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

Optimal. Leaf size=331 \[ \frac {(b c-a d)^2 \log (a+b x) (-5 a d f-4 b c f+9 b d e)}{162 b^{8/3} d^{7/3}}+\frac {(b c-a d)^2 (-5 a d f-4 b c f+9 b d e) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{54 b^{8/3} d^{7/3}}+\frac {(b c-a d)^2 (-5 a d f-4 b c f+9 b d e) \tan ^{-1}\left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{27 \sqrt {3} b^{8/3} d^{7/3}}+\frac {\sqrt [3]{a+b x} (c+d x)^{2/3} (b c-a d) (-5 a d f-4 b c f+9 b d e)}{27 b^2 d^2}+\frac {(a+b x)^{4/3} (c+d x)^{2/3} (-5 a d f-4 b c f+9 b d e)}{18 b^2 d}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3}}{3 b d} \]

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Rubi [A]  time = 0.23, antiderivative size = 331, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {80, 50, 59} \begin {gather*} \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 {(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 {(b c-a d)^2 (-5 a d f-4 b c f+9 b d e) \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^{7/3}}+\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} \end {gather*}

Antiderivative was successfully verified.

[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 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]

Rubi steps

\begin {align*} \int \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x) \, dx &=\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*}

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Mathematica [C]  time = 0.21, size = 103, normalized size = 0.31 \begin {gather*} \frac {(a+b x)^{4/3} (c+d x)^{2/3} \left (\frac {(-5 a d f-4 b c f+9 b d e) \, _2F_1\left (-\frac {2}{3},\frac {4}{3};\frac {7}{3};\frac {d (a+b x)}{a d-b c}\right )}{\left (\frac {b (c+d x)}{b c-a d}\right )^{2/3}}+4 b f (c+d x)\right )}{12 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.19, size = 465, normalized size = 1.40 \begin {gather*} \frac {(b c-a d)^2 (-5 a d f-4 b c f+9 b d e) \log \left (\sqrt [3]{b}-\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}\right )}{81 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 {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}}+b^{2/3}\right )}{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) \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^{8/3} d^{7/3}}+\frac {\sqrt [3]{a+b x} (b c-a d)^2 \left (-\frac {9 b^2 d^2 e (a+b x)}{c+d x}+\frac {22 b^2 c d f (a+b x)}{c+d x}-10 a b^2 d f-\frac {9 b d^3 e (a+b x)^2}{(c+d x)^2}+\frac {5 a d^3 f (a+b x)^2}{(c+d x)^2}-\frac {13 a b d^2 f (a+b x)}{c+d x}+\frac {4 b c d^2 f (a+b x)^2}{(c+d x)^2}-8 b^3 c f+18 b^3 d e\right )}{54 b^2 d^2 \sqrt [3]{c+d x} \left (b-\frac {d (a+b x)}{c+d x}\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*(a + b*x)^(1/3)*(18*b^3*d*e - 8*b^3*c*f - 10*a*b^2*d*f - (9*b*d^3*e*(a + b*x)^2)/(c + d*x)^2 +
(4*b*c*d^2*f*(a + b*x)^2)/(c + d*x)^2 + (5*a*d^3*f*(a + b*x)^2)/(c + d*x)^2 - (9*b^2*d^2*e*(a + b*x))/(c + d*x
) + (22*b^2*c*d*f*(a + b*x))/(c + d*x) - (13*a*b*d^2*f*(a + b*x))/(c + d*x)))/(54*b^2*d^2*(c + d*x)^(1/3)*(b -
(d*(a + b*x))/(c + d*x))^3) - ((b*c - a*d)^2*(9*b*d*e - 4*b*c*f - 5*a*d*f)*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^(8/3)*d^(7/3)) + ((b*c - a*d)^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)])/(81*b^(8/3)*d^(7/3)) - ((b*c - a*d)^2
*(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))

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fricas [B]  time = 1.37, size = 1196, normalized size = 3.61

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),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)]

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

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),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b x +a \right )^{\frac {1}{3}} \left (d x +c \right )^{\frac {2}{3}} \left (f x +e \right )\, dx \end {gather*}

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),x)

[Out]

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

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

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),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

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),x)

[Out]

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

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