3.5.36 \(\int \frac {x^3 (a+b x^3)^{2/3}}{c+d x^3} \, dx\)

Optimal. Leaf size=272 \[ \frac {\sqrt [3]{c} (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^2}-\frac {\sqrt [3]{c} (b c-a d)^{2/3} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^2}+\frac {(3 b c-2 a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 \sqrt [3]{b} d^2}-\frac {(3 b c-2 a d) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} \sqrt [3]{b} d^2}+\frac {\sqrt [3]{c} (b c-a d)^{2/3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} d^2}+\frac {x \left (a+b x^3\right )^{2/3}}{3 d} \]

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Rubi [C]  time = 0.06, 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} \begin {gather*} \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}} \end {gather*}

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 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])

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])

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*}

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Mathematica [C]  time = 0.59, size = 286, normalized size = 1.05 \begin {gather*} \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 {\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}+\frac {x^2 (b c-a d)^{2/3}}{\left (a x^3+b\right )^{2/3}}+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} \end {gather*}

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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IntegrateAlgebraic [C]  time = 4.96, size = 549, normalized size = 2.02 \begin {gather*} \frac {(2 a d-3 b c) \log \left (\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}+b^{2/3} x^2\right )}{18 \sqrt [3]{b} d^2}-\frac {i \left (\sqrt {3} \sqrt [3]{c} (b c-a d)^{2/3}-i \sqrt [3]{c} (b c-a d)^{2/3}\right ) \log \left (\left (\sqrt {3}+i\right ) c^{2/3} \left (a+b x^3\right )^{2/3}+\sqrt [3]{c} \left (-\sqrt {3} x+i x\right ) \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}-2 i x^2 (b c-a d)^{2/3}\right )}{12 d^2}+\frac {(3 b c-2 a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{9 \sqrt [3]{b} d^2}+\frac {\left (\sqrt [3]{c} (b c-a d)^{2/3}+i \sqrt {3} \sqrt [3]{c} (b c-a d)^{2/3}\right ) \log \left (2 x \sqrt [3]{b c-a d}+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{6 d^2}-\frac {(3 b c-2 a d) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{2 \sqrt [3]{a+b x^3}+\sqrt [3]{b} x}\right )}{3 \sqrt {3} \sqrt [3]{b} d^2}-\frac {\sqrt {-1+i \sqrt {3}} \sqrt [3]{c} (b c-a d)^{2/3} \tan ^{-1}\left (\frac {3 x \sqrt [3]{b c-a d}}{\sqrt {3} x \sqrt [3]{b c-a d}-\sqrt {3} \sqrt [3]{c} \sqrt [3]{a+b x^3}-3 i \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{\sqrt {6} d^2}+\frac {x \left (a+b x^3\right )^{2/3}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^3*(a + b*x^3)^(2/3))/(c + d*x^3),x]

[Out]

(x*(a + b*x^3)^(2/3))/(3*d) - ((3*b*c - 2*a*d)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))])/
(3*Sqrt[3]*b^(1/3)*d^2) - (Sqrt[-1 + I*Sqrt[3]]*c^(1/3)*(b*c - a*d)^(2/3)*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(Sqrt
[3]*(b*c - a*d)^(1/3)*x - (3*I)*c^(1/3)*(a + b*x^3)^(1/3) - Sqrt[3]*c^(1/3)*(a + b*x^3)^(1/3))])/(Sqrt[6]*d^2)
+ ((3*b*c - 2*a*d)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(9*b^(1/3)*d^2) + ((c^(1/3)*(b*c - a*d)^(2/3) + I*S
qrt[3]*c^(1/3)*(b*c - a*d)^(2/3))*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(6*d
^2) + ((-3*b*c + 2*a*d)*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(18*b^(1/3)*d^2) -
((I/12)*((-I)*c^(1/3)*(b*c - a*d)^(2/3) + Sqrt[3]*c^(1/3)*(b*c - a*d)^(2/3))*Log[(-2*I)*(b*c - a*d)^(2/3)*x^2
+ c^(1/3)*(b*c - a*d)^(1/3)*(I*x - Sqrt[3]*x)*(a + b*x^3)^(1/3) + (I + Sqrt[3])*c^(2/3)*(a + b*x^3)^(2/3)])/d
^2

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fricas [B]  time = 0.57, size = 1091, normalized size = 4.01

result too large to display

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{3}}{d x^{3} + c}\,{d x} \end {gather*}

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)

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maple [F]  time = 0.62, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{3}}{d \,x^{3}+c}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{3}}{d x^{3} + c}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3\,{\left (b\,x^3+a\right )}^{2/3}}{d\,x^3+c} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*x^3)^(2/3))/(c + d*x^3),x)

[Out]

int((x^3*(a + b*x^3)^(2/3))/(c + d*x^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (a + b x^{3}\right )^{\frac {2}{3}}}{c + d x^{3}}\, dx \end {gather*}

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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