3.3015 \(\int \frac {(e+f x)^3}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx\)

Optimal. Leaf size=587 \[ \frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} \left (28 a^2 d^2 f^2+3 b d f x (-7 a d f-8 b c f+15 b d e)-a b d f (108 d e-31 c f)+b^2 \left (40 c^2 f^2-135 c d e f+144 d^2 e^2\right )\right )}{54 b^3 d^3}+\frac {\log (c+d x) \left (14 a^3 d^3 f^3-6 a^2 b d^2 f^2 (9 d e-2 c f)+3 a b^2 d f \left (5 c^2 f^2-18 c d e f+27 d^2 e^2\right )-\left (b^3 \left (-40 c^3 f^3+135 c^2 d e f^2-162 c d^2 e^2 f+81 d^3 e^3\right )\right )\right )}{162 b^{10/3} d^{11/3}}+\frac {\left (14 a^3 d^3 f^3-6 a^2 b d^2 f^2 (9 d e-2 c f)+3 a b^2 d f \left (5 c^2 f^2-18 c d e f+27 d^2 e^2\right )-\left (b^3 \left (-40 c^3 f^3+135 c^2 d e f^2-162 c d^2 e^2 f+81 d^3 e^3\right )\right )\right ) \log \left (\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{54 b^{10/3} d^{11/3}}+\frac {\left (14 a^3 d^3 f^3-6 a^2 b d^2 f^2 (9 d e-2 c f)+3 a b^2 d f \left (5 c^2 f^2-18 c d e f+27 d^2 e^2\right )-\left (b^3 \left (-40 c^3 f^3+135 c^2 d e f^2-162 c d^2 e^2 f+81 d^3 e^3\right )\right )\right ) \tan ^{-1}\left (\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}+\frac {1}{\sqrt {3}}\right )}{27 \sqrt {3} b^{10/3} d^{11/3}}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)^2}{3 b d} \]

[Out]

1/3*f*(b*x+a)^(2/3)*(d*x+c)^(1/3)*(f*x+e)^2/b/d+1/54*f*(b*x+a)^(2/3)*(d*x+c)^(1/3)*(28*a^2*d^2*f^2-a*b*d*f*(-3
1*c*f+108*d*e)+b^2*(40*c^2*f^2-135*c*d*e*f+144*d^2*e^2)+3*b*d*f*(-7*a*d*f-8*b*c*f+15*b*d*e)*x)/b^3/d^3+1/162*(
14*a^3*d^3*f^3-6*a^2*b*d^2*f^2*(-2*c*f+9*d*e)+3*a*b^2*d*f*(5*c^2*f^2-18*c*d*e*f+27*d^2*e^2)-b^3*(-40*c^3*f^3+1
35*c^2*d*e*f^2-162*c*d^2*e^2*f+81*d^3*e^3))*ln(d*x+c)/b^(10/3)/d^(11/3)+1/54*(14*a^3*d^3*f^3-6*a^2*b*d^2*f^2*(
-2*c*f+9*d*e)+3*a*b^2*d*f*(5*c^2*f^2-18*c*d*e*f+27*d^2*e^2)-b^3*(-40*c^3*f^3+135*c^2*d*e*f^2-162*c*d^2*e^2*f+8
1*d^3*e^3))*ln(-1+d^(1/3)*(b*x+a)^(1/3)/b^(1/3)/(d*x+c)^(1/3))/b^(10/3)/d^(11/3)+1/81*(14*a^3*d^3*f^3-6*a^2*b*
d^2*f^2*(-2*c*f+9*d*e)+3*a*b^2*d*f*(5*c^2*f^2-18*c*d*e*f+27*d^2*e^2)-b^3*(-40*c^3*f^3+135*c^2*d*e*f^2-162*c*d^
2*e^2*f+81*d^3*e^3))*arctan(1/3*3^(1/2)+2/3*d^(1/3)*(b*x+a)^(1/3)/b^(1/3)/(d*x+c)^(1/3)*3^(1/2))/b^(10/3)/d^(1
1/3)*3^(1/2)

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Rubi [A]  time = 0.48, antiderivative size = 587, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {100, 147, 59} \[ \frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} \left (28 a^2 d^2 f^2+3 b d f x (-7 a d f-8 b c f+15 b d e)-a b d f (108 d e-31 c f)+b^2 \left (40 c^2 f^2-135 c d e f+144 d^2 e^2\right )\right )}{54 b^3 d^3}+\frac {\log (c+d x) \left (-6 a^2 b d^2 f^2 (9 d e-2 c f)+14 a^3 d^3 f^3+3 a b^2 d f \left (5 c^2 f^2-18 c d e f+27 d^2 e^2\right )+b^3 \left (-\left (135 c^2 d e f^2-40 c^3 f^3-162 c d^2 e^2 f+81 d^3 e^3\right )\right )\right )}{162 b^{10/3} d^{11/3}}+\frac {\left (-6 a^2 b d^2 f^2 (9 d e-2 c f)+14 a^3 d^3 f^3+3 a b^2 d f \left (5 c^2 f^2-18 c d e f+27 d^2 e^2\right )+b^3 \left (-\left (135 c^2 d e f^2-40 c^3 f^3-162 c d^2 e^2 f+81 d^3 e^3\right )\right )\right ) \log \left (\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{54 b^{10/3} d^{11/3}}+\frac {\left (-6 a^2 b d^2 f^2 (9 d e-2 c f)+14 a^3 d^3 f^3+3 a b^2 d f \left (5 c^2 f^2-18 c d e f+27 d^2 e^2\right )+b^3 \left (-\left (135 c^2 d e f^2-40 c^3 f^3-162 c d^2 e^2 f+81 d^3 e^3\right )\right )\right ) \tan ^{-1}\left (\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}+\frac {1}{\sqrt {3}}\right )}{27 \sqrt {3} b^{10/3} d^{11/3}}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)^2}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*(a + b*x)^(2/3)*(c + d*x)^(1/3)*(e + f*x)^2)/(3*b*d) + (f*(a + b*x)^(2/3)*(c + d*x)^(1/3)*(28*a^2*d^2*f^2 -
a*b*d*f*(108*d*e - 31*c*f) + b^2*(144*d^2*e^2 - 135*c*d*e*f + 40*c^2*f^2) + 3*b*d*f*(15*b*d*e - 8*b*c*f - 7*a
*d*f)*x))/(54*b^3*d^3) + ((14*a^3*d^3*f^3 - 6*a^2*b*d^2*f^2*(9*d*e - 2*c*f) + 3*a*b^2*d*f*(27*d^2*e^2 - 18*c*d
*e*f + 5*c^2*f^2) - b^3*(81*d^3*e^3 - 162*c*d^2*e^2*f + 135*c^2*d*e*f^2 - 40*c^3*f^3))*ArcTan[1/Sqrt[3] + (2*d
^(1/3)*(a + b*x)^(1/3))/(Sqrt[3]*b^(1/3)*(c + d*x)^(1/3))])/(27*Sqrt[3]*b^(10/3)*d^(11/3)) + ((14*a^3*d^3*f^3
- 6*a^2*b*d^2*f^2*(9*d*e - 2*c*f) + 3*a*b^2*d*f*(27*d^2*e^2 - 18*c*d*e*f + 5*c^2*f^2) - b^3*(81*d^3*e^3 - 162*
c*d^2*e^2*f + 135*c^2*d*e*f^2 - 40*c^3*f^3))*Log[c + d*x])/(162*b^(10/3)*d^(11/3)) + ((14*a^3*d^3*f^3 - 6*a^2*
b*d^2*f^2*(9*d*e - 2*c*f) + 3*a*b^2*d*f*(27*d^2*e^2 - 18*c*d*e*f + 5*c^2*f^2) - b^3*(81*d^3*e^3 - 162*c*d^2*e^
2*f + 135*c^2*d*e*f^2 - 40*c^3*f^3))*Log[-1 + (d^(1/3)*(a + b*x)^(1/3))/(b^(1/3)*(c + d*x)^(1/3))])/(54*b^(10/
3)*d^(11/3))

Rule 59

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

Rule 100

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

Rule 147

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

Rubi steps

\begin {align*} \int \frac {(e+f x)^3}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx &=\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)^2}{3 b d}+\frac {\int \frac {(e+f x) \left (\frac {1}{3} \left (9 b d e^2-f (2 b c e+a d e+6 a c f)\right )+\frac {1}{3} f (15 b d e-8 b c f-7 a d f) x\right )}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx}{3 b d}\\ &=\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)^2}{3 b d}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} \left (28 a^2 d^2 f^2-a b d f (108 d e-31 c f)+b^2 \left (144 d^2 e^2-135 c d e f+40 c^2 f^2\right )+3 b d f (15 b d e-8 b c f-7 a d f) x\right )}{54 b^3 d^3}-\frac {\left (14 a^3 d^3 f^3-6 a^2 b d^2 f^2 (9 d e-2 c f)+3 a b^2 d f \left (27 d^2 e^2-18 c d e f+5 c^2 f^2\right )-b^3 \left (81 d^3 e^3-162 c d^2 e^2 f+135 c^2 d e f^2-40 c^3 f^3\right )\right ) \int \frac {1}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx}{81 b^3 d^3}\\ &=\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)^2}{3 b d}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} \left (28 a^2 d^2 f^2-a b d f (108 d e-31 c f)+b^2 \left (144 d^2 e^2-135 c d e f+40 c^2 f^2\right )+3 b d f (15 b d e-8 b c f-7 a d f) x\right )}{54 b^3 d^3}+\frac {\left (14 a^3 d^3 f^3-6 a^2 b d^2 f^2 (9 d e-2 c f)+3 a b^2 d f \left (27 d^2 e^2-18 c d e f+5 c^2 f^2\right )-b^3 \left (81 d^3 e^3-162 c d^2 e^2 f+135 c^2 d e f^2-40 c^3 f^3\right )\right ) \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{27 \sqrt {3} b^{10/3} d^{11/3}}+\frac {\left (14 a^3 d^3 f^3-6 a^2 b d^2 f^2 (9 d e-2 c f)+3 a b^2 d f \left (27 d^2 e^2-18 c d e f+5 c^2 f^2\right )-b^3 \left (81 d^3 e^3-162 c d^2 e^2 f+135 c^2 d e f^2-40 c^3 f^3\right )\right ) \log (c+d x)}{162 b^{10/3} d^{11/3}}+\frac {\left (14 a^3 d^3 f^3-6 a^2 b d^2 f^2 (9 d e-2 c f)+3 a b^2 d f \left (27 d^2 e^2-18 c d e f+5 c^2 f^2\right )-b^3 \left (81 d^3 e^3-162 c d^2 e^2 f+135 c^2 d e f^2-40 c^3 f^3\right )\right ) \log \left (-1+\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{54 b^{10/3} d^{11/3}}\\ \end {align*}

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Mathematica [C]  time = 0.79, size = 291, normalized size = 0.50 \[ \frac {(a+b x)^{2/3} \left (\frac {f^2 (c+d x)^2 (-7 a d f-8 b c f+15 b d e) \, _2F_1\left (-\frac {4}{3},\frac {2}{3};\frac {5}{3};\frac {d (a+b x)}{a d-b c}\right )}{b d^2 \left (\frac {b (c+d x)}{b c-a d}\right )^{4/3}}+\frac {(d e-c f)^2 \left (\frac {b (c+d x)}{b c-a d}\right )^{2/3} (-a d f-8 b c f+9 b d e) \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {d (a+b x)}{a d-b c}\right )}{b d^2}+\frac {8 f (c+d x) (c f-d e) (a d f+2 b c f-3 b d e) \, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};\frac {d (a+b x)}{a d-b c}\right )}{b d^2 \sqrt [3]{\frac {b (c+d x)}{b c-a d}}}+2 f (c+d x) (e+f x)^2\right )}{6 b d (c+d x)^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(2/3)*(2*f*(c + d*x)*(e + f*x)^2 + (f^2*(15*b*d*e - 8*b*c*f - 7*a*d*f)*(c + d*x)^2*Hypergeometric2F
1[-4/3, 2/3, 5/3, (d*(a + b*x))/(-(b*c) + a*d)])/(b*d^2*((b*(c + d*x))/(b*c - a*d))^(4/3)) + (8*f*(-(d*e) + c*
f)*(-3*b*d*e + 2*b*c*f + a*d*f)*(c + d*x)*Hypergeometric2F1[-1/3, 2/3, 5/3, (d*(a + b*x))/(-(b*c) + a*d)])/(b*
d^2*((b*(c + d*x))/(b*c - a*d))^(1/3)) + ((d*e - c*f)^2*(9*b*d*e - 8*b*c*f - a*d*f)*((b*(c + d*x))/(b*c - a*d)
)^(2/3)*Hypergeometric2F1[2/3, 2/3, 5/3, (d*(a + b*x))/(-(b*c) + a*d)])/(b*d^2)))/(6*b*d*(c + d*x)^(2/3))

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fricas [A]  time = 1.87, size = 1481, normalized size = 2.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/162*(3*sqrt(1/3)*(81*b^4*d^4*e^3 - 81*(2*b^4*c*d^3 + a*b^3*d^4)*e^2*f + 27*(5*b^4*c^2*d^2 + 2*a*b^3*c*d^3
+ 2*a^2*b^2*d^4)*e*f^2 - (40*b^4*c^3*d + 15*a*b^3*c^2*d^2 + 12*a^2*b^2*c*d^3 + 14*a^3*b*d^4)*f^3)*sqrt(-(b*d^2
)^(1/3)/b)*log(-3*b*d^2*x - 2*b*c*d - a*d^2 + 3*(b*d^2)^(1/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3)*d + 3*sqrt(1/3)*
(2*(b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d - (b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (b*d^2)^(1/3)*(b*d*x
+ a*d))*sqrt(-(b*d^2)^(1/3)/b)) + 2*(81*b^3*d^3*e^3 - 81*(2*b^3*c*d^2 + a*b^2*d^3)*e^2*f + 27*(5*b^3*c^2*d + 2
*a*b^2*c*d^2 + 2*a^2*b*d^3)*e*f^2 - (40*b^3*c^3 + 15*a*b^2*c^2*d + 12*a^2*b*c*d^2 + 14*a^3*d^3)*f^3)*(b*d^2)^(
2/3)*log(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (b*d^2)^(2/3)*(b*x + a))/(b*x + a)) - (81*b^3*d^3*e^3 - 81*(2*
b^3*c*d^2 + a*b^2*d^3)*e^2*f + 27*(5*b^3*c^2*d + 2*a*b^2*c*d^2 + 2*a^2*b*d^3)*e*f^2 - (40*b^3*c^3 + 15*a*b^2*c
^2*d + 12*a^2*b*c*d^2 + 14*a^3*d^3)*f^3)*(b*d^2)^(2/3)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d + (b*d^2)^(2/3
)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (b*d^2)^(1/3)*(b*d*x + a*d))/(b*x + a)) - 3*(18*b^3*d^4*f^3*x^2 + 162*b^3*
d^4*e^2*f - 27*(5*b^3*c*d^3 + 4*a*b^2*d^4)*e*f^2 + (40*b^3*c^2*d^2 + 31*a*b^2*c*d^3 + 28*a^2*b*d^4)*f^3 + 3*(2
7*b^3*d^4*e*f^2 - (8*b^3*c*d^3 + 7*a*b^2*d^4)*f^3)*x)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b^4*d^5), 1/162*(6*sqr
t(1/3)*(81*b^4*d^4*e^3 - 81*(2*b^4*c*d^3 + a*b^3*d^4)*e^2*f + 27*(5*b^4*c^2*d^2 + 2*a*b^3*c*d^3 + 2*a^2*b^2*d^
4)*e*f^2 - (40*b^4*c^3*d + 15*a*b^3*c^2*d^2 + 12*a^2*b^2*c*d^3 + 14*a^3*b*d^4)*f^3)*sqrt((b*d^2)^(1/3)/b)*arct
an(sqrt(1/3)*(2*(b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (b*d^2)^(1/3)*(b*d*x + a*d))*sqrt((b*d^2)^(1/3
)/b)/(b*d^2*x + a*d^2)) - 2*(81*b^3*d^3*e^3 - 81*(2*b^3*c*d^2 + a*b^2*d^3)*e^2*f + 27*(5*b^3*c^2*d + 2*a*b^2*c
*d^2 + 2*a^2*b*d^3)*e*f^2 - (40*b^3*c^3 + 15*a*b^2*c^2*d + 12*a^2*b*c*d^2 + 14*a^3*d^3)*f^3)*(b*d^2)^(2/3)*log
(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (b*d^2)^(2/3)*(b*x + a))/(b*x + a)) + (81*b^3*d^3*e^3 - 81*(2*b^3*c*d^
2 + a*b^2*d^3)*e^2*f + 27*(5*b^3*c^2*d + 2*a*b^2*c*d^2 + 2*a^2*b*d^3)*e*f^2 - (40*b^3*c^3 + 15*a*b^2*c^2*d + 1
2*a^2*b*c*d^2 + 14*a^3*d^3)*f^3)*(b*d^2)^(2/3)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d + (b*d^2)^(2/3)*(b*x +
a)^(2/3)*(d*x + c)^(1/3) + (b*d^2)^(1/3)*(b*d*x + a*d))/(b*x + a)) + 3*(18*b^3*d^4*f^3*x^2 + 162*b^3*d^4*e^2*
f - 27*(5*b^3*c*d^3 + 4*a*b^2*d^4)*e*f^2 + (40*b^3*c^2*d^2 + 31*a*b^2*c*d^3 + 28*a^2*b*d^4)*f^3 + 3*(27*b^3*d^
4*e*f^2 - (8*b^3*c*d^3 + 7*a*b^2*d^4)*f^3)*x)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b^4*d^5)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{3}}{{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{3}}{\left (b x +a \right )^{\frac {1}{3}} \left (d x +c \right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{3}}{{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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