Integrand size = 30, antiderivative size = 153 \[ \int \frac {1}{\sqrt [3]{-a-b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{3 \sqrt {a}}\right )}{12 a^{5/6} d}-\frac {\sqrt {b} \arctan \left (\frac {\left (\sqrt [3]{a}+\sqrt [3]{-a-b x^2}\right )^2}{3 \sqrt [6]{a} \sqrt {b} x}\right )}{12 a^{5/6} d}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{a}+\sqrt [3]{-a-b x^2}\right )}{\sqrt {b} x}\right )}{4 \sqrt {3} a^{5/6} d} \] Output:
-1/12*b^(1/2)*arctan(1/3*b^(1/2)*x/a^(1/2))/a^(5/6)/d-1/12*b^(1/2)*arctan( 1/3*(a^(1/3)+(-b*x^2-a)^(1/3))^2/a^(1/6)/b^(1/2)/x)/a^(5/6)/d+1/12*b^(1/2) *arctanh(3^(1/2)*a^(1/6)*(a^(1/3)+(-b*x^2-a)^(1/3))/b^(1/2)/x)*3^(1/2)/a^( 5/6)/d
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 6.17 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\sqrt [3]{-a-b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\frac {27 a b x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {b x^2}{9 a}\right )}{d \sqrt [3]{-a-b x^2} \left (9 a+b x^2\right ) \left (27 a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {b x^2}{9 a}\right )-2 b x^2 \left (\operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {b x^2}{9 a}\right )+3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {4}{3},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {b x^2}{9 a}\right )\right )\right )} \] Input:
Integrate[1/((-a - b*x^2)^(1/3)*((9*a*d)/b + d*x^2)),x]
Output:
(27*a*b*x*AppellF1[1/2, 1/3, 1, 3/2, -((b*x^2)/a), -1/9*(b*x^2)/a])/(d*(-a - b*x^2)^(1/3)*(9*a + b*x^2)*(27*a*AppellF1[1/2, 1/3, 1, 3/2, -((b*x^2)/a ), -1/9*(b*x^2)/a] - 2*b*x^2*(AppellF1[3/2, 1/3, 2, 5/2, -((b*x^2)/a), -1/ 9*(b*x^2)/a] + 3*AppellF1[3/2, 4/3, 1, 5/2, -((b*x^2)/a), -1/9*(b*x^2)/a]) ))
Time = 0.20 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {306}
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.
| \(\displaystyle \int \frac {1}{\sqrt [3]{-a-b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx\) |
| \(\Big \downarrow \) 306 |
| \(\displaystyle -\frac {\sqrt {b} \arctan \left (\frac {\left (\sqrt [3]{-a-b x^2}+\sqrt [3]{a}\right )^2}{3 \sqrt [6]{a} \sqrt {b} x}\right )}{12 a^{5/6} d}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{3 \sqrt {a}}\right )}{12 a^{5/6} d}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{-a-b x^2}+\sqrt [3]{a}\right )}{\sqrt {b} x}\right )}{4 \sqrt {3} a^{5/6} d}\) |
Input:
Int[1/((-a - b*x^2)^(1/3)*((9*a*d)/b + d*x^2)),x]
Output:
-1/12*(Sqrt[b]*ArcTan[(Sqrt[b]*x)/(3*Sqrt[a])])/(a^(5/6)*d) - (Sqrt[b]*Arc Tan[(a^(1/3) + (-a - b*x^2)^(1/3))^2/(3*a^(1/6)*Sqrt[b]*x)])/(12*a^(5/6)*d ) + (Sqrt[b]*ArcTanh[(Sqrt[3]*a^(1/6)*(a^(1/3) + (-a - b*x^2)^(1/3)))/(Sqr t[b]*x)])/(4*Sqrt[3]*a^(5/6)*d)
Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Wit h[{q = Rt[b/a, 2]}, Simp[q*(ArcTan[q*(x/3)]/(12*Rt[a, 3]*d)), x] + (Simp[q* (ArcTan[(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*(ArcTanh[(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] && PosQ[b/a]
\[\int \frac {1}{\left (-b \,x^{2}-a \right )^{\frac {1}{3}} \left (\frac {9 a d}{b}+x^{2} d \right )}d x\]
Input:
int(1/(-b*x^2-a)^(1/3)/(9*a*d/b+x^2*d),x)
Output:
int(1/(-b*x^2-a)^(1/3)/(9*a*d/b+x^2*d),x)
Timed out. \[ \int \frac {1}{\sqrt [3]{-a-b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(1/(-b*x^2-a)^(1/3)/(9*a*d/b+d*x^2),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{\sqrt [3]{-a-b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\frac {b \int \frac {1}{9 a \sqrt [3]{- a - b x^{2}} + b x^{2} \sqrt [3]{- a - b x^{2}}}\, dx}{d} \] Input:
integrate(1/(-b*x**2-a)**(1/3)/(9*a*d/b+d*x**2),x)
Output:
b*Integral(1/(9*a*(-a - b*x**2)**(1/3) + b*x**2*(-a - b*x**2)**(1/3)), x)/ d
\[ \int \frac {1}{\sqrt [3]{-a-b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\int { \frac {1}{{\left (-b x^{2} - a\right )}^{\frac {1}{3}} {\left (d x^{2} + \frac {9 \, a d}{b}\right )}} \,d x } \] Input:
integrate(1/(-b*x^2-a)^(1/3)/(9*a*d/b+d*x^2),x, algorithm="maxima")
Output:
integrate(1/((-b*x^2 - a)^(1/3)*(d*x^2 + 9*a*d/b)), x)
\[ \int \frac {1}{\sqrt [3]{-a-b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\int { \frac {1}{{\left (-b x^{2} - a\right )}^{\frac {1}{3}} {\left (d x^{2} + \frac {9 \, a d}{b}\right )}} \,d x } \] Input:
integrate(1/(-b*x^2-a)^(1/3)/(9*a*d/b+d*x^2),x, algorithm="giac")
Output:
integrate(1/((-b*x^2 - a)^(1/3)*(d*x^2 + 9*a*d/b)), x)
Timed out. \[ \int \frac {1}{\sqrt [3]{-a-b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\int \frac {1}{{\left (-b\,x^2-a\right )}^{1/3}\,\left (d\,x^2+\frac {9\,a\,d}{b}\right )} \,d x \] Input:
int(1/((- a - b*x^2)^(1/3)*(d*x^2 + (9*a*d)/b)),x)
Output:
int(1/((- a - b*x^2)^(1/3)*(d*x^2 + (9*a*d)/b)), x)
\[ \int \frac {1}{\sqrt [3]{-a-b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=-\frac {\left (\int \frac {1}{9 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a +\left (b \,x^{2}+a \right )^{\frac {1}{3}} b \,x^{2}}d x \right ) b}{d} \] Input:
int(1/(-b*x^2-a)^(1/3)/(9*a*d/b+d*x^2),x)
Output:
( - int(1/(9*(a + b*x**2)**(1/3)*a + (a + b*x**2)**(1/3)*b*x**2),x)*b)/d