3.354 \(\int \frac {1}{x^4 \sqrt {a+\frac {b}{c+d x^2}}} \, dx\)

Optimal. Leaf size=435 \[ -\frac {a \sqrt {c} d^{3/2} \left (a c+a d x^2+b\right ) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 (a c+b)^2 \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}-\frac {d^{3/2} (b-a c) \left (a c+a d x^2+b\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 \sqrt {c} (a c+b)^2 \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}+\frac {d^2 x (b-a c) \left (a c+a d x^2+b\right )}{3 c (a c+b)^2 \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}-\frac {d (b-a c) \left (a c+a d x^2+b\right )}{3 c x (a c+b)^2 \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}+\frac {-a c-a d x^2-b}{3 x^3 (a c+b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}} \]

[Out]

1/3*(-a*d*x^2-a*c-b)/(a*c+b)/x^3/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)-1/3*(-a*c+b)*d*(a*d*x^2+a*c+b)/c/(a*c+b)^2/
x/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)+1/3*(-a*c+b)*d^2*x*(a*d*x^2+a*c+b)/c/(a*c+b)^2/(d*x^2+c)/((a*d*x^2+a*c+b)/
(d*x^2+c))^(1/2)-1/3*(-a*c+b)*d^(3/2)*(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))/(a*c+b)^2/(d*x^2+c)/c^(1/2)/((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)-1/3*a*d^(3/2)*(a*d*x^2+a*c+b)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(
1/2)*EllipticF(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(b/(a*c+b))^(1/2))*c^(1/2)/(a*c+b)^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 = 0.64, antiderivative size = 486, 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, 475, 583, 531, 418, 492, 411} \[ \frac {d^2 x (b-a c) \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b}}{3 c (a c+b)^2 \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}-\frac {a \sqrt {c} d^{3/2} \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 (a c+b)^2 \left (c+d x^2\right ) \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a+\frac {b}{c+d x^2}}}-\frac {d^{3/2} (b-a c) \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 \sqrt {c} (a c+b)^2 \left (c+d x^2\right ) \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a+\frac {b}{c+d x^2}}}-\frac {d (b-a c) \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b}}{3 c x (a c+b)^2 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b}}{3 x^3 (a c+b) \sqrt {a+\frac {b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[a + b/(c + d*x^2)]),x]

[Out]

-(Sqrt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)])/(3*(b + a*c)*x^3*Sqrt[a + b/(c + d*x^2)]) - ((b - a*c)*d*Sq
rt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)])/(3*c*(b + a*c)^2*x*Sqrt[a + b/(c + d*x^2)]) + ((b - a*c)*d^2*x*
Sqrt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)])/(3*c*(b + a*c)^2*(c + d*x^2)*Sqrt[a + b/(c + d*x^2)]) - ((b -
a*c)*d^(3/2)*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)])/(3*Sqrt[c]*(b + a*c)^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)]) - (a*Sqrt[c]*d^(3/2)*Sqrt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqr
t[c]], b/(b + a*c)])/(3*(b + a*c)^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 475

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*
x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x
], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] &&
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 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -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 {1}{x^4 \sqrt {a+\frac {b}{c+d x^2}}} \, dx &=\frac {\sqrt {b+a \left (c+d x^2\right )} \int \frac {\sqrt {c+d x^2}}{x^4 \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 {\sqrt {c+d x^2}}{x^4 \sqrt {b+a c+a d x^2}} \, dx}{\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {\sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 (b+a c) x^3 \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\sqrt {b+a \left (c+d x^2\right )} \int \frac {(b-a c) d-a d^2 x^2}{x^2 \sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{3 (b+a c) \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {\sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 (b+a c) x^3 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {(b-a c) d \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 c (b+a c)^2 x \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\sqrt {b+a \left (c+d x^2\right )} \int \frac {a c (b+a c) d^2-a (b-a c) d^3 x^2}{\sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{3 c (b+a c)^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {\sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 (b+a c) x^3 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {(b-a c) d \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 c (b+a c)^2 x \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\left (a d^2 \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}{3 (b+a c) \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\left (a (b-a c) d^3 \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}{3 c (b+a c)^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {\sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 (b+a c) x^3 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {(b-a c) d \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 c (b+a c)^2 x \sqrt {a+\frac {b}{c+d x^2}}}+\frac {(b-a c) d^2 x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 c (b+a c)^2 \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}-\frac {a \sqrt {c} d^{3/2} \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 )}{3 (b+a c)^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 ((b-a c) d^2 \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}{3 (b+a c)^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {\sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 (b+a c) x^3 \sqrt {a+\frac {b}{c+d x^2}}}-\frac {(b-a c) d \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 c (b+a c)^2 x \sqrt {a+\frac {b}{c+d x^2}}}+\frac {(b-a c) d^2 x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 c (b+a c)^2 \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}-\frac {(b-a c) d^{3/2} \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 )}{3 \sqrt {c} (b+a c)^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 {a \sqrt {c} d^{3/2} \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 )}{3 (b+a c)^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 = 1.02, size = 314, normalized size = 0.72 \[ \frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \left (-\left (c+d x^2\right ) \sqrt {\frac {a d}{a c+b}} \left (a^2 c \left (c^2-d^2 x^4\right )+a b \left (2 c^2+c d x^2+d^2 x^4\right )+b^2 \left (c+d x^2\right )\right )+2 i a b c d^2 x^3 \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )+i a c d^2 x^3 (a c-b) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )\right )}{3 c x^3 (a c+b)^2 \sqrt {\frac {a d}{a c+b}} \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[a + b/(c + d*x^2)]),x]

[Out]

(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(-(Sqrt[(a*d)/(b + a*c)]*(c + d*x^2)*(b^2*(c + d*x^2) + a^2*c*(c^2 - d^
2*x^4) + a*b*(2*c^2 + c*d*x^2 + d^2*x^4))) + I*a*c*(-b + a*c)*d^2*x^3*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)*a*b*c*d^2*x^3*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)]))/(3*c*(
b + a*c)^2*Sqrt[(a*d)/(b + a*c)]*x^3*(b + a*(c + d*x^2)))

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d x^{2} + c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{a d x^{6} + {\left (a c + b\right )} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + \frac {b}{d x^{2} + c}} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 596, normalized size = 1.37 \[ \frac {\left (\sqrt {-\frac {a d}{a c +b}}\, a^{2} c \,d^{3} x^{6}-\sqrt {-\frac {a d}{a c +b}}\, a b \,d^{3} x^{6}+\sqrt {-\frac {a d}{a c +b}}\, a^{2} c^{2} d^{2} x^{4}-\sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} c^{2} d^{2} x^{3} \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )-2 \sqrt {-\frac {a d}{a c +b}}\, a b c \,d^{2} x^{4}+\sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b c \,d^{2} x^{3} \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )-2 \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b c \,d^{2} x^{3} \EllipticF \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )-\sqrt {-\frac {a d}{a c +b}}\, a^{2} c^{3} d \,x^{2}-\sqrt {-\frac {a d}{a c +b}}\, b^{2} d^{2} x^{4}-3 \sqrt {-\frac {a d}{a c +b}}\, a b \,c^{2} d \,x^{2}-\sqrt {-\frac {a d}{a c +b}}\, a^{2} c^{4}-2 \sqrt {-\frac {a d}{a c +b}}\, b^{2} c d \,x^{2}-2 \sqrt {-\frac {a d}{a c +b}}\, a b \,c^{3}-\sqrt {-\frac {a d}{a c +b}}\, b^{2} c^{2}\right ) \left (d \,x^{2}+c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{3 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, \left (a c +b \right )^{2} \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, c \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((-1/(a*c+b)*a*d)^(1/2)*a^2*c*d^3*x^6-(-1/(a*c+b)*a*d)^(1/2)*a*b*d^3*x^6-((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))*x^3*a^2*c^2*d^2+(-1/(a*c+b)*a*d)^(1
/2)*a^2*c^2*d^2*x^4-2*((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))*x^3*a*b*c*d^2+((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))*x^3*a*b*c*d^2-2*(-1/(a*c+b)*a*d)^(1/2)*a*b*c*d^2*x^4-(-1/(a*c+b)*a*d)^(1/2)*b^
2*d^2*x^4-(-1/(a*c+b)*a*d)^(1/2)*a^2*c^3*d*x^2-3*(-1/(a*c+b)*a*d)^(1/2)*x^2*a*b*c^2*d-2*(-1/(a*c+b)*a*d)^(1/2)
*b^2*c*d*x^2-(-1/(a*c+b)*a*d)^(1/2)*a^2*c^4-2*(-1/(a*c+b)*a*d)^(1/2)*a*b*c^3-(-1/(a*c+b)*a*d)^(1/2)*b^2*c^2)*(
d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/(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)/c/x^3/(a*c+b)^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 \[ \int \frac {1}{\sqrt {a + \frac {b}{d x^{2} + c}} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(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 \[ \int \frac {1}{x^{4} \sqrt {\frac {a c + a d x^{2} + b}{c + d x^{2}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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