3.266 \(\int \frac {(a+b x^3+c x^6)^p}{x^3} \, dx\)

Optimal. Leaf size=138 \[ -\frac {\left (\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^3+c x^6\right )^p F_1\left (-\frac {2}{3};-p,-p;\frac {1}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1385, 510} \[ -\frac {\left (\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^3+c x^6\right )^p F_1\left (-\frac {2}{3};-p,-p;\frac {1}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3 + c*x^6)^p/x^3,x]

[Out]

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

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 1385

Int[((d_.)*(x_))^(m_.)*((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]))^FracPart[p]), Int[(d*x)^m*(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, d, m, n, p}, x] && EqQ[n2, 2*n]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 166, normalized size = 1.20 \[ -\frac {\left (\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{\sqrt {b^2-4 a c}+b}\right )^{-p} \left (a+b x^3+c x^6\right )^p F_1\left (-\frac {2}{3};-p,-p;\frac {1}{3};-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{\sqrt {b^2-4 a c}-b}\right )}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{6} + b x^{3} + a\right )}^{p}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3 + c*x^6)^p/x^3,x)

[Out]

int((a + b*x^3 + c*x^6)^p/x^3, x)

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

[Out]

Timed out

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