\(\int \frac {(d x)^m}{\sqrt {a+b x^3+c x^6}} \, dx\) [254]

Optimal result

Integrand size = 22, antiderivative size = 157 \[ \int \frac {(d x)^m}{\sqrt {a+b x^3+c x^6}} \, dx=\frac {(d x)^{1+m} \sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1+m}{3},\frac {1}{2},\frac {1}{2},\frac {4+m}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{d (1+m) \sqrt {a+b x^3+c x^6}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1399, 524} \[ \int \frac {(d x)^m}{\sqrt {a+b x^3+c x^6}} \, dx=\frac {(d x)^{m+1} \sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {m+1}{3},\frac {1}{2},\frac {1}{2},\frac {m+4}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1) \sqrt {a+b x^3+c x^6}} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 1.74 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.15 \[ \int \frac {(d x)^m}{\sqrt {a+b x^3+c x^6}} \, dx=\frac {x (d x)^m \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1+m}{3},\frac {1}{2},\frac {1}{2},\frac {4+m}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )}{(1+m) \sqrt {a+b x^3+c x^6}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

int((d*x)^m/(c*x^6+b*x^3+a)^(1/2),x)

[Out]

int((d*x)^m/(c*x^6+b*x^3+a)^(1/2),x)

Fricas [F]

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

[In]

integrate((d*x)^m/(c*x^6+b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

integral((d*x)^m/sqrt(c*x^6 + b*x^3 + a), x)

Sympy [F]

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

[In]

integrate((d*x)**m/(c*x**6+b*x**3+a)**(1/2),x)

[Out]

Integral((d*x)**m/sqrt(a + b*x**3 + c*x**6), x)

Maxima [F]

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

[In]

integrate((d*x)^m/(c*x^6+b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((d*x)^m/(c*x^6+b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int((d*x)^m/(a + b*x^3 + c*x^6)^(1/2),x)

[Out]

int((d*x)^m/(a + b*x^3 + c*x^6)^(1/2), x)