3.1103 \(\int \frac {1}{\sqrt {d x} (a+b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=148 \[ \frac {2 \sqrt {d x} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {1}{4};\frac {3}{2},\frac {3}{2};\frac {5}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{a d \sqrt {a+b x^2+c x^4}} \]

[Out]

2*AppellF1(1/4,3/2,3/2,5/4,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))*(d*x)^(1/2)*(1+2*c
*x^2/(b-(-4*a*c+b^2)^(1/2)))^(1/2)*(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2)/a/d/(c*x^4+b*x^2+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1141, 510} \[ \frac {2 \sqrt {d x} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {1}{4};\frac {3}{2},\frac {3}{2};\frac {5}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{a d \sqrt {a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d*x]*(a + b*x^2 + c*x^4)^(3/2)),x]

[Out]

(2*Sqrt[d*x]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[
1/4, 3/2, 3/2, 5/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(a*d*Sqrt[a + b*x
^2 + c*x^4])

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1141

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^2 +
c*x^4)^FracPart[p])/((1 + (2*c*x^2)/(b + Rt[b^2 - 4*a*c, 2]))^FracPart[p]*(1 + (2*c*x^2)/(b - Rt[b^2 - 4*a*c,
2]))^FracPart[p]), Int[(d*x)^m*(1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c
]))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {d x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx &=\frac {\left (\sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}\right ) \int \frac {1}{\sqrt {d x} \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{3/2} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{3/2}} \, dx}{a \sqrt {a+b x^2+c x^4}}\\ &=\frac {2 \sqrt {d x} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} F_1\left (\frac {1}{4};\frac {3}{2},\frac {3}{2};\frac {5}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{a d \sqrt {a+b x^2+c x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.47, size = 366, normalized size = 2.47 \[ \frac {x \left (b c x^2 \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {5}{4};\frac {1}{2},\frac {1}{2};\frac {9}{4};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}\right )-5 \left (b^2-6 a c\right ) \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {1}{4};\frac {1}{2},\frac {1}{2};\frac {5}{4};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}\right )-5 \left (-2 a c+b^2+b c x^2\right )\right )}{5 a \sqrt {d x} \left (4 a c-b^2\right ) \sqrt {a+b x^2+c x^4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sqrt[d*x]*(a + b*x^2 + c*x^4)^(3/2)),x]

[Out]

(x*(-5*(b^2 - 2*a*c + b*c*x^2) - 5*(b^2 - 6*a*c)*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]
)]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[1/4, 1/2, 1/2, 5/4, (-2*c*x^2)/(b
+ Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + b*c*x^2*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b
- Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[5/4, 1/2, 1/2,
9/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]))/(5*a*(-b^2 + 4*a*c)*Sqrt[d*x]*S
qrt[a + b*x^2 + c*x^4])

________________________________________________________________________________________

fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + b x^{2} + a} \sqrt {d x}}{c^{2} d x^{9} + 2 \, b c d x^{7} + {\left (b^{2} + 2 \, a c\right )} d x^{5} + 2 \, a b d x^{3} + a^{2} d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)^(1/2)/(c*x^4+b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2 + a)*sqrt(d*x)/(c^2*d*x^9 + 2*b*c*d*x^7 + (b^2 + 2*a*c)*d*x^5 + 2*a*b*d*x^3 + a^2*
d*x), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {d x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)^(1/2)/(c*x^4+b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((c*x^4 + b*x^2 + a)^(3/2)*sqrt(d*x)), x)

________________________________________________________________________________________

maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {d x}\, \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*x)^(1/2)/(c*x^4+b*x^2+a)^(3/2),x)

[Out]

int(1/(d*x)^(1/2)/(c*x^4+b*x^2+a)^(3/2),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {d x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)^(1/2)/(c*x^4+b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((c*x^4 + b*x^2 + a)^(3/2)*sqrt(d*x)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d*x)^(1/2)*(a + b*x^2 + c*x^4)^(3/2)),x)

[Out]

int(1/((d*x)^(1/2)*(a + b*x^2 + c*x^4)^(3/2)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {d x} \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)**(1/2)/(c*x**4+b*x**2+a)**(3/2),x)

[Out]

Integral(1/(sqrt(d*x)*(a + b*x**2 + c*x**4)**(3/2)), x)

________________________________________________________________________________________