3.685 \(\int \frac{x^3 (a+b x^3)^{2/3}}{c+d x^3} \, dx\)
Optimal. Leaf size=272 \[ \frac{\sqrt [3]{c} (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^2}-\frac{\sqrt [3]{c} (b c-a d)^{2/3} \log \left (\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^2}+\frac{(3 b c-2 a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 \sqrt [3]{b} d^2}-\frac{(3 b c-2 a d) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{3 \sqrt{3} \sqrt [3]{b} d^2}+\frac{\sqrt [3]{c} (b c-a d)^{2/3} \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} d^2}+\frac{x \left (a+b x^3\right )^{2/3}}{3 d} \]
[Out]
(x*(a + b*x^3)^(2/3))/(3*d) - ((3*b*c - 2*a*d)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[
3]*b^(1/3)*d^2) + (c^(1/3)*(b*c - a*d)^(2/3)*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/
Sqrt[3]])/(Sqrt[3]*d^2) + (c^(1/3)*(b*c - a*d)^(2/3)*Log[c + d*x^3])/(6*d^2) - (c^(1/3)*(b*c - a*d)^(2/3)*Log[
((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)])/(2*d^2) + ((3*b*c - 2*a*d)*Log[-(b^(1/3)*x) + (a + b*x^3)^
(1/3)])/(6*b^(1/3)*d^2)
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Rubi [C] time = 0.0591927, antiderivative size = 64, normalized size of antiderivative =
0.24, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules
used = {511, 510} \[ \frac{x^4 \left (a+b x^3\right )^{2/3} F_1\left (\frac{4}{3};-\frac{2}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{4 c \left (\frac{b x^3}{a}+1\right )^{2/3}} \]
Warning: Unable to verify antiderivative.
[In]
Int[(x^3*(a + b*x^3)^(2/3))/(c + d*x^3),x]
[Out]
(x^4*(a + b*x^3)^(2/3)*AppellF1[4/3, -2/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])/(4*c*(1 + (b*x^3)/a)^(2/3))
Rule 511
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && !(IntegerQ[
p] || GtQ[a, 0])
Rule 510
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])
Rubi steps
\begin{align*} \int \frac{x^3 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx &=\frac{\left (a+b x^3\right )^{2/3} \int \frac{x^3 \left (1+\frac{b x^3}{a}\right )^{2/3}}{c+d x^3} \, dx}{\left (1+\frac{b x^3}{a}\right )^{2/3}}\\ &=\frac{x^4 \left (a+b x^3\right )^{2/3} F_1\left (\frac{4}{3};-\frac{2}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{4 c \left (1+\frac{b x^3}{a}\right )^{2/3}}\\ \end{align*}
Mathematica [C] time = 0.49689, size = 286, normalized size = 1.05 \[ \frac{\frac{3 x^4 \sqrt [3]{\frac{b x^3}{a}+1} (2 a d-3 b c) F_1\left (\frac{4}{3};\frac{1}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{c \sqrt [3]{a+b x^3}}+\frac{2 \left (-a \sqrt [3]{c} \log \left (\frac{x^2 (b c-a d)^{2/3}}{\left (a x^3+b\right )^{2/3}}+\frac{\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}+c^{2/3}\right )+6 x \left (a+b x^3\right )^{2/3} \sqrt [3]{b c-a d}+2 a \sqrt [3]{c} \log \left (\sqrt [3]{c}-\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}\right )-2 \sqrt{3} a \sqrt [3]{c} \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a x^3+b}}+1}{\sqrt{3}}\right )\right )}{\sqrt [3]{b c-a d}}}{36 d} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(x^3*(a + b*x^3)^(2/3))/(c + d*x^3),x]
[Out]
((3*(-3*b*c + 2*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)])/(c*(a +
b*x^3)^(1/3)) + (2*(6*(b*c - a*d)^(1/3)*x*(a + b*x^3)^(2/3) - 2*Sqrt[3]*a*c^(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*c^(1/3)*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(b + a*x^3)^(
1/3)] - a*c^(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)]))/(b*c - a*d)^(1/3))/(36*d)
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Maple [F] time = 0.049, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3}}{d{x}^{3}+c} \left ( b{x}^{3}+a \right ) ^{{\frac{2}{3}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*(b*x^3+a)^(2/3)/(d*x^3+c),x)
[Out]
int(x^3*(b*x^3+a)^(2/3)/(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{2}{3}} x^{3}}{d x^{3} + c}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="maxima")
[Out]
integrate((b*x^3 + a)^(2/3)*x^3/(d*x^3 + c), x)
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Fricas [B] time = 2.36371, size = 2546, normalized size = 9.36 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")
[Out]
[1/18*(6*(b*x^3 + a)^(2/3)*b*d*x - 3*sqrt(1/3)*(3*b^2*c - 2*a*b*d)*sqrt((-b)^(1/3)/b)*log(3*b*x^3 - 3*(b*x^3 +
a)^(1/3)*(-b)^(2/3)*x^2 - 3*sqrt(1/3)*((-b)^(1/3)*b*x^3 - (b*x^3 + a)^(1/3)*b*x^2 + 2*(b*x^3 + a)^(2/3)*(-b)^
(2/3)*x)*sqrt((-b)^(1/3)/b) + 2*a) + 6*sqrt(3)*(-b^2*c^3 + 2*a*b*c^2*d - a^2*c*d^2)^(1/3)*b*arctan(1/3*(sqrt(3
)*(b*c - a*d)*x - 2*sqrt(3)*(-b^2*c^3 + 2*a*b*c^2*d - a^2*c*d^2)^(1/3)*(b*x^3 + a)^(1/3))/((b*c - a*d)*x)) + 2
*(3*b*c - 2*a*d)*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) - (3*b*c - 2*a*d)*(-b)^(2/3)*log(((-b)^(
2/3)*x^2 - (b*x^3 + a)^(1/3)*(-b)^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) + 6*(-b^2*c^3 + 2*a*b*c^2*d - a^2*c*d^2)^(
1/3)*b*log(((-b^2*c^3 + 2*a*b*c^2*d - a^2*c*d^2)^(2/3)*x - (b*x^3 + a)^(1/3)*(b*c^2 - a*c*d))/x) - 3*(-b^2*c^3
+ 2*a*b*c^2*d - a^2*c*d^2)^(1/3)*b*log(((-b^2*c^3 + 2*a*b*c^2*d - a^2*c*d^2)^(1/3)*(b*c - a*d)*x^2 - (-b^2*c^
3 + 2*a*b*c^2*d - a^2*c*d^2)^(2/3)*(b*x^3 + a)^(1/3)*x - (b*x^3 + a)^(2/3)*(b*c^2 - a*c*d))/x^2))/(b*d^2), 1/1
8*(6*(b*x^3 + a)^(2/3)*b*d*x + 6*sqrt(1/3)*(3*b^2*c - 2*a*b*d)*sqrt(-(-b)^(1/3)/b)*arctan(-sqrt(1/3)*((-b)^(1/
3)*x - 2*(b*x^3 + a)^(1/3))*sqrt(-(-b)^(1/3)/b)/x) + 6*sqrt(3)*(-b^2*c^3 + 2*a*b*c^2*d - a^2*c*d^2)^(1/3)*b*ar
ctan(1/3*(sqrt(3)*(b*c - a*d)*x - 2*sqrt(3)*(-b^2*c^3 + 2*a*b*c^2*d - a^2*c*d^2)^(1/3)*(b*x^3 + a)^(1/3))/((b*
c - a*d)*x)) + 2*(3*b*c - 2*a*d)*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) - (3*b*c - 2*a*d)*(-b)^(
2/3)*log(((-b)^(2/3)*x^2 - (b*x^3 + a)^(1/3)*(-b)^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) + 6*(-b^2*c^3 + 2*a*b*c^2*
d - a^2*c*d^2)^(1/3)*b*log(((-b^2*c^3 + 2*a*b*c^2*d - a^2*c*d^2)^(2/3)*x - (b*x^3 + a)^(1/3)*(b*c^2 - a*c*d))/
x) - 3*(-b^2*c^3 + 2*a*b*c^2*d - a^2*c*d^2)^(1/3)*b*log(((-b^2*c^3 + 2*a*b*c^2*d - a^2*c*d^2)^(1/3)*(b*c - a*d
)*x^2 - (-b^2*c^3 + 2*a*b*c^2*d - a^2*c*d^2)^(2/3)*(b*x^3 + a)^(1/3)*x - (b*x^3 + a)^(2/3)*(b*c^2 - a*c*d))/x^
2))/(b*d^2)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \left (a + b x^{3}\right )^{\frac{2}{3}}}{c + d x^{3}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**3*(b*x**3+a)**(2/3)/(d*x**3+c),x)
[Out]
Integral(x**3*(a + b*x**3)**(2/3)/(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{2}{3}} x^{3}}{d x^{3} + c}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")
[Out]
integrate((b*x^3 + a)^(2/3)*x^3/(d*x^3 + c), x)