3.87 \(\int \frac{(a+b x^3)^{5/3}}{c+d x^3} \, dx\)

Optimal. Leaf size=273 \[ \frac{b^{2/3} (3 b c-5 a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 d^2}-\frac{b^{2/3} (3 b c-5 a d) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{3 \sqrt{3} d^2}+\frac{(b c-a d)^{5/3} \log \left (c+d x^3\right )}{6 c^{2/3} d^2}-\frac{(b c-a d)^{5/3} \log \left (\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} d^2}+\frac{(b c-a d)^{5/3} \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} c^{2/3} d^2}+\frac{b x \left (a+b x^3\right )^{2/3}}{3 d} \]

[Out]

(b*x*(a + b*x^3)^(2/3))/(3*d) - (b^(2/3)*(3*b*c - 5*a*d)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]
)/(3*Sqrt[3]*d^2) + ((b*c - a*d)^(5/3)*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3
]])/(Sqrt[3]*c^(2/3)*d^2) + ((b*c - a*d)^(5/3)*Log[c + d*x^3])/(6*c^(2/3)*d^2) - ((b*c - a*d)^(5/3)*Log[((b*c
- a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)])/(2*c^(2/3)*d^2) + (b^(2/3)*(3*b*c - 5*a*d)*Log[-(b^(1/3)*x) + (a
+ b*x^3)^(1/3)])/(6*d^2)

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Rubi [C]  time = 0.027235, antiderivative size = 60, normalized size of antiderivative = 0.22, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {430, 429} \[ \frac{a x \left (a+b x^3\right )^{2/3} F_1\left (\frac{1}{3};-\frac{5}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{c \left (\frac{b x^3}{a}+1\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*x^3)^(5/3)/(c + d*x^3),x]

[Out]

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

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])

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])

Rubi steps

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

Mathematica [C]  time = 0.476772, size = 443, normalized size = 1.62 \[ \frac{2 \sqrt [3]{c} \left (3 a^2 d \sqrt [3]{a+b x^3} \log \left (\frac{x^2 (b c-a d)^{2/3}}{\left (a x^3+b\right )^{2/3}}+\frac{\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}+c^{2/3}\right )+6 b^2 c^{2/3} x^4 \sqrt [3]{b c-a d}-a b c \sqrt [3]{a+b x^3} \log \left (\frac{x^2 (b c-a d)^{2/3}}{\left (a x^3+b\right )^{2/3}}+\frac{\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}+c^{2/3}\right )+6 a b c^{2/3} x \sqrt [3]{b c-a d}+2 a \sqrt [3]{a+b x^3} (b c-3 a d) \log \left (\sqrt [3]{c}-\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}\right )+2 \sqrt{3} a \sqrt [3]{a+b x^3} (3 a d-b c) \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a x^3+b}}+1}{\sqrt{3}}\right )\right )+3 b x^4 \sqrt [3]{\frac{b x^3}{a}+1} \sqrt [3]{b c-a d} (5 a d-3 b c) F_1\left (\frac{4}{3};\frac{1}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{36 c d \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(5/3)/(c + d*x^3),x]

[Out]

(3*b*(b*c - a*d)^(1/3)*(-3*b*c + 5*a*d)*x^4*(1 + (b*x^3)/a)^(1/3)*AppellF1[4/3, 1/3, 1, 7/3, -((b*x^3)/a), -((
d*x^3)/c)] + 2*c^(1/3)*(6*a*b*c^(2/3)*(b*c - a*d)^(1/3)*x + 6*b^2*c^(2/3)*(b*c - a*d)^(1/3)*x^4 + 2*Sqrt[3]*a*
(-(b*c) + 3*a*d)*(a + b*x^3)^(1/3)*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(b + a*x^3)^(1/3)))/Sqrt[3]] +
2*a*(b*c - 3*a*d)*(a + b*x^3)^(1/3)*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(b + a*x^3)^(1/3)] - a*b*c*(a + b*x^3
)^(1/3)*Log[c^(2/3) + ((b*c - a*d)^(2/3)*x^2)/(b + a*x^3)^(2/3) + (c^(1/3)*(b*c - a*d)^(1/3)*x)/(b + a*x^3)^(1
/3)] + 3*a^2*d*(a + b*x^3)^(1/3)*Log[c^(2/3) + ((b*c - a*d)^(2/3)*x^2)/(b + a*x^3)^(2/3) + (c^(1/3)*(b*c - a*d
)^(1/3)*x)/(b + a*x^3)^(1/3)]))/(36*c*d*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/3))

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Maple [F]  time = 0.427, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{d{x}^{3}+c} \left ( b{x}^{3}+a \right ) ^{{\frac{5}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(5/3)/(d*x^3+c),x)

[Out]

int((b*x^3+a)^(5/3)/(d*x^3+c),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{5}{3}}}{d x^{3} + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(5/3)/(d*x^3 + c), x)

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Fricas [B]  time = 6.45721, size = 1258, normalized size = 4.61 \begin{align*} \frac{6 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}} b d x + 6 \, \sqrt{3}{\left (b c - a d\right )} \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac{1}{3}} \arctan \left (-\frac{\sqrt{3}{\left (b c - a d\right )} x + 2 \, \sqrt{3}{\left (b x^{3} + a\right )}^{\frac{1}{3}} c \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac{1}{3}}}{3 \,{\left (b c - a d\right )} x}\right ) + 2 \, \sqrt{3} \left (-b^{2}\right )^{\frac{1}{3}}{\left (3 \, b c - 5 \, a d\right )} \arctan \left (-\frac{\sqrt{3} b x - 2 \, \sqrt{3}{\left (b x^{3} + a\right )}^{\frac{1}{3}} \left (-b^{2}\right )^{\frac{1}{3}}}{3 \, b x}\right ) - 6 \,{\left (b c - a d\right )} \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac{1}{3}} \log \left (\frac{c x \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac{2}{3}} -{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (b c - a d\right )}}{x}\right ) - 2 \, \left (-b^{2}\right )^{\frac{1}{3}}{\left (3 \, b c - 5 \, a d\right )} \log \left (-\frac{\left (-b^{2}\right )^{\frac{2}{3}} x -{\left (b x^{3} + a\right )}^{\frac{1}{3}} b}{x}\right ) + \left (-b^{2}\right )^{\frac{1}{3}}{\left (3 \, b c - 5 \, a d\right )} \log \left (-\frac{\left (-b^{2}\right )^{\frac{1}{3}} b x^{2} -{\left (b x^{3} + a\right )}^{\frac{1}{3}} \left (-b^{2}\right )^{\frac{2}{3}} x -{\left (b x^{3} + a\right )}^{\frac{2}{3}} b}{x^{2}}\right ) + 3 \,{\left (b c - a d\right )} \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac{1}{3}} \log \left (-\frac{{\left (b c - a d\right )} x^{2} \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac{1}{3}} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} c x \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac{2}{3}} +{\left (b x^{3} + a\right )}^{\frac{2}{3}}{\left (b c - a d\right )}}{x^{2}}\right )}{18 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(6*(b*x^3 + a)^(2/3)*b*d*x + 6*sqrt(3)*(b*c - a*d)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)*arctan(-1/
3*(sqrt(3)*(b*c - a*d)*x + 2*sqrt(3)*(b*x^3 + a)^(1/3)*c*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3))/((b*c -
a*d)*x)) + 2*sqrt(3)*(-b^2)^(1/3)*(3*b*c - 5*a*d)*arctan(-1/3*(sqrt(3)*b*x - 2*sqrt(3)*(b*x^3 + a)^(1/3)*(-b^2
)^(1/3))/(b*x)) - 6*(b*c - a*d)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)*log((c*x*((b^2*c^2 - 2*a*b*c*d + a
^2*d^2)/c^2)^(2/3) - (b*x^3 + a)^(1/3)*(b*c - a*d))/x) - 2*(-b^2)^(1/3)*(3*b*c - 5*a*d)*log(-((-b^2)^(2/3)*x -
(b*x^3 + a)^(1/3)*b)/x) + (-b^2)^(1/3)*(3*b*c - 5*a*d)*log(-((-b^2)^(1/3)*b*x^2 - (b*x^3 + a)^(1/3)*(-b^2)^(2
/3)*x - (b*x^3 + a)^(2/3)*b)/x^2) + 3*(b*c - a*d)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)*log(-((b*c - a*d
)*x^2*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3) + (b*x^3 + a)^(1/3)*c*x*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2
)^(2/3) + (b*x^3 + a)^(2/3)*(b*c - a*d))/x^2))/d^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{3}\right )^{\frac{5}{3}}}{c + d x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(5/3)/(d*x**3+c),x)

[Out]

Integral((a + b*x**3)**(5/3)/(c + d*x**3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{5}{3}}}{d x^{3} + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(5/3)/(d*x^3 + c), x)