\(\int \frac {1}{\sqrt {d+e x^3} (a+b x^3+c x^6)} \, dx\) [19]

Optimal result

Integrand size = 26, antiderivative size = 194 \[ \int \frac {1}{\sqrt {d+e x^3} \left (a+b x^3+c x^6\right )} \, dx=-\frac {2 c x \sqrt {1+\frac {e x^3}{d}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {e x^3}{d},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \sqrt {d+e x^3}}-\frac {2 c x \sqrt {1+\frac {e x^3}{d}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {e x^3}{d},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) \sqrt {d+e x^3}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(4385\) vs. \(2(194)=388\).

Time = 16.20 (sec) , antiderivative size = 4385, normalized size of antiderivative = 22.60 \[ \int \frac {1}{\sqrt {d+e x^3} \left (a+b x^3+c x^6\right )} \, dx=\text {Result too large to show} \] Input:

Integrate[1/(Sqrt[d + e*x^3]*(a + b*x^3 + c*x^6)),x]
 

Output:

(4*(((-1)^(1/3)*d^(1/3))/e^(1/3) + ((-1)^(2/3)*d^(1/3))/e^(1/3))*Sqrt[(d^( 
1/3)/e^(1/3) + x)/(d^(1/3)/e^(1/3) + ((-1)^(1/3)*d^(1/3))/e^(1/3))]*Sqrt[( 
(-(((-1)^(2/3)*d^(1/3))/e^(1/3)) - x)*(-(((-1)^(1/3)*d^(1/3))/e^(1/3)) + x 
))/(((-1)^(1/3)*d^(1/3))/e^(1/3) + ((-1)^(2/3)*d^(1/3))/e^(1/3))^2]*Ellipt 
icPi[(2*(d^(1/3) + (-1)^(1/3)*d^(1/3)))/(2*d^(1/3) + 2^(2/3)*((-b - Sqrt[b 
^2 - 4*a*c])/c)^(1/3)*e^(1/3)), ArcSin[Sqrt[-(((-1)^(2/3)*((-1)^(1/3)*d^(1 
/3) - e^(1/3)*x))/((1 + (-1)^(1/3))*d^(1/3)))]], (-1)^(1/3)])/(c*(-((-1/2) 
^(1/3)*(-(b/c) - Sqrt[b^2 - 4*a*c]/c)^(1/3)) - (-(b/c) - Sqrt[b^2 - 4*a*c] 
/c)^(1/3)/2^(1/3))*(-((-1/2)^(1/3)*(-(b/c) - Sqrt[b^2 - 4*a*c]/c)^(1/3)) - 
 ((-1)^(2/3)*(-(b/c) - Sqrt[b^2 - 4*a*c]/c)^(1/3))/2^(1/3))*(-((-1/2)^(1/3 
)*(-(b/c) - Sqrt[b^2 - 4*a*c]/c)^(1/3)) + (-1/2)^(1/3)*(-(b/c) + Sqrt[b^2 
- 4*a*c]/c)^(1/3))*(-((-1/2)^(1/3)*(-(b/c) - Sqrt[b^2 - 4*a*c]/c)^(1/3)) - 
 (-(b/c) + Sqrt[b^2 - 4*a*c]/c)^(1/3)/2^(1/3))*(-((-1/2)^(1/3)*(-(b/c) - S 
qrt[b^2 - 4*a*c]/c)^(1/3)) - ((-1)^(2/3)*(-(b/c) + Sqrt[b^2 - 4*a*c]/c)^(1 
/3))/2^(1/3))*(-((-1)^(1/3)*2^(2/3)*((-b - Sqrt[b^2 - 4*a*c])/c)^(1/3)) - 
(2*(-1)^(1/3)*d^(1/3))/e^(1/3))*Sqrt[d + e*x^3]) + (4*(((-1)^(1/3)*d^(1/3) 
)/e^(1/3) + ((-1)^(2/3)*d^(1/3))/e^(1/3))*Sqrt[(d^(1/3)/e^(1/3) + x)/(d^(1 
/3)/e^(1/3) + ((-1)^(1/3)*d^(1/3))/e^(1/3))]*Sqrt[((-(((-1)^(2/3)*d^(1/3)) 
/e^(1/3)) - x)*(-(((-1)^(1/3)*d^(1/3))/e^(1/3)) + x))/(((-1)^(1/3)*d^(1/3) 
)/e^(1/3) + ((-1)^(2/3)*d^(1/3))/e^(1/3))^2]*EllipticPi[(-2*((-1)^(1/3)...
 

Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1758, 937, 936}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\sqrt {d+e x^3} \left (a+b x^3+c x^6\right )} \, dx\)

\(\Big \downarrow \) 1758

\(\displaystyle \frac {2 c \int \frac {1}{\left (2 c x^3+b-\sqrt {b^2-4 a c}\right ) \sqrt {e x^3+d}}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {1}{\left (2 c x^3+b+\sqrt {b^2-4 a c}\right ) \sqrt {e x^3+d}}dx}{\sqrt {b^2-4 a c}}\)

\(\Big \downarrow \) 937

\(\displaystyle \frac {2 c \sqrt {\frac {e x^3}{d}+1} \int \frac {1}{\left (2 c x^3+b-\sqrt {b^2-4 a c}\right ) \sqrt {\frac {e x^3}{d}+1}}dx}{\sqrt {b^2-4 a c} \sqrt {d+e x^3}}-\frac {2 c \sqrt {\frac {e x^3}{d}+1} \int \frac {1}{\left (2 c x^3+b+\sqrt {b^2-4 a c}\right ) \sqrt {\frac {e x^3}{d}+1}}dx}{\sqrt {b^2-4 a c} \sqrt {d+e x^3}}\)

\(\Big \downarrow \) 936

\(\displaystyle \frac {2 c x \sqrt {\frac {e x^3}{d}+1} \operatorname {AppellF1}\left (\frac {1}{3},1,\frac {1}{2},\frac {4}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {e x^3}{d}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) \sqrt {d+e x^3}}-\frac {2 c x \sqrt {\frac {e x^3}{d}+1} \operatorname {AppellF1}\left (\frac {1}{3},1,\frac {1}{2},\frac {4}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},-\frac {e x^3}{d}\right )}{\sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right ) \sqrt {d+e x^3}}\)

Input:

Int[1/(Sqrt[d + e*x^3]*(a + b*x^3 + c*x^6)),x]
 

Output:

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

Defintions of rubi rules used

rule 936

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) 
], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] 
 && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 937

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) 
  Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q 
}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 1758

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_ 
)), x_Symbol] :> With[{r = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/r)   Int[(d + e*x 
^n)^q/(b - r + 2*c*x^n), x], x] - Simp[2*(c/r)   Int[(d + e*x^n)^q/(b + r + 
 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && Ne 
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[q]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 6.

Time = 0.34 (sec) , antiderivative size = 776, normalized size of antiderivative = 4.00

method	result	size
default	\(\text {Expression too large to display}\)	\(776\)
elliptic	\(\text {Expression too large to display}\)	\(776\)

Input:

int(1/(e*x^3+d)^(1/2)/(c*x^6+b*x^3+a),x,method=_RETURNVERBOSE)
 

Output:

-1/3*I/e^2*2^(1/2)*sum(1/_alpha^2/(2*_alpha^3*c+b)/(a*e^2-b*d*e+c*d^2)*(-d 
*e^2)^(1/3)*(1/2*I*e*(2*x+1/e*(-I*3^(1/2)*(-d*e^2)^(1/3)+(-d*e^2)^(1/3)))/ 
(-d*e^2)^(1/3))^(1/2)*(e*(x-1/e*(-d*e^2)^(1/3))/(-3*(-d*e^2)^(1/3)+I*3^(1/ 
2)*(-d*e^2)^(1/3)))^(1/2)*(-1/2*I*e*(2*x+1/e*(I*3^(1/2)*(-d*e^2)^(1/3)+(-d 
*e^2)^(1/3)))/(-d*e^2)^(1/3))^(1/2)/(e*x^3+d)^(1/2)*(2*e^2*(_alpha^5*c*e+_ 
alpha^2*b*e-_alpha^2*c*d)+I*(-d*e^2)^(1/3)*3^(1/2)*_alpha^4*c*e^2-I*(-d*e^ 
2)^(2/3)*3^(1/2)*_alpha^3*c*e-(-d*e^2)^(1/3)*_alpha^4*c*e^2-(-d*e^2)^(2/3) 
*_alpha^3*c*e+I*(-d*e^2)^(1/3)*3^(1/2)*_alpha*b*e^2-I*(-d*e^2)^(1/3)*3^(1/ 
2)*_alpha*c*d*e-I*(-d*e^2)^(2/3)*3^(1/2)*b*e+I*(-d*e^2)^(2/3)*3^(1/2)*c*d- 
(-d*e^2)^(1/3)*_alpha*b*e^2+(-d*e^2)^(1/3)*_alpha*c*d*e-(-d*e^2)^(2/3)*b*e 
+(-d*e^2)^(2/3)*c*d)*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2 
*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2),1/2/e*(2*I*(- 
d*e^2)^(1/3)*3^(1/2)*_alpha^5*c*e^2-I*(-d*e^2)^(2/3)*3^(1/2)*_alpha^4*c*e+ 
I*3^(1/2)*_alpha^3*c*d*e^2-3*c*(-d*e^2)^(2/3)*_alpha^4*e+2*I*(-d*e^2)^(1/3 
)*3^(1/2)*_alpha^2*b*e^2-2*I*(-d*e^2)^(1/3)*3^(1/2)*_alpha^2*c*d*e-I*(-d*e 
^2)^(2/3)*3^(1/2)*_alpha*b*e+I*(-d*e^2)^(2/3)*3^(1/2)*_alpha*c*d-3*c*_alph 
a^3*e^2*d+I*3^(1/2)*b*d*e^2-I*3^(1/2)*c*d^2*e-3*(-d*e^2)^(2/3)*_alpha*b*e+ 
3*(-d*e^2)^(2/3)*_alpha*c*d-3*b*d*e^2+3*d^2*e*c)/(a*e^2-b*d*e+c*d^2),(I*3^ 
(1/2)/e*(-d*e^2)^(1/3)/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/ 
3)))^(1/2)),_alpha=RootOf(_Z^6*c+_Z^3*b+a))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {d+e x^3} \left (a+b x^3+c x^6\right )} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{\sqrt {d+e x^3} \left (a+b x^3+c x^6\right )} \, dx=\int \frac {1}{\sqrt {d + e x^{3}} \left (a + b x^{3} + c x^{6}\right )}\, dx \] Input:

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

Output:

Integral(1/(sqrt(d + e*x**3)*(a + b*x**3 + c*x**6)), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {d+e x^3} \left (a+b x^3+c x^6\right )} \, dx=\int { \frac {1}{{\left (c x^{6} + b x^{3} + a\right )} \sqrt {e x^{3} + d}} \,d x } \] Input:

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

Output:

integrate(1/((c*x^6 + b*x^3 + a)*sqrt(e*x^3 + d)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {d+e x^3} \left (a+b x^3+c x^6\right )} \, dx=\int { \frac {1}{{\left (c x^{6} + b x^{3} + a\right )} \sqrt {e x^{3} + d}} \,d x } \] Input:

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

Output:

integrate(1/((c*x^6 + b*x^3 + a)*sqrt(e*x^3 + d)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {d+e x^3} \left (a+b x^3+c x^6\right )} \, dx=\int \frac {1}{\sqrt {e\,x^3+d}\,\left (c\,x^6+b\,x^3+a\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {d+e x^3} \left (a+b x^3+c x^6\right )} \, dx=\int \frac {\sqrt {e \,x^{3}+d}}{c e \,x^{9}+b e \,x^{6}+c d \,x^{6}+a e \,x^{3}+b d \,x^{3}+a d}d x \] Input:

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

Output:

int(sqrt(d + e*x**3)/(a*d + a*e*x**3 + b*d*x**3 + b*e*x**6 + c*d*x**6 + c* 
e*x**9),x)