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

Optimal. Leaf size=141 \[ -\frac {a (6 b c-a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{18 b^{4/3}}+\frac {a (6 b c-a d) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} b^{4/3}}+\frac {x \left (a+b x^3\right )^{2/3} (6 b c-a d)}{18 b}+\frac {d x \left (a+b x^3\right )^{5/3}}{6 b} \]

[Out]

1/18*(-a*d+6*b*c)*x*(b*x^3+a)^(2/3)/b+1/6*d*x*(b*x^3+a)^(5/3)/b-1/18*a*(-a*d+6*b*c)*ln(-b^(1/3)*x+(b*x^3+a)^(1
/3))/b^(4/3)+1/27*a*(-a*d+6*b*c)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(4/3)*3^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {388, 195, 239} \[ -\frac {a (6 b c-a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{18 b^{4/3}}+\frac {a (6 b c-a d) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} b^{4/3}}+\frac {x \left (a+b x^3\right )^{2/3} (6 b c-a d)}{18 b}+\frac {d x \left (a+b x^3\right )^{5/3}}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(2/3)*(c + d*x^3),x]

[Out]

((6*b*c - a*d)*x*(a + b*x^3)^(2/3))/(18*b) + (d*x*(a + b*x^3)^(5/3))/(6*b) + (a*(6*b*c - a*d)*ArcTan[(1 + (2*b
^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(9*Sqrt[3]*b^(4/3)) - (a*(6*b*c - a*d)*Log[-(b^(1/3)*x) + (a + b*x^3)^(
1/3)])/(18*b^(4/3))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 239

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + (2*Rt[b, 3]*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rubi steps

\begin {align*} \int \left (a+b x^3\right )^{2/3} \left (c+d x^3\right ) \, dx &=\frac {d x \left (a+b x^3\right )^{5/3}}{6 b}-\frac {(-6 b c+a d) \int \left (a+b x^3\right )^{2/3} \, dx}{6 b}\\ &=\frac {(6 b c-a d) x \left (a+b x^3\right )^{2/3}}{18 b}+\frac {d x \left (a+b x^3\right )^{5/3}}{6 b}+\frac {(a (6 b c-a d)) \int \frac {1}{\sqrt [3]{a+b x^3}} \, dx}{9 b}\\ &=\frac {(6 b c-a d) x \left (a+b x^3\right )^{2/3}}{18 b}+\frac {d x \left (a+b x^3\right )^{5/3}}{6 b}+\frac {a (6 b c-a d) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{4/3}}-\frac {a (6 b c-a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{18 b^{4/3}}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 72, normalized size = 0.51 \[ \frac {x \left (a+b x^3\right )^{2/3} \left (\frac {(6 b c-a d) \, _2F_1\left (-\frac {2}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{\left (\frac {b x^3}{a}+1\right )^{2/3}}+d \left (a+b x^3\right )\right )}{6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x^3)^(2/3)*(d*(a + b*x^3) + ((6*b*c - a*d)*Hypergeometric2F1[-2/3, 1/3, 4/3, -((b*x^3)/a)])/(1 + (b*
x^3)/a)^(2/3)))/(6*b)

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fricas [A]  time = 0.87, size = 424, normalized size = 3.01 \[ \left [-\frac {3 \, \sqrt {\frac {1}{3}} {\left (6 \, a b^{2} c - a^{2} b d\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (b^{\frac {4}{3}} x^{3} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} - 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{\frac {2}{3}} x\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} + 2 \, a\right ) + 2 \, {\left (6 \, a b c - a^{2} d\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - {\left (6 \, a b c - a^{2} d\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 3 \, {\left (3 \, b^{2} d x^{4} + 2 \, {\left (3 \, b^{2} c + a b d\right )} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b^{2}}, -\frac {2 \, {\left (6 \, a b c - a^{2} d\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - {\left (6 \, a b c - a^{2} d\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + \frac {6 \, \sqrt {\frac {1}{3}} {\left (6 \, a b^{2} c - a^{2} b d\right )} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (b^{\frac {1}{3}} x + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{b^{\frac {1}{3}} x}\right )}{b^{\frac {1}{3}}} - 3 \, {\left (3 \, b^{2} d x^{4} + 2 \, {\left (3 \, b^{2} c + a b d\right )} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (d \,x^{3}+c \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.44, size = 322, normalized size = 2.28 \[ -\frac {1}{9} \, {\left (\frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {1}{3}}} + \frac {2 \, a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {1}{3}}} + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a}{{\left (b - \frac {b x^{3} + a}{x^{3}}\right )} x^{2}}\right )} c + \frac {1}{54} \, {\left (\frac {2 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {4}{3}}} - \frac {a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {4}{3}}} + \frac {2 \, a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {4}{3}}} + \frac {3 \, {\left (\frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b}{x^{2}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2}}{x^{5}}\right )}}{b^{3} - \frac {2 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b}{x^{6}}}\right )} d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*(2*sqrt(3)*a*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(1/3) - a*log(b^(2/3) + (b*x
^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(1/3) + 2*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(1/3) + 3
*(b*x^3 + a)^(2/3)*a/((b - (b*x^3 + a)/x^3)*x^2))*c + 1/54*(2*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x
^3 + a)^(1/3)/x)/b^(1/3))/b^(4/3) - a^2*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(
4/3) + 2*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(4/3) + 3*((b*x^3 + a)^(2/3)*a^2*b/x^2 + 2*(b*x^3 + a)^(5/3
)*a^2/x^5)/(b^3 - 2*(b*x^3 + a)*b^2/x^3 + (b*x^3 + a)^2*b/x^6))*d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,x^3+a\right )}^{2/3}\,\left (d\,x^3+c\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(2/3)*(c + d*x^3),x)

[Out]

int((a + b*x^3)^(2/3)*(c + d*x^3), x)

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sympy [C]  time = 5.35, size = 82, normalized size = 0.58 \[ \frac {a^{\frac {2}{3}} c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {a^{\frac {2}{3}} d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(2/3)*c*x*gamma(1/3)*hyper((-2/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3)) + a**(2/3)*d*x**4*
gamma(4/3)*hyper((-2/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3))

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