\(\int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x) \, dx\) [3002]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 24, antiderivative size = 331 \[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x) \, dx=\frac {(b c-a d) (9 b d e-4 b c f-5 a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{27 b^2 d^2}+\frac {(9 b d e-4 b c f-5 a d f) (a+b x)^{4/3} (c+d x)^{2/3}}{18 b^2 d}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3}}{3 b d}+\frac {(b c-a d)^2 (9 b d e-4 b c f-5 a d f) \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{27 \sqrt {3} b^{8/3} d^{7/3}}+\frac {(b c-a d)^2 (9 b d e-4 b c f-5 a d f) \log (a+b x)}{162 b^{8/3} d^{7/3}}+\frac {(b c-a d)^2 (9 b d e-4 b c f-5 a d f) \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{54 b^{8/3} d^{7/3}} \]

[Out]

1/27*(-a*d+b*c)*(-5*a*d*f-4*b*c*f+9*b*d*e)*(b*x+a)^(1/3)*(d*x+c)^(2/3)/b^2/d^2+1/18*(-5*a*d*f-4*b*c*f+9*b*d*e)
*(b*x+a)^(4/3)*(d*x+c)^(2/3)/b^2/d+1/3*f*(b*x+a)^(4/3)*(d*x+c)^(5/3)/b/d+1/162*(-a*d+b*c)^2*(-5*a*d*f-4*b*c*f+
9*b*d*e)*ln(b*x+a)/b^(8/3)/d^(7/3)+1/54*(-a*d+b*c)^2*(-5*a*d*f-4*b*c*f+9*b*d*e)*ln(-1+b^(1/3)*(d*x+c)^(1/3)/d^
(1/3)/(b*x+a)^(1/3))/b^(8/3)/d^(7/3)+1/81*(-a*d+b*c)^2*(-5*a*d*f-4*b*c*f+9*b*d*e)*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^(8/3)/d^(7/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {81, 52, 61} \[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x) \, dx=\frac {(b c-a d)^2 \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right ) (-5 a d f-4 b c f+9 b d e)}{27 \sqrt {3} b^{8/3} d^{7/3}}+\frac {(b c-a d)^2 \log (a+b x) (-5 a d f-4 b c f+9 b d e)}{162 b^{8/3} d^{7/3}}+\frac {(b c-a d)^2 (-5 a d f-4 b c f+9 b d e) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{54 b^{8/3} d^{7/3}}+\frac {\sqrt [3]{a+b x} (c+d x)^{2/3} (b c-a d) (-5 a d f-4 b c f+9 b d e)}{27 b^2 d^2}+\frac {(a+b x)^{4/3} (c+d x)^{2/3} (-5 a d f-4 b c f+9 b d e)}{18 b^2 d}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3}}{3 b d} \]

[In]

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

[Out]

((b*c - a*d)*(9*b*d*e - 4*b*c*f - 5*a*d*f)*(a + b*x)^(1/3)*(c + d*x)^(2/3))/(27*b^2*d^2) + ((9*b*d*e - 4*b*c*f
 - 5*a*d*f)*(a + b*x)^(4/3)*(c + d*x)^(2/3))/(18*b^2*d) + (f*(a + b*x)^(4/3)*(c + d*x)^(5/3))/(3*b*d) + ((b*c
- a*d)^2*(9*b*d*e - 4*b*c*f - 5*a*d*f)*ArcTan[1/Sqrt[3] + (2*b^(1/3)*(c + d*x)^(1/3))/(Sqrt[3]*d^(1/3)*(a + b*
x)^(1/3))])/(27*Sqrt[3]*b^(8/3)*d^(7/3)) + ((b*c - a*d)^2*(9*b*d*e - 4*b*c*f - 5*a*d*f)*Log[a + b*x])/(162*b^(
8/3)*d^(7/3)) + ((b*c - a*d)^2*(9*b*d*e - 4*b*c*f - 5*a*d*f)*Log[-1 + (b^(1/3)*(c + d*x)^(1/3))/(d^(1/3)*(a +
b*x)^(1/3))])/(54*b^(8/3)*d^(7/3))

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]

Rubi steps \begin{align*} \text {integral}& = \frac {f (a+b x)^{4/3} (c+d x)^{5/3}}{3 b d}+\frac {\left (3 b d e-\left (\frac {4 b c}{3}+\frac {5 a d}{3}\right ) f\right ) \int \sqrt [3]{a+b x} (c+d x)^{2/3} \, dx}{3 b d} \\ & = \frac {(9 b d e-4 b c f-5 a d f) (a+b x)^{4/3} (c+d x)^{2/3}}{18 b^2 d}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3}}{3 b d}+\frac {((b c-a d) (9 b d e-4 b c f-5 a d f)) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx}{27 b^2 d} \\ & = \frac {(b c-a d) (9 b d e-4 b c f-5 a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{27 b^2 d^2}+\frac {(9 b d e-4 b c f-5 a d f) (a+b x)^{4/3} (c+d x)^{2/3}}{18 b^2 d}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3}}{3 b d}-\frac {\left ((b c-a d)^2 (9 b d e-4 b c f-5 a d f)\right ) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}} \, dx}{81 b^2 d^2} \\ & = \frac {(b c-a d) (9 b d e-4 b c f-5 a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{27 b^2 d^2}+\frac {(9 b d e-4 b c f-5 a d f) (a+b x)^{4/3} (c+d x)^{2/3}}{18 b^2 d}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3}}{3 b d}+\frac {(b c-a d)^2 (9 b d e-4 b c f-5 a d f) \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 )}{27 \sqrt {3} b^{8/3} d^{7/3}}+\frac {(b c-a d)^2 (9 b d e-4 b c f-5 a d f) \log (a+b x)}{162 b^{8/3} d^{7/3}}+\frac {(b c-a d)^2 (9 b d e-4 b c f-5 a d f) \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{54 b^{8/3} d^{7/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 319, normalized size of antiderivative = 0.96 \[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x) \, dx=\frac {(b c-a d)^2 \left (\frac {3 b^{2/3} \sqrt [3]{d} \sqrt [3]{a+b x} (c+d x)^{2/3} \left (-5 a^2 d^2 f+a b d (9 d e+4 c f+3 d f x)+b^2 \left (-8 c^2 f+6 c d (3 e+f x)+9 d^2 x (3 e+2 f x)\right )\right )}{(b c-a d)^2}-2 \sqrt {3} (9 b d e-4 b c f-5 a d f) \arctan \left (\frac {1+\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}}{\sqrt {3}}\right )+2 (9 b d e-4 b c f-5 a d f) \log \left (\sqrt [3]{b}-\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}\right )-(9 b d e-4 b c f-5 a d f) \log \left (b^{2/3}+\frac {d^{2/3} (a+b x)^{2/3}}{(c+d x)^{2/3}}+\frac {\sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}\right )\right )}{162 b^{8/3} d^{7/3}} \]

[In]

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

[Out]

((b*c - a*d)^2*((3*b^(2/3)*d^(1/3)*(a + b*x)^(1/3)*(c + d*x)^(2/3)*(-5*a^2*d^2*f + a*b*d*(9*d*e + 4*c*f + 3*d*
f*x) + b^2*(-8*c^2*f + 6*c*d*(3*e + f*x) + 9*d^2*x*(3*e + 2*f*x))))/(b*c - a*d)^2 - 2*Sqrt[3]*(9*b*d*e - 4*b*c
*f - 5*a*d*f)*ArcTan[(1 + (2*d^(1/3)*(a + b*x)^(1/3))/(b^(1/3)*(c + d*x)^(1/3)))/Sqrt[3]] + 2*(9*b*d*e - 4*b*c
*f - 5*a*d*f)*Log[b^(1/3) - (d^(1/3)*(a + b*x)^(1/3))/(c + d*x)^(1/3)] - (9*b*d*e - 4*b*c*f - 5*a*d*f)*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)]))/(162*b
^(8/3)*d^(7/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 575 vs. \(2 (275) = 550\).

Time = 0.28 (sec) , antiderivative size = 1196, normalized size of antiderivative = 3.61 \[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

[-1/162*(3*sqrt(1/3)*(9*(b^4*c^2*d^2 - 2*a*b^3*c*d^3 + a^2*b^2*d^4)*e - (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)*f)*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)) + (-b^2*d)^(2/3)*(9*(b^3*c^2*d - 2*
a*b^2*c*d^2 + a^2*b*d^3)*e - (4*b^3*c^3 - 3*a*b^2*c^2*d - 6*a^2*b*c*d^2 + 5*a^3*d^3)*f)*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))
 - 2*(-b^2*d)^(2/3)*(9*(b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e - (4*b^3*c^3 - 3*a*b^2*c^2*d - 6*a^2*b*c*d^2
+ 5*a^3*d^3)*f)*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*(18*b^4*d^
3*f*x^2 + 9*(2*b^4*c*d^2 + a*b^3*d^3)*e - (8*b^4*c^2*d - 4*a*b^3*c*d^2 + 5*a^2*b^2*d^3)*f + 3*(9*b^4*d^3*e + (
2*b^4*c*d^2 + a*b^3*d^3)*f)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^4*d^3), -1/162*(6*sqrt(1/3)*(9*(b^4*c^2*d^2
 - 2*a*b^3*c*d^3 + a^2*b^2*d^4)*e - (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)*f)*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)) + (-b^2*d)^(2/3)*(9*(b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e
 - (4*b^3*c^3 - 3*a*b^2*c^2*d - 6*a^2*b*c*d^2 + 5*a^3*d^3)*f)*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)) - 2*(-b^2*d)^(2/3)*(9*(b^
3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e - (4*b^3*c^3 - 3*a*b^2*c^2*d - 6*a^2*b*c*d^2 + 5*a^3*d^3)*f)*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*(18*b^4*d^3*f*x^2 + 9*(2*b^4*c*d^2 +
 a*b^3*d^3)*e - (8*b^4*c^2*d - 4*a*b^3*c*d^2 + 5*a^2*b^2*d^3)*f + 3*(9*b^4*d^3*e + (2*b^4*c*d^2 + a*b^3*d^3)*f
)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^4*d^3)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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