\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^p}{x^4} \, dx\) [136]

Optimal result

Integrand size = 24, antiderivative size = 64 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^4} \, dx=\frac {b \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p \operatorname {Hypergeometric2F1}\left (2,1+2 p,2 (1+p),1+\frac {b x^3}{a}\right )}{3 a^2 (1+2 p)} \]

[Out]

1/3*b*(b*x^3+a)*(b^2*x^6+2*a*b*x^3+a^2)^p*hypergeom([2, 1+2*p],[2+2*p],1+b*x^3/a)/a^2/(1+2*p)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1370, 272, 67} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^4} \, dx=\frac {b \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p \operatorname {Hypergeometric2F1}\left (2,2 p+1,2 (p+1),\frac {b x^3}{a}+1\right )}{3 a^2 (2 p+1)} \]

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^p/x^4,x]

[Out]

(b*(a + b*x^3)*(a^2 + 2*a*b*x^3 + b^2*x^6)^p*Hypergeometric2F1[2, 1 + 2*p, 2*(1 + p), 1 + (b*x^3)/a])/(3*a^2*(
1 + 2*p))

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1370

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

Rubi steps \begin{align*} \text {integral}& = \left (\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \int \frac {\left (1+\frac {b x^3}{a}\right )^{2 p}}{x^4} \, dx \\ & = \frac {1}{3} \left (\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \text {Subst}\left (\int \frac {\left (1+\frac {b x}{a}\right )^{2 p}}{x^2} \, dx,x,x^3\right ) \\ & = \frac {b \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p \, _2F_1\left (2,1+2 p;2 (1+p);1+\frac {b x^3}{a}\right )}{3 a^2 (1+2 p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^4} \, dx=\frac {b \left (a+b x^3\right ) \left (\left (a+b x^3\right )^2\right )^p \operatorname {Hypergeometric2F1}\left (2,1+2 p,2+2 p,1+\frac {b x^3}{a}\right )}{3 a^2 (1+2 p)} \]

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^p/x^4,x]

[Out]

(b*(a + b*x^3)*((a + b*x^3)^2)^p*Hypergeometric2F1[2, 1 + 2*p, 2 + 2*p, 1 + (b*x^3)/a])/(3*a^2*(1 + 2*p))

Maple [F]

\[\int \frac {\left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p}}{x^{4}}d x\]

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^p/x^4,x)

[Out]

int((b^2*x^6+2*a*b*x^3+a^2)^p/x^4,x)

Fricas [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^4} \, dx=\int { \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{x^{4}} \,d x } \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^p/x^4,x, algorithm="fricas")

[Out]

integral((b^2*x^6 + 2*a*b*x^3 + a^2)^p/x^4, x)

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^4} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{p}}{x^{4}}\, dx \]

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**p/x**4,x)

[Out]

Integral(((a + b*x**3)**2)**p/x**4, x)

Maxima [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^4} \, dx=\int { \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{x^{4}} \,d x } \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^p/x^4,x, algorithm="maxima")

[Out]

integrate((b^2*x^6 + 2*a*b*x^3 + a^2)^p/x^4, x)

Giac [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^4} \, dx=\int { \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{x^{4}} \,d x } \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^p/x^4,x, algorithm="giac")

[Out]

integrate((b^2*x^6 + 2*a*b*x^3 + a^2)^p/x^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^4} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^p}{x^4} \,d x \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^p/x^4,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^p/x^4, x)