3.2.53 \(\int \frac {1}{\sqrt [3]{-a-b x^2} (\frac {9 a d}{b}+d x^2)} \, dx\) [153]

Optimal. Leaf size=153 \[ -\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x}{3 \sqrt {a}}\right )}{12 a^{5/6} d}-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\left (\sqrt [3]{a}+\sqrt [3]{-a-b x^2}\right )^2}{3 \sqrt [6]{a} \sqrt {b} x}\right )}{12 a^{5/6} d}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{a}+\sqrt [3]{-a-b x^2}\right )}{\sqrt {b} x}\right )}{4 \sqrt {3} a^{5/6} d} \]

[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)

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Rubi [A]
time = 0.02, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {403} \begin {gather*} -\frac {\sqrt {b} \text {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} \text {ArcTan}\left (\frac {\sqrt {b} x}{3 \sqrt {a}}\right )}{12 a^{5/6} d}+\frac {\sqrt {b} \tanh ^{-1}\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(Sqrt[b]*ArcTan[(Sqrt[b]*x)/(3*Sqrt[a])])/(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)))/(S
qrt[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*} \int \frac {1}{\sqrt [3]{-a-b x^2} \left (\frac {9 a d}{b}+d x^2\right )} \, dx &=-\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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 6.91, size = 172, normalized size = 1.12 \begin {gather*} \frac {27 a b x F_1\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 F_1\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 (F_1\left (\frac {3}{2};\frac {1}{3},2;\frac {5}{2};-\frac {b x^2}{a},-\frac {b x^2}{9 a}\right )+3 F_1\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 )} \end {gather*}

Warning: Unable to verify antiderivative.

[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*
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])))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (-b \,x^{2}-a \right )^{\frac {1}{3}} \left (\frac {9 a d}{b}+d \,x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {b \int \frac {1}{9 a \sqrt [3]{- a - b x^{2}} + b x^{2} \sqrt [3]{- a - b x^{2}}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (-b\,x^2-a\right )}^{1/3}\,\left (d\,x^2+\frac {9\,a\,d}{b}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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