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

Optimal. Leaf size=571 \[ \frac {(b c-a d)^2 \log (a+b x) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right )}{486 b^{11/3} d^{10/3}}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right ) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{162 b^{11/3} d^{10/3}}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right ) \tan ^{-1}\left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{81 \sqrt {3} b^{11/3} d^{10/3}}+\frac {(a+b x)^{4/3} (c+d x)^{2/3} \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right )}{54 b^3 d^2}+\frac {\sqrt [3]{a+b x} (c+d x)^{2/3} (b c-a d) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right )}{81 b^3 d^3}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (-8 a d f-7 b c f+15 b d e)}{36 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d} \]

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Rubi [A]  time = 0.56, antiderivative size = 571, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {90, 80, 50, 59} \begin {gather*} \frac {\sqrt [3]{a+b x} (c+d x)^{2/3} (b c-a d) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right )}{81 b^3 d^3}+\frac {(a+b x)^{4/3} (c+d x)^{2/3} \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right )}{54 b^3 d^2}+\frac {(b c-a d)^2 \log (a+b x) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right )}{486 b^{11/3} d^{10/3}}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 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 )}{162 b^{11/3} d^{10/3}}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (7 c^2 f^2-24 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 )}{81 \sqrt {3} b^{11/3} d^{10/3}}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (-8 a d f-7 b c f+15 b d e)}{36 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)*(10*a^2*d^2*f^2 - 10*a*b*d*f*(3*d*e - c*f) + b^2*(27*d^2*e^2 - 24*c*d*e*f + 7*c^2*f^2))*(a + b*x)
^(1/3)*(c + d*x)^(2/3))/(81*b^3*d^3) + ((10*a^2*d^2*f^2 - 10*a*b*d*f*(3*d*e - c*f) + b^2*(27*d^2*e^2 - 24*c*d*
e*f + 7*c^2*f^2))*(a + b*x)^(4/3)*(c + d*x)^(2/3))/(54*b^3*d^2) + (f*(15*b*d*e - 7*b*c*f - 8*a*d*f)*(a + b*x)^
(4/3)*(c + d*x)^(5/3))/(36*b^2*d^2) + (f*(a + b*x)^(4/3)*(c + d*x)^(5/3)*(e + f*x))/(4*b*d) + ((b*c - a*d)^2*(
10*a^2*d^2*f^2 - 10*a*b*d*f*(3*d*e - c*f) + b^2*(27*d^2*e^2 - 24*c*d*e*f + 7*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))])/(81*Sqrt[3]*b^(11/3)*d^(10/3)) + ((b*c - a*d)^2*(1
0*a^2*d^2*f^2 - 10*a*b*d*f*(3*d*e - c*f) + b^2*(27*d^2*e^2 - 24*c*d*e*f + 7*c^2*f^2))*Log[a + b*x])/(486*b^(11
/3)*d^(10/3)) + ((b*c - a*d)^2*(10*a^2*d^2*f^2 - 10*a*b*d*f*(3*d*e - c*f) + b^2*(27*d^2*e^2 - 24*c*d*e*f + 7*c
^2*f^2))*Log[-1 + (b^(1/3)*(c + d*x)^(1/3))/(d^(1/3)*(a + b*x)^(1/3))])/(162*b^(11/3)*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 \sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)^2 \, dx &=\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d}+\frac {\int \sqrt [3]{a+b x} (c+d x)^{2/3} \left (\frac {1}{3} \left (12 b d e^2-f (4 b c e+5 a d e+3 a c f)\right )+\frac {1}{3} f (15 b d e-7 b c f-8 a d f) x\right ) \, dx}{4 b d}\\ &=\frac {f (15 b d e-7 b c f-8 a d f) (a+b x)^{4/3} (c+d x)^{5/3}}{36 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d}+\frac {\left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \int \sqrt [3]{a+b x} (c+d x)^{2/3} \, dx}{27 b^2 d^2}\\ &=\frac {\left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) (a+b x)^{4/3} (c+d x)^{2/3}}{54 b^3 d^2}+\frac {f (15 b d e-7 b c f-8 a d f) (a+b x)^{4/3} (c+d x)^{5/3}}{36 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d}+\frac {\left ((b c-a d) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right )\right ) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx}{81 b^3 d^2}\\ &=\frac {(b c-a d) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{81 b^3 d^3}+\frac {\left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) (a+b x)^{4/3} (c+d x)^{2/3}}{54 b^3 d^2}+\frac {f (15 b d e-7 b c f-8 a d f) (a+b x)^{4/3} (c+d x)^{5/3}}{36 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d}-\frac {\left ((b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right )\right ) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}} \, dx}{243 b^3 d^3}\\ &=\frac {(b c-a d) \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{81 b^3 d^3}+\frac {\left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) (a+b x)^{4/3} (c+d x)^{2/3}}{54 b^3 d^2}+\frac {f (15 b d e-7 b c f-8 a d f) (a+b x)^{4/3} (c+d x)^{5/3}}{36 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{5/3} (e+f x)}{4 b d}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\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 )}{81 \sqrt {3} b^{11/3} d^{10/3}}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \log (a+b x)}{486 b^{11/3} d^{10/3}}+\frac {(b c-a d)^2 \left (10 a^2 d^2 f^2-10 a b d f (3 d e-c f)+b^2 \left (27 d^2 e^2-24 c d e f+7 c^2 f^2\right )\right ) \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{162 b^{11/3} d^{10/3}}\\ \end {align*}

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Mathematica [C]  time = 0.34, size = 179, normalized size = 0.31 \begin {gather*} \frac {(a+b x)^{4/3} (c+d x)^{2/3} \left (\frac {\left (10 a^2 d^2 f^2+10 a b d f (c f-3 d e)+b^2 \left (7 c^2 f^2-24 c d e f+27 d^2 e^2\right )\right ) \, _2F_1\left (-\frac {2}{3},\frac {4}{3};\frac {7}{3};\frac {d (a+b x)}{a d-b c}\right )}{b^2 d \left (\frac {b (c+d x)}{b c-a d}\right )^{2/3}}+\frac {f (c+d x) (-8 a d f-7 b c f+15 b d e)}{b d}+9 f (c+d x) (e+f x)\right )}{36 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(4/3)*(c + d*x)^(2/3)*((f*(15*b*d*e - 7*b*c*f - 8*a*d*f)*(c + d*x))/(b*d) + 9*f*(c + d*x)*(e + f*x)
+ ((10*a^2*d^2*f^2 + 10*a*b*d*f*(-3*d*e + c*f) + b^2*(27*d^2*e^2 - 24*c*d*e*f + 7*c^2*f^2))*Hypergeometric2F1
[-2/3, 4/3, 7/3, (d*(a + b*x))/(-(b*c) + a*d)])/(b^2*d*((b*(c + d*x))/(b*c - a*d))^(2/3))))/(36*b*d)

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IntegrateAlgebraic [A]  time = 1.90, size = 946, normalized size = 1.66 \begin {gather*} \frac {\sqrt [3]{a+b x} \left (108 d^2 e^2 b^5+28 c^2 f^2 b^5-96 c d e f b^5+40 a c d f^2 b^4-120 a d^2 e f b^4-\frac {162 d^3 e^2 (a+b x) b^4}{c+d x}-\frac {105 c^2 d f^2 (a+b x) b^4}{c+d x}+\frac {360 c d^2 e f (a+b x) b^4}{c+d x}+40 a^2 d^2 f^2 b^3-\frac {150 a c d^2 f^2 (a+b x) b^3}{c+d x}-\frac {36 a d^3 e f (a+b x) b^3}{c+d x}+\frac {144 c^2 d^2 f^2 (a+b x)^2 b^3}{(c+d x)^2}-\frac {216 c d^3 e f (a+b x)^2 b^3}{(c+d x)^2}+\frac {93 a^2 d^3 f^2 (a+b x) b^2}{c+d x}-\frac {72 a c d^3 f^2 (a+b x)^2 b^2}{(c+d x)^2}+\frac {216 a d^4 e f (a+b x)^2 b^2}{(c+d x)^2}+\frac {54 d^5 e^2 (a+b x)^3 b^2}{(c+d x)^3}+\frac {14 c^2 d^3 f^2 (a+b x)^3 b^2}{(c+d x)^3}-\frac {48 c d^4 e f (a+b x)^3 b^2}{(c+d x)^3}-\frac {72 a^2 d^4 f^2 (a+b x)^2 b}{(c+d x)^2}+\frac {20 a c d^4 f^2 (a+b x)^3 b}{(c+d x)^3}-\frac {60 a d^5 e f (a+b x)^3 b}{(c+d x)^3}+\frac {20 a^2 d^5 f^2 (a+b x)^3}{(c+d x)^3}\right ) (b c-a d)^2}{324 b^3 d^3 \sqrt [3]{c+d x} \left (b-\frac {d (a+b x)}{c+d x}\right )^4}-\frac {\left (27 d^2 e^2 b^2+7 c^2 f^2 b^2-24 c d e f b^2+10 a c d f^2 b-30 a d^2 e f b+10 a^2 d^2 f^2\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 ) (b c-a d)^2}{81 \sqrt {3} b^{11/3} d^{10/3}}+\frac {\left (27 d^2 e^2 b^2+7 c^2 f^2 b^2-24 c d e f b^2+10 a c d f^2 b-30 a d^2 e f b+10 a^2 d^2 f^2\right ) \log \left (\sqrt [3]{b}-\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}\right ) (b c-a d)^2}{243 b^{11/3} d^{10/3}}-\frac {\left (27 d^2 e^2 b^2+7 c^2 f^2 b^2-24 c d e f b^2+10 a c d f^2 b-30 a d^2 e f b+10 a^2 d^2 f^2\right ) \log \left (b^{2/3}+\frac {\sqrt [3]{d} \sqrt [3]{a+b x} \sqrt [3]{b}}{\sqrt [3]{c+d x}}+\frac {d^{2/3} (a+b x)^{2/3}}{(c+d x)^{2/3}}\right ) (b c-a d)^2}{486 b^{11/3} d^{10/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*(a + b*x)^(1/3)*(108*b^5*d^2*e^2 - 96*b^5*c*d*e*f - 120*a*b^4*d^2*e*f + 28*b^5*c^2*f^2 + 40*a*b
^4*c*d*f^2 + 40*a^2*b^3*d^2*f^2 + (54*b^2*d^5*e^2*(a + b*x)^3)/(c + d*x)^3 - (48*b^2*c*d^4*e*f*(a + b*x)^3)/(c
+ d*x)^3 - (60*a*b*d^5*e*f*(a + b*x)^3)/(c + d*x)^3 + (14*b^2*c^2*d^3*f^2*(a + b*x)^3)/(c + d*x)^3 + (20*a*b*
c*d^4*f^2*(a + b*x)^3)/(c + d*x)^3 + (20*a^2*d^5*f^2*(a + b*x)^3)/(c + d*x)^3 - (216*b^3*c*d^3*e*f*(a + b*x)^2
)/(c + d*x)^2 + (216*a*b^2*d^4*e*f*(a + b*x)^2)/(c + d*x)^2 + (144*b^3*c^2*d^2*f^2*(a + b*x)^2)/(c + d*x)^2 -
(72*a*b^2*c*d^3*f^2*(a + b*x)^2)/(c + d*x)^2 - (72*a^2*b*d^4*f^2*(a + b*x)^2)/(c + d*x)^2 - (162*b^4*d^3*e^2*(
a + b*x))/(c + d*x) + (360*b^4*c*d^2*e*f*(a + b*x))/(c + d*x) - (36*a*b^3*d^3*e*f*(a + b*x))/(c + d*x) - (105*
b^4*c^2*d*f^2*(a + b*x))/(c + d*x) - (150*a*b^3*c*d^2*f^2*(a + b*x))/(c + d*x) + (93*a^2*b^2*d^3*f^2*(a + b*x)
)/(c + d*x)))/(324*b^3*d^3*(c + d*x)^(1/3)*(b - (d*(a + b*x))/(c + d*x))^4) - ((b*c - a*d)^2*(27*b^2*d^2*e^2 -
24*b^2*c*d*e*f - 30*a*b*d^2*e*f + 7*b^2*c^2*f^2 + 10*a*b*c*d*f^2 + 10*a^2*d^2*f^2)*ArcTan[1/Sqrt[3] + (2*d^(1
/3)*(a + b*x)^(1/3))/(Sqrt[3]*b^(1/3)*(c + d*x)^(1/3))])/(81*Sqrt[3]*b^(11/3)*d^(10/3)) + ((b*c - a*d)^2*(27*b
^2*d^2*e^2 - 24*b^2*c*d*e*f - 30*a*b*d^2*e*f + 7*b^2*c^2*f^2 + 10*a*b*c*d*f^2 + 10*a^2*d^2*f^2)*Log[b^(1/3) -
(d^(1/3)*(a + b*x)^(1/3))/(c + d*x)^(1/3)])/(243*b^(11/3)*d^(10/3)) - ((b*c - a*d)^2*(27*b^2*d^2*e^2 - 24*b^2*
c*d*e*f - 30*a*b*d^2*e*f + 7*b^2*c^2*f^2 + 10*a*b*c*d*f^2 + 10*a^2*d^2*f^2)*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)])/(486*b^(11/3)*d^(10/3))

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fricas [A]  time = 2.00, size = 1857, normalized size = 3.25

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)^2,x, algorithm="fricas")

[Out]

[1/972*(6*sqrt(1/3)*(27*(b^5*c^2*d^3 - 2*a*b^4*c*d^4 + a^2*b^3*d^5)*e^2 - 6*(4*b^5*c^3*d^2 - 3*a*b^4*c^2*d^3 -
6*a^2*b^3*c*d^4 + 5*a^3*b^2*d^5)*e*f + (7*b^5*c^4*d - 4*a*b^4*c^3*d^2 - 3*a^2*b^3*c^2*d^3 - 10*a^3*b^2*c*d^4
+ 10*a^4*b*d^5)*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)) - 2*(b^2*d)^(2/3)*(27*(b^4*c^2*d^2 - 2*a*b^3*c
*d^3 + a^2*b^2*d^4)*e^2 - 6*(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)*e*f + (7*b^4*c^4 -
4*a*b^3*c^3*d - 3*a^2*b^2*c^2*d^2 - 10*a^3*b*c*d^3 + 10*a^4*d^4)*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)) + 4*(b^2*d)^(2/3)*
(27*(b^4*c^2*d^2 - 2*a*b^3*c*d^3 + a^2*b^2*d^4)*e^2 - 6*(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)*e*f + (7*b^4*c^4 - 4*a*b^3*c^3*d - 3*a^2*b^2*c^2*d^2 - 10*a^3*b*c*d^3 + 10*a^4*d^4)*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*(81*b^5*d^4*f^2*x^3 + 54*(2*b^5*c*d^3
+ a*b^4*d^4)*e^2 - 12*(8*b^5*c^2*d^2 - 4*a*b^4*c*d^3 + 5*a^2*b^3*d^4)*e*f + (28*b^5*c^3*d - 9*a*b^4*c^2*d^2 -
12*a^2*b^3*c*d^3 + 20*a^3*b^2*d^4)*f^2 + 9*(24*b^5*d^4*e*f + (2*b^5*c*d^3 + a*b^4*d^4)*f^2)*x^2 + 3*(54*b^5*d^
4*e^2 + 12*(2*b^5*c*d^3 + a*b^4*d^4)*e*f - (7*b^5*c^2*d^2 - 2*a*b^4*c*d^3 + 4*a^2*b^3*d^4)*f^2)*x)*(b*x + a)^(
1/3)*(d*x + c)^(2/3))/(b^5*d^4), -1/972*(12*sqrt(1/3)*(27*(b^5*c^2*d^3 - 2*a*b^4*c*d^4 + a^2*b^3*d^5)*e^2 - 6*
(4*b^5*c^3*d^2 - 3*a*b^4*c^2*d^3 - 6*a^2*b^3*c*d^4 + 5*a^3*b^2*d^5)*e*f + (7*b^5*c^4*d - 4*a*b^4*c^3*d^2 - 3*a
^2*b^3*c^2*d^3 - 10*a^3*b^2*c*d^4 + 10*a^4*b*d^5)*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)) + 2*(
b^2*d)^(2/3)*(27*(b^4*c^2*d^2 - 2*a*b^3*c*d^3 + a^2*b^2*d^4)*e^2 - 6*(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)*e*f + (7*b^4*c^4 - 4*a*b^3*c^3*d - 3*a^2*b^2*c^2*d^2 - 10*a^3*b*c*d^3 + 10*a^4*d^4)*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)) - 4*(b^2*d)^(2/3)*(27*(b^4*c^2*d^2 - 2*a*b^3*c*d^3 + a^2*b^2*d^4)*e^2 - 6*(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)*e*f + (7*b^4*c^4 - 4*a*b^3*c^3*d - 3*a^2*b^2*c^2*d^2 - 10*a
^3*b*c*d^3 + 10*a^4*d^4)*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*(81*b^5*d^4*f^2*x^3 + 54*(2*b^5*c*d^3 + a*b^4*d^4)*e^2 - 12*(8*b^5*c^2*d^2 - 4*a*b^4*c*d^3 + 5*a^2*b^3*d^4)
*e*f + (28*b^5*c^3*d - 9*a*b^4*c^2*d^2 - 12*a^2*b^3*c*d^3 + 20*a^3*b^2*d^4)*f^2 + 9*(24*b^5*d^4*e*f + (2*b^5*c
*d^3 + a*b^4*d^4)*f^2)*x^2 + 3*(54*b^5*d^4*e^2 + 12*(2*b^5*c*d^3 + a*b^4*d^4)*e*f - (7*b^5*c^2*d^2 - 2*a*b^4*c
*d^3 + 4*a^2*b^3*d^4)*f^2)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^5*d^4)]

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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 )}^{2}\,{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)^2,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.06, 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 )^{2}\, 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)^2,x)

[Out]

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

[Out]

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

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

[Out]

int((e + f*x)^2*(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 )^{2}\, 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)**2,x)

[Out]

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

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