3.216 \(\int (a+b x^3+c x^6)^{3/2} \, dx\)

Optimal. Leaf size=136 \[ \frac {a x \sqrt {a+b x^3+c x^6} F_1\left (\frac {1}{3};-\frac {3}{2},-\frac {3}{2};\frac {4}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1}} \]

[Out]

a*x*AppellF1(1/3,-3/2,-3/2,4/3,-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 = 0.07, antiderivative size = 136, 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} \[ \frac {a x \sqrt {a+b x^3+c x^6} F_1\left (\frac {1}{3};-\frac {3}{2},-\frac {3}{2};\frac {4}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3 + c*x^6)^(3/2),x]

[Out]

(a*x*Sqrt[a + b*x^3 + c*x^6]*AppellF1[1/3, -3/2, -3/2, 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 \left (a+b x^3+c x^6\right )^{3/2} \, dx &=\frac {\left (a \sqrt {a+b x^3+c x^6}\right ) \int \left (1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}\right )^{3/2} \left (1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )^{3/2} \, 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 {a x \sqrt {a+b x^3+c x^6} F_1\left (\frac {1}{3};-\frac {3}{2},-\frac {3}{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 = 0.67, size = 408, normalized size = 3.00 \[ \frac {x \left (8 \left (364 a^2 c+27 a b^2+548 a b c x^3+476 a c^2 x^6+27 b^3 x^3+211 b^2 c x^6+296 b c^2 x^9+112 c^3 x^{12}\right )-27 b x^3 \left (5 b^2-44 a c\right ) \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {4}{3};\frac {1}{2},\frac {1}{2};\frac {7}{3};-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{\sqrt {b^2-4 a c}-b}\right )-216 a \left (b^2-28 a c\right ) \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {1}{3};\frac {1}{2},\frac {1}{2};\frac {4}{3};-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{\sqrt {b^2-4 a c}-b}\right )\right )}{8960 c \sqrt {a+b x^3+c x^6}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3 + c*x^6)^(3/2),x]

[Out]

(x*(8*(27*a*b^2 + 364*a^2*c + 27*b^3*x^3 + 548*a*b*c*x^3 + 211*b^2*c*x^6 + 476*a*c^2*x^6 + 296*b*c^2*x^9 + 112
*c^3*x^12) - 216*a*(b^2 - 28*a*c)*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sq
rt[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])] - 27*b*(5*b^2 - 44*a*c)*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])]))/(8960*c*Sqrt[a + b*x^3 + c*x
^6])

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,x^6+b\,x^3+a\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3 + c*x^6)^(3/2),x)

[Out]

int((a + b*x^3 + c*x^6)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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