3.3.63 \(\int x^5 \sqrt {\frac {e (a+b x^2)}{c+d x^2}} \, dx\) [263]

Optimal. Leaf size=244 \[ \frac {\left (11 b^2 c^2-2 a b c d-a^2 d^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{16 b^2 d^3}-\frac {(3 b c+a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{8 b d^3}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac {(b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{16 b^{5/2} d^{7/2}} \]

[Out]

1/6*(e*(b*x^2+a)/(d*x^2+c))^(3/2)*(d*x^2+c)^3/b/d^2/e-1/16*(-a*d+b*c)*(a^2*d^2+2*a*b*c*d+5*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))*e^(1/2)/b^(5/2)/d^(7/2)+1/16*(-a^2*d^2-2*a*b*c*d+11*b^2*c
^2)*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^2/d^3-1/8*(a*d+3*b*c)*(d*x^2+c)^2*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/
b/d^3

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Rubi [A]
time = 0.30, antiderivative size = 244, 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, 474, 466, 393, 214} \begin {gather*} \frac {\left (c+d x^2\right ) \left (-a^2 d^2-2 a b c d+11 b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{16 b^2 d^3}-\frac {\sqrt {e} (b c-a d) \left (a^2 d^2+2 a b c d+5 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^{5/2} d^{7/2}}-\frac {\left (c+d x^2\right )^2 (a d+3 b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 b d^3}+\frac {\left (c+d x^2\right )^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{6 b d^2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((11*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(16*b^2*d^3) - ((3*b*c + a*
d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^2)/(8*b*d^3) + (((e*(a + b*x^2))/(c + d*x^2))^(3/2)*(c + d*x^
2)^3)/(6*b*d^2*e) - ((b*c - a*d)*(5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[e]*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2
))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(16*b^(5/2)*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 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 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 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 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx &=((b c-a d) e) \text {Subst}\left (\int \frac {x^2 \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 )^{3/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac {(b c-a d) \text {Subst}\left (\int \frac {x^2 \left (-3 \left (2 a^2 d^2 e^2-(b c e-a d e)^2\right )+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 {(3 b c+a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{8 b d^3}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \left (c+d x^2\right )^3}{6 b d^2 e}+\frac {(b c-a d) \text {Subst}\left (\int \frac {3 d (b c-a d) (3 b c+a d) e^2+24 b c^2 d^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 )}{24 b d^4}\\ &=\frac {\left (11 b^2 c^2-2 a b c d-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^2 d^3}-\frac {(3 b c+a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{8 b d^3}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac {\left ((b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) e\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^2 d^3}\\ &=\frac {\left (11 b^2 c^2-2 a b c d-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^2 d^3}-\frac {(3 b c+a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{8 b d^3}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac {(b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {e} \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^{5/2} d^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 1.64, size = 198, normalized size = 0.81 \begin {gather*} \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (-b \sqrt {d} \left (c+d x^2\right ) \left (3 a^2 d^2-2 a b d \left (-2 c+d x^2\right )+b^2 \left (-15 c^2+10 c d x^2-8 d^2 x^4\right )\right )-\frac {3 (b c-a d)^{3/2} \left (5 b^2 c^2+2 a b c d+a^2 d^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 )}{\sqrt {a+b x^2}}\right )}{48 b^3 d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(-(b*Sqrt[d]*(c + d*x^2)*(3*a^2*d^2 - 2*a*b*d*(-2*c + d*x^2) + b^2*(-15*c^2
+ 10*c*d*x^2 - 8*d^2*x^4))) - (3*(b*c - a*d)^(3/2)*(5*b^2*c^2 + 2*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]])/Sqrt[a + b*x^2]))/(48*b^3*d^(7/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(526\) vs. \(2(216)=432\).
time = 0.10, size = 527, normalized size = 2.16

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

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

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Maxima [A]
time = 0.51, size = 391, normalized size = 1.60 \begin {gather*} \frac {1}{96} \, {\left (\frac {2 \, {\left (3 \, {\left (11 \, b^{3} c^{3} d^{2} - 13 \, a b^{2} c^{2} d^{3} + 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 (5 \, b^{4} c^{3} d - 3 \, a b^{3} c^{2} d^{2} - 3 \, 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 (5 \, b^{5} c^{3} - 3 \, a b^{4} c^{2} d - 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^{5} d^{3} - \frac {3 \, {\left (b x^{2} + a\right )} b^{4} d^{4}}{d x^{2} + c} + \frac {3 \, {\left (b x^{2} + a\right )}^{2} b^{3} d^{5}}{{\left (d x^{2} + c\right )}^{2}} - \frac {{\left (b x^{2} + a\right )}^{3} b^{2} d^{6}}{{\left (d x^{2} + c\right )}^{3}}} + \frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - 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^{2} d^{3}}\right )} e^{\frac {1}{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 528, normalized size = 2.16 \begin {gather*} \left [-\frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {b d} e^{\frac {1}{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} + 15 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} - 3 \, a^{2} b c d^{3} - 2 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{4} + {\left (5 \, b^{3} c^{2} d^{2} - 2 \, 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 {1}{2}}}{192 \, b^{3} d^{4}}, \frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - 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 {1}{2}} + 2 \, {\left (8 \, b^{3} d^{4} x^{6} + 15 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} - 3 \, a^{2} b c d^{3} - 2 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{4} + {\left (5 \, b^{3} c^{2} d^{2} - 2 \, 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 {1}{2}}}{96 \, b^{3} d^{4}}\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*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*sqrt(b*d)*e^(1/2)*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)*sqr
t(b*d)*sqrt((b*x^2 + a)/(d*x^2 + c))) - 4*(8*b^3*d^4*x^6 + 15*b^3*c^3*d - 4*a*b^2*c^2*d^2 - 3*a^2*b*c*d^3 - 2*
(b^3*c*d^3 - a*b^2*d^4)*x^4 + (5*b^3*c^2*d^2 - 2*a*b^2*c*d^3 - 3*a^2*b*d^4)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c))
*e^(1/2))/(b^3*d^4), 1/96*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - 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))*e^(1/2) + 2*(8*b^3*d^4*x^6 + 15
*b^3*c^3*d - 4*a*b^2*c^2*d^2 - 3*a^2*b*c*d^3 - 2*(b^3*c*d^3 - a*b^2*d^4)*x^4 + (5*b^3*c^2*d^2 - 2*a*b^2*c*d^3
- 3*a^2*b*d^4)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c))*e^(1/2))/(b^3*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**5*(e*(b*x**2+a)/(d*x**2+c))**(1/2),x)

[Out]

Timed out

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Giac [A]
time = 5.27, size = 211, normalized size = 0.86 \begin {gather*} \frac {1}{96} \, {\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}}{d} - \frac {5 \, b^{2} c d - a b d^{2}}{b^{2} d^{3}}\right )} + \frac {15 \, b^{2} c^{2} - 4 \, a b c d - 3 \, a^{2} d^{2}}{b^{2} d^{3}}\right )} + \frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - 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^{2} d^{3}}\right )} e^{\frac {1}{2}} \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/d - (5*b^2*c*d - a*b*d^2)/(b^2*d^3)) + (15*b^2*c
^2 - 4*a*b*c*d - 3*a^2*d^2)/(b^2*d^3)) + 3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - 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^2*d^3))*e^(1/2)*sgn
(d*x^2 + c)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int 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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