\(\int \frac {1}{\sqrt [3]{-2+b x^2} (-\frac {18 d}{b}+d x^2)} \, dx\) [365]

Optimal result

Integrand size = 26, antiderivative size = 147 \[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\frac {\sqrt {b} \arctan \left (\frac {\sqrt [6]{2} \sqrt {3} \left (\sqrt [3]{2}+\sqrt [3]{-2+b x^2}\right )}{\sqrt {b} x}\right )}{4\ 2^{5/6} \sqrt {3} d}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{3 \sqrt {2}}\right )}{12\ 2^{5/6} d}-\frac {\sqrt {b} \text {arctanh}\left (\frac {\left (\sqrt [3]{2}+\sqrt [3]{-2+b x^2}\right )^2}{3 \sqrt [6]{2} \sqrt {b} x}\right )}{12\ 2^{5/6} d} \] Output:

1/24*b^(1/2)*arctan(2^(1/6)*3^(1/2)*(2^(1/3)+(b*x^2-2)^(1/3))/b^(1/2)/x)*2 
^(1/6)*3^(1/2)/d+1/24*b^(1/2)*arctanh(1/6*b^(1/2)*x*2^(1/2))*2^(1/6)/d-1/2 
4*b^(1/2)*arctanh(1/6*(2^(1/3)+(b*x^2-2)^(1/3))^2*2^(5/6)/b^(1/2)/x)*2^(1/ 
6)/d
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 5.20 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\frac {27 b x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},\frac {b x^2}{2},\frac {b x^2}{18}\right )}{d \left (-18+b x^2\right ) \sqrt [3]{-2+b x^2} \left (27 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},\frac {b x^2}{2},\frac {b x^2}{18}\right )+b x^2 \left (\operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},2,\frac {5}{2},\frac {b x^2}{2},\frac {b x^2}{18}\right )+3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {4}{3},1,\frac {5}{2},\frac {b x^2}{2},\frac {b x^2}{18}\right )\right )\right )} \] Input:

Integrate[1/((-2 + b*x^2)^(1/3)*((-18*d)/b + d*x^2)),x]
 

Output:

(27*b*x*AppellF1[1/2, 1/3, 1, 3/2, (b*x^2)/2, (b*x^2)/18])/(d*(-18 + b*x^2 
)*(-2 + b*x^2)^(1/3)*(27*AppellF1[1/2, 1/3, 1, 3/2, (b*x^2)/2, (b*x^2)/18] 
 + b*x^2*(AppellF1[3/2, 1/3, 2, 5/2, (b*x^2)/2, (b*x^2)/18] + 3*AppellF1[3 
/2, 4/3, 1, 5/2, (b*x^2)/2, (b*x^2)/18])))
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {307}

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.

\(\displaystyle \int \frac {1}{\sqrt [3]{b x^2-2} \left (d x^2-\frac {18 d}{b}\right )} \, dx\)

\(\Big \downarrow \) 307

\(\displaystyle \frac {\sqrt {b} \arctan \left (\frac {\sqrt [6]{2} \sqrt {3} \left (\sqrt [3]{b x^2-2}+\sqrt [3]{2}\right )}{\sqrt {b} x}\right )}{4\ 2^{5/6} \sqrt {3} d}-\frac {\sqrt {b} \text {arctanh}\left (\frac {\left (\sqrt [3]{b x^2-2}+\sqrt [3]{2}\right )^2}{3 \sqrt [6]{2} \sqrt {b} x}\right )}{12\ 2^{5/6} d}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{3 \sqrt {2}}\right )}{12\ 2^{5/6} d}\)

Input:

Int[1/((-2 + b*x^2)^(1/3)*((-18*d)/b + d*x^2)),x]
 

Output:

(Sqrt[b]*ArcTan[(2^(1/6)*Sqrt[3]*(2^(1/3) + (-2 + b*x^2)^(1/3)))/(Sqrt[b]* 
x)])/(4*2^(5/6)*Sqrt[3]*d) + (Sqrt[b]*ArcTanh[(Sqrt[b]*x)/(3*Sqrt[2])])/(1 
2*2^(5/6)*d) - (Sqrt[b]*ArcTanh[(2^(1/3) + (-2 + b*x^2)^(1/3))^2/(3*2^(1/6 
)*Sqrt[b]*x)])/(12*2^(5/6)*d)
 

Defintions of rubi rules used

rule 307

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Wit 
h[{q = Rt[-b/a, 2]}, Simp[(-q)*(ArcTanh[q*(x/3)]/(12*Rt[a, 3]*d)), x] + (Si 
mp[q*(ArcTanh[(Rt[a, 3] - (a + b*x^2)^(1/3))^2/(3*Rt[a, 3]^2*q*x)]/(12*Rt[a 
, 3]*d)), x] - Simp[q*(ArcTan[(Sqrt[3]*(Rt[a, 3] - (a + b*x^2)^(1/3)))/(Rt[ 
a, 3]*q*x)]/(4*Sqrt[3]*Rt[a, 3]*d)), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[ 
b*c - a*d, 0] && EqQ[b*c - 9*a*d, 0] && NegQ[b/a]
 

Maple [F]

\[\int \frac {1}{\left (b \,x^{2}-2\right )^{\frac {1}{3}} \left (-\frac {18 d}{b}+x^{2} d \right )}d x\]

Input:

int(1/(b*x^2-2)^(1/3)/(-18*d/b+x^2*d),x)
 

Output:

int(1/(b*x^2-2)^(1/3)/(-18*d/b+x^2*d),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x^2-2)^(1/3)/(-18*d/b+d*x^2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\frac {b \int \frac {1}{b x^{2} \sqrt [3]{b x^{2} - 2} - 18 \sqrt [3]{b x^{2} - 2}}\, dx}{d} \] Input:

integrate(1/(b*x**2-2)**(1/3)/(-18*d/b+d*x**2),x)
 

Output:

b*Integral(1/(b*x**2*(b*x**2 - 2)**(1/3) - 18*(b*x**2 - 2)**(1/3)), x)/d
 

Maxima [F]

\[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} - 2\right )}^{\frac {1}{3}} {\left (d x^{2} - \frac {18 \, d}{b}\right )}} \,d x } \] Input:

integrate(1/(b*x^2-2)^(1/3)/(-18*d/b+d*x^2),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 - 2)^(1/3)*(d*x^2 - 18*d/b)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} - 2\right )}^{\frac {1}{3}} {\left (d x^{2} - \frac {18 \, d}{b}\right )}} \,d x } \] Input:

integrate(1/(b*x^2-2)^(1/3)/(-18*d/b+d*x^2),x, algorithm="giac")
 

Output:

integrate(1/((b*x^2 - 2)^(1/3)*(d*x^2 - 18*d/b)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\int -\frac {1}{\left (\frac {18\,d}{b}-d\,x^2\right )\,{\left (b\,x^2-2\right )}^{1/3}} \,d x \] Input:

int(-1/(((18*d)/b - d*x^2)*(b*x^2 - 2)^(1/3)),x)
 

Output:

int(-1/(((18*d)/b - d*x^2)*(b*x^2 - 2)^(1/3)), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\frac {\left (\int \frac {1}{\left (b \,x^{2}-2\right )^{\frac {1}{3}} b \,x^{2}-18 \left (b \,x^{2}-2\right )^{\frac {1}{3}}}d x \right ) b}{d} \] Input:

int(1/(b*x^2-2)^(1/3)/(-18*d/b+d*x^2),x)
 

Output:

(int(1/((b*x**2 - 2)**(1/3)*b*x**2 - 18*(b*x**2 - 2)**(1/3)),x)*b)/d