\(\int \frac {(a+b x^3+c x^6)^{3/2}}{x^3} \, dx\) [218]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 20, antiderivative size = 141 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^3} \, dx=-\frac {a \sqrt {a+b x^3+c x^6} \operatorname {AppellF1}\left (-\frac {2}{3},-\frac {3}{2},-\frac {3}{2},\frac {1}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{2 x^2 \sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1399, 524} \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^3} \, dx=-\frac {a \sqrt {a+b x^3+c x^6} \operatorname {AppellF1}\left (-\frac {2}{3},-\frac {3}{2},-\frac {3}{2},\frac {1}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{2 x^2 \sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1}} \]

[In]

Int[(a + b*x^3 + c*x^6)^(3/2)/x^3,x]

[Out]

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

Rule 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 1399

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(379\) vs. \(2(141)=282\).

Time = 10.35 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.69 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^3} \, dx=\frac {8 \left (-28 a^2-11 a b x^3+17 b^2 x^6-20 a c x^6+25 b c x^9+8 c^2 x^{12}\right )+648 a b x^3 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )+27 \left (b^2+8 a c\right ) x^6 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},\frac {1}{2},\frac {7}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )}{448 x^2 \sqrt {a+b x^3+c x^6}} \]

[In]

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

[Out]

(8*(-28*a^2 - 11*a*b*x^3 + 17*b^2*x^6 - 20*a*c*x^6 + 25*b*c*x^9 + 8*c^2*x^12) + 648*a*b*x^3*Sqrt[(b - Sqrt[b^2
 - 4*a*c] + 2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])]*
AppellF1[1/3, 1/2, 1/2, 4/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])] + 27*(b^2
 + 8*a*c)*x^6*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*
c*x^3)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[4/3, 1/2, 1/2, 7/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b
 + Sqrt[b^2 - 4*a*c])])/(448*x^2*Sqrt[a + b*x^3 + c*x^6])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((a + b*x**3 + c*x**6)**(3/2)/x**3, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + b*x^3 + c*x^6)^(3/2)/x^3,x)

[Out]

int((a + b*x^3 + c*x^6)^(3/2)/x^3, x)