3.362 \(\int \frac {x^2}{(a+\frac {b}{c+d x^2})^{3/2}} \, dx\)

Optimal. Leaf size=409 \[ \frac {\sqrt {c} (a c+8 b) \left (a c+a d x^2+b\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 a^3 d^{3/2} \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}-\frac {x (a c+8 b) \left (a c+a d x^2+b\right )}{3 a^3 d \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}-\frac {c^{3/2} (a c+4 b) \left (a c+a d x^2+b\right ) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 a^2 d^{3/2} (a c+b) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}+\frac {4 x \left (a c+a d x^2+b\right )}{3 a^2 d \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}-\frac {x \left (c+d x^2\right )}{a d \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}} \]

[Out]

-x*(d*x^2+c)/a/d/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)+4/3*x*(a*d*x^2+a*c+b)/a^2/d/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/
2)-1/3*(a*c+8*b)*x*(a*d*x^2+a*c+b)/a^3/d/(d*x^2+c)/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)-1/3*c^(3/2)*(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)*EllipticF(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(b/(a*c+b))
^(1/2))/a^2/(a*c+b)/d^(3/2)/(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)+1/3*(a*c+8*b)*(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^(3/2)/(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 = 0.66, antiderivative size = 475, normalized size of antiderivative = 1.16, 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, 467, 528, 531, 418, 492, 411} \[ -\frac {c^{3/2} (a c+4 b) \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 a^2 d^{3/2} (a c+b) \left (c+d x^2\right ) \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\sqrt {c} (a c+8 b) \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 a^3 d^{3/2} \left (c+d x^2\right ) \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a+\frac {b}{c+d x^2}}}+\frac {4 x \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b}}{3 a^2 d \sqrt {a+\frac {b}{c+d x^2}}}-\frac {x (a c+8 b) \sqrt {a c+a d x^2+b} \sqrt {a \left (c+d x^2\right )+b}}{3 a^3 d \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}-\frac {x \left (c+d x^2\right ) \sqrt {a \left (c+d x^2\right )+b}}{a d \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b/(c + d*x^2))^(3/2),x]

[Out]

-((x*(c + d*x^2)*Sqrt[b + a*(c + d*x^2)])/(a*d*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])) + (4*x*Sqrt[b
+ a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)])/(3*a^2*d*Sqrt[a + b/(c + d*x^2)]) - ((8*b + a*c)*x*Sqrt[b + a*c + a
*d*x^2]*Sqrt[b + a*(c + d*x^2)])/(3*a^3*d*(c + d*x^2)*Sqrt[a + b/(c + d*x^2)]) + (Sqrt[c]*(8*b + a*c)*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*a^3*d^(3/2)*(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 + a*c
)*Sqrt[b + a*c + a*d*x^2]*Sqrt[b + a*(c + d*x^2)]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(3*a^2*
(b + a*c)*d^(3/2)*(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 467

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*n*(p + 1)), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && 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 528

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

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 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^2}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx &=\frac {\sqrt {b+a \left (c+d x^2\right )} \int \frac {x^2 \left (c+d x^2\right )^{3/2}}{\left (b+a \left (c+d x^2\right )\right )^{3/2}} \, 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^2 \left (c+d x^2\right )^{3/2}}{\left (b+a c+a d x^2\right )^{3/2}} \, dx}{\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {x \left (c+d x^2\right ) \sqrt {b+a \left (c+d x^2\right )}}{a d \sqrt {b+a c+a 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} \left (c+4 d x^2\right )}{\sqrt {b+a c+a d x^2}} \, dx}{a d \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {x \left (c+d x^2\right ) \sqrt {b+a \left (c+d x^2\right )}}{a d \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}+\frac {4 x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 a^2 d \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\sqrt {b+a \left (c+d x^2\right )} \int \frac {-c (4 b+a c) d-(8 b+a c) d^2 x^2}{\sqrt {c+d x^2} \sqrt {b+a c+a d x^2}} \, dx}{3 a^2 d^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {x \left (c+d x^2\right ) \sqrt {b+a \left (c+d x^2\right )}}{a d \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}+\frac {4 x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 a^2 d \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\left ((8 b+a c) \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 a^2 \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}-\frac {\left (c (4 b+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}{3 a^2 d \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {x \left (c+d x^2\right ) \sqrt {b+a \left (c+d x^2\right )}}{a d \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}+\frac {4 x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 a^2 d \sqrt {a+\frac {b}{c+d x^2}}}-\frac {(8 b+a c) x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 a^3 d \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}-\frac {c^{3/2} (4 b+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 )}{3 a^2 (b+a c) d^{3/2} \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 (8 b+a c) \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 a^3 d \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}}\\ &=-\frac {x \left (c+d x^2\right ) \sqrt {b+a \left (c+d x^2\right )}}{a d \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}+\frac {4 x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 a^2 d \sqrt {a+\frac {b}{c+d x^2}}}-\frac {(8 b+a c) x \sqrt {b+a c+a d x^2} \sqrt {b+a \left (c+d x^2\right )}}{3 a^3 d \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}+\frac {\sqrt {c} (8 b+a c) \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 a^3 d^{3/2} \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+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 )}{3 a^2 (b+a c) d^{3/2} \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 = 0.60, size = 255, normalized size = 0.62 \[ \frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \left (x \left (c+d x^2\right ) \sqrt {\frac {a d}{a c+b}} \left (a c+a d x^2+4 b\right )-4 i b c \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )+i c (a c+8 b) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )\right )}{3 a^2 d \sqrt {\frac {a d}{a c+b}} \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b/(c + d*x^2))^(3/2),x]

[Out]

(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 + a*c + a*d*x^2) + I*c*(8*b +
a*c)*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)] - (4*I)*b*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)]))/(3*a^2*d*Sqrt[(a*d)/(b + a*c)]*(b + a*(c + d*x^2)))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d^{2} x^{6} + 2 \, c d x^{4} + c^{2} x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{a^{2} d^{2} x^{4} + a^{2} c^{2} + 2 \, {\left (a^{2} c + a b\right )} d x^{2} + 2 \, a b c + b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b/(d*x^2+c))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (a + \frac {b}{d x^{2} + c}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b/(d*x^2+c))^(3/2),x, algorithm="giac")

[Out]

integrate(x^2/(a + b/(d*x^2 + c))^(3/2), x)

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maple [A]  time = 0.04, size = 667, normalized size = 1.63 \[ \frac {\left (\sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {-\frac {a d}{a c +b}}\, a \,d^{2} x^{5}+2 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {-\frac {a d}{a c +b}}\, a c d \,x^{3}+\sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {-\frac {a d}{a c +b}}\, b d \,x^{3}+3 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, b d \,x^{3}+\sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {-\frac {a d}{a c +b}}\, a \,c^{2} x -\sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a \,c^{2} \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+\sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {-\frac {a d}{a c +b}}\, b c x +3 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, b c x -8 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b c \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+4 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b c \EllipticF \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )\right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{3 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, \left (a d \,x^{2}+a c +b \right ) a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a+b/(d*x^2+c))^(3/2),x)

[Out]

1/3*(((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(-1/(a*c+b)*a*d)^(1/2)*x^5*a*d^2+2*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(-
1/(a*c+b)*a*d)^(1/2)*x^3*a*c*d+((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(-1/(a*c+b)*a*d)^(1/2)*x^3*b*d-((d*x^2+c)*(a*
d*x^2+a*c+b))^(1/2)*((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))*a*c^2+3*(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)*x^3*b*d+((d*x
^2+c)*(a*d*x^2+a*c+b))^(1/2)*(-1/(a*c+b)*a*d)^(1/2)*x*a*c^2+4*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*((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))*b*c-8*((d*x^2+c)
*(a*d*x^2+a*c+b))^(1/2)*((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))*b*c+((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(-1/(a*c+b)*a*d)^(1/2)*x*b*c+3*(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)*x*b*c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/a^2/d/(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)/(a*d*x^2+a*c+b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (a + \frac {b}{d x^{2} + c}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b/(d*x^2+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(x^2/(a + b/(d*x^2 + c))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2}{{\left (a+\frac {b}{d\,x^2+c}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b/(c + d*x^2))^(3/2),x)

[Out]

int(x^2/(a + b/(c + d*x^2))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (\frac {a c + a d x^{2} + b}{c + d x^{2}}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a+b/(d*x**2+c))**(3/2),x)

[Out]

Integral(x**2/((a*c + a*d*x**2 + b)/(c + d*x**2))**(3/2), x)

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