3.1095 \(\int \frac {(a+b x^2+c x^4)^{3/2}}{\sqrt {d x}} \, dx\)

Optimal. Leaf size=146 \[ \frac {2 a \sqrt {d x} \sqrt {a+b x^2+c x^4} F_1\left (\frac {1}{4};-\frac {3}{2},-\frac {3}{2};\frac {5}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 146, 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} \[ \frac {2 a \sqrt {d x} \sqrt {a+b x^2+c x^4} F_1\left (\frac {1}{4};-\frac {3}{2},-\frac {3}{2};\frac {5}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [B]  time = 0.71, size = 415, normalized size = 2.84 \[ \frac {2 x \left (5 \left (51 a^2 c+4 a b^2+76 a b c x^2+66 a c^2 x^4+4 b^3 x^2+29 b^2 c x^4+40 b c^2 x^6+15 c^3 x^8\right )-4 b x^2 \left (3 b^2-28 a c\right ) \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {5}{4};\frac {1}{2},\frac {1}{2};\frac {9}{4};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}\right )-20 a \left (b^2-36 a c\right ) \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {1}{4};\frac {1}{2},\frac {1}{2};\frac {5}{4};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}\right )\right )}{975 c \sqrt {d x} \sqrt {a+b x^2+c x^4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*x*(5*(4*a*b^2 + 51*a^2*c + 4*b^3*x^2 + 76*a*b*c*x^2 + 29*b^2*c*x^4 + 66*a*c^2*x^4 + 40*b*c^2*x^6 + 15*c^3*x
^8) - 20*a*(b^2 - 36*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])] - 4*b*(3*b^2 - 28*a*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])]))/(975*c*Sqrt[d*x]*Sqrt[a + b*x^2 + c*x
^4])

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {d x}}{d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{\sqrt {d x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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