\(\int \frac {1}{\sqrt [3]{a+b x^2} (\frac {9 a d}{b}+d x^2)} \, dx\) [150]

Optimal result

Integrand size = 27, antiderivative size = 151 \[ \int \frac {1}{\sqrt [3]{a+b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{3 \sqrt {a}}\right )}{12 a^{5/6} d}+\frac {\sqrt {b} \arctan \left (\frac {\left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}{3 \sqrt [6]{a} \sqrt {b} x}\right )}{12 a^{5/6} d}-\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\sqrt {b} x}\right )}{4 \sqrt {3} a^{5/6} d} \]

[Out]

1/12*arctan(1/3*(a^(1/3)-(b*x^2+a)^(1/3))^2/a^(1/6)/x/b^(1/2))*b^(1/2)/a^(5/6)/d+1/12*arctan(1/3*x*b^(1/2)/a^(
1/2))*b^(1/2)/a^(5/6)/d-1/12*arctanh(a^(1/6)*(a^(1/3)-(b*x^2+a)^(1/3))*3^(1/2)/x/b^(1/2))*b^(1/2)/a^(5/6)/d*3^
(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {403} \[ \int \frac {1}{\sqrt [3]{a+b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\frac {\sqrt {b} \arctan \left (\frac {\left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}{3 \sqrt [6]{a} \sqrt {b} x}\right )}{12 a^{5/6} d}+\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{3 \sqrt {a}}\right )}{12 a^{5/6} d}-\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\sqrt {b} x}\right )}{4 \sqrt {3} a^{5/6} d} \]

[In]

Int[1/((a + b*x^2)^(1/3)*((9*a*d)/b + d*x^2)),x]

[Out]

(Sqrt[b]*ArcTan[(Sqrt[b]*x)/(3*Sqrt[a])])/(12*a^(5/6)*d) + (Sqrt[b]*ArcTan[(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)))/(Sqrt[b
]*x)])/(4*Sqrt[3]*a^(5/6)*d)

Rule 403

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x}{3 \sqrt {a}}\right )}{12 a^{5/6} d}+\frac {\sqrt {b} \tan ^{-1}\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} \tanh ^{-1}\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} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 5.64 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\sqrt [3]{a+b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\frac {27 a b x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {b x^2}{9 a}\right )}{d \sqrt [3]{a+b x^2} \left (9 a+b x^2\right ) \left (27 a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {b x^2}{9 a}\right )-2 b x^2 \left (\operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {b x^2}{9 a}\right )+3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {4}{3},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {b x^2}{9 a}\right )\right )\right )} \]

[In]

Integrate[1/((a + b*x^2)^(1/3)*((9*a*d)/b + d*x^2)),x]

[Out]

(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*A
ppellF1[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])))

Maple [F]

\[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{3}} \left (\frac {9 a d}{b}+d \,x^{2}\right )}d x\]

[In]

int(1/(b*x^2+a)^(1/3)/(9*a*d/b+d*x^2),x)

[Out]

int(1/(b*x^2+a)^(1/3)/(9*a*d/b+d*x^2),x)

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [3]{a+b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\text {Timed out} \]

[In]

integrate(1/(b*x^2+a)^(1/3)/(9*a*d/b+d*x^2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{\sqrt [3]{a+b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\frac {b \int \frac {1}{9 a \sqrt [3]{a + b x^{2}} + b x^{2} \sqrt [3]{a + b x^{2}}}\, dx}{d} \]

[In]

integrate(1/(b*x**2+a)**(1/3)/(9*a*d/b+d*x**2),x)

[Out]

b*Integral(1/(9*a*(a + b*x**2)**(1/3) + b*x**2*(a + b*x**2)**(1/3)), x)/d

Maxima [F]

\[ \int \frac {1}{\sqrt [3]{a+b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (d x^{2} + \frac {9 \, a d}{b}\right )}} \,d x } \]

[In]

integrate(1/(b*x^2+a)^(1/3)/(9*a*d/b+d*x^2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(1/3)*(d*x^2 + 9*a*d/b)), x)

Giac [F]

\[ \int \frac {1}{\sqrt [3]{a+b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (d x^{2} + \frac {9 \, a d}{b}\right )}} \,d x } \]

[In]

integrate(1/(b*x^2+a)^(1/3)/(9*a*d/b+d*x^2),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(1/3)*(d*x^2 + 9*a*d/b)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [3]{a+b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{1/3}\,\left (d\,x^2+\frac {9\,a\,d}{b}\right )} \,d x \]

[In]

int(1/((a + b*x^2)^(1/3)*(d*x^2 + (9*a*d)/b)),x)

[Out]

int(1/((a + b*x^2)^(1/3)*(d*x^2 + (9*a*d)/b)), x)