3.86 \(\int \frac{(a+b x^3)^{8/3}}{c+d x^3} \, dx\)

Optimal. Leaf size=331 \[ -\frac{b^{2/3} \left (20 a^2 d^2-24 a b c d+9 b^2 c^2\right ) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{18 d^3}+\frac{b^{2/3} \left (20 a^2 d^2-24 a b c d+9 b^2 c^2\right ) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{9 \sqrt{3} d^3}-\frac{(b c-a d)^{8/3} \log \left (c+d x^3\right )}{6 c^{2/3} d^3}+\frac{(b c-a d)^{8/3} \log \left (\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} d^3}-\frac{(b c-a d)^{8/3} \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} c^{2/3} d^3}-\frac{b x \left (a+b x^3\right )^{2/3} (6 b c-11 a d)}{18 d^2}+\frac{b x \left (a+b x^3\right )^{5/3}}{6 d} \]

[Out]

-(b*(6*b*c - 11*a*d)*x*(a + b*x^3)^(2/3))/(18*d^2) + (b*x*(a + b*x^3)^(5/3))/(6*d) + (b^(2/3)*(9*b^2*c^2 - 24*
a*b*c*d + 20*a^2*d^2)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(9*Sqrt[3]*d^3) - ((b*c - a*d)^(8
/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^3) - ((b*c -
a*d)^(8/3)*Log[c + d*x^3])/(6*c^(2/3)*d^3) + ((b*c - a*d)^(8/3)*Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^
3)^(1/3)])/(2*c^(2/3)*d^3) - (b^(2/3)*(9*b^2*c^2 - 24*a*b*c*d + 20*a^2*d^2)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/
3)])/(18*d^3)

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Rubi [C]  time = 0.0275217, antiderivative size = 62, normalized size of antiderivative = 0.19, 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} \[ \frac{a^2 x \left (a+b x^3\right )^{2/3} F_1\left (\frac{1}{3};-\frac{8}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{c \left (\frac{b x^3}{a}+1\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*x^3)^(8/3)/(c + d*x^3),x]

[Out]

(a^2*x*(a + b*x^3)^(2/3)*AppellF1[1/3, -8/3, 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 )^{8/3}}{c+d x^3} \, dx &=\frac{\left (a^2 \left (a+b x^3\right )^{2/3}\right ) \int \frac{\left (1+\frac{b x^3}{a}\right )^{8/3}}{c+d x^3} \, dx}{\left (1+\frac{b x^3}{a}\right )^{2/3}}\\ &=\frac{a^2 x \left (a+b x^3\right )^{2/3} F_1\left (\frac{1}{3};-\frac{8}{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 = 0.788875, size = 655, normalized size = 1.98 \[ \frac{3 b x^4 \sqrt [3]{\frac{b x^3}{a}+1} \sqrt [3]{b c-a d} \left (20 a^2 d^2-24 a b c d+9 b^2 c^2\right ) F_1\left (\frac{4}{3};\frac{1}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )+2 \sqrt [3]{c} \left (-2 a \sqrt [3]{a+b x^3} \left (9 a^2 d^2-7 a b c d+3 b^2 c^2\right ) \log \left (\sqrt [3]{c}-\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}\right )+2 \sqrt{3} a \sqrt [3]{a+b x^3} \left (9 a^2 d^2-7 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a x^3+b}}+1}{\sqrt{3}}\right )+9 a^3 d^2 \sqrt [3]{a+b x^3} \log \left (\frac{x^2 (b c-a d)^{2/3}}{\left (a x^3+b\right )^{2/3}}+\frac{\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}+c^{2/3}\right )-7 a^2 b c d \sqrt [3]{a+b x^3} \log \left (\frac{x^2 (b c-a d)^{2/3}}{\left (a x^3+b\right )^{2/3}}+\frac{\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}+c^{2/3}\right )+42 a^2 b c^{2/3} d x \sqrt [3]{b c-a d}+9 b^3 c^{2/3} d x^7 \sqrt [3]{b c-a d}-18 b^3 c^{5/3} x^4 \sqrt [3]{b c-a d}+51 a b^2 c^{2/3} d x^4 \sqrt [3]{b c-a d}+3 a b^2 c^2 \sqrt [3]{a+b x^3} \log \left (\frac{x^2 (b c-a d)^{2/3}}{\left (a x^3+b\right )^{2/3}}+\frac{\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}+c^{2/3}\right )-18 a b^2 c^{5/3} x \sqrt [3]{b c-a d}\right )}{108 c d^2 \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(8/3)/(c + d*x^3),x]

[Out]

(3*b*(b*c - a*d)^(1/3)*(9*b^2*c^2 - 24*a*b*c*d + 20*a^2*d^2)*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)*(-18*a*b^2*c^(5/3)*(b*c - a*d)^(1/3)*x + 42*a^2*b*c^(2/3)*d*(b*c -
a*d)^(1/3)*x - 18*b^3*c^(5/3)*(b*c - a*d)^(1/3)*x^4 + 51*a*b^2*c^(2/3)*d*(b*c - a*d)^(1/3)*x^4 + 9*b^3*c^(2/3
)*d*(b*c - a*d)^(1/3)*x^7 + 2*Sqrt[3]*a*(3*b^2*c^2 - 7*a*b*c*d + 9*a^2*d^2)*(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*(3*b^2*c^2 - 7*a*b*c*d + 9*a^2*d^2)*(a + b*x^3
)^(1/3)*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(b + a*x^3)^(1/3)] + 3*a*b^2*c^2*(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)] - 7*a^2*b*c*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)] + 9*a^3*d^2*(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)]))/(108*c*d^2*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/3))

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Maple [F]  time = 0.426, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{d{x}^{3}+c} \left ( b{x}^{3}+a \right ) ^{{\frac{8}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(8/3)/(d*x^3+c),x)

[Out]

int((b*x^3+a)^(8/3)/(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{8}{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)^(8/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(8/3)/(d*x^3 + c), x)

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Fricas [B]  time = 61.6469, size = 1486, normalized size = 4.49 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(8/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

-1/54*(18*sqrt(3)*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)*arctan(-1/3*(sqr
t(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)*(9*b^2*c^2 - 24*a*b*c*d + 20*a^2*d^2)*(-b^2)^(1/3)*arctan(-1/3*(sqrt(3)*b*x - 2*sqrt(3)*(b*x^3
+ a)^(1/3)*(-b^2)^(1/3))/(b*x)) - 18*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*((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*(9*b^2*c^2 - 2
4*a*b*c*d + 20*a^2*d^2)*(-b^2)^(1/3)*log(-((-b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b)/x) + (9*b^2*c^2 - 24*a*b*c*d
+ 20*a^2*d^2)*(-b^2)^(1/3)*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) + 9*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*((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) - 3*(3*b^2*d^2*x^4 - 2*(3*b^2*c*d - 7*a*b*d^2)*x)*(b*x^3 + a)^(2/3))/d^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(8/3)/(d*x**3+c),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{8}{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)^(8/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(8/3)/(d*x^3 + c), x)