3.5.35 \(\int \frac {x^6 (a+b x^3)^{2/3}}{c+d x^3} \, dx\)

Optimal. Leaf size=334 \[ -\frac {\left (-a^2 d^2-6 a b c d+9 b^2 c^2\right ) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{18 b^{4/3} d^3}+\frac {\left (-a^2 d^2-6 a b c d+9 b^2 c^2\right ) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} b^{4/3} d^3}-\frac {c^{4/3} (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^3}+\frac {c^{4/3} (b c-a d)^{2/3} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^3}-\frac {c^{4/3} (b c-a d)^{2/3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} d^3}-\frac {x \left (a+b x^3\right )^{2/3} (3 b c-a d)}{9 b d^2}+\frac {x^4 \left (a+b x^3\right )^{2/3}}{6 d} \]

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Rubi [C]  time = 0.06, antiderivative size = 64, normalized size of antiderivative = 0.19, 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^7 \left (a+b x^3\right )^{2/3} F_1\left (\frac {7}{3};-\frac {2}{3},1;\frac {10}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{7 c \left (\frac {b x^3}{a}+1\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[(x^6*(a + b*x^3)^(2/3))/(c + d*x^3),x]

[Out]

(x^7*(a + b*x^3)^(2/3)*AppellF1[7/3, -2/3, 1, 10/3, -((b*x^3)/a), -((d*x^3)/c)])/(7*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^6 \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^6 \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^7 \left (a+b x^3\right )^{2/3} F_1\left (\frac {7}{3};-\frac {2}{3},1;\frac {10}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{7 c \left (1+\frac {b x^3}{a}\right )^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.76, size = 525, normalized size = 1.57 \begin {gather*} \frac {3 x^4 \sqrt [3]{\frac {b x^3}{a}+1} \sqrt [3]{b c-a d} \left (-a^2 d^2-6 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 c \left (-a^2 \sqrt [3]{c} d \sqrt [3]{a+b x^3} \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 a^2 d x \sqrt [3]{b c-a d}+9 b^2 d x^7 \sqrt [3]{b c-a d}-18 b^2 c x^4 \sqrt [3]{b c-a d}+3 a b c^{4/3} \sqrt [3]{a+b x^3} \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 )+15 a b d x^4 \sqrt [3]{b c-a d}+2 a \sqrt [3]{c} \sqrt [3]{a+b x^3} (a d-3 b 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} \sqrt [3]{a+b x^3} (a d-3 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 )-18 a b c x \sqrt [3]{b c-a d}\right )}{108 b c d^2 \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^6*(a + b*x^3)^(2/3))/(c + d*x^3),x]

[Out]

(3*(b*c - a*d)^(1/3)*(9*b^2*c^2 - 6*a*b*c*d - 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*(-18*a*b*c*(b*c - a*d)^(1/3)*x + 6*a^2*d*(b*c - a*d)^(1/3)*x - 18*b^2*c*(b*c -
a*d)^(1/3)*x^4 + 15*a*b*d*(b*c - a*d)^(1/3)*x^4 + 9*b^2*d*(b*c - a*d)^(1/3)*x^7 - 2*Sqrt[3]*a*c^(1/3)*(-3*b*c
+ 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*c^(1
/3)*(-3*b*c + 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)] + 3*a*b*c^(4/3)*(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)] - a^2*c^(1/3)*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)]))/(108*b*c*d^2*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/3))

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IntegrateAlgebraic [C]  time = 6.45, size = 608, normalized size = 1.82 \begin {gather*} \frac {\left (a^2 d^2+6 a b c d-9 b^2 c^2\right ) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{27 b^{4/3} d^3}+\frac {\left (-a^2 d^2-6 a b c d+9 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{2 \sqrt [3]{a+b x^3}+\sqrt [3]{b} x}\right )}{9 \sqrt {3} b^{4/3} d^3}+\frac {\left (-a^2 d^2-6 a b c d+9 b^2 c^2\right ) \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 )}{54 b^{4/3} d^3}-\frac {i \left (\sqrt {3} c^{4/3} (b c-a d)^{2/3}-i c^{4/3} (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^3}+\frac {\sqrt {\frac {1}{6} \left (-1+i \sqrt {3}\right )} c^{4/3} (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 )}{d^3}+\frac {\left (c^{4/3} (b c-a d)^{2/3}+i \sqrt {3} c^{4/3} (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^3}+\frac {\left (a+b x^3\right )^{2/3} \left (2 a d x-6 b c x+3 b d x^4\right )}{18 b d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^6*(a + b*x^3)^(2/3))/(c + d*x^3),x]

[Out]

((a + b*x^3)^(2/3)*(-6*b*c*x + 2*a*d*x + 3*b*d*x^4))/(18*b*d^2) + ((9*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*ArcTan[(S
qrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))])/(9*Sqrt[3]*b^(4/3)*d^3) + (Sqrt[(-1 + I*Sqrt[3])/6]*c^(4
/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))])/d^3 + ((-9*b^2*c^2 + 6*a*b*c*d + a^2*d^2)*Log[-(b^(1/3)*x) + (a +
b*x^3)^(1/3)])/(27*b^(4/3)*d^3) - ((I/6)*((-I)*c^(4/3)*(b*c - a*d)^(2/3) + Sqrt[3]*c^(4/3)*(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)])/d^3 + ((9*b^2*c^2 - 6*a*b*c*d - a^2*d
^2)*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(54*b^(4/3)*d^3) + ((c^(4/3)*(b*c - a*
d)^(2/3) + I*Sqrt[3]*c^(4/3)*(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)])/(12*d^3)

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fricas [B]  time = 1.60, size = 1164, normalized size = 3.49

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

[-1/54*(18*sqrt(3)*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)^(1/3)*b^2*c*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)) - 18*(b^2*c^3 - 2*a*b*c^2*d
+ a^2*c*d^2)^(1/3)*b^2*c*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) + 9*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)^(1/3)*b^2*c*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) + 3*sqrt(1/3)*(9*b^3*c^2 - 6*a*b^2*c*d - a^2*b*d^2)*sqrt(-1/b^(2/3))*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*b
^(2/3)*x^2 - 3*sqrt(1/3)*(b^(4/3)*x^3 + (b*x^3 + a)^(1/3)*b*x^2 - 2*(b*x^3 + a)^(2/3)*b^(2/3)*x)*sqrt(-1/b^(2/
3)) + 2*a) + 2*(9*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - (9*b^2*c^2
- 6*a*b*c*d - a^2*d^2)*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) - 3*(3
*b^2*d^2*x^4 - 2*(3*b^2*c*d - a*b*d^2)*x)*(b*x^3 + a)^(2/3))/(b^2*d^3), -1/54*(18*sqrt(3)*(b^2*c^3 - 2*a*b*c^2
*d + a^2*c*d^2)^(1/3)*b^2*c*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)) - 18*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)^(1/3)*b^2*c*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) + 9*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*
d^2)^(1/3)*b^2*c*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) + 2*(9*b^2*c^2 - 6*a*b*c*d - a^2*
d^2)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - (9*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*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*sqrt(1/3)*(9*b^3*c^2 - 6*a*b^2*c*d - a^2*b*d^2
)*arctan(sqrt(1/3)*(b^(1/3)*x + 2*(b*x^3 + a)^(1/3))/(b^(1/3)*x))/b^(1/3) - 3*(3*b^2*d^2*x^4 - 2*(3*b^2*c*d -
a*b*d^2)*x)*(b*x^3 + a)^(2/3))/(b^2*d^3)]

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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^{6}}{d x^{3} + c}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(2/3)*x^6/(d*x^3 + c), x)

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maple [F]  time = 0.58, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{6}}{d \,x^{3}+c}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(b*x^3+a)^(2/3)/(d*x^3+c),x)

[Out]

int(x^6*(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^{6}}{d x^{3} + c}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(2/3)*x^6/(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^6\,{\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^6*(a + b*x^3)^(2/3))/(c + d*x^3),x)

[Out]

int((x^6*(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^{6} \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**6*(b*x**3+a)**(2/3)/(d*x**3+c),x)

[Out]

Integral(x**6*(a + b*x**3)**(2/3)/(c + d*x**3), x)

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