3.56 \(\int (a+b x^3)^{5/3} (c+d x^3) \, dx\)

Optimal. Leaf size=174 \[ -\frac {5 a^2 (9 b c-a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{162 b^{4/3}}+\frac {5 a^2 (9 b c-a d) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{81 \sqrt {3} b^{4/3}}+\frac {x \left (a+b x^3\right )^{5/3} (9 b c-a d)}{54 b}+\frac {5 a x \left (a+b x^3\right )^{2/3} (9 b c-a d)}{162 b}+\frac {d x \left (a+b x^3\right )^{8/3}}{9 b} \]

[Out]

5/162*a*(-a*d+9*b*c)*x*(b*x^3+a)^(2/3)/b+1/54*(-a*d+9*b*c)*x*(b*x^3+a)^(5/3)/b+1/9*d*x*(b*x^3+a)^(8/3)/b-5/162
*a^2*(-a*d+9*b*c)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(4/3)+5/243*a^2*(-a*d+9*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 = 0.06, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {388, 195, 239} \[ -\frac {5 a^2 (9 b c-a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{162 b^{4/3}}+\frac {5 a^2 (9 b c-a d) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{81 \sqrt {3} b^{4/3}}+\frac {x \left (a+b x^3\right )^{5/3} (9 b c-a d)}{54 b}+\frac {5 a x \left (a+b x^3\right )^{2/3} (9 b c-a d)}{162 b}+\frac {d x \left (a+b x^3\right )^{8/3}}{9 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(5/3)*(c + d*x^3),x]

[Out]

(5*a*(9*b*c - a*d)*x*(a + b*x^3)^(2/3))/(162*b) + ((9*b*c - a*d)*x*(a + b*x^3)^(5/3))/(54*b) + (d*x*(a + b*x^3
)^(8/3))/(9*b) + (5*a^2*(9*b*c - a*d)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(81*Sqrt[3]*b^(4/
3)) - (5*a^2*(9*b*c - a*d)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(162*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 )^{5/3} \left (c+d x^3\right ) \, dx &=\frac {d x \left (a+b x^3\right )^{8/3}}{9 b}-\frac {(-9 b c+a d) \int \left (a+b x^3\right )^{5/3} \, dx}{9 b}\\ &=\frac {(9 b c-a d) x \left (a+b x^3\right )^{5/3}}{54 b}+\frac {d x \left (a+b x^3\right )^{8/3}}{9 b}+\frac {(5 a (9 b c-a d)) \int \left (a+b x^3\right )^{2/3} \, dx}{54 b}\\ &=\frac {5 a (9 b c-a d) x \left (a+b x^3\right )^{2/3}}{162 b}+\frac {(9 b c-a d) x \left (a+b x^3\right )^{5/3}}{54 b}+\frac {d x \left (a+b x^3\right )^{8/3}}{9 b}+\frac {\left (5 a^2 (9 b c-a d)\right ) \int \frac {1}{\sqrt [3]{a+b x^3}} \, dx}{81 b}\\ &=\frac {5 a (9 b c-a d) x \left (a+b x^3\right )^{2/3}}{162 b}+\frac {(9 b c-a d) x \left (a+b x^3\right )^{5/3}}{54 b}+\frac {d x \left (a+b x^3\right )^{8/3}}{9 b}+\frac {5 a^2 (9 b c-a d) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{81 \sqrt {3} b^{4/3}}-\frac {5 a^2 (9 b c-a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{162 b^{4/3}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 75, normalized size = 0.43 \[ \frac {x \left (a+b x^3\right )^{2/3} \left (d \left (a+b x^3\right )^2-\frac {a (a d-9 b c) \, _2F_1\left (-\frac {5}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{\left (\frac {b x^3}{a}+1\right )^{2/3}}\right )}{9 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(5/3)*(c + d*x^3),x]

[Out]

(x*(a + b*x^3)^(2/3)*(d*(a + b*x^3)^2 - (a*(-9*b*c + a*d)*Hypergeometric2F1[-5/3, 1/3, 4/3, -((b*x^3)/a)])/(1
+ (b*x^3)/a)^(2/3)))/(9*b)

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fricas [A]  time = 0.71, size = 482, normalized size = 2.77 \[ \left [-\frac {15 \, \sqrt {\frac {1}{3}} {\left (9 \, a^{2} b^{2} c - a^{3} b d\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (b^{\frac {4}{3}} x^{3} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} - 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{\frac {2}{3}} x\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} + 2 \, a\right ) + 10 \, {\left (9 \, a^{2} b c - a^{3} d\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - 5 \, {\left (9 \, a^{2} b c - a^{3} d\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 3 \, {\left (18 \, b^{3} d x^{7} + 3 \, {\left (9 \, b^{3} c + 11 \, a b^{2} d\right )} x^{4} + 2 \, {\left (36 \, a b^{2} c + 5 \, a^{2} b d\right )} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{486 \, b^{2}}, -\frac {10 \, {\left (9 \, a^{2} b c - a^{3} d\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - 5 \, {\left (9 \, a^{2} b c - a^{3} d\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + \frac {30 \, \sqrt {\frac {1}{3}} {\left (9 \, a^{2} b^{2} c - a^{3} b d\right )} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (b^{\frac {1}{3}} x + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{b^{\frac {1}{3}} x}\right )}{b^{\frac {1}{3}}} - 3 \, {\left (18 \, b^{3} d x^{7} + 3 \, {\left (9 \, b^{3} c + 11 \, a b^{2} d\right )} x^{4} + 2 \, {\left (36 \, a b^{2} c + 5 \, a^{2} b d\right )} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{486 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/3)*(d*x^3+c),x, algorithm="fricas")

[Out]

[-1/486*(15*sqrt(1/3)*(9*a^2*b^2*c - a^3*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*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) +
10*(9*a^2*b*c - a^3*d)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - 5*(9*a^2*b*c - a^3*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*(18*b^3*d*x^7 + 3*(9*b^3*c + 11*a*b^2*d
)*x^4 + 2*(36*a*b^2*c + 5*a^2*b*d)*x)*(b*x^3 + a)^(2/3))/b^2, -1/486*(10*(9*a^2*b*c - a^3*d)*b^(2/3)*log(-(b^(
1/3)*x - (b*x^3 + a)^(1/3))/x) - 5*(9*a^2*b*c - a^3*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) + 30*sqrt(1/3)*(9*a^2*b^2*c - a^3*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*(18*b^3*d*x^7 + 3*(9*b^3*c + 11*a*b^2*d)*x^4 + 2*(36*a*b^2*c + 5*a^2*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 \[ \int {\left (b x^{3} + a\right )}^{\frac {5}{3}} {\left (d x^{3} + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/3)*(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(5/3)*(d*x^3 + c), x)

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (d \,x^{3}+c \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(5/3)*(d*x^3+c),x)

[Out]

int((b*x^3+a)^(5/3)*(d*x^3+c),x)

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maxima [B]  time = 1.68, size = 406, normalized size = 2.33 \[ -\frac {1}{54} \, {\left (\frac {10 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {5 \, a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {1}{3}}} + \frac {10 \, a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {1}{3}}} + \frac {3 \, {\left (\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b}{x^{2}} - \frac {8 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2}}{x^{5}}\right )}}{b^{2} - \frac {2 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2}}{x^{6}}}\right )} c + \frac {1}{486} \, {\left (\frac {10 \, \sqrt {3} a^{3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {4}{3}}} - \frac {5 \, a^{3} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {4}{3}}} + \frac {10 \, a^{3} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {4}{3}}} + \frac {3 \, {\left (\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{3} b^{2}}{x^{2}} - \frac {13 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{3} b}{x^{5}} - \frac {10 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} a^{3}}{x^{8}}\right )}}{b^{4} - \frac {3 \, {\left (b x^{3} + a\right )} b^{3}}{x^{3}} + \frac {3 \, {\left (b x^{3} + a\right )}^{2} b^{2}}{x^{6}} - \frac {{\left (b x^{3} + a\right )}^{3} b}{x^{9}}}\right )} d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/3)*(d*x^3+c),x, algorithm="maxima")

[Out]

-1/54*(10*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(1/3) - 5*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^(1/3) + 10*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/
b^(1/3) + 3*(5*(b*x^3 + a)^(2/3)*a^2*b/x^2 - 8*(b*x^3 + a)^(5/3)*a^2/x^5)/(b^2 - 2*(b*x^3 + a)*b/x^3 + (b*x^3
+ a)^2/x^6))*c + 1/486*(10*sqrt(3)*a^3*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(4/3) -
5*a^3*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) + 10*a^3*log(-b^(1/3) + (b*x
^3 + a)^(1/3)/x)/b^(4/3) + 3*(5*(b*x^3 + a)^(2/3)*a^3*b^2/x^2 - 13*(b*x^3 + a)^(5/3)*a^3*b/x^5 - 10*(b*x^3 + a
)^(8/3)*a^3/x^8)/(b^4 - 3*(b*x^3 + a)*b^3/x^3 + 3*(b*x^3 + a)^2*b^2/x^6 - (b*x^3 + a)^3*b/x^9))*d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,x^3+a\right )}^{5/3}\,\left (d\,x^3+c\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(5/3)*(c + d*x^3),x)

[Out]

int((a + b*x^3)^(5/3)*(c + d*x^3), x)

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sympy [C]  time = 10.42, size = 170, normalized size = 0.98 \[ \frac {a^{\frac {5}{3}} c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {a^{\frac {5}{3}} d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {a^{\frac {2}{3}} b c x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {a^{\frac {2}{3}} b d x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(5/3)*(d*x**3+c),x)

[Out]

a**(5/3)*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**(5/3)*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)) + a**(2/3)*b*c*x**4*gamma(4/3)*
hyper((-2/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3)) + a**(2/3)*b*d*x**7*gamma(7/3)*hyper((-2/3
, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(10/3))

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