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

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

[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 = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1960, 413, 385, 199, 208} \[ \frac {\left (c+d x^2\right ) \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{16 b^3 d^2 e}+\frac {(b c-a d) \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{16 b^{7/2} d^{5/2} \sqrt {e}}-\frac {\left (c+d x^2\right )^2 (5 a d+3 b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{24 b^2 d^2 e}-\frac {\left (c+d x^2\right )^3 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{6 b d e (b c-a d)} \]

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 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 385

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 413

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 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 \frac {x^5}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx &=((b c-a d) e) \operatorname {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) \operatorname {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 ) \operatorname {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 ) \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 )}{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 = 0.39, size = 224, normalized size = 0.80 \[ \frac {\sqrt {a+b x^2} \left (3 \sqrt {b c-a d} \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )+\sqrt {d} \sqrt {a+b x^2} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \left (15 a^2 d^2-2 a b d \left (2 c+5 d x^2\right )+b^2 \left (-3 c^2+2 c d x^2+8 d^2 x^4\right )\right )\right )}{48 b^3 d^{5/2} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \]

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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fricas [A]  time = 0.48, size = 545, normalized size = 1.94 \[ \left [-\frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {b d e} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e - 4 \, {\left (2 \, b d^{2} x^{4} + b c^{2} + a c d + {\left (3 \, b c d + a d^{2}\right )} x^{2}\right )} \sqrt {b d e} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}\right ) - 4 \, {\left (8 \, b^{3} d^{4} x^{6} - 3 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} + 15 \, a^{2} b c d^{3} + 10 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{4} - {\left (b^{3} c^{2} d^{2} + 14 \, a b^{2} c d^{3} - 15 \, a^{2} b d^{4}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{192 \, b^{4} d^{3} e}, -\frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {-b d e} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-b d e} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (b^{2} d e x^{2} + a b d e\right )}}\right ) - 2 \, {\left (8 \, b^{3} d^{4} x^{6} - 3 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} + 15 \, a^{2} b c d^{3} + 10 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{4} - {\left (b^{3} c^{2} d^{2} + 14 \, a b^{2} c d^{3} - 15 \, a^{2} b d^{4}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{96 \, b^{4} d^{3} e}\right ] \]

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

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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^5/(e*(b*x^2+a)/(d*x^2+c))^(1/2),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(t_nostep*d+c)]Unable to divide, perhaps due to rounding error%%%{%%%{1,[0,1,0]%%%},[2,0]%%%}+%%%{%%{[-2,0]:[
1,0,%%%{-1,[1,1,1]%%%}]%%},[1,1]%%%}+%%%{%%%{1,[1,0,1]%%%},[0,2]%%%} / %%%{%%%{1,[0,2,0]%%%},[2,0]%%%}+%%%{%%{
[%%%{-2,[0,1,0]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,1]%%%}+%%%{%%%{1,[1,1,1]%%%},[0,2]%%%} Error: Bad Argume
nt Value

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maple [B]  time = 0.05, size = 527, normalized size = 1.88 \[ \frac {\left (b \,x^{2}+a \right ) \left (-15 a^{3} d^{3} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+9 a^{2} b c \,d^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 a \,b^{2} c^{2} d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 b^{3} c^{3} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-36 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b \,d^{2} x^{2}-12 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} c d \,x^{2}+30 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{2} d^{2}-24 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b c d -6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} c^{2}+16 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {b d}\, b d \right )}{96 \sqrt {\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, b^{3} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.27, size = 413, normalized size = 1.47 \[ \frac {1}{96} \, e {\left (\frac {2 \, {\left (3 \, {\left (b^{3} c^{3} d^{2} + a b^{2} c^{2} d^{3} + 3 \, a^{2} b c d^{4} - 5 \, a^{3} d^{5}\right )} \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {5}{2}} + 8 \, {\left (b^{4} c^{3} d - 3 \, a b^{3} c^{2} d^{2} - 3 \, a^{2} b^{2} c d^{3} + 5 \, a^{3} b d^{4}\right )} \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {3}{2}} e - 3 \, {\left (b^{5} c^{3} + a b^{4} c^{2} d - 13 \, a^{2} b^{3} c d^{2} + 11 \, a^{3} b^{2} d^{3}\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} e^{2}\right )}}{b^{6} d^{2} e^{4} - \frac {3 \, {\left (b x^{2} + a\right )} b^{5} d^{3} e^{4}}{d x^{2} + c} + \frac {3 \, {\left (b x^{2} + a\right )}^{2} b^{4} d^{4} e^{4}}{{\left (d x^{2} + c\right )}^{2}} - \frac {{\left (b x^{2} + a\right )}^{3} b^{3} d^{5} e^{4}}{{\left (d x^{2} + c\right )}^{3}}} - \frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left (\frac {d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {b d e}}{d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {b d e}}\right )}{\sqrt {b d e} b^{3} d^{2} e}\right )} \]

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*e*(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)*e/(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)*e/(d*x^2 + c))^(3/2)*e - 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)*e/(d*x^2 + c))*e^2)/(b^6*d^2*e^4 - 3*(b*x^
2 + a)*b^5*d^3*e^4/(d*x^2 + c) + 3*(b*x^2 + a)^2*b^4*d^4*e^4/(d*x^2 + c)^2 - (b*x^2 + a)^3*b^3*d^5*e^4/(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)*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)*b^3*d^2*e))

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

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

[Out]

Timed out

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