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

Optimal. Leaf size=219 \[ \frac{x \left (a+b x^3\right )^{2/3} \left (2 a^2 d^2-9 a b c d+27 b^2 c^2\right )}{81 b^2}-\frac{a \left (2 a^2 d^2-9 a b c d+27 b^2 c^2\right ) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{81 b^{7/3}}+\frac{2 a \left (2 a^2 d^2-9 a b c d+27 b^2 c^2\right ) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{81 \sqrt{3} b^{7/3}}+\frac{2 d x \left (a+b x^3\right )^{5/3} (3 b c-a d)}{27 b^2}+\frac{d x \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )}{9 b} \]

[Out]

((27*b^2*c^2 - 9*a*b*c*d + 2*a^2*d^2)*x*(a + b*x^3)^(2/3))/(81*b^2) + (2*d*(3*b*c - a*d)*x*(a + b*x^3)^(5/3))/
(27*b^2) + (d*x*(a + b*x^3)^(5/3)*(c + d*x^3))/(9*b) + (2*a*(27*b^2*c^2 - 9*a*b*c*d + 2*a^2*d^2)*ArcTan[(1 + (
2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(81*Sqrt[3]*b^(7/3)) - (a*(27*b^2*c^2 - 9*a*b*c*d + 2*a^2*d^2)*Log[-
(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(81*b^(7/3))

________________________________________________________________________________________

Rubi [A]  time = 0.166311, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {416, 388, 195, 239} \[ \frac{x \left (a+b x^3\right )^{2/3} \left (2 a^2 d^2-9 a b c d+27 b^2 c^2\right )}{81 b^2}-\frac{a \left (2 a^2 d^2-9 a b c d+27 b^2 c^2\right ) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{81 b^{7/3}}+\frac{2 a \left (2 a^2 d^2-9 a b c d+27 b^2 c^2\right ) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{81 \sqrt{3} b^{7/3}}+\frac{2 d x \left (a+b x^3\right )^{5/3} (3 b c-a d)}{27 b^2}+\frac{d x \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )}{9 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((27*b^2*c^2 - 9*a*b*c*d + 2*a^2*d^2)*x*(a + b*x^3)^(2/3))/(81*b^2) + (2*d*(3*b*c - a*d)*x*(a + b*x^3)^(5/3))/
(27*b^2) + (d*x*(a + b*x^3)^(5/3)*(c + d*x^3))/(9*b) + (2*a*(27*b^2*c^2 - 9*a*b*c*d + 2*a^2*d^2)*ArcTan[(1 + (
2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(81*Sqrt[3]*b^(7/3)) - (a*(27*b^2*c^2 - 9*a*b*c*d + 2*a^2*d^2)*Log[-
(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(81*b^(7/3))

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, 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]

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 )^{2/3} \left (c+d x^3\right )^2 \, dx &=\frac{d x \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )}{9 b}+\frac{\int \left (a+b x^3\right )^{2/3} \left (c (9 b c-a d)+4 d (3 b c-a d) x^3\right ) \, dx}{9 b}\\ &=\frac{2 d (3 b c-a d) x \left (a+b x^3\right )^{5/3}}{27 b^2}+\frac{d x \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )}{9 b}-\frac{(4 a d (3 b c-a d)-6 b c (9 b c-a d)) \int \left (a+b x^3\right )^{2/3} \, dx}{54 b^2}\\ &=\frac{\left (27 b^2 c^2-9 a b c d+2 a^2 d^2\right ) x \left (a+b x^3\right )^{2/3}}{81 b^2}+\frac{2 d (3 b c-a d) x \left (a+b x^3\right )^{5/3}}{27 b^2}+\frac{d x \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )}{9 b}+\frac{\left (2 a \left (27 b^2 c^2-9 a b c d+2 a^2 d^2\right )\right ) \int \frac{1}{\sqrt [3]{a+b x^3}} \, dx}{81 b^2}\\ &=\frac{\left (27 b^2 c^2-9 a b c d+2 a^2 d^2\right ) x \left (a+b x^3\right )^{2/3}}{81 b^2}+\frac{2 d (3 b c-a d) x \left (a+b x^3\right )^{5/3}}{27 b^2}+\frac{d x \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )}{9 b}+\frac{2 a \left (27 b^2 c^2-9 a b c d+2 a^2 d^2\right ) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt{3}}\right )}{81 \sqrt{3} b^{7/3}}-\frac{a \left (27 b^2 c^2-9 a b c d+2 a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{81 b^{7/3}}\\ \end{align*}

Mathematica [C]  time = 4.1492, size = 179, normalized size = 0.82 \[ \frac{x \left (a+b x^3\right )^{2/3} \left (9 b x^3 \text{Gamma}\left (\frac{1}{3}\right ) \left (c+d x^3\right )^2 \text{HypergeometricPFQ}\left (\left \{\frac{1}{3},\frac{4}{3},2\right \},\left \{1,\frac{13}{3}\right \},-\frac{b x^3}{a}\right )+3 b x^3 \text{Gamma}\left (\frac{1}{3}\right ) \left (11 c^2+16 c d x^3+5 d^2 x^6\right ) \, _2F_1\left (\frac{1}{3},\frac{4}{3};\frac{13}{3};-\frac{b x^3}{a}\right )-20 a \text{Gamma}\left (-\frac{2}{3}\right ) \left (14 c^2+7 c d x^3+2 d^2 x^6\right ) \, _2F_1\left (-\frac{2}{3},\frac{1}{3};\frac{10}{3};-\frac{b x^3}{a}\right )\right )}{420 a \text{Gamma}\left (\frac{1}{3}\right ) \left (\frac{b x^3}{a}+1\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(a + b*x^3)^(2/3)*(-20*a*(14*c^2 + 7*c*d*x^3 + 2*d^2*x^6)*Gamma[-2/3]*Hypergeometric2F1[-2/3, 1/3, 10/3, -(
(b*x^3)/a)] + 3*b*x^3*(11*c^2 + 16*c*d*x^3 + 5*d^2*x^6)*Gamma[1/3]*Hypergeometric2F1[1/3, 4/3, 13/3, -((b*x^3)
/a)] + 9*b*x^3*(c + d*x^3)^2*Gamma[1/3]*HypergeometricPFQ[{1/3, 4/3, 2}, {1, 13/3}, -((b*x^3)/a)]))/(420*a*(1
+ (b*x^3)/a)^(2/3)*Gamma[1/3])

________________________________________________________________________________________

Maple [F]  time = 0.219, size = 0, normalized size = 0. \begin{align*} \int \left ( b{x}^{3}+a \right ) ^{{\frac{2}{3}}} \left ( d{x}^{3}+c \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.8091, size = 1519, normalized size = 6.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/243*(3*sqrt(1/3)*(27*a*b^3*c^2 - 9*a^2*b^2*c*d + 2*a^3*b*d^2)*sqrt((-b)^(1/3)/b)*log(3*b*x^3 - 3*(b*x^3 + a
)^(1/3)*(-b)^(2/3)*x^2 - 3*sqrt(1/3)*((-b)^(1/3)*b*x^3 - (b*x^3 + a)^(1/3)*b*x^2 + 2*(b*x^3 + a)^(2/3)*(-b)^(2
/3)*x)*sqrt((-b)^(1/3)/b) + 2*a) - 2*(27*a*b^2*c^2 - 9*a^2*b*c*d + 2*a^3*d^2)*(-b)^(2/3)*log(((-b)^(1/3)*x + (
b*x^3 + a)^(1/3))/x) + (27*a*b^2*c^2 - 9*a^2*b*c*d + 2*a^3*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*(9*b^3*d^2*x^7 + 3*(9*b^3*c*d + a*b^2*d^2)*x^4 + (27*b^3*c^2 +
18*a*b^2*c*d - 4*a^2*b*d^2)*x)*(b*x^3 + a)^(2/3))/b^3, -1/243*(6*sqrt(1/3)*(27*a*b^3*c^2 - 9*a^2*b^2*c*d + 2*
a^3*b*d^2)*sqrt(-(-b)^(1/3)/b)*arctan(-sqrt(1/3)*((-b)^(1/3)*x - 2*(b*x^3 + a)^(1/3))*sqrt(-(-b)^(1/3)/b)/x) +
2*(27*a*b^2*c^2 - 9*a^2*b*c*d + 2*a^3*d^2)*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) - (27*a*b^2*c
^2 - 9*a^2*b*c*d + 2*a^3*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*(9*b^3*d^2*x^7 + 3*(9*b^3*c*d + a*b^2*d^2)*x^4 + (27*b^3*c^2 + 18*a*b^2*c*d - 4*a^2*b*d^2)*x)*(b
*x^3 + a)^(2/3))/b^3]

________________________________________________________________________________________

Sympy [C]  time = 4.88447, size = 131, normalized size = 0.6 \begin{align*} \frac{a^{\frac{2}{3}} c^{2} 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{2 a^{\frac{2}{3}} c 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}} d^{2} 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)**(2/3)*(d*x**3+c)**2,x)

[Out]

a**(2/3)*c**2*x*gamma(1/3)*hyper((-2/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3)) + 2*a**(2/3)*c*
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)*d**2*x**7*gam
ma(7/3)*hyper((-2/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(10/3))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{3} + a\right )}^{\frac{2}{3}}{\left (d x^{3} + c\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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