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*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))

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Rubi [A]  time = 0.0592267, antiderivative size = 174, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [C]  time = 0.065927, 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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Maple [F]  time = 0.214, size = 0, normalized size = 0. \begin{align*} \int \left ( b{x}^{3}+a \right ) ^{{\frac{5}{3}}} \left ( d{x}^{3}+c \right ) \, dx \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 2.06421, size = 1222, normalized size = 7.02 \begin{align*} \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 ] \end{align*}

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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Sympy [C]  time = 6.70869, size = 170, normalized size = 0.98 \begin{align*} \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 )} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{3} + a\right )}^{\frac{5}{3}}{\left (d x^{3} + c\right )}\,{d x} \end{align*}

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)