3.35 \(\int \frac {(a+b x^3)^{4/3}}{a-b x^3} \, dx\)

Optimal. Leaf size=464 \[ -\frac {\sqrt [3]{2} a^{2/3} \log \left (2^{2/3}-\frac {\sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 \sqrt [3]{b}}+\frac {\sqrt [3]{2} a^{2/3} \log \left (\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}-\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+1\right )}{3 \sqrt [3]{b}}-\frac {2 \sqrt [3]{2} a^{2/3} \log \left (\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+1\right )}{3 \sqrt [3]{b}}+\frac {a^{2/3} \log \left (\frac {\left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}+\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+2 \sqrt [3]{2}\right )}{3\ 2^{2/3} \sqrt [3]{b}}-\frac {2 \sqrt [3]{2} a^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\sqrt [3]{2} a^{2/3} \tan ^{-1}\left (\frac {\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {a x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 \left (a+b x^3\right )^{2/3}}-\frac {1}{2} x \sqrt [3]{a+b x^3} \]

[Out]

-1/2*x*(b*x^3+a)^(1/3)-1/2*a*x*(1+b*x^3/a)^(2/3)*hypergeom([1/3, 2/3],[4/3],-b*x^3/a)/(b*x^3+a)^(2/3)-1/3*2^(1
/3)*a^(2/3)*ln(2^(2/3)+(-a^(1/3)-b^(1/3)*x)/(b*x^3+a)^(1/3))/b^(1/3)+1/3*2^(1/3)*a^(2/3)*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))/b^(1/3)-2/3*2^(1/3)*a^(2/3)*ln(1+2^
(1/3)*(a^(1/3)+b^(1/3)*x)/(b*x^3+a)^(1/3))/b^(1/3)+1/6*a^(2/3)*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)/b^(1/3)-2/3*2^(1/3)*a^(2/3)*arctan(1/3*(1-2*2^(1/3)*(
a^(1/3)+b^(1/3)*x)/(b*x^3+a)^(1/3))*3^(1/2))/b^(1/3)*3^(1/2)-1/3*2^(1/3)*a^(2/3)*arctan(1/3*(1+2^(1/3)*(a^(1/3
)+b^(1/3)*x)/(b*x^3+a)^(1/3))*3^(1/2))/b^(1/3)*3^(1/2)

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Rubi [C]  time = 0.03, antiderivative size = 55, 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} \[ \frac {x \sqrt [3]{a+b x^3} F_1\left (\frac {1}{3};1,-\frac {4}{3};\frac {4}{3};\frac {b x^3}{a},-\frac {b x^3}{a}\right )}{\sqrt [3]{\frac {b x^3}{a}+1}} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*x^3)^(4/3)/(a - b*x^3),x]

[Out]

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

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

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Mathematica [C]  time = 0.12, size = 217, normalized size = 0.47 \[ \frac {x \left (\frac {48 a^3 F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )}{\left (a-b x^3\right ) \left (b x^3 \left (3 F_1\left (\frac {4}{3};\frac {2}{3},2;\frac {7}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )-2 F_1\left (\frac {4}{3};\frac {5}{3},1;\frac {7}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )\right )+4 a F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )\right )}+5 b x^3 \left (\frac {b x^3}{a}+1\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )-4 \left (a+b x^3\right )\right )}{8 \left (a+b x^3\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(4/3)/(a - b*x^3),x]

[Out]

(x*(-4*(a + b*x^3) + 5*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] + (48*a
^3*AppellF1[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])))))/(8*(a + b*x^3)^(2/3))

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fricas [F]  time = 34.20, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{b x^{3} - a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)/(-b*x^3+a),x, algorithm="fricas")

[Out]

integral(-(b*x^3 + a)^(4/3)/(b*x^3 - a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{b x^{3} - a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)/(-b*x^3+a),x, algorithm="giac")

[Out]

integrate(-(b*x^3 + a)^(4/3)/(b*x^3 - a), x)

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maple [F]  time = 0.61, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}}}{-b \,x^{3}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(4/3)/(-b*x^3+a),x)

[Out]

int((b*x^3+a)^(4/3)/(-b*x^3+a),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{b x^{3} - a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)/(-b*x^3+a),x, algorithm="maxima")

[Out]

-integrate((b*x^3 + a)^(4/3)/(b*x^3 - a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^3+a\right )}^{4/3}}{a-b\,x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(4/3)/(a - b*x^3),x)

[Out]

int((a + b*x^3)^(4/3)/(a - b*x^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {a \sqrt [3]{a + b x^{3}}}{- a + b x^{3}}\, dx - \int \frac {b x^{3} \sqrt [3]{a + b x^{3}}}{- a + b x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(4/3)/(-b*x**3+a),x)

[Out]

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

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