2
*(b*d)^(1/2))/(e*(b*x^2+a)/(d*x^2+c))^(1/2)/((d*x^2+c)*(b*x^2+a))^(1/2)/b^3/d^2/(b*d)^(1/2)

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Maxima [A]
time = 0.52, size = 387, normalized size = 1.38 \begin {gather*} \frac {1}{96} \, {\left (\frac {2 \, {\left (3 \, {\left (b^{3} c^{3} d^{2} + a b^{2} c^{2} d^{3} + 3 \, a^{2} b c d^{4} - 5 \, a^{3} d^{5}\right )} \left (\frac {b x^{2} + a}{d x^{2} + c}\right )^{\frac {5}{2}} + 8 \, {\left (b^{4} c^{3} d - 3 \, a b^{3} c^{2} d^{2} - 3 \, a^{2} b^{2} c d^{3} + 5 \, a^{3} b d^{4}\right )} \left (\frac {b x^{2} + a}{d x^{2} + c}\right )^{\frac {3}{2}} - 3 \, {\left (b^{5} c^{3} + a b^{4} c^{2} d - 13 \, a^{2} b^{3} c d^{2} + 11 \, a^{3} b^{2} d^{3}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )}}{b^{6} d^{2} - \frac {3 \, {\left (b x^{2} + a\right )} b^{5} d^{3}}{d x^{2} + c} + \frac {3 \, {\left (b x^{2} + a\right )}^{2} b^{4} d^{4}}{{\left (d x^{2} + c\right )}^{2}} - \frac {{\left (b x^{2} + a\right )}^{3} b^{3} d^{5}}{{\left (d x^{2} + c\right )}^{3}}} - \frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, 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^{3} d^{2}}\right )} e^{\left (-\frac {1}{2}\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))^(1/2),x, algorithm="maxima")

[Out]

1/96*(2*(3*(b^3*c^3*d^2 + a*b^2*c^2*d^3 + 3*a^2*b*c*d^4 - 5*a^3*d^5)*((b*x^2 + a)/(d*x^2 + c))^(5/2) + 8*(b^4*
c^3*d - 3*a*b^3*c^2*d^2 - 3*a^2*b^2*c*d^3 + 5*a^3*b*d^4)*((b*x^2 + a)/(d*x^2 + c))^(3/2) - 3*(b^5*c^3 + a*b^4*
c^2*d - 13*a^2*b^3*c*d^2 + 11*a^3*b^2*d^3)*sqrt((b*x^2 + a)/(d*x^2 + c)))/(b^6*d^2 - 3*(b*x^2 + a)*b^5*d^3/(d*
x^2 + c) + 3*(b*x^2 + a)^2*b^4*d^4/(d*x^2 + c)^2 - (b*x^2 + a)^3*b^3*d^5/(d*x^2 + c)^3) - 3*(b^3*c^3 + a*b^2*c
^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*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)*b^3*d^2))*e^(-1/2)

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Fricas [A]
time = 0.39, size = 520, normalized size = 1.85 \begin {gather*} \left [-\frac {{\left (3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {b d} \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} - 3 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} + 15 \, a^{2} b c d^{3} + 10 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{4} - {\left (b^{3} c^{2} d^{2} + 14 \, a b^{2} c d^{3} - 15 \, a^{2} b d^{4}\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {1}{2}\right )}}{192 \, b^{4} d^{3}}, -\frac {{\left (3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, 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 ) - 2 \, {\left (8 \, b^{3} d^{4} x^{6} - 3 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} + 15 \, a^{2} b c d^{3} + 10 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{4} - {\left (b^{3} c^{2} d^{2} + 14 \, a b^{2} c d^{3} - 15 \, a^{2} b d^{4}\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {1}{2}\right )}}{96 \, b^{4} d^{3}}\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))^(1/2),x, algorithm="fricas")

[Out]

[-1/192*(3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(b*d)*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)*s
qrt((b*x^2 + a)/(d*x^2 + c))) - 4*(8*b^3*d^4*x^6 - 3*b^3*c^3*d - 4*a*b^2*c^2*d^2 + 15*a^2*b*c*d^3 + 10*(b^3*c*
d^3 - a*b^2*d^4)*x^4 - (b^3*c^2*d^2 + 14*a*b^2*c*d^3 - 15*a^2*b*d^4)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c)))*e^(-1
/2)/(b^4*d^3), -1/96*(3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*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)) - 2*(8*b^3*d^4*x^6 - 3*b^3*c^3*d - 4
*a*b^2*c^2*d^2 + 15*a^2*b*c*d^3 + 10*(b^3*c*d^3 - a*b^2*d^4)*x^4 - (b^3*c^2*d^2 + 14*a*b^2*c*d^3 - 15*a^2*b*d^
4)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c)))*e^(-1/2)/(b^4*d^3)]

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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))**(1/2),x)

[Out]

Timed out

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Giac [A]
time = 3.76, size = 210, normalized size = 0.75 \begin {gather*} \frac {{\left (2 \, \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c} {\left (2 \, x^{2} {\left (\frac {4 \, x^{2}}{b} + \frac {b^{2} c d - 5 \, a b d^{2}}{b^{3} d^{2}}\right )} - \frac {3 \, b^{2} c^{2} + 4 \, a b c d - 15 \, a^{2} d^{2}}{b^{3} d^{2}}\right )} - \frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left ({\left | -b c - a d - 2 \, {\left (\sqrt {b d} x^{2} - \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c}\right )} \sqrt {b d} \right |}\right )}{\sqrt {b d} b^{3} d^{2}}\right )} e^{\left (-\frac {1}{2}\right )}}{96 \, \mathrm {sgn}\left (d x^{2} + c\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))^(1/2),x, algorithm="giac")

[Out]

1/96*(2*sqrt(b*d*x^4 + b*c*x^2 + a*d*x^2 + a*c)*(2*x^2*(4*x^2/b + (b^2*c*d - 5*a*b*d^2)/(b^3*d^2)) - (3*b^2*c^
2 + 4*a*b*c*d - 15*a^2*d^2)/(b^3*d^2)) - 3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*log(abs(-b*c -
a*d - 2*(sqrt(b*d)*x^2 - sqrt(b*d*x^4 + b*c*x^2 + a*d*x^2 + a*c))*sqrt(b*d)))/(sqrt(b*d)*b^3*d^2))*e^(-1/2)/sg
n(d*x^2 + c)

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

[Out]

int(x^5/((e*(a + b*x^2))/(c + d*x^2))^(1/2), x)

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