3.70 \(\int (a+b x^3)^{5/3} (c+d x^3)^2 \, dx\)

Optimal. Leaf size=262 \[ \frac {x \left (a+b x^3\right )^{5/3} \left (a^2 d^2-6 a b c d+27 b^2 c^2\right )}{162 b^2}+\frac {5 a x \left (a+b x^3\right )^{2/3} \left (a^2 d^2-6 a b c d+27 b^2 c^2\right )}{486 b^2}-\frac {5 a^2 \left (a^2 d^2-6 a b c d+27 b^2 c^2\right ) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{486 b^{7/3}}+\frac {5 a^2 \left (a^2 d^2-6 a b c d+27 b^2 c^2\right ) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{243 \sqrt {3} b^{7/3}}+\frac {d x \left (a+b x^3\right )^{8/3} (15 b c-4 a d)}{108 b^2}+\frac {d x \left (a+b x^3\right )^{8/3} \left (c+d x^3\right )}{12 b} \]

[Out]

5/486*a*(a^2*d^2-6*a*b*c*d+27*b^2*c^2)*x*(b*x^3+a)^(2/3)/b^2+1/162*(a^2*d^2-6*a*b*c*d+27*b^2*c^2)*x*(b*x^3+a)^
(5/3)/b^2+1/108*d*(-4*a*d+15*b*c)*x*(b*x^3+a)^(8/3)/b^2+1/12*d*x*(b*x^3+a)^(8/3)*(d*x^3+c)/b-5/486*a^2*(a^2*d^
2-6*a*b*c*d+27*b^2*c^2)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(7/3)+5/729*a^2*(a^2*d^2-6*a*b*c*d+27*b^2*c^2)*arctan
(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(7/3)*3^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {416, 388, 195, 239} \[ \frac {x \left (a+b x^3\right )^{5/3} \left (a^2 d^2-6 a b c d+27 b^2 c^2\right )}{162 b^2}+\frac {5 a x \left (a+b x^3\right )^{2/3} \left (a^2 d^2-6 a b c d+27 b^2 c^2\right )}{486 b^2}-\frac {5 a^2 \left (a^2 d^2-6 a b c d+27 b^2 c^2\right ) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{486 b^{7/3}}+\frac {5 a^2 \left (a^2 d^2-6 a b c d+27 b^2 c^2\right ) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{243 \sqrt {3} b^{7/3}}+\frac {d x \left (a+b x^3\right )^{8/3} (15 b c-4 a d)}{108 b^2}+\frac {d x \left (a+b x^3\right )^{8/3} \left (c+d x^3\right )}{12 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(5/3)*(c + d*x^3)^2,x]

[Out]

(5*a*(27*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*x*(a + b*x^3)^(2/3))/(486*b^2) + ((27*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*x
*(a + b*x^3)^(5/3))/(162*b^2) + (d*(15*b*c - 4*a*d)*x*(a + b*x^3)^(8/3))/(108*b^2) + (d*x*(a + b*x^3)^(8/3)*(c
+ d*x^3))/(12*b) + (5*a^2*(27*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqr
t[3]])/(243*Sqrt[3]*b^(7/3)) - (5*a^2*(27*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)]
)/(486*b^(7/3))

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]

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]

Rubi steps

\begin {align*} \int \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2 \, dx &=\frac {d x \left (a+b x^3\right )^{8/3} \left (c+d x^3\right )}{12 b}+\frac {\int \left (a+b x^3\right )^{5/3} \left (c (12 b c-a d)+d (15 b c-4 a d) x^3\right ) \, dx}{12 b}\\ &=\frac {d (15 b c-4 a d) x \left (a+b x^3\right )^{8/3}}{108 b^2}+\frac {d x \left (a+b x^3\right )^{8/3} \left (c+d x^3\right )}{12 b}+\frac {\left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) \int \left (a+b x^3\right )^{5/3} \, dx}{27 b^2}\\ &=\frac {\left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) x \left (a+b x^3\right )^{5/3}}{162 b^2}+\frac {d (15 b c-4 a d) x \left (a+b x^3\right )^{8/3}}{108 b^2}+\frac {d x \left (a+b x^3\right )^{8/3} \left (c+d x^3\right )}{12 b}+\frac {\left (5 a \left (27 b^2 c^2-6 a b c d+a^2 d^2\right )\right ) \int \left (a+b x^3\right )^{2/3} \, dx}{162 b^2}\\ &=\frac {5 a \left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) x \left (a+b x^3\right )^{2/3}}{486 b^2}+\frac {\left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) x \left (a+b x^3\right )^{5/3}}{162 b^2}+\frac {d (15 b c-4 a d) x \left (a+b x^3\right )^{8/3}}{108 b^2}+\frac {d x \left (a+b x^3\right )^{8/3} \left (c+d x^3\right )}{12 b}+\frac {\left (5 a^2 \left (27 b^2 c^2-6 a b c d+a^2 d^2\right )\right ) \int \frac {1}{\sqrt [3]{a+b x^3}} \, dx}{243 b^2}\\ &=\frac {5 a \left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) x \left (a+b x^3\right )^{2/3}}{486 b^2}+\frac {\left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) x \left (a+b x^3\right )^{5/3}}{162 b^2}+\frac {d (15 b c-4 a d) x \left (a+b x^3\right )^{8/3}}{108 b^2}+\frac {d x \left (a+b x^3\right )^{8/3} \left (c+d x^3\right )}{12 b}+\frac {5 a^2 \left (27 b^2 c^2-6 a b c d+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 )}{243 \sqrt {3} b^{7/3}}-\frac {5 a^2 \left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{486 b^{7/3}}\\ \end {align*}

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Mathematica [A]  time = 5.19, size = 238, normalized size = 0.91 \[ \frac {10 a^2 \left (a^2 d^2-6 a b c d+27 b^2 c^2\right ) \left (\log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )-2 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )\right )+3 \sqrt [3]{b} x \left (a+b x^3\right )^{2/3} \left (-20 a^3 d^2+15 a^2 b d \left (8 c+d x^3\right )+18 a b^2 \left (24 c^2+22 c d x^3+7 d^2 x^6\right )+27 b^3 x^3 \left (6 c^2+8 c d x^3+3 d^2 x^6\right )\right )}{2916 b^{7/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(5/3)*(c + d*x^3)^2,x]

[Out]

(3*b^(1/3)*x*(a + b*x^3)^(2/3)*(-20*a^3*d^2 + 15*a^2*b*d*(8*c + d*x^3) + 27*b^3*x^3*(6*c^2 + 8*c*d*x^3 + 3*d^2
*x^6) + 18*a*b^2*(24*c^2 + 22*c*d*x^3 + 7*d^2*x^6)) + 10*a^2*(27*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*(2*Sqrt[3]*Arc
Tan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]] - 2*Log[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)] + Log[1 + (b^(2/
3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)]))/(2916*b^(7/3))

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fricas [A]  time = 0.85, size = 717, normalized size = 2.74 \[ \left [\frac {30 \, \sqrt {\frac {1}{3}} {\left (27 \, a^{2} b^{3} c^{2} - 6 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (\left (-b\right )^{\frac {1}{3}} b x^{3} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} + 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b\right )^{\frac {2}{3}} x\right )} \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} + 2 \, a\right ) - 20 \, {\left (27 \, a^{2} b^{2} c^{2} - 6 \, a^{3} b c d + a^{4} d^{2}\right )} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + 10 \, {\left (27 \, a^{2} b^{2} c^{2} - 6 \, a^{3} b c d + a^{4} d^{2}\right )} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 3 \, {\left (81 \, b^{4} d^{2} x^{10} + 18 \, {\left (12 \, b^{4} c d + 7 \, a b^{3} d^{2}\right )} x^{7} + 3 \, {\left (54 \, b^{4} c^{2} + 132 \, a b^{3} c d + 5 \, a^{2} b^{2} d^{2}\right )} x^{4} + 4 \, {\left (108 \, a b^{3} c^{2} + 30 \, a^{2} b^{2} c d - 5 \, a^{3} b d^{2}\right )} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{2916 \, b^{3}}, -\frac {60 \, \sqrt {\frac {1}{3}} {\left (27 \, a^{2} b^{3} c^{2} - 6 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (\left (-b\right )^{\frac {1}{3}} x - 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}}}{x}\right ) + 20 \, {\left (27 \, a^{2} b^{2} c^{2} - 6 \, a^{3} b c d + a^{4} d^{2}\right )} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - 10 \, {\left (27 \, a^{2} b^{2} c^{2} - 6 \, a^{3} b c d + a^{4} d^{2}\right )} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 3 \, {\left (81 \, b^{4} d^{2} x^{10} + 18 \, {\left (12 \, b^{4} c d + 7 \, a b^{3} d^{2}\right )} x^{7} + 3 \, {\left (54 \, b^{4} c^{2} + 132 \, a b^{3} c d + 5 \, a^{2} b^{2} d^{2}\right )} x^{4} + 4 \, {\left (108 \, a b^{3} c^{2} + 30 \, a^{2} b^{2} c d - 5 \, a^{3} b d^{2}\right )} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{2916 \, b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/3)*(d*x^3+c)^2,x, algorithm="fricas")

[Out]

[1/2916*(30*sqrt(1/3)*(27*a^2*b^3*c^2 - 6*a^3*b^2*c*d + a^4*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) - 20*(27*a^2*b^2*c^2 - 6*a^3*b*c*d + a^4*d^2)*(-b)^(2/3)*log(((-b)^(1/3)*x
+ (b*x^3 + a)^(1/3))/x) + 10*(27*a^2*b^2*c^2 - 6*a^3*b*c*d + a^4*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*(81*b^4*d^2*x^10 + 18*(12*b^4*c*d + 7*a*b^3*d^2)*x^7 + 3
*(54*b^4*c^2 + 132*a*b^3*c*d + 5*a^2*b^2*d^2)*x^4 + 4*(108*a*b^3*c^2 + 30*a^2*b^2*c*d - 5*a^3*b*d^2)*x)*(b*x^3
+ a)^(2/3))/b^3, -1/2916*(60*sqrt(1/3)*(27*a^2*b^3*c^2 - 6*a^3*b^2*c*d + a^4*b*d^2)*sqrt(-(-b)^(1/3)/b)*arcta
n(-sqrt(1/3)*((-b)^(1/3)*x - 2*(b*x^3 + a)^(1/3))*sqrt(-(-b)^(1/3)/b)/x) + 20*(27*a^2*b^2*c^2 - 6*a^3*b*c*d +
a^4*d^2)*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) - 10*(27*a^2*b^2*c^2 - 6*a^3*b*c*d + a^4*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*(81*b^4*d^2*x^10 +
18*(12*b^4*c*d + 7*a*b^3*d^2)*x^7 + 3*(54*b^4*c^2 + 132*a*b^3*c*d + 5*a^2*b^2*d^2)*x^4 + 4*(108*a*b^3*c^2 + 3
0*a^2*b^2*c*d - 5*a^3*b*d^2)*x)*(b*x^3 + a)^(2/3))/b^3]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{\frac {5}{3}} {\left (d x^{3} + c\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/3)*(d*x^3+c)^2,x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(5/3)*(d*x^3 + c)^2, x)

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (d \,x^{3}+c \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(5/3)*(d*x^3+c)^2,x)

[Out]

int((b*x^3+a)^(5/3)*(d*x^3+c)^2,x)

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maxima [B]  time = 1.29, size = 672, normalized size = 2.56 \[ -\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^{2} + \frac {1}{243} \, {\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 )} c d - \frac {1}{2916} \, {\left (\frac {20 \, \sqrt {3} a^{4} \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 {7}{3}}} - \frac {10 \, a^{4} \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 {7}{3}}} + \frac {20 \, a^{4} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {7}{3}}} + \frac {3 \, {\left (\frac {10 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{4} b^{3}}{x^{2}} - \frac {36 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{4} b^{2}}{x^{5}} - \frac {75 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} a^{4} b}{x^{8}} + \frac {20 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}} a^{4}}{x^{11}}\right )}}{b^{6} - \frac {4 \, {\left (b x^{3} + a\right )} b^{5}}{x^{3}} + \frac {6 \, {\left (b x^{3} + a\right )}^{2} b^{4}}{x^{6}} - \frac {4 \, {\left (b x^{3} + a\right )}^{3} b^{3}}{x^{9}} + \frac {{\left (b x^{3} + a\right )}^{4} b^{2}}{x^{12}}}\right )} d^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/3)*(d*x^3+c)^2,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^2 + 1/243*(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))*c*d - 1/2916
*(20*sqrt(3)*a^4*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(7/3) - 10*a^4*log(b^(2/3) +
(b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(7/3) + 20*a^4*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(7
/3) + 3*(10*(b*x^3 + a)^(2/3)*a^4*b^3/x^2 - 36*(b*x^3 + a)^(5/3)*a^4*b^2/x^5 - 75*(b*x^3 + a)^(8/3)*a^4*b/x^8
+ 20*(b*x^3 + a)^(11/3)*a^4/x^11)/(b^6 - 4*(b*x^3 + a)*b^5/x^3 + 6*(b*x^3 + a)^2*b^4/x^6 - 4*(b*x^3 + a)^3*b^3
/x^9 + (b*x^3 + a)^4*b^2/x^12))*d^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (b\,x^3+a\right )}^{5/3}\,{\left (d\,x^3+c\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(5/3)*(c + d*x^3)^2,x)

[Out]

int((a + b*x^3)^(5/3)*(c + d*x^3)^2, x)

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sympy [C]  time = 13.13, size = 270, normalized size = 1.03 \[ \frac {a^{\frac {5}{3}} c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {2 a^{\frac {5}{3}} c d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {a^{\frac {5}{3}} d^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} + \frac {a^{\frac {2}{3}} b c^{2} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {2 a^{\frac {2}{3}} b c d x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} + \frac {a^{\frac {2}{3}} b d^{2} x^{10} \Gamma \left (\frac {10}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {10}{3} \\ \frac {13}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {13}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(5/3)*(d*x**3+c)**2,x)

[Out]

a**(5/3)*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**(5/3)*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**(5/3)*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)) + a**(2/3)*b*c**2*x**4*gamma(4/3
)*hyper((-2/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3)) + 2*a**(2/3)*b*c*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)) + a**(2/3)*b*d**2*x**10*gamma(10/3)*hyper((-2
/3, 10/3), (13/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(13/3))

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