3.3008 \(\int \frac {\sqrt [3]{a+b x} (e+f x)^2}{\sqrt [3]{c+d x}} \, dx\)

Optimal. Leaf size=475 \[ \frac {\sqrt [3]{a+b x} (c+d x)^{2/3} \left (5 a^2 d^2 f^2-2 a b d f (9 d e-4 c f)+b^2 \left (14 c^2 f^2-36 c d e f+27 d^2 e^2\right )\right )}{27 b^2 d^3}+\frac {(b c-a d) \log (a+b x) \left (5 a^2 d^2 f^2-2 a b d f (9 d e-4 c f)+b^2 \left (14 c^2 f^2-36 c d e f+27 d^2 e^2\right )\right )}{162 b^{8/3} d^{10/3}}+\frac {(b c-a d) \left (5 a^2 d^2 f^2-2 a b d f (9 d e-4 c f)+b^2 \left (14 c^2 f^2-36 c d e f+27 d^2 e^2\right )\right ) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{54 b^{8/3} d^{10/3}}+\frac {(b c-a d) \left (5 a^2 d^2 f^2-2 a b d f (9 d e-4 c f)+b^2 \left (14 c^2 f^2-36 c d e f+27 d^2 e^2\right )\right ) \tan ^{-1}\left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{27 \sqrt {3} b^{8/3} d^{10/3}}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (-5 a d f-7 b c f+12 b d e)}{18 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (e+f x)}{3 b d} \]

[Out]

1/27*(5*a^2*d^2*f^2-2*a*b*d*f*(-4*c*f+9*d*e)+b^2*(14*c^2*f^2-36*c*d*e*f+27*d^2*e^2))*(b*x+a)^(1/3)*(d*x+c)^(2/
3)/b^2/d^3+1/18*f*(-5*a*d*f-7*b*c*f+12*b*d*e)*(b*x+a)^(4/3)*(d*x+c)^(2/3)/b^2/d^2+1/3*f*(b*x+a)^(4/3)*(d*x+c)^
(2/3)*(f*x+e)/b/d+1/162*(-a*d+b*c)*(5*a^2*d^2*f^2-2*a*b*d*f*(-4*c*f+9*d*e)+b^2*(14*c^2*f^2-36*c*d*e*f+27*d^2*e
^2))*ln(b*x+a)/b^(8/3)/d^(10/3)+1/54*(-a*d+b*c)*(5*a^2*d^2*f^2-2*a*b*d*f*(-4*c*f+9*d*e)+b^2*(14*c^2*f^2-36*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^(8/3)/d^(10/3)+1/81*(-a*d+b*c)*(5*a^2*d
^2*f^2-2*a*b*d*f*(-4*c*f+9*d*e)+b^2*(14*c^2*f^2-36*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^(8/3)/d^(10/3)*3^(1/2)

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Rubi [A]  time = 0.40, antiderivative size = 475, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {90, 80, 50, 59} \[ \frac {\sqrt [3]{a+b x} (c+d x)^{2/3} \left (5 a^2 d^2 f^2-2 a b d f (9 d e-4 c f)+b^2 \left (14 c^2 f^2-36 c d e f+27 d^2 e^2\right )\right )}{27 b^2 d^3}+\frac {(b c-a d) \log (a+b x) \left (5 a^2 d^2 f^2-2 a b d f (9 d e-4 c f)+b^2 \left (14 c^2 f^2-36 c d e f+27 d^2 e^2\right )\right )}{162 b^{8/3} d^{10/3}}+\frac {(b c-a d) \left (5 a^2 d^2 f^2-2 a b d f (9 d e-4 c f)+b^2 \left (14 c^2 f^2-36 c d e f+27 d^2 e^2\right )\right ) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{54 b^{8/3} d^{10/3}}+\frac {(b c-a d) \left (5 a^2 d^2 f^2-2 a b d f (9 d e-4 c f)+b^2 \left (14 c^2 f^2-36 c d e f+27 d^2 e^2\right )\right ) \tan ^{-1}\left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{27 \sqrt {3} b^{8/3} d^{10/3}}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (-5 a d f-7 b c f+12 b d e)}{18 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (e+f x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a^2*d^2*f^2 - 2*a*b*d*f*(9*d*e - 4*c*f) + b^2*(27*d^2*e^2 - 36*c*d*e*f + 14*c^2*f^2))*(a + b*x)^(1/3)*(c +
d*x)^(2/3))/(27*b^2*d^3) + (f*(12*b*d*e - 7*b*c*f - 5*a*d*f)*(a + b*x)^(4/3)*(c + d*x)^(2/3))/(18*b^2*d^2) +
(f*(a + b*x)^(4/3)*(c + d*x)^(2/3)*(e + f*x))/(3*b*d) + ((b*c - a*d)*(5*a^2*d^2*f^2 - 2*a*b*d*f*(9*d*e - 4*c*f
) + b^2*(27*d^2*e^2 - 36*c*d*e*f + 14*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))])/(27*Sqrt[3]*b^(8/3)*d^(10/3)) + ((b*c - a*d)*(5*a^2*d^2*f^2 - 2*a*b*d*f*(9*d*e - 4*c*f) +
b^2*(27*d^2*e^2 - 36*c*d*e*f + 14*c^2*f^2))*Log[a + b*x])/(162*b^(8/3)*d^(10/3)) + ((b*c - a*d)*(5*a^2*d^2*f^
2 - 2*a*b*d*f*(9*d*e - 4*c*f) + b^2*(27*d^2*e^2 - 36*c*d*e*f + 14*c^2*f^2))*Log[-1 + (b^(1/3)*(c + d*x)^(1/3))
/(d^(1/3)*(a + b*x)^(1/3))])/(54*b^(8/3)*d^(10/3))

Rule 50

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 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 80

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 90

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*} \int \frac {\sqrt [3]{a+b x} (e+f x)^2}{\sqrt [3]{c+d x}} \, dx &=\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (e+f x)}{3 b d}+\frac {\int \frac {\sqrt [3]{a+b x} \left (\frac {1}{3} \left (9 b d e^2-f (4 b c e+2 a d e+3 a c f)\right )+\frac {1}{3} f (12 b d e-7 b c f-5 a d f) x\right )}{\sqrt [3]{c+d x}} \, dx}{3 b d}\\ &=\frac {f (12 b d e-7 b c f-5 a d f) (a+b x)^{4/3} (c+d x)^{2/3}}{18 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (e+f x)}{3 b d}+\frac {\left (\frac {5 a^2 d f^2}{b}-2 a f (9 d e-4 c f)+b \left (27 d e^2-36 c e f+\frac {14 c^2 f^2}{d}\right )\right ) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx}{27 b d}\\ &=\frac {\left (\frac {5 a^2 d f^2}{b}-2 a f (9 d e-4 c f)+b \left (27 d e^2-36 c e f+\frac {14 c^2 f^2}{d}\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{27 b d^2}+\frac {f (12 b d e-7 b c f-5 a d f) (a+b x)^{4/3} (c+d x)^{2/3}}{18 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (e+f x)}{3 b d}-\frac {\left ((b c-a d) \left (\frac {5 a^2 d f^2}{b}-2 a f (9 d e-4 c f)+b \left (27 d e^2-36 c e f+\frac {14 c^2 f^2}{d}\right )\right )\right ) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}} \, dx}{81 b d^2}\\ &=\frac {\left (\frac {5 a^2 d f^2}{b}-2 a f (9 d e-4 c f)+b \left (27 d e^2-36 c e f+\frac {14 c^2 f^2}{d}\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{27 b d^2}+\frac {f (12 b d e-7 b c f-5 a d f) (a+b x)^{4/3} (c+d x)^{2/3}}{18 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (e+f x)}{3 b d}+\frac {(b c-a d) \left (\frac {5 a^2 d f^2}{b}-2 a f (9 d e-4 c f)+b \left (27 d e^2-36 c e f+\frac {14 c^2 f^2}{d}\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 )}{27 \sqrt {3} b^{5/3} d^{7/3}}+\frac {(b c-a d) \left (\frac {5 a^2 d f^2}{b}-2 a f (9 d e-4 c f)+b \left (27 d e^2-36 c e f+\frac {14 c^2 f^2}{d}\right )\right ) \log (a+b x)}{162 b^{5/3} d^{7/3}}+\frac {(b c-a d) \left (\frac {5 a^2 d f^2}{b}-2 a f (9 d e-4 c f)+b \left (27 d e^2-36 c e f+\frac {14 c^2 f^2}{d}\right )\right ) \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{54 b^{5/3} d^{7/3}}\\ \end {align*}

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Mathematica [C]  time = 0.28, size = 175, normalized size = 0.37 \[ \frac {(a+b x)^{4/3} \left (2 \sqrt [3]{\frac {b (c+d x)}{b c-a d}} \left (5 a^2 d^2 f^2+2 a b d f (4 c f-9 d e)+b^2 \left (14 c^2 f^2-36 c d e f+27 d^2 e^2\right )\right ) \, _2F_1\left (\frac {1}{3},\frac {4}{3};\frac {7}{3};\frac {d (a+b x)}{a d-b c}\right )-4 b f (c+d x) (5 a d f+7 b c f-12 b d e)+24 b^2 d f (c+d x) (e+f x)\right )}{72 b^3 d^2 \sqrt [3]{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(4/3)*(-4*b*f*(-12*b*d*e + 7*b*c*f + 5*a*d*f)*(c + d*x) + 24*b^2*d*f*(c + d*x)*(e + f*x) + 2*(5*a^2
*d^2*f^2 + 2*a*b*d*f*(-9*d*e + 4*c*f) + b^2*(27*d^2*e^2 - 36*c*d*e*f + 14*c^2*f^2))*((b*(c + d*x))/(b*c - a*d)
)^(1/3)*Hypergeometric2F1[1/3, 4/3, 7/3, (d*(a + b*x))/(-(b*c) + a*d)]))/(72*b^3*d^2*(c + d*x)^(1/3))

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fricas [A]  time = 1.26, size = 1400, normalized size = 2.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/162*(3*sqrt(1/3)*(27*(b^4*c*d^3 - a*b^3*d^4)*e^2 - 18*(2*b^4*c^2*d^2 - a*b^3*c*d^3 - a^2*b^2*d^4)*e*f + (1
4*b^4*c^3*d - 6*a*b^3*c^2*d^2 - 3*a^2*b^2*c*d^3 - 5*a^3*b*d^4)*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)) + (-b^2*d)^(2/3)*(27*(b^3*c*d^2 - a*b^2*d^3)*e^2 - 18*(2*b^3*c^2*d - a*b^2*c*d^2 - a^2*b*d^3)*e*f + (14*b^
3*c^3 - 6*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 5*a^3*d^3)*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)) - 2*(-b^2*d)^(2/3)*(27*(b^3*c*d
^2 - a*b^2*d^3)*e^2 - 18*(2*b^3*c^2*d - a*b^2*c*d^2 - a^2*b*d^3)*e*f + (14*b^3*c^3 - 6*a*b^2*c^2*d - 3*a^2*b*c
*d^2 - 5*a^3*d^3)*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*(18
*b^4*d^3*f^2*x^2 + 54*b^4*d^3*e^2 - 18*(4*b^4*c*d^2 - a*b^3*d^3)*e*f + (28*b^4*c^2*d - 5*a*b^3*c*d^2 - 5*a^2*b
^2*d^3)*f^2 + 3*(18*b^4*d^3*e*f - (7*b^4*c*d^2 - a*b^3*d^3)*f^2)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^4*d^4)
, -1/162*(6*sqrt(1/3)*(27*(b^4*c*d^3 - a*b^3*d^4)*e^2 - 18*(2*b^4*c^2*d^2 - a*b^3*c*d^3 - a^2*b^2*d^4)*e*f + (
14*b^4*c^3*d - 6*a*b^3*c^2*d^2 - 3*a^2*b^2*c*d^3 - 5*a^3*b*d^4)*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)) + (-b^2*d)^(2/3)*(27*(b^3*c*d^2 - a*b^2*d^3)*e^2 - 18*(2*b^3*c^2*d - a*b^2*c*d^2 - a^2*b*d^3)*e
*f + (14*b^3*c^3 - 6*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 5*a^3*d^3)*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)) - 2*(-b^2*d)^(2/3)*(
27*(b^3*c*d^2 - a*b^2*d^3)*e^2 - 18*(2*b^3*c^2*d - a*b^2*c*d^2 - a^2*b*d^3)*e*f + (14*b^3*c^3 - 6*a*b^2*c^2*d
- 3*a^2*b*c*d^2 - 5*a^3*d^3)*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*(18*b^4*d^3*f^2*x^2 + 54*b^4*d^3*e^2 - 18*(4*b^4*c*d^2 - a*b^3*d^3)*e*f + (28*b^4*c^2*d - 5*a*b^3*c*d^
2 - 5*a^2*b^2*d^3)*f^2 + 3*(18*b^4*d^3*e*f - (7*b^4*c*d^2 - a*b^3*d^3)*f^2)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3)
)/(b^4*d^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{\frac {1}{3}} \left (f x +e \right )^{2}}{\left (d x +c \right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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