3.3006 \(\int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx\)
Optimal. Leaf size=325 \[ -\frac {\sqrt [3]{a+b x} (c+d x)^{2/3} (b c-a d)}{6 (e+f x) (b e-a f) (d e-c f)}+\frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (e+f x)^2 (d e-c f)}-\frac {(b c-a d)^2 \log (e+f x)}{18 (b e-a f)^{5/3} (d e-c f)^{4/3}}+\frac {(b c-a d)^2 \log \left (\frac {\sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt [3]{d e-c f}}-\sqrt [3]{a+b x}\right )}{6 (b e-a f)^{5/3} (d e-c f)^{4/3}}+\frac {(b c-a d)^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt {3} \sqrt [3]{a+b x} \sqrt [3]{d e-c f}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3} (b e-a f)^{5/3} (d e-c f)^{4/3}} \]
[Out]
1/2*(b*x+a)^(1/3)*(d*x+c)^(5/3)/(-c*f+d*e)/(f*x+e)^2-1/6*(-a*d+b*c)*(b*x+a)^(1/3)*(d*x+c)^(2/3)/(-a*f+b*e)/(-c
*f+d*e)/(f*x+e)-1/18*(-a*d+b*c)^2*ln(f*x+e)/(-a*f+b*e)^(5/3)/(-c*f+d*e)^(4/3)+1/6*(-a*d+b*c)^2*ln(-(b*x+a)^(1/
3)+(-a*f+b*e)^(1/3)*(d*x+c)^(1/3)/(-c*f+d*e)^(1/3))/(-a*f+b*e)^(5/3)/(-c*f+d*e)^(4/3)+1/9*(-a*d+b*c)^2*arctan(
1/3*3^(1/2)+2/3*(-a*f+b*e)^(1/3)*(d*x+c)^(1/3)/(-c*f+d*e)^(1/3)/(b*x+a)^(1/3)*3^(1/2))/(-a*f+b*e)^(5/3)/(-c*f+
d*e)^(4/3)*3^(1/2)
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Rubi [A] time = 0.23, antiderivative size = 325, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used =
{94, 91} \[ -\frac {\sqrt [3]{a+b x} (c+d x)^{2/3} (b c-a d)}{6 (e+f x) (b e-a f) (d e-c f)}+\frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (e+f x)^2 (d e-c f)}-\frac {(b c-a d)^2 \log (e+f x)}{18 (b e-a f)^{5/3} (d e-c f)^{4/3}}+\frac {(b c-a d)^2 \log \left (\frac {\sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt [3]{d e-c f}}-\sqrt [3]{a+b x}\right )}{6 (b e-a f)^{5/3} (d e-c f)^{4/3}}+\frac {(b c-a d)^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt {3} \sqrt [3]{a+b x} \sqrt [3]{d e-c f}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3} (b e-a f)^{5/3} (d e-c f)^{4/3}} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*x)^(1/3)*(c + d*x)^(2/3))/(e + f*x)^3,x]
[Out]
((a + b*x)^(1/3)*(c + d*x)^(5/3))/(2*(d*e - c*f)*(e + f*x)^2) - ((b*c - a*d)*(a + b*x)^(1/3)*(c + d*x)^(2/3))/
(6*(b*e - a*f)*(d*e - c*f)*(e + f*x)) + ((b*c - a*d)^2*ArcTan[1/Sqrt[3] + (2*(b*e - a*f)^(1/3)*(c + d*x)^(1/3)
)/(Sqrt[3]*(d*e - c*f)^(1/3)*(a + b*x)^(1/3))])/(3*Sqrt[3]*(b*e - a*f)^(5/3)*(d*e - c*f)^(4/3)) - ((b*c - a*d)
^2*Log[e + f*x])/(18*(b*e - a*f)^(5/3)*(d*e - c*f)^(4/3)) + ((b*c - a*d)^2*Log[-(a + b*x)^(1/3) + ((b*e - a*f)
^(1/3)*(c + d*x)^(1/3))/(d*e - c*f)^(1/3)])/(6*(b*e - a*f)^(5/3)*(d*e - c*f)^(4/3))
Rule 91
Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)*((e_.) + (f_.)*(x_))), x_Symbol] :> With[{q = Rt[
(d*e - c*f)/(b*e - a*f), 3]}, -Simp[(Sqrt[3]*q*ArcTan[1/Sqrt[3] + (2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/
3))])/(d*e - c*f), x] + (Simp[(q*Log[e + f*x])/(2*(d*e - c*f)), x] - Simp[(3*q*Log[q*(a + b*x)^(1/3) - (c + d*
x)^(1/3)])/(2*(d*e - c*f)), x])] /; FreeQ[{a, b, c, d, e, f}, x]
Rule 94
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && !(SumSimplerQ[p, 1] && !SumSimplerQ[m, 1])
Rubi steps
\begin {align*} \int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx &=\frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (d e-c f) (e+f x)^2}-\frac {(b c-a d) \int \frac {(c+d x)^{2/3}}{(a+b x)^{2/3} (e+f x)^2} \, dx}{6 (d e-c f)}\\ &=\frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (d e-c f) (e+f x)^2}-\frac {(b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{6 (b e-a f) (d e-c f) (e+f x)}-\frac {(b c-a d)^2 \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)} \, dx}{9 (b e-a f) (d e-c f)}\\ &=\frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (d e-c f) (e+f x)^2}-\frac {(b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{6 (b e-a f) (d e-c f) (e+f x)}+\frac {(b c-a d)^2 \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d e-c f} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} (b e-a f)^{5/3} (d e-c f)^{4/3}}-\frac {(b c-a d)^2 \log (e+f x)}{18 (b e-a f)^{5/3} (d e-c f)^{4/3}}+\frac {(b c-a d)^2 \log \left (-\sqrt [3]{a+b x}+\frac {\sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt [3]{d e-c f}}\right )}{6 (b e-a f)^{5/3} (d e-c f)^{4/3}}\\ \end {align*}
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Mathematica [C] time = 0.15, size = 142, normalized size = 0.44 \[ \frac {\sqrt [3]{a+b x} \left (\frac {(e+f x) (b c-a d) \left (2 (e+f x) (b c-a d) \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )+(c+d x) (b e-a f)\right )}{(b e-a f)^2}-3 (c+d x)^2\right )}{6 \sqrt [3]{c+d x} (e+f x)^2 (c f-d e)} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*x)^(1/3)*(c + d*x)^(2/3))/(e + f*x)^3,x]
[Out]
((a + b*x)^(1/3)*(-3*(c + d*x)^2 + ((b*c - a*d)*(e + f*x)*((b*e - a*f)*(c + d*x) + 2*(b*c - a*d)*(e + f*x)*Hyp
ergeometric2F1[1/3, 1, 4/3, ((d*e - c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))]))/(b*e - a*f)^2))/(6*(-(d*e) + c*
f)*(c + d*x)^(1/3)*(e + f*x)^2)
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fricas [B] time = 1.65, size = 3992, normalized size = 12.28 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(1/3)*(d*x+c)^(2/3)/(f*x+e)^3,x, algorithm="fricas")
[Out]
[1/18*(3*sqrt(1/3)*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e^4 - (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d
^3)*e^3*f + (a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*e^2*f^2 + ((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e^2*f^2
- (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*e*f^3 + (a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*f^4)*x^2 +
2*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e^3*f - (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*e^2*f^2 + (
a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*e*f^3)*x)*sqrt((-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a
*b*c + a^2*d)*e*f^2)^(1/3)/(d*e - c*f))*log(-(3*a^2*c*f^2 + (b^2*c + 2*a*b*d)*e^2 - 2*(2*a*b*c + a^2*d)*e*f +
3*(-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)*(b*e - a*f)*(b*x + a)^(1/
3)*(d*x + c)^(2/3) + (3*b^2*d*e^2 - 2*(b^2*c + 2*a*b*d)*e*f + (2*a*b*c + a^2*d)*f^2)*x - 3*sqrt(1/3)*(2*(b*d*e
^2 + a*c*f^2 - (b*c + a*d)*e*f)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*
e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) + (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*
a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)*(b*c*e - a*c*f + (b*d*e - a*d*f)*x))*sqrt((-b^2*d*e^3 + a^2*c*f^
3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)/(d*e - c*f)))/(f*x + e)) - (-b^2*d*e^3 + a^2*c*f^
3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*x^2 + 2*(b^2
*c^2 - 2*a*b*c*d + a^2*d^2)*e*f*x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2)*log(((b*d*e^2 + a*c*f^2 - (b*c + a*d)
*e*f)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*
e*f^2)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) - (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c +
a^2*d)*e*f^2)^(1/3)*(b*c*e - a*c*f + (b*d*e - a*d*f)*x))/(d*x + c)) + 2*(-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a
*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*x^2 + 2*(b^2*c^2 - 2*a*b*c*d
+ a^2*d^2)*e*f*x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2)*log(((b*d*e^2 + a*c*f^2 - (b*c + a*d)*e*f)*(b*x + a)^
(1/3)*(d*x + c)^(2/3) - (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*(d*
x + c))/(d*x + c)) + 3*(3*a^3*c^2*f^4 + (2*b^3*c*d + a*b^2*d^2)*e^4 - 2*(b^3*c^2 + 4*a*b^2*c*d + a^2*b*d^2)*e^
3*f + (7*a*b^2*c^2 + 10*a^2*b*c*d + a^3*d^2)*e^2*f^2 - 4*(2*a^2*b*c^2 + a^3*c*d)*e*f^3 + (3*b^3*d^2*e^4 - 4*(b
^3*c*d + 2*a*b^2*d^2)*e^3*f + (b^3*c^2 + 10*a*b^2*c*d + 7*a^2*b*d^2)*e^2*f^2 - 2*(a*b^2*c^2 + 4*a^2*b*c*d + a^
3*d^2)*e*f^3 + (a^2*b*c^2 + 2*a^3*c*d)*f^4)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^3*d^2*e^7 - a^3*c^2*e^2*f^5
- (2*b^3*c*d + 3*a*b^2*d^2)*e^6*f + (b^3*c^2 + 6*a*b^2*c*d + 3*a^2*b*d^2)*e^5*f^2 - (3*a*b^2*c^2 + 6*a^2*b*c*
d + a^3*d^2)*e^4*f^3 + (3*a^2*b*c^2 + 2*a^3*c*d)*e^3*f^4 + (b^3*d^2*e^5*f^2 - a^3*c^2*f^7 - (2*b^3*c*d + 3*a*b
^2*d^2)*e^4*f^3 + (b^3*c^2 + 6*a*b^2*c*d + 3*a^2*b*d^2)*e^3*f^4 - (3*a*b^2*c^2 + 6*a^2*b*c*d + a^3*d^2)*e^2*f^
5 + (3*a^2*b*c^2 + 2*a^3*c*d)*e*f^6)*x^2 + 2*(b^3*d^2*e^6*f - a^3*c^2*e*f^6 - (2*b^3*c*d + 3*a*b^2*d^2)*e^5*f^
2 + (b^3*c^2 + 6*a*b^2*c*d + 3*a^2*b*d^2)*e^4*f^3 - (3*a*b^2*c^2 + 6*a^2*b*c*d + a^3*d^2)*e^3*f^4 + (3*a^2*b*c
^2 + 2*a^3*c*d)*e^2*f^5)*x), -1/18*(6*sqrt(1/3)*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e^4 - (b^3*c^3 - a*b^
2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*e^3*f + (a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*e^2*f^2 + ((b^3*c^2*d - 2*a*b
^2*c*d^2 + a^2*b*d^3)*e^2*f^2 - (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*e*f^3 + (a*b^2*c^3 - 2*a^2*b*c
^2*d + a^3*c*d^2)*f^4)*x^2 + 2*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e^3*f - (b^3*c^3 - a*b^2*c^2*d - a^2*b
*c*d^2 + a^3*d^3)*e^2*f^2 + (a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*e*f^3)*x)*sqrt(-(-b^2*d*e^3 + a^2*c*f^3 +
(b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)/(d*e - c*f))*arctan(sqrt(1/3)*(2*(-b^2*d*e^3 + a^2*c*
f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) - (-b^2*d*e^3 +
a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)*(b*c*e - a*c*f + (b*d*e - a*d*f)*x))*sqr
t(-(-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)/(d*e - c*f))/(b^2*c*e^2
- 2*a*b*c*e*f + a^2*c*f^2 + (b^2*d*e^2 - 2*a*b*d*e*f + a^2*d*f^2)*x)) + (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a
*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*x^2 + 2*(b^2*c^2 - 2*a*b*c*d
+ a^2*d^2)*e*f*x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2)*log(((b*d*e^2 + a*c*f^2 - (b*c + a*d)*e*f)*(b*x + a)^
(2/3)*(d*x + c)^(1/3) + (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*(b*
x + a)^(1/3)*(d*x + c)^(2/3) - (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1
/3)*(b*c*e - a*c*f + (b*d*e - a*d*f)*x))/(d*x + c)) - 2*(-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2
*a*b*c + a^2*d)*e*f^2)^(2/3)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*x^2 + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e*f*
x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2)*log(((b*d*e^2 + a*c*f^2 - (b*c + a*d)*e*f)*(b*x + a)^(1/3)*(d*x + c)^
(2/3) - (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*(d*x + c))/(d*x + c
)) - 3*(3*a^3*c^2*f^4 + (2*b^3*c*d + a*b^2*d^2)*e^4 - 2*(b^3*c^2 + 4*a*b^2*c*d + a^2*b*d^2)*e^3*f + (7*a*b^2*c
^2 + 10*a^2*b*c*d + a^3*d^2)*e^2*f^2 - 4*(2*a^2*b*c^2 + a^3*c*d)*e*f^3 + (3*b^3*d^2*e^4 - 4*(b^3*c*d + 2*a*b^2
*d^2)*e^3*f + (b^3*c^2 + 10*a*b^2*c*d + 7*a^2*b*d^2)*e^2*f^2 - 2*(a*b^2*c^2 + 4*a^2*b*c*d + a^3*d^2)*e*f^3 + (
a^2*b*c^2 + 2*a^3*c*d)*f^4)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^3*d^2*e^7 - a^3*c^2*e^2*f^5 - (2*b^3*c*d +
3*a*b^2*d^2)*e^6*f + (b^3*c^2 + 6*a*b^2*c*d + 3*a^2*b*d^2)*e^5*f^2 - (3*a*b^2*c^2 + 6*a^2*b*c*d + a^3*d^2)*e^4
*f^3 + (3*a^2*b*c^2 + 2*a^3*c*d)*e^3*f^4 + (b^3*d^2*e^5*f^2 - a^3*c^2*f^7 - (2*b^3*c*d + 3*a*b^2*d^2)*e^4*f^3
+ (b^3*c^2 + 6*a*b^2*c*d + 3*a^2*b*d^2)*e^3*f^4 - (3*a*b^2*c^2 + 6*a^2*b*c*d + a^3*d^2)*e^2*f^5 + (3*a^2*b*c^2
+ 2*a^3*c*d)*e*f^6)*x^2 + 2*(b^3*d^2*e^6*f - a^3*c^2*e*f^6 - (2*b^3*c*d + 3*a*b^2*d^2)*e^5*f^2 + (b^3*c^2 + 6
*a*b^2*c*d + 3*a^2*b*d^2)*e^4*f^3 - (3*a*b^2*c^2 + 6*a^2*b*c*d + a^3*d^2)*e^3*f^4 + (3*a^2*b*c^2 + 2*a^3*c*d)*
e^2*f^5)*x)]
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{{\left (f x + e\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(1/3)*(d*x+c)^(2/3)/(f*x+e)^3,x, algorithm="giac")
[Out]
integrate((b*x + a)^(1/3)*(d*x + c)^(2/3)/(f*x + e)^3, x)
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maple [F] time = 0.23, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{\frac {1}{3}} \left (d x +c \right )^{\frac {2}{3}}}{\left (f x +e \right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x+a)^(1/3)*(d*x+c)^(2/3)/(f*x+e)^3,x)
[Out]
int((b*x+a)^(1/3)*(d*x+c)^(2/3)/(f*x+e)^3,x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{{\left (f x + e\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(1/3)*(d*x+c)^(2/3)/(f*x+e)^3,x, algorithm="maxima")
[Out]
integrate((b*x + a)^(1/3)*(d*x + c)^(2/3)/(f*x + e)^3, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,x\right )}^{1/3}\,{\left (c+d\,x\right )}^{2/3}}{{\left (e+f\,x\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((a + b*x)^(1/3)*(c + d*x)^(2/3))/(e + f*x)^3,x)
[Out]
int(((a + b*x)^(1/3)*(c + d*x)^(2/3))/(e + f*x)^3, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt [3]{a + b x} \left (c + d x\right )^{\frac {2}{3}}}{\left (e + f x\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)**(1/3)*(d*x+c)**(2/3)/(f*x+e)**3,x)
[Out]
Integral((a + b*x)**(1/3)*(c + d*x)**(2/3)/(e + f*x)**3, x)
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