3.246 \(\int \frac {x^{9/2} (A+B x^2)}{\sqrt {b x^2+c x^4}} \, dx\)

Optimal. Leaf size=204 \[ -\frac {5 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (9 b B-11 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{231 c^{13/4} \sqrt {b x^2+c x^4}}+\frac {10 b \sqrt {b x^2+c x^4} (9 b B-11 A c)}{231 c^3 \sqrt {x}}-\frac {2 x^{3/2} \sqrt {b x^2+c x^4} (9 b B-11 A c)}{77 c^2}+\frac {2 B x^{7/2} \sqrt {b x^2+c x^4}}{11 c} \]

[Out]

-2/77*(-11*A*c+9*B*b)*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c^2+2/11*B*x^(7/2)*(c*x^4+b*x^2)^(1/2)/c+10/231*b*(-11*A*c+9
*B*b)*(c*x^4+b*x^2)^(1/2)/c^3/x^(1/2)-5/231*b^(7/4)*(-11*A*c+9*B*b)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^
2)^(1/2)/cos(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)/c^(13/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2039, 2024, 2032, 329, 220} \[ -\frac {5 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (9 b B-11 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{231 c^{13/4} \sqrt {b x^2+c x^4}}-\frac {2 x^{3/2} \sqrt {b x^2+c x^4} (9 b B-11 A c)}{77 c^2}+\frac {10 b \sqrt {b x^2+c x^4} (9 b B-11 A c)}{231 c^3 \sqrt {x}}+\frac {2 B x^{7/2} \sqrt {b x^2+c x^4}}{11 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^(9/2)*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]

[Out]

(10*b*(9*b*B - 11*A*c)*Sqrt[b*x^2 + c*x^4])/(231*c^3*Sqrt[x]) - (2*(9*b*B - 11*A*c)*x^(3/2)*Sqrt[b*x^2 + c*x^4
])/(77*c^2) + (2*B*x^(7/2)*Sqrt[b*x^2 + c*x^4])/(11*c) - (5*b^(7/4)*(9*b*B - 11*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*S
qrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(231*c^(13/4)*Sq
rt[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 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

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 2039

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

Rubi steps

\begin {align*} \int \frac {x^{9/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx &=\frac {2 B x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {\left (2 \left (\frac {9 b B}{2}-\frac {11 A c}{2}\right )\right ) \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx}{11 c}\\ &=-\frac {2 (9 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c^2}+\frac {2 B x^{7/2} \sqrt {b x^2+c x^4}}{11 c}+\frac {(5 b (9 b B-11 A c)) \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx}{77 c^2}\\ &=\frac {10 b (9 b B-11 A c) \sqrt {b x^2+c x^4}}{231 c^3 \sqrt {x}}-\frac {2 (9 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c^2}+\frac {2 B x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {\left (5 b^2 (9 b B-11 A c)\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{231 c^3}\\ &=\frac {10 b (9 b B-11 A c) \sqrt {b x^2+c x^4}}{231 c^3 \sqrt {x}}-\frac {2 (9 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c^2}+\frac {2 B x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {\left (5 b^2 (9 b B-11 A c) x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{231 c^3 \sqrt {b x^2+c x^4}}\\ &=\frac {10 b (9 b B-11 A c) \sqrt {b x^2+c x^4}}{231 c^3 \sqrt {x}}-\frac {2 (9 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c^2}+\frac {2 B x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {\left (10 b^2 (9 b B-11 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 )}{231 c^3 \sqrt {b x^2+c x^4}}\\ &=\frac {10 b (9 b B-11 A c) \sqrt {b x^2+c x^4}}{231 c^3 \sqrt {x}}-\frac {2 (9 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c^2}+\frac {2 B x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {5 b^{7/4} (9 b B-11 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 )}{231 c^{13/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.15, size = 122, normalized size = 0.60 \[ \frac {2 x^{3/2} \left (\left (b+c x^2\right ) \left (-b c \left (55 A+27 B x^2\right )+3 c^2 x^2 \left (11 A+7 B x^2\right )+45 b^2 B\right )+5 b^2 \sqrt {\frac {c x^2}{b}+1} (11 A c-9 b B) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^2}{b}\right )\right )}{231 c^3 \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(9/2)*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]

[Out]

(2*x^(3/2)*((b + c*x^2)*(45*b^2*B + 3*c^2*x^2*(11*A + 7*B*x^2) - b*c*(55*A + 27*B*x^2)) + 5*b^2*(-9*b*B + 11*A
*c)*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(231*c^3*Sqrt[x^2*(b + c*x^2)])

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{4} + A x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{c x^{2} + b}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} x^{\frac {9}{2}}}{\sqrt {c x^{4} + b x^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*x^(9/2)/sqrt(c*x^4 + b*x^2), x)

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maple [A]  time = 0.07, size = 274, normalized size = 1.34 \[ \frac {\left (42 B \,c^{4} x^{7}+66 A \,c^{4} x^{5}-12 B b \,c^{3} x^{5}-44 A b \,c^{3} x^{3}+36 B \,b^{2} c^{2} x^{3}-110 A \,b^{2} c^{2} x +90 B \,b^{3} c x +55 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A \,b^{2} c \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-45 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B \,b^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {x}}{231 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x)

[Out]

1/231/(c*x^4+b*x^2)^(1/2)*x^(1/2)*(42*B*c^4*x^7+55*A*(-b*c)^(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)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-
1/(-b*c)^(1/2)*c*x)^(1/2)*b^2*c+66*A*c^4*x^5-45*B*(-b*c)^(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)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-1/(
-b*c)^(1/2)*c*x)^(1/2)*b^3-12*B*b*c^3*x^5-44*A*b*c^3*x^3+36*B*x^3*b^2*c^2-110*A*b^2*c^2*x+90*B*x*b^3*c)/c^4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} x^{\frac {9}{2}}}{\sqrt {c x^{4} + b x^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*x^(9/2)/sqrt(c*x^4 + b*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^{9/2}\,\left (B\,x^2+A\right )}{\sqrt {c\,x^4+b\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(9/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(1/2),x)

[Out]

int((x^(9/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(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**(9/2)*(B*x**2+A)/(c*x**4+b*x**2)**(1/2),x)

[Out]

Timed out

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