\(\int \frac {\sqrt [3]{a+b x} (e+f x)}{\sqrt [3]{c+d x}} \, dx\) [3009]

Optimal result

Integrand size = 24, antiderivative size = 273 \[ \int \frac {\sqrt [3]{a+b x} (e+f x)}{\sqrt [3]{c+d x}} \, dx=\frac {(3 b d e-2 b c f-a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 b d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 b d}+\frac {(b c-a d) (3 b d e-2 b c f-a d f) \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} b^{5/3} d^{7/3}}+\frac {(b c-a d) (3 b d e-2 b c f-a d f) \log (a+b x)}{18 b^{5/3} d^{7/3}}+\frac {(b c-a d) (3 b d e-2 b c f-a d f) \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{6 b^{5/3} d^{7/3}} \]

[Out]

1/3*(-a*d*f-2*b*c*f+3*b*d*e)*(b*x+a)^(1/3)*(d*x+c)^(2/3)/b/d^2+1/2*f*(b*x+a)^(4/3)*(d*x+c)^(2/3)/b/d+1/18*(-a*
d+b*c)*(-a*d*f-2*b*c*f+3*b*d*e)*ln(b*x+a)/b^(5/3)/d^(7/3)+1/6*(-a*d+b*c)*(-a*d*f-2*b*c*f+3*b*d*e)*ln(-1+b^(1/3
)*(d*x+c)^(1/3)/d^(1/3)/(b*x+a)^(1/3))/b^(5/3)/d^(7/3)+1/9*(-a*d+b*c)*(-a*d*f-2*b*c*f+3*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^(5/3)/d^(7/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {81, 52, 61} \[ \int \frac {\sqrt [3]{a+b x} (e+f x)}{\sqrt [3]{c+d x}} \, dx=\frac {(b c-a d) \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right ) (-a d f-2 b c f+3 b d e)}{3 \sqrt {3} b^{5/3} d^{7/3}}+\frac {(b c-a d) \log (a+b x) (-a d f-2 b c f+3 b d e)}{18 b^{5/3} d^{7/3}}+\frac {(b c-a d) (-a d f-2 b c f+3 b d e) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{6 b^{5/3} d^{7/3}}+\frac {\sqrt [3]{a+b x} (c+d x)^{2/3} (-a d f-2 b c f+3 b d e)}{3 b d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 b d} \]

[In]

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

[Out]

((3*b*d*e - 2*b*c*f - a*d*f)*(a + b*x)^(1/3)*(c + d*x)^(2/3))/(3*b*d^2) + (f*(a + b*x)^(4/3)*(c + d*x)^(2/3))/
(2*b*d) + ((b*c - a*d)*(3*b*d*e - 2*b*c*f - 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))])/(3*Sqrt[3]*b^(5/3)*d^(7/3)) + ((b*c - a*d)*(3*b*d*e - 2*b*c*f - a*d*f)*Log[a + b*x])/(
18*b^(5/3)*d^(7/3)) + ((b*c - a*d)*(3*b*d*e - 2*b*c*f - a*d*f)*Log[-1 + (b^(1/3)*(c + d*x)^(1/3))/(d^(1/3)*(a
+ b*x)^(1/3))])/(6*b^(5/3)*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)^{2/3}}{2 b d}+\frac {\left (2 b d e-\left (\frac {4 b c}{3}+\frac {2 a d}{3}\right ) f\right ) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx}{2 b d} \\ & = \frac {(3 b d e-2 b c f-a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 b d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 b d}-\frac {((b c-a d) (3 b d e-2 b c f-a d f)) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}} \, dx}{9 b d^2} \\ & = \frac {(3 b d e-2 b c f-a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 b d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 b d}+\frac {(b c-a d) (3 b d e-2 b c f-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 )}{3 \sqrt {3} b^{5/3} d^{7/3}}+\frac {(b c-a d) (3 b d e-2 b c f-a d f) \log (a+b x)}{18 b^{5/3} d^{7/3}}+\frac {(b c-a d) (3 b d e-2 b c f-a d f) \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{6 b^{5/3} d^{7/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 9.54 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt [3]{a+b x} (e+f x)}{\sqrt [3]{c+d x}} \, dx=\frac {\sqrt [3]{a+b x} \left (3 b^{2/3} \sqrt [3]{d (a+b x)} (c+d x)^{2/3} (a d f+b (6 d e-4 c f+3 d f x))-2 \sqrt {3} (b c-a d) (-3 b d e+2 b c f+a d f) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}{2 \sqrt [3]{d (a+b x)}+\sqrt [3]{b} \sqrt [3]{c+d x}}\right )-2 (b c-a d) (-3 b d e+2 b c f+a d f) \log \left (\sqrt [3]{d (a+b x)}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )+(b c-a d) (-3 b d e+2 b c f+a d f) \log \left ((d (a+b x))^{2/3}+\sqrt [3]{b} \sqrt [3]{d (a+b x)} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}\right )\right )}{18 b^{5/3} d^2 \sqrt [3]{d (a+b x)}} \]

[In]

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

[Out]

((a + b*x)^(1/3)*(3*b^(2/3)*(d*(a + b*x))^(1/3)*(c + d*x)^(2/3)*(a*d*f + b*(6*d*e - 4*c*f + 3*d*f*x)) - 2*Sqrt
[3]*(b*c - a*d)*(-3*b*d*e + 2*b*c*f + a*d*f)*ArcTan[(Sqrt[3]*b^(1/3)*(c + d*x)^(1/3))/(2*(d*(a + b*x))^(1/3) +
b^(1/3)*(c + d*x)^(1/3))] - 2*(b*c - a*d)*(-3*b*d*e + 2*b*c*f + a*d*f)*Log[(d*(a + b*x))^(1/3) - b^(1/3)*(c +
d*x)^(1/3)] + (b*c - a*d)*(-3*b*d*e + 2*b*c*f + a*d*f)*Log[(d*(a + b*x))^(2/3) + b^(1/3)*(d*(a + b*x))^(1/3)*
(c + d*x)^(1/3) + b^(2/3)*(c + d*x)^(2/3)]))/(18*b^(5/3)*d^2*(d*(a + b*x))^(1/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 892, normalized size of antiderivative = 3.27 \[ \int \frac {\sqrt [3]{a+b x} (e+f x)}{\sqrt [3]{c+d x}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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