3.263 \(\int \frac {x^{5/2} (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=137 \[ \frac {x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (A c+b B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 b^{5/4} c^{5/4} \sqrt {b x^2+c x^4}}-\frac {x^{3/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}} \]

[Out]

-(-A*c+B*b)*x^(3/2)/b/c/(c*x^4+b*x^2)^(1/2)+1/2*(A*c+B*b)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/c
os(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x
*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/b^(5/4)/c^(5/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2037, 2032, 329, 220} \[ \frac {x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (A c+b B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 b^{5/4} c^{5/4} \sqrt {b x^2+c x^4}}-\frac {x^{3/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(((b*B - A*c)*x^(3/2))/(b*c*Sqrt[b*x^2 + c*x^4])) + ((b*B + A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sq
rt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(2*b^(5/4)*c^(5/4)*Sqrt[b*x^2 + c*x
^4])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 2037

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> -Si
mp[(e^(j - 1)*(b*c - a*d)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*b*n*(p + 1)), x] - Dist[(e^j*(a*
d*(m + j*p + 1) - b*c*(m + n + p*(j + n) + 1)))/(a*b*n*(p + 1)), Int[(e*x)^(m - j)*(a*x^j + b*x^(j + n))^(p +
1), x], x] /; FreeQ[{a, b, c, d, e, j, m, n}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && Lt
Q[p, -1] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])

Rubi steps

\begin {align*} \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac {(b B-A c) x^{3/2}}{b c \sqrt {b x^2+c x^4}}+\frac {(b B+A c) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{2 b c}\\ &=-\frac {(b B-A c) x^{3/2}}{b c \sqrt {b x^2+c x^4}}+\frac {\left ((b B+A c) x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{2 b c \sqrt {b x^2+c x^4}}\\ &=-\frac {(b B-A c) x^{3/2}}{b c \sqrt {b x^2+c x^4}}+\frac {\left ((b B+A c) x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{b c \sqrt {b x^2+c x^4}}\\ &=-\frac {(b B-A c) x^{3/2}}{b c \sqrt {b x^2+c x^4}}+\frac {(b B+A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 b^{5/4} c^{5/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 76, normalized size = 0.55 \[ \frac {x^{3/2} \left (\sqrt {\frac {c x^2}{b}+1} (A c+b B) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^2}{b}\right )+A c-b B\right )}{b c \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x^(3/2)*(-(b*B) + A*c + (b*B + A*c)*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(b*c
*Sqrt[x^2*(b + c*x^2)])

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )} \sqrt {x}}{c^{2} x^{6} + 2 \, b c x^{4} + b^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)*sqrt(x)/(c^2*x^6 + 2*b*c*x^4 + b^2*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} x^{\frac {5}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*x^(5/2)/(c*x^4 + b*x^2)^(3/2), x)

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maple [A]  time = 0.09, size = 222, normalized size = 1.62 \[ \frac {\left (c \,x^{2}+b \right ) \left (2 A \,c^{2} x -2 B b c x +\sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-b c}\, A c \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+\sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-b c}\, B b \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )\right ) x^{\frac {5}{2}}}{2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b \,c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)

[Out]

1/2/(c*x^4+b*x^2)^(3/2)*x^(5/2)*(c*x^2+b)*(A*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-1/(-b*c)^(1/2)
*c*x)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(
1/2)*(-b*c)^(1/2)*c+B*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-1/(-b*c)^(1/2)*c*x)^(1/2)*EllipticF((
(c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-b*c)^(1/2)*b+2*A
*c^2*x-2*B*b*c*x)/b/c^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} x^{\frac {5}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*x^(5/2)/(c*x^4 + b*x^2)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{5/2}\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)

[Out]

int((x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2), 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**(5/2)*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Timed out

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