3.270 \(\int \frac{(a+b x^3+c x^6)^p}{x^7} \, dx\)

Optimal. Leaf size=168 \[ -\frac{2^{2 p-1} \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p F_1\left (2 (1-p);-p,-p;3-2 p;-\frac{b-\sqrt{b^2-4 a c}}{2 c x^3},-\frac{b+\sqrt{b^2-4 a c}}{2 c x^3}\right )}{3 (1-p) x^6} \]

[Out]

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

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Rubi [A]  time = 0.126097, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1357, 758, 133} \[ -\frac{2^{2 p-1} \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p F_1\left (2 (1-p);-p,-p;3-2 p;-\frac{b-\sqrt{b^2-4 a c}}{2 c x^3},-\frac{b+\sqrt{b^2-4 a c}}{2 c x^3}\right )}{3 (1-p) x^6} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3 + c*x^6)^p/x^7,x]

[Out]

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

Rule 1357

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 758

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, -Dist[((1/(d + e*x))^(2*p)*(a + b*x + c*x^2)^p)/(e*((e*(b - q + 2*c*x))/(2*c*(d + e*x)))^p*((e*(b + q +
2*c*x))/(2*c*(d + e*x)))^p), Subst[Int[x^(-m - 2*(p + 1))*Simp[1 - (d - (e*(b - q))/(2*c))*x, x]^p*Simp[1 - (d
- (e*(b + q))/(2*c))*x, x]^p, x], x, 1/(d + e*x)], x]] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] &&  !IntegerQ[p] && ILtQ[m, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+b x^3+c x^6\right )^p}{x^7} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{\left (a+b x+c x^2\right )^p}{x^3} \, dx,x,x^3\right )\\ &=-\left (\frac{1}{3} \left (2^{2 p} \left (\frac{1}{x^3}\right )^{2 p} \left (\frac{b-\sqrt{b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (\frac{b+\sqrt{b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p\right ) \operatorname{Subst}\left (\int x^{3-2 (1+p)} \left (1+\frac{\left (b-\sqrt{b^2-4 a c}\right ) x}{2 c}\right )^p \left (1+\frac{\left (b+\sqrt{b^2-4 a c}\right ) x}{2 c}\right )^p \, dx,x,\frac{1}{x^3}\right )\right )\\ &=-\frac{2^{-1+2 p} \left (\frac{b-\sqrt{b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (\frac{b+\sqrt{b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p F_1\left (2 (1-p);-p,-p;3-2 p;-\frac{b-\sqrt{b^2-4 a c}}{2 c x^3},-\frac{b+\sqrt{b^2-4 a c}}{2 c x^3}\right )}{3 (1-p) x^6}\\ \end{align*}

Mathematica [A]  time = 0.233864, size = 164, normalized size = 0.98 \[ \frac{2^{2 p-1} \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p F_1\left (2-2 p;-p,-p;3-2 p;-\frac{b+\sqrt{b^2-4 a c}}{2 c x^3},\frac{\sqrt{b^2-4 a c}-b}{2 c x^3}\right )}{3 (p-1) x^6} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3 + c*x^6)^p/x^7,x]

[Out]

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

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{6}+b{x}^{3}+a \right ) ^{p}}{{x}^{7}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^6+b*x^3+a)^p/x^7,x)

[Out]

int((c*x^6+b*x^3+a)^p/x^7,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{6} + b x^{3} + a\right )}^{p}}{x^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)^p/x^7,x, algorithm="maxima")

[Out]

integrate((c*x^6 + b*x^3 + a)^p/x^7, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c x^{6} + b x^{3} + a\right )}^{p}}{x^{7}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)^p/x^7,x, algorithm="fricas")

[Out]

integral((c*x^6 + b*x^3 + a)^p/x^7, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**6+b*x**3+a)**p/x**7,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{6} + b x^{3} + a\right )}^{p}}{x^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)^p/x^7,x, algorithm="giac")

[Out]

integrate((c*x^6 + b*x^3 + a)^p/x^7, x)