3.38 \(\int \frac {1}{(a-b x^3) (a+b x^3)^{5/3}} \, dx\)

Optimal. Leaf size=473 \[ -\frac {\log \left (2^{2/3}-\frac {\sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{12\ 2^{2/3} a^{7/3} \sqrt [3]{b}}+\frac {\log \left (\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}-\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+1\right )}{12\ 2^{2/3} a^{7/3} \sqrt [3]{b}}-\frac {\log \left (\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+1\right )}{6\ 2^{2/3} a^{7/3} \sqrt [3]{b}}+\frac {\log \left (\frac {\left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}+\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+2 \sqrt [3]{2}\right )}{24\ 2^{2/3} a^{7/3} \sqrt [3]{b}}-\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{2\ 2^{2/3} \sqrt {3} a^{7/3} \sqrt [3]{b}}-\frac {\tan ^{-1}\left (\frac {\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{4\ 2^{2/3} \sqrt {3} a^{7/3} \sqrt [3]{b}}+\frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 a^2 \left (a+b x^3\right )^{2/3}}+\frac {x}{4 a^2 \left (a+b x^3\right )^{2/3}} \]

[Out]

1/4*x/a^2/(b*x^3+a)^(2/3)+1/2*x*(1+b*x^3/a)^(2/3)*hypergeom([1/3, 2/3],[4/3],-b*x^3/a)/a^2/(b*x^3+a)^(2/3)-1/2
4*ln(2^(2/3)+(-a^(1/3)-b^(1/3)*x)/(b*x^3+a)^(1/3))*2^(1/3)/a^(7/3)/b^(1/3)+1/24*ln(1+2^(2/3)*(a^(1/3)+b^(1/3)*
x)^2/(b*x^3+a)^(2/3)-2^(1/3)*(a^(1/3)+b^(1/3)*x)/(b*x^3+a)^(1/3))*2^(1/3)/a^(7/3)/b^(1/3)-1/12*ln(1+2^(1/3)*(a
^(1/3)+b^(1/3)*x)/(b*x^3+a)^(1/3))*2^(1/3)/a^(7/3)/b^(1/3)+1/48*ln(2*2^(1/3)+(a^(1/3)+b^(1/3)*x)^2/(b*x^3+a)^(
2/3)+2^(2/3)*(a^(1/3)+b^(1/3)*x)/(b*x^3+a)^(1/3))*2^(1/3)/a^(7/3)/b^(1/3)-1/12*arctan(1/3*(1-2*2^(1/3)*(a^(1/3
)+b^(1/3)*x)/(b*x^3+a)^(1/3))*3^(1/2))*2^(1/3)/a^(7/3)/b^(1/3)*3^(1/2)-1/24*arctan(1/3*(1+2^(1/3)*(a^(1/3)+b^(
1/3)*x)/(b*x^3+a)^(1/3))*3^(1/2))*2^(1/3)/a^(7/3)/b^(1/3)*3^(1/2)

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Rubi [C]  time = 0.03, antiderivative size = 58, normalized size of antiderivative = 0.12, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {430, 429} \[ \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} F_1\left (\frac {1}{3};1,\frac {5}{3};\frac {4}{3};\frac {b x^3}{a},-\frac {b x^3}{a}\right )}{a^2 \left (a+b x^3\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Int[1/((a - b*x^3)*(a + b*x^3)^(5/3)),x]

[Out]

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

Rule 429

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 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[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])

Rubi steps

\begin {align*} \int \frac {1}{\left (a-b x^3\right ) \left (a+b x^3\right )^{5/3}} \, dx &=\frac {\left (1+\frac {b x^3}{a}\right )^{2/3} \int \frac {1}{\left (a-b x^3\right ) \left (1+\frac {b x^3}{a}\right )^{5/3}} \, dx}{a \left (a+b x^3\right )^{2/3}}\\ &=\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} F_1\left (\frac {1}{3};1,\frac {5}{3};\frac {4}{3};\frac {b x^3}{a},-\frac {b x^3}{a}\right )}{a^2 \left (a+b x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 213, normalized size = 0.45 \[ \frac {x \left (-\frac {b x^3 \left (\frac {b x^3}{a}+1\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )}{a^3}+\frac {4}{a^2}+\frac {48 F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )}{\left (a-b x^3\right ) \left (b x^3 \left (3 F_1\left (\frac {4}{3};\frac {2}{3},2;\frac {7}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )-2 F_1\left (\frac {4}{3};\frac {5}{3},1;\frac {7}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )\right )+4 a F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )\right )}\right )}{16 \left (a+b x^3\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a - b*x^3)*(a + b*x^3)^(5/3)),x]

[Out]

(x*(4/a^2 - (b*x^3*(1 + (b*x^3)/a)^(2/3)*AppellF1[4/3, 2/3, 1, 7/3, -((b*x^3)/a), (b*x^3)/a])/a^3 + (48*Appell
F1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), (b*x^3)/a])/((a - b*x^3)*(4*a*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), (b*x
^3)/a] + b*x^3*(3*AppellF1[4/3, 2/3, 2, 7/3, -((b*x^3)/a), (b*x^3)/a] - 2*AppellF1[4/3, 5/3, 1, 7/3, -((b*x^3)
/a), (b*x^3)/a])))))/(16*(a + b*x^3)^(2/3))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^3+a)/(b*x^3+a)^(5/3),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {1}{{\left (b x^{3} + a\right )}^{\frac {5}{3}} {\left (b x^{3} - a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^3+a)/(b*x^3+a)^(5/3),x, algorithm="giac")

[Out]

integrate(-1/((b*x^3 + a)^(5/3)*(b*x^3 - a)), x)

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maple [F]  time = 0.68, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{\frac {5}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-b*x^3+a)/(b*x^3+a)^(5/3),x)

[Out]

int(1/(-b*x^3+a)/(b*x^3+a)^(5/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {1}{{\left (b x^{3} + a\right )}^{\frac {5}{3}} {\left (b x^{3} - a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^3+a)/(b*x^3+a)^(5/3),x, algorithm="maxima")

[Out]

-integrate(1/((b*x^3 + a)^(5/3)*(b*x^3 - a)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (b\,x^3+a\right )}^{5/3}\,\left (a-b\,x^3\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^3)^(5/3)*(a - b*x^3)),x)

[Out]

int(1/((a + b*x^3)^(5/3)*(a - b*x^3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {1}{- a^{2} \left (a + b x^{3}\right )^{\frac {2}{3}} + b^{2} x^{6} \left (a + b x^{3}\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x**3+a)/(b*x**3+a)**(5/3),x)

[Out]

-Integral(1/(-a**2*(a + b*x**3)**(2/3) + b**2*x**6*(a + b*x**3)**(2/3)), x)

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