3.220 \(\int x^{5/2} (A+B x^2) \sqrt {b x^2+c x^4} \, dx\)

Optimal. Leaf size=243 \[ -\frac {2 b^{11/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (3 b B-5 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{231 c^{13/4} \sqrt {b x^2+c x^4}}+\frac {4 b^2 \sqrt {b x^2+c x^4} (3 b B-5 A c)}{231 c^3 \sqrt {x}}-\frac {4 b x^{3/2} \sqrt {b x^2+c x^4} (3 b B-5 A c)}{385 c^2}-\frac {2 x^{7/2} \sqrt {b x^2+c x^4} (3 b B-5 A c)}{55 c}+\frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c} \]

[Out]

2/15*B*x^(3/2)*(c*x^4+b*x^2)^(3/2)/c-4/385*b*(-5*A*c+3*B*b)*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c^2-2/55*(-5*A*c+3*B*b
)*x^(7/2)*(c*x^4+b*x^2)^(1/2)/c+4/231*b^2*(-5*A*c+3*B*b)*(c*x^4+b*x^2)^(1/2)/c^3/x^(1/2)-2/231*b^(11/4)*(-5*A*
c+3*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(s
in(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)

________________________________________________________________________________________

Rubi [A]  time = 0.39, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2039, 2021, 2024, 2032, 329, 220} \[ \frac {4 b^2 \sqrt {b x^2+c x^4} (3 b B-5 A c)}{231 c^3 \sqrt {x}}-\frac {2 b^{11/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (3 b B-5 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{231 c^{13/4} \sqrt {b x^2+c x^4}}-\frac {4 b x^{3/2} \sqrt {b x^2+c x^4} (3 b B-5 A c)}{385 c^2}-\frac {2 x^{7/2} \sqrt {b x^2+c x^4} (3 b B-5 A c)}{55 c}+\frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b^2*(3*b*B - 5*A*c)*Sqrt[b*x^2 + c*x^4])/(231*c^3*Sqrt[x]) - (4*b*(3*b*B - 5*A*c)*x^(3/2)*Sqrt[b*x^2 + c*x^
4])/(385*c^2) - (2*(3*b*B - 5*A*c)*x^(7/2)*Sqrt[b*x^2 + c*x^4])/(55*c) + (2*B*x^(3/2)*(b*x^2 + c*x^4)^(3/2))/(
15*c) - (2*b^(11/4)*(3*b*B - 5*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*Elliptic
F[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(231*c^(13/4)*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 2021

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

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 x^{5/2} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx &=\frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {\left (2 \left (\frac {9 b B}{2}-\frac {15 A c}{2}\right )\right ) \int x^{5/2} \sqrt {b x^2+c x^4} \, dx}{15 c}\\ &=-\frac {2 (3 b B-5 A c) x^{7/2} \sqrt {b x^2+c x^4}}{55 c}+\frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {(2 b (3 b B-5 A c)) \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx}{55 c}\\ &=-\frac {4 b (3 b B-5 A c) x^{3/2} \sqrt {b x^2+c x^4}}{385 c^2}-\frac {2 (3 b B-5 A c) x^{7/2} \sqrt {b x^2+c x^4}}{55 c}+\frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}+\frac {\left (2 b^2 (3 b B-5 A c)\right ) \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx}{77 c^2}\\ &=\frac {4 b^2 (3 b B-5 A c) \sqrt {b x^2+c x^4}}{231 c^3 \sqrt {x}}-\frac {4 b (3 b B-5 A c) x^{3/2} \sqrt {b x^2+c x^4}}{385 c^2}-\frac {2 (3 b B-5 A c) x^{7/2} \sqrt {b x^2+c x^4}}{55 c}+\frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {\left (2 b^3 (3 b B-5 A c)\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{231 c^3}\\ &=\frac {4 b^2 (3 b B-5 A c) \sqrt {b x^2+c x^4}}{231 c^3 \sqrt {x}}-\frac {4 b (3 b B-5 A c) x^{3/2} \sqrt {b x^2+c x^4}}{385 c^2}-\frac {2 (3 b B-5 A c) x^{7/2} \sqrt {b x^2+c x^4}}{55 c}+\frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {\left (2 b^3 (3 b B-5 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 {4 b^2 (3 b B-5 A c) \sqrt {b x^2+c x^4}}{231 c^3 \sqrt {x}}-\frac {4 b (3 b B-5 A c) x^{3/2} \sqrt {b x^2+c x^4}}{385 c^2}-\frac {2 (3 b B-5 A c) x^{7/2} \sqrt {b x^2+c x^4}}{55 c}+\frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {\left (4 b^3 (3 b B-5 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 {4 b^2 (3 b B-5 A c) \sqrt {b x^2+c x^4}}{231 c^3 \sqrt {x}}-\frac {4 b (3 b B-5 A c) x^{3/2} \sqrt {b x^2+c x^4}}{385 c^2}-\frac {2 (3 b B-5 A c) x^{7/2} \sqrt {b x^2+c x^4}}{55 c}+\frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac {2 b^{11/4} (3 b B-5 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*}

________________________________________________________________________________________

Mathematica [C]  time = 0.18, size = 136, normalized size = 0.56 \[ \frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (\left (b+c x^2\right ) \sqrt {\frac {c x^2}{b}+1} \left (-3 b c \left (25 A+21 B x^2\right )+7 c^2 x^2 \left (15 A+11 B x^2\right )+45 b^2 B\right )+15 b^2 (5 A c-3 b B) \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {c x^2}{b}\right )\right )}{1155 c^3 \sqrt {x} \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x^2*(b + c*x^2)]*((b + c*x^2)*Sqrt[1 + (c*x^2)/b]*(45*b^2*B + 7*c^2*x^2*(15*A + 11*B*x^2) - 3*b*c*(25*
A + 21*B*x^2)) + 15*b^2*(-3*b*B + 5*A*c)*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x^2)/b)]))/(1155*c^3*Sqrt[x]*S
qrt[1 + (c*x^2)/b])

________________________________________________________________________________________

fricas [F]  time = 1.08, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B x^{4} + A x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}, x\right ) \]

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)^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

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)^(1/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.20, size = 307, normalized size = 1.26 \[ \frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (77 B \,c^{5} x^{9}+105 A \,c^{5} x^{7}+91 B b \,c^{4} x^{7}+135 A b \,c^{4} x^{5}-4 B \,b^{2} c^{3} x^{5}-20 A \,b^{2} c^{3} x^{3}+12 B \,b^{3} c^{2} x^{3}-50 A \,b^{3} c^{2} x +30 B \,b^{4} c x +25 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A \,b^{3} c \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-15 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B \,b^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )\right )}{1155 \left (c \,x^{2}+b \right ) c^{4} x^{\frac {3}{2}}} \]

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)^(1/2),x)

[Out]

2/1155*(c*x^4+b*x^2)^(1/2)/x^(3/2)/(c*x^2+b)*(77*B*x^9*c^5+105*A*x^7*c^5+91*B*x^7*b*c^4+25*A*(-b*c)^(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)*(-x*c/(-b*c)^(1/2))^(1/2)
*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b^3*c+135*A*x^5*b*c^4-15*B*(-b*c)^(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)*(-x*c/(-b*c)^(1/2))^(1/2)*E
llipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b^4-4*B*x^5*b^2*c^3-20*A*x^3*b^2*c^3+12*B*x^3*b^
3*c^2-50*A*x*b^3*c^2+30*B*x*b^4*c)/c^4

________________________________________________________________________________________

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

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)^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)**(1/2),x)

[Out]

Integral(x**(5/2)*sqrt(x**2*(b + c*x**2))*(A + B*x**2), x)

________________________________________________________________________________________