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

Optimal. Leaf size=203 \[ -\frac {5 c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (7 b B-9 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{42 b^{13/4} \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4} (7 b B-9 A c)}{21 b^3 x^{5/2}}+\frac {7 b B-9 A c}{7 b^2 \sqrt {x} \sqrt {b x^2+c x^4}}-\frac {2 A}{7 b x^{5/2} \sqrt {b x^2+c x^4}} \]

[Out]

-2/7*A/b/x^(5/2)/(c*x^4+b*x^2)^(1/2)+1/7*(-9*A*c+7*B*b)/b^2/x^(1/2)/(c*x^4+b*x^2)^(1/2)-5/21*(-9*A*c+7*B*b)*(c
*x^4+b*x^2)^(1/2)/b^3/x^(5/2)-5/42*c^(3/4)*(-9*A*c+7*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^(13/4)/(c*x^4+b*x^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.33, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2038, 2023, 2025, 2032, 329, 220} \[ -\frac {5 c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (7 b B-9 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{42 b^{13/4} \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4} (7 b B-9 A c)}{21 b^3 x^{5/2}}+\frac {7 b B-9 A c}{7 b^2 \sqrt {x} \sqrt {b x^2+c x^4}}-\frac {2 A}{7 b x^{5/2} \sqrt {b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(7*b*x^(5/2)*Sqrt[b*x^2 + c*x^4]) + (7*b*B - 9*A*c)/(7*b^2*Sqrt[x]*Sqrt[b*x^2 + c*x^4]) - (5*(7*b*B - 9
*A*c)*Sqrt[b*x^2 + c*x^4])/(21*b^3*x^(5/2)) - (5*c^(3/4)*(7*b*B - 9*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x
^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(42*b^(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 2023

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(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]) && LtQ[p, -1]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*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]) && LtQ[m + j*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 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {A+B x^2}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac {2 A}{7 b x^{5/2} \sqrt {b x^2+c x^4}}-\frac {\left (2 \left (-\frac {7 b B}{2}+\frac {9 A c}{2}\right )\right ) \int \frac {\sqrt {x}}{\left (b x^2+c x^4\right )^{3/2}} \, dx}{7 b}\\ &=-\frac {2 A}{7 b x^{5/2} \sqrt {b x^2+c x^4}}+\frac {7 b B-9 A c}{7 b^2 \sqrt {x} \sqrt {b x^2+c x^4}}+\frac {(5 (7 b B-9 A c)) \int \frac {1}{x^{3/2} \sqrt {b x^2+c x^4}} \, dx}{14 b^2}\\ &=-\frac {2 A}{7 b x^{5/2} \sqrt {b x^2+c x^4}}+\frac {7 b B-9 A c}{7 b^2 \sqrt {x} \sqrt {b x^2+c x^4}}-\frac {5 (7 b B-9 A c) \sqrt {b x^2+c x^4}}{21 b^3 x^{5/2}}-\frac {(5 c (7 b B-9 A c)) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{42 b^3}\\ &=-\frac {2 A}{7 b x^{5/2} \sqrt {b x^2+c x^4}}+\frac {7 b B-9 A c}{7 b^2 \sqrt {x} \sqrt {b x^2+c x^4}}-\frac {5 (7 b B-9 A c) \sqrt {b x^2+c x^4}}{21 b^3 x^{5/2}}-\frac {\left (5 c (7 b B-9 A c) x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{42 b^3 \sqrt {b x^2+c x^4}}\\ &=-\frac {2 A}{7 b x^{5/2} \sqrt {b x^2+c x^4}}+\frac {7 b B-9 A c}{7 b^2 \sqrt {x} \sqrt {b x^2+c x^4}}-\frac {5 (7 b B-9 A c) \sqrt {b x^2+c x^4}}{21 b^3 x^{5/2}}-\frac {\left (5 c (7 b B-9 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 )}{21 b^3 \sqrt {b x^2+c x^4}}\\ &=-\frac {2 A}{7 b x^{5/2} \sqrt {b x^2+c x^4}}+\frac {7 b B-9 A c}{7 b^2 \sqrt {x} \sqrt {b x^2+c x^4}}-\frac {5 (7 b B-9 A c) \sqrt {b x^2+c x^4}}{21 b^3 x^{5/2}}-\frac {5 c^{3/4} (7 b B-9 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 )}{42 b^{13/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 79, normalized size = 0.39 \[ \frac {2 x^2 \sqrt {\frac {c x^2}{b}+1} (9 A c-7 b B) \, _2F_1\left (-\frac {3}{4},\frac {3}{2};\frac {1}{4};-\frac {c x^2}{b}\right )-6 A b}{21 b^2 x^{5/2} \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*A*b + 2*(-7*b*B + 9*A*c)*x^2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[-3/4, 3/2, 1/4, -((c*x^2)/b)])/(21*b^2*
x^(5/2)*Sqrt[x^2*(b + c*x^2)])

________________________________________________________________________________________

fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )} \sqrt {x}}{c^{2} x^{10} + 2 \, b c x^{8} + b^{2} x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^(3/2)/(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^10 + 2*b*c*x^8 + b^2*x^6), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.09, size = 254, normalized size = 1.25 \[ \frac {\left (c \,x^{2}+b \right ) \left (90 A \,c^{2} x^{4}-70 B b c \,x^{4}+45 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A c \,x^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-35 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B b \,x^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+36 A b c \,x^{2}-28 B \,b^{2} x^{2}-12 A \,b^{2}\right )}{42 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{3} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/42/(c*x^4+b*x^2)^(3/2)/x^(1/2)*(c*x^2+b)*(45*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)*(-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))*x^3*c-35*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)*(-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))*x^3*b+90*A*c^2*x^4-70*B*b*c*x^4+36*A*b*c*x^2-28*B*b^2*x^2-12*A*b^2)/b^3

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________