\(\int \frac {(a+b x^3+c x^6)^p}{x^7} \, dx\) [270]

Optimal result

Integrand size = 18, antiderivative size = 168 \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^7} \, dx=-\frac {2^{-1+2 p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (2 (1-p),-p,-p,3-2 p,-\frac {b-\sqrt {b^2-4 a c}}{2 c x^3},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^3}\right )}{3 (1-p) x^6} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1371, 772, 138} \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^7} \, dx=-\frac {2^{2 p-1} \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (2 (1-p),-p,-p,3-2 p,-\frac {b-\sqrt {b^2-4 a c}}{2 c x^3},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^3}\right )}{3 (1-p) x^6} \]

[In]

Int[(a + b*x^3 + c*x^6)^p/x^7,x]

[Out]

-1/3*(2^(-1 + 2*p)*(a + b*x^3 + c*x^6)^p*AppellF1[2*(1 - p), -p, -p, 3 - 2*p, -1/2*(b - Sqrt[b^2 - 4*a*c])/(c*
x^3), -1/2*(b + Sqrt[b^2 - 4*a*c])/(c*x^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 138

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

Rule 772

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 1371

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]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {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 ) \text {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] (verified)

Time = 0.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^7} \, dx=\frac {2^{-1+2 p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (2-2 p,-p,-p,3-2 p,-\frac {b+\sqrt {b^2-4 a c}}{2 c x^3},\frac {-b+\sqrt {b^2-4 a c}}{2 c x^3}\right )}{3 (-1+p) x^6} \]

[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, -1/2*(b + Sqrt[b^2 - 4*a*c])/(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)

Maple [F]

\[\int \frac {\left (c \,x^{6}+b \,x^{3}+a \right )^{p}}{x^{7}}d x\]

[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)

Fricas [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^7} \, dx=\int { \frac {{\left (c x^{6} + b x^{3} + a\right )}^{p}}{x^{7}} \,d x } \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^7} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^7} \, dx=\int { \frac {{\left (c x^{6} + b x^{3} + a\right )}^{p}}{x^{7}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^7} \, dx=\int { \frac {{\left (c x^{6} + b x^{3} + a\right )}^{p}}{x^{7}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^7} \, dx=\int \frac {{\left (c\,x^6+b\,x^3+a\right )}^p}{x^7} \,d x \]

[In]

int((a + b*x^3 + c*x^6)^p/x^7,x)

[Out]

int((a + b*x^3 + c*x^6)^p/x^7, x)