\(\int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)^2 \, dx\) [3001]

Optimal result

Integrand size = 26, antiderivative size = 571 \[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)^2 \, dx=\frac {(b c-a d) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{81 b^3 d^3}+\frac {\left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) (a+b x)^{4/3} (c+d x)^{2/3}}{54 b^3 d^2}+\frac {f (15 b d e-7 b c f-8 a d f) (a+b x)^{4/3} (c+d x)^{5/3}}{36 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{81 \sqrt {3} b^{11/3} d^{10/3}}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \log (a+b x)}{486 b^{11/3} d^{10/3}}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{162 b^{11/3} d^{10/3}} \]

[Out]

1/81*(-a*d+b*c)*(10*a^2*d^2*f^2-10*a*b*d*f*(-c*f+3*d*e)+b^2*(7*c^2*f^2-24*c*d*e*f+27*d^2*e^2))*(b*x+a)^(1/3)*(
d*x+c)^(2/3)/b^3/d^3+1/54*(10*a^2*d^2*f^2-10*a*b*d*f*(-c*f+3*d*e)+b^2*(7*c^2*f^2-24*c*d*e*f+27*d^2*e^2))*(b*x+
a)^(4/3)*(d*x+c)^(2/3)/b^3/d^2+1/36*f*(-8*a*d*f-7*b*c*f+15*b*d*e)*(b*x+a)^(4/3)*(d*x+c)^(5/3)/b^2/d^2+1/4*f*(b
*x+a)^(4/3)*(d*x+c)^(5/3)*(f*x+e)/b/d+1/486*(-a*d+b*c)^2*(10*a^2*d^2*f^2-10*a*b*d*f*(-c*f+3*d*e)+b^2*(7*c^2*f^
2-24*c*d*e*f+27*d^2*e^2))*ln(b*x+a)/b^(11/3)/d^(10/3)+1/162*(-a*d+b*c)^2*(10*a^2*d^2*f^2-10*a*b*d*f*(-c*f+3*d*
e)+b^2*(7*c^2*f^2-24*c*d*e*f+27*d^2*e^2))*ln(-1+b^(1/3)*(d*x+c)^(1/3)/d^(1/3)/(b*x+a)^(1/3))/b^(11/3)/d^(10/3)
+1/243*(-a*d+b*c)^2*(10*a^2*d^2*f^2-10*a*b*d*f*(-c*f+3*d*e)+b^2*(7*c^2*f^2-24*c*d*e*f+27*d^2*e^2))*arctan(1/3*
3^(1/2)+2/3*b^(1/3)*(d*x+c)^(1/3)/d^(1/3)/(b*x+a)^(1/3)*3^(1/2))/b^(11/3)/d^(10/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 571, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {92, 81, 52, 61} \[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)^2 \, dx=\frac {(b c-a d)^2 \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right ) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right )}{81 \sqrt {3} b^{11/3} d^{10/3}}+\frac {(b c-a d)^2 \log (a+b x) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right )}{486 b^{11/3} d^{10/3}}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right ) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{162 b^{11/3} d^{10/3}}+\frac {(a+b x)^{4/3} (c+d x)^{2/3} \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right )}{54 b^3 d^2}+\frac {\sqrt [3]{a+b x} (c+d x)^{2/3} (b c-a d) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right )}{81 b^3 d^3}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (-8 a d f-7 b c f+15 b d e)}{36 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d} \]

[In]

Int[(a + b*x)^(1/3)*(c + d*x)^(2/3)*(e + f*x)^2,x]

[Out]

((b*c - a*d)*(10*a^2*d^2*f^2 - 10*a*b*d*f*(3*d*e - c*f) + b^2*(27*d^2*e^2 - 24*c*d*e*f + 7*c^2*f^2))*(a + b*x)
^(1/3)*(c + d*x)^(2/3))/(81*b^3*d^3) + ((10*a^2*d^2*f^2 - 10*a*b*d*f*(3*d*e - c*f) + b^2*(27*d^2*e^2 - 24*c*d*
e*f + 7*c^2*f^2))*(a + b*x)^(4/3)*(c + d*x)^(2/3))/(54*b^3*d^2) + (f*(15*b*d*e - 7*b*c*f - 8*a*d*f)*(a + b*x)^
(4/3)*(c + d*x)^(5/3))/(36*b^2*d^2) + (f*(a + b*x)^(4/3)*(c + d*x)^(5/3)*(e + f*x))/(4*b*d) + ((b*c - a*d)^2*(
10*a^2*d^2*f^2 - 10*a*b*d*f*(3*d*e - c*f) + b^2*(27*d^2*e^2 - 24*c*d*e*f + 7*c^2*f^2))*ArcTan[1/Sqrt[3] + (2*b
^(1/3)*(c + d*x)^(1/3))/(Sqrt[3]*d^(1/3)*(a + b*x)^(1/3))])/(81*Sqrt[3]*b^(11/3)*d^(10/3)) + ((b*c - a*d)^2*(1
0*a^2*d^2*f^2 - 10*a*b*d*f*(3*d*e - c*f) + b^2*(27*d^2*e^2 - 24*c*d*e*f + 7*c^2*f^2))*Log[a + b*x])/(486*b^(11
/3)*d^(10/3)) + ((b*c - a*d)^2*(10*a^2*d^2*f^2 - 10*a*b*d*f*(3*d*e - c*f) + b^2*(27*d^2*e^2 - 24*c*d*e*f + 7*c
^2*f^2))*Log[-1 + (b^(1/3)*(c + d*x)^(1/3))/(d^(1/3)*(a + b*x)^(1/3))])/(162*b^(11/3)*d^(10/3))

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 61

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, Simp[(-Sqrt
[3])*(q/d)*ArcTan[2*q*((a + b*x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3))) + 1/Sqrt[3]], x] + (-Simp[3*(q/(2*d))*Log[q*
((a + b*x)^(1/3)/(c + d*x)^(1/3)) - 1], x] - Simp[(q/(2*d))*Log[c + d*x], x])] /; FreeQ[{a, b, c, d}, x] && Ne
Q[b*c - a*d, 0] && PosQ[d/b]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d}+\frac {\int \sqrt [3]{a+b x} (c+d x)^{2/3} \left (\frac {1}{3} \left (12 b d e^2-f (4 b c e+5 a d e+3 a c f)\right )+\frac {1}{3} f (15 b d e-7 b c f-8 a d f) x\right ) \, dx}{4 b d} \\ & = \frac {f (15 b d e-7 b c f-8 a d f) (a+b x)^{4/3} (c+d x)^{5/3}}{36 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d}+\frac {\left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \int \sqrt [3]{a+b x} (c+d x)^{2/3} \, dx}{27 b^2 d^2} \\ & = \frac {\left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) (a+b x)^{4/3} (c+d x)^{2/3}}{54 b^3 d^2}+\frac {f (15 b d e-7 b c f-8 a d f) (a+b x)^{4/3} (c+d x)^{5/3}}{36 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d}+\frac {\left ((b c-a d) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right )\right ) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx}{81 b^3 d^2} \\ & = \frac {(b c-a d) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{81 b^3 d^3}+\frac {\left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) (a+b x)^{4/3} (c+d x)^{2/3}}{54 b^3 d^2}+\frac {f (15 b d e-7 b c f-8 a d f) (a+b x)^{4/3} (c+d x)^{5/3}}{36 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d}-\frac {\left ((b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right )\right ) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}} \, dx}{243 b^3 d^3} \\ & = \frac {(b c-a d) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{81 b^3 d^3}+\frac {\left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) (a+b x)^{4/3} (c+d x)^{2/3}}{54 b^3 d^2}+\frac {f (15 b d e-7 b c f-8 a d f) (a+b x)^{4/3} (c+d x)^{5/3}}{36 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{81 \sqrt {3} b^{11/3} d^{10/3}}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \log (a+b x)}{486 b^{11/3} d^{10/3}}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{162 b^{11/3} d^{10/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.75 (sec) , antiderivative size = 528, normalized size of antiderivative = 0.92 \[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)^2 \, dx=\frac {(b c-a d)^2 \left (\frac {3 b^{2/3} \sqrt [3]{d} \sqrt [3]{a+b x} (c+d x)^{2/3} \left (20 a^3 d^3 f^2-12 a^2 b d^2 f (5 d e+c f+d f x)+3 a b^2 d \left (-3 c^2 f^2+2 c d f (8 e+f x)+3 d^2 \left (6 e^2+4 e f x+f^2 x^2\right )\right )+b^3 \left (28 c^3 f^2-3 c^2 d f (32 e+7 f x)+18 c d^2 \left (6 e^2+4 e f x+f^2 x^2\right )+27 d^3 x \left (6 e^2+8 e f x+3 f^2 x^2\right )\right )\right )}{(b c-a d)^2}-4 \sqrt {3} \left (10 a^2 d^2 f^2+10 a b d f (-3 d e+c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}}{\sqrt {3}}\right )+4 \left (10 a^2 d^2 f^2+10 a b d f (-3 d e+c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \log \left (\sqrt [3]{b}-\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}\right )-2 \left (10 a^2 d^2 f^2+10 a b d f (-3 d e+c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \log \left (b^{2/3}+\frac {d^{2/3} (a+b x)^{2/3}}{(c+d x)^{2/3}}+\frac {\sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}\right )\right )}{972 b^{11/3} d^{10/3}} \]

[In]

Integrate[(a + b*x)^(1/3)*(c + d*x)^(2/3)*(e + f*x)^2,x]

[Out]

((b*c - a*d)^2*((3*b^(2/3)*d^(1/3)*(a + b*x)^(1/3)*(c + d*x)^(2/3)*(20*a^3*d^3*f^2 - 12*a^2*b*d^2*f*(5*d*e + c
*f + d*f*x) + 3*a*b^2*d*(-3*c^2*f^2 + 2*c*d*f*(8*e + f*x) + 3*d^2*(6*e^2 + 4*e*f*x + f^2*x^2)) + b^3*(28*c^3*f
^2 - 3*c^2*d*f*(32*e + 7*f*x) + 18*c*d^2*(6*e^2 + 4*e*f*x + f^2*x^2) + 27*d^3*x*(6*e^2 + 8*e*f*x + 3*f^2*x^2))
))/(b*c - a*d)^2 - 4*Sqrt[3]*(10*a^2*d^2*f^2 + 10*a*b*d*f*(-3*d*e + c*f) + b^2*(27*d^2*e^2 - 24*c*d*e*f + 7*c^
2*f^2))*ArcTan[(1 + (2*d^(1/3)*(a + b*x)^(1/3))/(b^(1/3)*(c + d*x)^(1/3)))/Sqrt[3]] + 4*(10*a^2*d^2*f^2 + 10*a
*b*d*f*(-3*d*e + c*f) + b^2*(27*d^2*e^2 - 24*c*d*e*f + 7*c^2*f^2))*Log[b^(1/3) - (d^(1/3)*(a + b*x)^(1/3))/(c
+ d*x)^(1/3)] - 2*(10*a^2*d^2*f^2 + 10*a*b*d*f*(-3*d*e + c*f) + b^2*(27*d^2*e^2 - 24*c*d*e*f + 7*c^2*f^2))*Log
[b^(2/3) + (d^(2/3)*(a + b*x)^(2/3))/(c + d*x)^(2/3) + (b^(1/3)*d^(1/3)*(a + b*x)^(1/3))/(c + d*x)^(1/3)]))/(9
72*b^(11/3)*d^(10/3))

Maple [F]

\[\int \left (b x +a \right )^{\frac {1}{3}} \left (d x +c \right )^{\frac {2}{3}} \left (f x +e \right )^{2}d x\]

[In]

int((b*x+a)^(1/3)*(d*x+c)^(2/3)*(f*x+e)^2,x)

[Out]

int((b*x+a)^(1/3)*(d*x+c)^(2/3)*(f*x+e)^2,x)

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 1857, normalized size of antiderivative = 3.25 \[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)^2 \, dx=\text {Too large to display} \]

[In]

integrate((b*x+a)^(1/3)*(d*x+c)^(2/3)*(f*x+e)^2,x, algorithm="fricas")

[Out]

[1/972*(6*sqrt(1/3)*(27*(b^5*c^2*d^3 - 2*a*b^4*c*d^4 + a^2*b^3*d^5)*e^2 - 6*(4*b^5*c^3*d^2 - 3*a*b^4*c^2*d^3 -
6*a^2*b^3*c*d^4 + 5*a^3*b^2*d^5)*e*f + (7*b^5*c^4*d - 4*a*b^4*c^3*d^2 - 3*a^2*b^3*c^2*d^3 - 10*a^3*b^2*c*d^4
+ 10*a^4*b*d^5)*f^2)*sqrt(-(b^2*d)^(1/3)/d)*log(3*b^2*d*x + b^2*c + 2*a*b*d - 3*(b^2*d)^(1/3)*(b*x + a)^(1/3)*
(d*x + c)^(2/3)*b - 3*sqrt(1/3)*(2*(b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (b^2*d)^(2/3)*(b*x + a)^(1/3)*(d*x +
c)^(2/3) - (b^2*d)^(1/3)*(b*d*x + b*c))*sqrt(-(b^2*d)^(1/3)/d)) - 2*(b^2*d)^(2/3)*(27*(b^4*c^2*d^2 - 2*a*b^3*c
*d^3 + a^2*b^2*d^4)*e^2 - 6*(4*b^4*c^3*d - 3*a*b^3*c^2*d^2 - 6*a^2*b^2*c*d^3 + 5*a^3*b*d^4)*e*f + (7*b^4*c^4 -
4*a*b^3*c^3*d - 3*a^2*b^2*c^2*d^2 - 10*a^3*b*c*d^3 + 10*a^4*d^4)*f^2)*log(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*
d + (b^2*d)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) + (b^2*d)^(1/3)*(b*d*x + b*c))/(d*x + c)) + 4*(b^2*d)^(2/3)*
(27*(b^4*c^2*d^2 - 2*a*b^3*c*d^3 + a^2*b^2*d^4)*e^2 - 6*(4*b^4*c^3*d - 3*a*b^3*c^2*d^2 - 6*a^2*b^2*c*d^3 + 5*a
^3*b*d^4)*e*f + (7*b^4*c^4 - 4*a*b^3*c^3*d - 3*a^2*b^2*c^2*d^2 - 10*a^3*b*c*d^3 + 10*a^4*d^4)*f^2)*log(((b*x +
a)^(1/3)*(d*x + c)^(2/3)*b*d - (b^2*d)^(2/3)*(d*x + c))/(d*x + c)) + 3*(81*b^5*d^4*f^2*x^3 + 54*(2*b^5*c*d^3
+ a*b^4*d^4)*e^2 - 12*(8*b^5*c^2*d^2 - 4*a*b^4*c*d^3 + 5*a^2*b^3*d^4)*e*f + (28*b^5*c^3*d - 9*a*b^4*c^2*d^2 -
12*a^2*b^3*c*d^3 + 20*a^3*b^2*d^4)*f^2 + 9*(24*b^5*d^4*e*f + (2*b^5*c*d^3 + a*b^4*d^4)*f^2)*x^2 + 3*(54*b^5*d^
4*e^2 + 12*(2*b^5*c*d^3 + a*b^4*d^4)*e*f - (7*b^5*c^2*d^2 - 2*a*b^4*c*d^3 + 4*a^2*b^3*d^4)*f^2)*x)*(b*x + a)^(
1/3)*(d*x + c)^(2/3))/(b^5*d^4), -1/972*(12*sqrt(1/3)*(27*(b^5*c^2*d^3 - 2*a*b^4*c*d^4 + a^2*b^3*d^5)*e^2 - 6*
(4*b^5*c^3*d^2 - 3*a*b^4*c^2*d^3 - 6*a^2*b^3*c*d^4 + 5*a^3*b^2*d^5)*e*f + (7*b^5*c^4*d - 4*a*b^4*c^3*d^2 - 3*a
^2*b^3*c^2*d^3 - 10*a^3*b^2*c*d^4 + 10*a^4*b*d^5)*f^2)*sqrt((b^2*d)^(1/3)/d)*arctan(sqrt(1/3)*(2*(b^2*d)^(2/3)
*(b*x + a)^(1/3)*(d*x + c)^(2/3) + (b^2*d)^(1/3)*(b*d*x + b*c))*sqrt((b^2*d)^(1/3)/d)/(b^2*d*x + b^2*c)) + 2*(
b^2*d)^(2/3)*(27*(b^4*c^2*d^2 - 2*a*b^3*c*d^3 + a^2*b^2*d^4)*e^2 - 6*(4*b^4*c^3*d - 3*a*b^3*c^2*d^2 - 6*a^2*b^
2*c*d^3 + 5*a^3*b*d^4)*e*f + (7*b^4*c^4 - 4*a*b^3*c^3*d - 3*a^2*b^2*c^2*d^2 - 10*a^3*b*c*d^3 + 10*a^4*d^4)*f^2
)*log(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d + (b^2*d)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) + (b^2*d)^(1/3)*(b*
d*x + b*c))/(d*x + c)) - 4*(b^2*d)^(2/3)*(27*(b^4*c^2*d^2 - 2*a*b^3*c*d^3 + a^2*b^2*d^4)*e^2 - 6*(4*b^4*c^3*d
- 3*a*b^3*c^2*d^2 - 6*a^2*b^2*c*d^3 + 5*a^3*b*d^4)*e*f + (7*b^4*c^4 - 4*a*b^3*c^3*d - 3*a^2*b^2*c^2*d^2 - 10*a
^3*b*c*d^3 + 10*a^4*d^4)*f^2)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d - (b^2*d)^(2/3)*(d*x + c))/(d*x + c)) -
3*(81*b^5*d^4*f^2*x^3 + 54*(2*b^5*c*d^3 + a*b^4*d^4)*e^2 - 12*(8*b^5*c^2*d^2 - 4*a*b^4*c*d^3 + 5*a^2*b^3*d^4)
*e*f + (28*b^5*c^3*d - 9*a*b^4*c^2*d^2 - 12*a^2*b^3*c*d^3 + 20*a^3*b^2*d^4)*f^2 + 9*(24*b^5*d^4*e*f + (2*b^5*c
*d^3 + a*b^4*d^4)*f^2)*x^2 + 3*(54*b^5*d^4*e^2 + 12*(2*b^5*c*d^3 + a*b^4*d^4)*e*f - (7*b^5*c^2*d^2 - 2*a*b^4*c
*d^3 + 4*a^2*b^3*d^4)*f^2)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^5*d^4)]

Sympy [F]

\[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)^2 \, dx=\int \sqrt [3]{a + b x} \left (c + d x\right )^{\frac {2}{3}} \left (e + f x\right )^{2}\, dx \]

[In]

integrate((b*x+a)**(1/3)*(d*x+c)**(2/3)*(f*x+e)**2,x)

[Out]

Integral((a + b*x)**(1/3)*(c + d*x)**(2/3)*(e + f*x)**2, x)

Maxima [F]

\[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)^2 \, dx=\int { {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} {\left (f x + e\right )}^{2} \,d x } \]

[In]

integrate((b*x+a)^(1/3)*(d*x+c)^(2/3)*(f*x+e)^2,x, algorithm="maxima")

[Out]

integrate((b*x + a)^(1/3)*(d*x + c)^(2/3)*(f*x + e)^2, x)

Giac [F]

\[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)^2 \, dx=\int { {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} {\left (f x + e\right )}^{2} \,d x } \]

[In]

integrate((b*x+a)^(1/3)*(d*x+c)^(2/3)*(f*x+e)^2,x, algorithm="giac")

[Out]

integrate((b*x + a)^(1/3)*(d*x + c)^(2/3)*(f*x + e)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)^2 \, dx=\int {\left (e+f\,x\right )}^2\,{\left (a+b\,x\right )}^{1/3}\,{\left (c+d\,x\right )}^{2/3} \,d x \]

[In]

int((e + f*x)^2*(a + b*x)^(1/3)*(c + d*x)^(2/3),x)

[Out]

int((e + f*x)^2*(a + b*x)^(1/3)*(c + d*x)^(2/3), x)