3.350 \(\int \frac {x^4}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx\)

Optimal. Leaf size=443 \[ \frac {c^{3/2} (3 a c+4 b) \left (a c+a d x^2+b\right ) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 a^2 d^{5/2} \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}-\frac {x (3 a c+4 b) \left (a c+a d x^2+b\right )}{15 a^2 d^2 \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}-\frac {\sqrt {c} \left (3 a^2 c^2+13 a b c+8 b^2\right ) \left (a c+a d x^2+b\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 a^3 d^{5/2} \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}+\frac {x \left (3 a^2 c^2+13 a b c+8 b^2\right ) \left (a c+a d x^2+b\right )}{15 a^3 d^2 \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}+\frac {x^3 \left (a c+a d x^2+b\right )}{5 a d \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}} \]

[Out]

-1/15*(3*a*c+4*b)*x*(a*d*x^2+a*c+b)/a^2/d^2/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)+1/5*x^3*(a*d*x^2+a*c+b)/a/d/((a*
d*x^2+a*c+b)/(d*x^2+c))^(1/2)+1/15*(3*a^2*c^2+13*a*b*c+8*b^2)*x*(a*d*x^2+a*c+b)/a^3/d^2/(d*x^2+c)/((a*d*x^2+a*
c+b)/(d*x^2+c))^(1/2)+1/15*c^(3/2)*(3*a*c+4*b)*(a*d*x^2+a*c+b)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*Ellipti
cF(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(b/(a*c+b))^(1/2))/a^2/d^(5/2)/(d*x^2+c)/((a*d*x^2+a*c+b)/(d*x^2+c))^(1
/2)/(c*(a*d*x^2+a*c+b)/(a*c+b)/(d*x^2+c))^(1/2)-1/15*(3*a^2*c^2+13*a*b*c+8*b^2)*(a*d*x^2+a*c+b)*(1/(1+d*x^2/c)
)^(1/2)*(1+d*x^2/c)^(1/2)*EllipticE(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(b/(a*c+b))^(1/2))*c^(1/2)/a^3/d^(5/2)
/(d*x^2+c)/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/(c*(a*d*x^2+a*c+b)/(a*c+b)/(d*x^2+c))^(1/2)

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Rubi [A]  time = 0.71, antiderivative size = 498, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {6722, 1975, 478, 582, 531, 418, 492, 411} \[ \frac {x \left (3 a^2 c^2+13 a b c+8 b^2\right ) \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b}}{15 a^3 d^2 \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\sqrt {c} \left (3 a^2 c^2+13 a b c+8 b^2\right ) \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 a^3 d^{5/2} \left (c+d x^2\right ) \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a+\frac {b}{c+d x^2}}}+\frac {c^{3/2} (3 a c+4 b) \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 a^2 d^{5/2} \left (c+d x^2\right ) \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a+\frac {b}{c+d x^2}}}-\frac {x (3 a c+4 b) \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b}}{15 a^2 d^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {x^3 \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b}}{5 a d \sqrt {a+\frac {b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[a + b/(c + d*x^2)],x]

[Out]

-((4*b + 3*a*c)*x*Sqrt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)])/(15*a^2*d^2*Sqrt[a + b/(c + d*x^2)]) + (x^3
*Sqrt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)])/(5*a*d*Sqrt[a + b/(c + d*x^2)]) + ((8*b^2 + 13*a*b*c + 3*a^2
*c^2)*x*Sqrt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)])/(15*a^3*d^2*(c + d*x^2)*Sqrt[a + b/(c + d*x^2)]) - (S
qrt[c]*(8*b^2 + 13*a*b*c + 3*a^2*c^2)*Sqrt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d
]*x)/Sqrt[c]], b/(b + a*c)])/(15*a^3*d^(5/2)*(c + d*x^2)*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]
*Sqrt[a + b/(c + d*x^2)]) + (c^(3/2)*(4*b + 3*a*c)*Sqrt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)]*EllipticF[A
rcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(15*a^2*d^(5/2)*(c + d*x^2)*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*
(c + d*x^2))]*Sqrt[a + b/(c + d*x^2)])

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 478

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

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 531

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

Rule 582

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

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rubi steps

\begin {align*} \int \frac {x^4}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx &=\frac {\sqrt {b+a \left (c+d x^2\right )} \int \frac {x^4 \sqrt {c+d x^2}}{\sqrt {b+a \left (c+d x^2\right )}} \, dx}{\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=\frac {\sqrt {b+a \left (c+d x^2\right )} \int \frac {x^4 \sqrt {c+d x^2}}{\sqrt {b+a c+a d x^2}} \, dx}{\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=\frac {x^3 \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{5 a d \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\sqrt {b+a \left (c+d x^2\right )} \int \frac {x^2 \left (3 c (b+a c)+(4 b+3 a c) d x^2\right )}{\sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{5 a d \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {(4 b+3 a c) x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{15 a^2 d^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {x^3 \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{5 a d \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\sqrt {b+a \left (c+d x^2\right )} \int \frac {c (b+a c) (4 b+3 a c) d+\left (8 b^2+13 a b c+3 a^2 c^2\right ) d^2 x^2}{\sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{15 a^2 d^3 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {(4 b+3 a c) x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{15 a^2 d^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {x^3 \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{5 a d \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (c (b+a c) (4 b+3 a c) \sqrt {b+a \left (c+d x^2\right )}\right ) \int \frac {1}{\sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{15 a^2 d^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (\left (8 b^2+13 a b c+3 a^2 c^2\right ) \sqrt {b+a \left (c+d x^2\right )}\right ) \int \frac {x^2}{\sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{15 a^2 d \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {(4 b+3 a c) x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{15 a^2 d^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {x^3 \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{5 a d \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (8 b^2+13 a b c+3 a^2 c^2\right ) x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{15 a^3 d^2 \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}+\frac {c^{3/2} (4 b+3 a c) \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 a^2 d^{5/2} \left (c+d x^2\right ) \sqrt {\frac {c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\left (c \left (8 b^2+13 a b c+3 a^2 c^2\right ) \sqrt {b+a \left (c+d x^2\right )}\right ) \int \frac {\sqrt {b+a c+a d x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{15 a^3 d^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {(4 b+3 a c) x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{15 a^2 d^2 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {x^3 \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{5 a d \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (8 b^2+13 a b c+3 a^2 c^2\right ) x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{15 a^3 d^2 \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\sqrt {c} \left (8 b^2+13 a b c+3 a^2 c^2\right ) \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 a^3 d^{5/2} \left (c+d x^2\right ) \sqrt {\frac {c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt {a+\frac {b}{c+d x^2}}}+\frac {c^{3/2} (4 b+3 a c) \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{15 a^2 d^{5/2} \left (c+d x^2\right ) \sqrt {\frac {c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt {a+\frac {b}{c+d x^2}}}\\ \end {align*}

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Mathematica [C]  time = 0.79, size = 297, normalized size = 0.67 \[ -\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \left (x \left (c+d x^2\right ) \sqrt {\frac {a d}{a c+b}} \left (3 a^2 \left (c^2-d^2 x^4\right )+a b \left (7 c+d x^2\right )+4 b^2\right )+i c \left (3 a^2 c^2+13 a b c+8 b^2\right ) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )-2 i b c (3 a c+2 b) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )\right )}{15 a^2 d^2 \sqrt {\frac {a d}{a c+b}} \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[a + b/(c + d*x^2)],x]

[Out]

-1/15*(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(Sqrt[(a*d)/(b + a*c)]*x*(c + d*x^2)*(4*b^2 + a*b*(7*c + d*x^2) +
3*a^2*(c^2 - d^2*x^4)) + I*c*(8*b^2 + 13*a*b*c + 3*a^2*c^2)*Sqrt[(b + a*c + a*d*x^2)/(b + a*c)]*Sqrt[1 + (d*x
^2)/c]*EllipticE[I*ArcSinh[Sqrt[(a*d)/(b + a*c)]*x], 1 + b/(a*c)] - (2*I)*b*c*(2*b + 3*a*c)*Sqrt[(b + a*c + a*
d*x^2)/(b + a*c)]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[(a*d)/(b + a*c)]*x], 1 + b/(a*c)]))/(a^2*d^2*Sq
rt[(a*d)/(b + a*c)]*(b + a*(c + d*x^2)))

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d x^{6} + c x^{4}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{a d x^{2} + a c + b}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b/(d*x^2+c))^(1/2),x, algorithm="fricas")

[Out]

integral((d*x^6 + c*x^4)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a*d*x^2 + a*c + b), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {a + \frac {b}{d x^{2} + c}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b/(d*x^2+c))^(1/2),x, algorithm="giac")

[Out]

integrate(x^4/sqrt(a + b/(d*x^2 + c)), x)

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maple [A]  time = 0.04, size = 665, normalized size = 1.50 \[ \frac {\left (3 \sqrt {-\frac {a d}{a c +b}}\, a^{2} d^{3} x^{7}+3 \sqrt {-\frac {a d}{a c +b}}\, a^{2} c \,d^{2} x^{5}-\sqrt {-\frac {a d}{a c +b}}\, a b \,d^{2} x^{5}-3 \sqrt {-\frac {a d}{a c +b}}\, a^{2} c^{2} d \,x^{3}-8 \sqrt {-\frac {a d}{a c +b}}\, a b c d \,x^{3}-3 \sqrt {-\frac {a d}{a c +b}}\, a^{2} c^{3} x +3 \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} c^{3} \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )-4 \sqrt {-\frac {a d}{a c +b}}\, b^{2} d \,x^{3}-7 \sqrt {-\frac {a d}{a c +b}}\, a b \,c^{2} x +13 \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b \,c^{2} \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )-6 \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b \,c^{2} \EllipticF \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )-4 \sqrt {-\frac {a d}{a c +b}}\, b^{2} c x +8 \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )-4 \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c \EllipticF \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )\right ) \left (d \,x^{2}+c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{15 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, a^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a+b/(d*x^2+c))^(1/2),x)

[Out]

1/15*(3*(-1/(a*c+b)*a*d)^(1/2)*a^2*d^3*x^7+3*(-1/(a*c+b)*a*d)^(1/2)*a^2*c*d^2*x^5-(-1/(a*c+b)*a*d)^(1/2)*a*b*d
^2*x^5-3*(-1/(a*c+b)*a*d)^(1/2)*a^2*c^2*d*x^3-8*(-1/(a*c+b)*a*d)^(1/2)*a*b*c*d*x^3+3*((a*d*x^2+a*c+b)/(a*c+b))
^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*a^2*c^3-4*(-1/(a*c+b)*a*d)^
(1/2)*b^2*d*x^3-3*(-1/(a*c+b)*a*d)^(1/2)*a^2*c^3*x-6*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*Ellip
ticF((-1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*a*b*c^2+13*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1
/2)*EllipticE((-1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*a*b*c^2-7*(-1/(a*c+b)*a*d)^(1/2)*a*b*c^2*x-4*((a*d
*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*b^2*c+8
*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*b
^2*c-4*(-1/(a*c+b)*a*d)^(1/2)*b^2*c*x)*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/d^2/a^2/(a*d^2*x^4+2*a*c*d*
x^2+b*d*x^2+a*c^2+b*c)^(1/2)/(-1/(a*c+b)*a*d)^(1/2)/((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {a + \frac {b}{d x^{2} + c}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b/(d*x^2+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/sqrt(a + b/(d*x^2 + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4}{\sqrt {a+\frac {b}{d\,x^2+c}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + b/(c + d*x^2))^(1/2),x)

[Out]

int(x^4/(a + b/(c + d*x^2))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {\frac {a c + a d x^{2} + b}{c + d x^{2}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(a+b/(d*x**2+c))**(1/2),x)

[Out]

Integral(x**4/sqrt((a*c + a*d*x**2 + b)/(c + d*x**2)), x)

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