3.198 \(\int \sqrt {a+b x^3+c x^6} \, dx\)

Optimal. Leaf size=135 \[ \frac {x \sqrt {a+b x^3+c x^6} F_1\left (\frac {1}{3};-\frac {1}{2},-\frac {1}{2};\frac {4}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1}} \]

[Out]

x*AppellF1(1/3,-1/2,-1/2,4/3,-2*c*x^3/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^3/(b+(-4*a*c+b^2)^(1/2)))*(c*x^6+b*x^3+a)^
(1/2)/(1+2*c*x^3/(b-(-4*a*c+b^2)^(1/2)))^(1/2)/(1+2*c*x^3/(b+(-4*a*c+b^2)^(1/2)))^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1348, 429} \[ \frac {x \sqrt {a+b x^3+c x^6} F_1\left (\frac {1}{3};-\frac {1}{2},-\frac {1}{2};\frac {4}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^3 + c*x^6],x]

[Out]

(x*Sqrt[a + b*x^3 + c*x^6]*AppellF1[1/3, -1/2, -1/2, 4/3, (-2*c*x^3)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^3)/(b +
Sqrt[b^2 - 4*a*c])])/(Sqrt[1 + (2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])])

Rule 429

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

Rule 1348

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n + c*x^(2*n))
^FracPart[p])/((1 + (2*c*x^n)/(b + Rt[b^2 - 4*a*c, 2]))^FracPart[p]*(1 + (2*c*x^n)/(b - Rt[b^2 - 4*a*c, 2]))^F
racPart[p]), Int[(1 + (2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]))^p, x], x] /
; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \sqrt {a+b x^3+c x^6} \, dx &=\frac {\sqrt {a+b x^3+c x^6} \int \sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}} \, dx}{\sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}}}\\ &=\frac {x \sqrt {a+b x^3+c x^6} F_1\left (\frac {1}{3};-\frac {1}{2},-\frac {1}{2};\frac {4}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}}}\\ \end {align*}

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Mathematica [B]  time = 0.34, size = 335, normalized size = 2.48 \[ \frac {x \left (3 b x^3 \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {4}{3};\frac {1}{2},\frac {1}{2};\frac {7}{3};-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{\sqrt {b^2-4 a c}-b}\right )+24 a \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {1}{3};\frac {1}{2},\frac {1}{2};\frac {4}{3};-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{\sqrt {b^2-4 a c}-b}\right )+8 \left (a+b x^3+c x^6\right )\right )}{32 \sqrt {a+b x^3+c x^6}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[a + b*x^3 + c*x^6],x]

[Out]

(x*(8*(a + b*x^3 + c*x^6) + 24*a*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqr
t[b^2 - 4*a*c] + 2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[1/3, 1/2, 1/2, 4/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a
*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])] + 3*b*x^3*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b - Sqrt[b^2 - 4*a
*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[4/3, 1/2, 1/2, 7/3, (-2*c*x^3)/
(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])]))/(32*Sqrt[a + b*x^3 + c*x^6])

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {c x^{6} + b x^{3} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^6 + b*x^3 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c x^{6} + b x^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^6 + b*x^3 + a), x)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \sqrt {c \,x^{6}+b \,x^{3}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^6+b*x^3+a)^(1/2),x)

[Out]

int((c*x^6+b*x^3+a)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c x^{6} + b x^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^6 + b*x^3 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {c\,x^6+b\,x^3+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3 + c*x^6)^(1/2),x)

[Out]

int((a + b*x^3 + c*x^6)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b x^{3} + c x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**6+b*x**3+a)**(1/2),x)

[Out]

Integral(sqrt(a + b*x**3 + c*x**6), x)

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