\(\int \frac {(d x)^m}{(a^2+2 a b x^3+b^2 x^6)^{3/2}} \, dx\) [122]

Optimal result

Integrand size = 28, antiderivative size = 73 \[ \int \frac {(d x)^m}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {(d x)^{1+m} \left (a+b x^3\right ) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )}{a^3 d (1+m) \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[Out]

(d*x)^(1+m)*(b*x^3+a)*hypergeom([3, 1/3+1/3*m],[4/3+1/3*m],-b*x^3/a)/a^3/d/(1+m)/((b*x^3+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1369, 371} \[ \int \frac {(d x)^m}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {\left (a+b x^3\right ) (d x)^{m+1} \operatorname {Hypergeometric2F1}\left (3,\frac {m+1}{3},\frac {m+4}{3},-\frac {b x^3}{a}\right )}{a^3 d (m+1) \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[In]

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

[Out]

((d*x)^(1 + m)*(a + b*x^3)*Hypergeometric2F1[3, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/(a^3*d*(1 + m)*Sqrt[a^2 +
2*a*b*x^3 + b^2*x^6])

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1369

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int \frac {(d x)^m}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {x (d x)^m \left (a+b x^3\right ) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )}{a^3 (1+m) \sqrt {\left (a+b x^3\right )^2}} \]

[In]

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

[Out]

(x*(d*x)^m*(a + b*x^3)*Hypergeometric2F1[3, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/(a^3*(1 + m)*Sqrt[(a + b*x^3)
^2])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*(d*x)^m/(b^4*x^12 + 4*a*b^3*x^9 + 6*a^2*b^2*x^6 + 4*a^3*b*x^3 + a^4),
x)

Sympy [F]

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

[In]

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

[Out]

Integral((d*x)**m/((a + b*x**3)**2)**(3/2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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