3.5.86 \(\int \frac {x^9}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=322 \[ \frac {(4 a d+3 b c) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 b^{7/3} d^2}-\frac {(4 a d+3 b c) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{7/3} d^2}+\frac {x \left (a+b x^3\right )^{2/3} (b c-4 a d)}{3 b^2 d (b c-a d)}+\frac {c^{7/3} \log \left (c+d x^3\right )}{6 d^2 (b c-a d)^{4/3}}-\frac {c^{7/3} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^2 (b c-a d)^{4/3}}+\frac {c^{7/3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} d^2 (b c-a d)^{4/3}}+\frac {a x^4}{b \sqrt [3]{a+b x^3} (b c-a d)} \]

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Rubi [C]  time = 0.06, antiderivative size = 67, normalized size of antiderivative = 0.21, 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^{10} \sqrt [3]{\frac {b x^3}{a}+1} F_1\left (\frac {10}{3};\frac {4}{3},1;\frac {13}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{10 a c \sqrt [3]{a+b x^3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[x^9/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

(x^10*(1 + (b*x^3)/a)^(1/3)*AppellF1[10/3, 4/3, 1, 13/3, -((b*x^3)/a), -((d*x^3)/c)])/(10*a*c*(a + b*x^3)^(1/3
))

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^9}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx &=\frac {\sqrt [3]{1+\frac {b x^3}{a}} \int \frac {x^9}{\left (1+\frac {b x^3}{a}\right )^{4/3} \left (c+d x^3\right )} \, dx}{a \sqrt [3]{a+b x^3}}\\ &=\frac {x^{10} \sqrt [3]{1+\frac {b x^3}{a}} F_1\left (\frac {10}{3};\frac {4}{3},1;\frac {13}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{10 a c \sqrt [3]{a+b x^3}}\\ \end {align*}

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Mathematica [C]  time = 1.08, size = 504, 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 (-4 a^2 d^2+a b c d+3 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 (-4 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 )+24 a^2 d x \sqrt [3]{b c-a d}-6 b^2 c x^4 \sqrt [3]{b c-a d}+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 )+6 a b d x^4 \sqrt [3]{b c-a d}+2 a \sqrt [3]{c} \sqrt [3]{a+b x^3} (4 a d-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} (4 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 )-6 a b c x \sqrt [3]{b c-a d}\right )}{36 b^2 c d \sqrt [3]{a+b x^3} (b c-a d)^{4/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[x^9/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

-1/36*(3*(b*c - a*d)^(1/3)*(3*b^2*c^2 + a*b*c*d - 4*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*(-6*a*b*c*(b*c - a*d)^(1/3)*x + 24*a^2*d*(b*c - a*d)^(1/3)*x - 6*b^2*c*(
b*c - a*d)^(1/3)*x^4 + 6*a*b*d*(b*c - a*d)^(1/3)*x^4 - 2*Sqrt[3]*a*c^(1/3)*(-(b*c) + 4*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)*(-(b*c) + 4*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^(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)] - 4*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)]))/(b^2*c*d*(b*c - a*d)^(4/3)*(a + b*x^3)^(1/3))

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IntegrateAlgebraic [C]  time = 9.27, size = 565, normalized size = 1.75 \begin {gather*} -\frac {4 a^2 d x-a b c x+a b d x^4-b^2 c x^4}{3 b^2 d \sqrt [3]{a+b x^3} (b c-a d)}+\frac {(4 a d+3 b c) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{9 b^{7/3} d^2}-\frac {(4 a d+3 b c) \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} b^{7/3} d^2}+\frac {(-4 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 b^{7/3} d^2}+\frac {\left (c^{7/3}+i \sqrt {3} c^{7/3}\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 (b c-a d)^{4/3}}-\frac {\sqrt {\frac {1}{6} \left (-1+i \sqrt {3}\right )} c^{7/3} \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^2 (b c-a d)^{4/3}}-\frac {i \left (\sqrt {3} c^{7/3}-i c^{7/3}\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 (b c-a d)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^9/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

-1/3*(-(a*b*c*x) + 4*a^2*d*x - b^2*c*x^4 + a*b*d*x^4)/(b^2*d*(b*c - a*d)*(a + b*x^3)^(1/3)) - ((3*b*c + 4*a*d)
*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))])/(3*Sqrt[3]*b^(7/3)*d^2) - (Sqrt[(-1 + I*Sqrt[3
])/6]*c^(7/3)*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^2*(b*c - a*d)^(4/3)) + ((3*b*c + 4*a*d)*Log[-(b^(1/3)*x) + (a + b*x^3)
^(1/3)])/(9*b^(7/3)*d^2) + ((c^(7/3) + I*Sqrt[3]*c^(7/3))*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*(b*c - a*d)^(4/3)) + ((-3*b*c - 4*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^(7/3)*d^2) - ((I/12)*((-I)*c^(7/3) + Sqrt[3]*c^(7/3))*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*(b*c - a*d)^(4/3))

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fricas [B]  time = 1.36, size = 1329, normalized size = 4.13

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

[1/18*(3*sqrt(1/3)*(3*a*b^3*c^2 + a^2*b^2*c*d - 4*a^3*b*d^2 + (3*b^4*c^2 + a*b^3*c*d - 4*a^2*b^2*d^2)*x^3)*sqr
t(-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) - 6*sqrt(3)*(b^4*c^2*x^3 + a*b^3*c^2)*(c/(b*c - a*
d))^(1/3)*arctan(1/3*(sqrt(3)*x + 2*sqrt(3)*(b*x^3 + a)^(1/3)*(c/(b*c - a*d))^(1/3))/x) + 2*(3*a*b^2*c^2 + a^2
*b*c*d - 4*a^3*d^2 + (3*b^3*c^2 + a*b^2*c*d - 4*a^2*b*d^2)*x^3)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x
) - (3*a*b^2*c^2 + a^2*b*c*d - 4*a^3*d^2 + (3*b^3*c^2 + a*b^2*c*d - 4*a^2*b*d^2)*x^3)*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^4*c^2*x^3 + a*b^3*c^2)*(c/(b*c - a*d))^(1/3)*l
og(-((b*c - a*d)*x*(c/(b*c - a*d))^(2/3) - (b*x^3 + a)^(1/3)*c)/x) + 3*(b^4*c^2*x^3 + a*b^3*c^2)*(c/(b*c - a*d
))^(1/3)*log(((b*c - a*d)*x^2*(c/(b*c - a*d))^(1/3) + (b*x^3 + a)^(1/3)*(b*c - a*d)*x*(c/(b*c - a*d))^(2/3) +
(b*x^3 + a)^(2/3)*c)/x^2) + 6*((b^3*c*d - a*b^2*d^2)*x^4 + (a*b^2*c*d - 4*a^2*b*d^2)*x)*(b*x^3 + a)^(2/3))/(a*
b^4*c*d^2 - a^2*b^3*d^3 + (b^5*c*d^2 - a*b^4*d^3)*x^3), -1/18*(6*sqrt(3)*(b^4*c^2*x^3 + a*b^3*c^2)*(c/(b*c - a
*d))^(1/3)*arctan(1/3*(sqrt(3)*x + 2*sqrt(3)*(b*x^3 + a)^(1/3)*(c/(b*c - a*d))^(1/3))/x) - 2*(3*a*b^2*c^2 + a^
2*b*c*d - 4*a^3*d^2 + (3*b^3*c^2 + a*b^2*c*d - 4*a^2*b*d^2)*x^3)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/
x) + (3*a*b^2*c^2 + a^2*b*c*d - 4*a^3*d^2 + (3*b^3*c^2 + a*b^2*c*d - 4*a^2*b*d^2)*x^3)*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^4*c^2*x^3 + a*b^3*c^2)*(c/(b*c - a*d))^(1/3)*
log(-((b*c - a*d)*x*(c/(b*c - a*d))^(2/3) - (b*x^3 + a)^(1/3)*c)/x) - 3*(b^4*c^2*x^3 + a*b^3*c^2)*(c/(b*c - a*
d))^(1/3)*log(((b*c - a*d)*x^2*(c/(b*c - a*d))^(1/3) + (b*x^3 + a)^(1/3)*(b*c - a*d)*x*(c/(b*c - a*d))^(2/3) +
(b*x^3 + a)^(2/3)*c)/x^2) - 6*sqrt(1/3)*(3*a*b^3*c^2 + a^2*b^2*c*d - 4*a^3*b*d^2 + (3*b^4*c^2 + a*b^3*c*d - 4
*a^2*b^2*d^2)*x^3)*arctan(sqrt(1/3)*(b^(1/3)*x + 2*(b*x^3 + a)^(1/3))/(b^(1/3)*x))/b^(1/3) - 6*((b^3*c*d - a*b
^2*d^2)*x^4 + (a*b^2*c*d - 4*a^2*b*d^2)*x)*(b*x^3 + a)^(2/3))/(a*b^4*c*d^2 - a^2*b^3*d^3 + (b^5*c*d^2 - a*b^4*
d^3)*x^3)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{9}}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate(x^9/((b*x^3 + a)^(4/3)*(d*x^3 + c)), x)

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maple [F]  time = 0.45, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{9}}{\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (d \,x^{3}+c \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x)

[Out]

int(x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{9}}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate(x^9/((b*x^3 + a)^(4/3)*(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^9}{{\left (b\,x^3+a\right )}^{4/3}\,\left (d\,x^3+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/((a + b*x^3)^(4/3)*(c + d*x^3)),x)

[Out]

int(x^9/((a + b*x^3)^(4/3)*(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^{9}}{\left (a + b x^{3}\right )^{\frac {4}{3}} \left (c + d x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b*x**3+a)**(4/3)/(d*x**3+c),x)

[Out]

Integral(x**9/((a + b*x**3)**(4/3)*(c + d*x**3)), x)

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