3.26.22 \(\int \frac {(a+b x)^{4/3} (e+f x)^2}{(c+d x)^{4/3}} \, dx\)

Optimal. Leaf size=562 \[ -\frac {4 \sqrt [3]{a+b x} (c+d x)^{2/3} \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-\left (b^2 \left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right )\right )}{27 b d^4}+\frac {(a+b x)^{4/3} (c+d x)^{2/3} \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-\left (b^2 \left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right )\right )}{9 b d^3 (b c-a d)}-\frac {2 (b c-a d) \log (a+b x) \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-\left (b^2 \left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right )\right )}{81 b^{5/3} d^{13/3}}-\frac {2 (b c-a d) \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-\left (b^2 \left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right )\right ) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{27 b^{5/3} d^{13/3}}-\frac {4 (b c-a d) \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-\left (b^2 \left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right )\right ) \tan ^{-1}\left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{27 \sqrt {3} b^{5/3} d^{13/3}}+\frac {3 (a+b x)^{7/3} (d e-c f)^2}{d^2 \sqrt [3]{c+d x} (b c-a d)}+\frac {f^2 (a+b x)^{7/3} (c+d x)^{2/3}}{3 b d^2} \]

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Rubi [A]  time = 0.53, antiderivative size = 562, 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 = {89, 80, 50, 59} \begin {gather*} \frac {(a+b x)^{4/3} (c+d x)^{2/3} \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)+b^2 \left (-\left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right )\right )}{9 b d^3 (b c-a d)}-\frac {4 \sqrt [3]{a+b x} (c+d x)^{2/3} \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)+b^2 \left (-\left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right )\right )}{27 b d^4}-\frac {2 (b c-a d) \log (a+b x) \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)+b^2 \left (-\left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right )\right )}{81 b^{5/3} d^{13/3}}-\frac {2 (b c-a d) \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)+b^2 \left (-\left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right )\right ) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{27 b^{5/3} d^{13/3}}-\frac {4 (b c-a d) \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)+b^2 \left (-\left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\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^{5/3} d^{13/3}}+\frac {3 (a+b x)^{7/3} (d e-c f)^2}{d^2 \sqrt [3]{c+d x} (b c-a d)}+\frac {f^2 (a+b x)^{7/3} (c+d x)^{2/3}}{3 b d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(4/3)*(e + f*x)^2)/(c + d*x)^(4/3),x]

[Out]

(3*(d*e - c*f)^2*(a + b*x)^(7/3))/(d^2*(b*c - a*d)*(c + d*x)^(1/3)) - (4*(a^2*d^2*f^2 - a*b*d*f*(9*d*e - 7*c*f
) - b^2*(27*d^2*e^2 - 63*c*d*e*f + 35*c^2*f^2))*(a + b*x)^(1/3)*(c + d*x)^(2/3))/(27*b*d^4) + ((a^2*d^2*f^2 -
a*b*d*f*(9*d*e - 7*c*f) - b^2*(27*d^2*e^2 - 63*c*d*e*f + 35*c^2*f^2))*(a + b*x)^(4/3)*(c + d*x)^(2/3))/(9*b*d^
3*(b*c - a*d)) + (f^2*(a + b*x)^(7/3)*(c + d*x)^(2/3))/(3*b*d^2) - (4*(b*c - a*d)*(a^2*d^2*f^2 - a*b*d*f*(9*d*
e - 7*c*f) - b^2*(27*d^2*e^2 - 63*c*d*e*f + 35*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^(5/3)*d^(13/3)) - (2*(b*c - a*d)*(a^2*d^2*f^2 - a*b*d*f*(9*d*e - 7
*c*f) - b^2*(27*d^2*e^2 - 63*c*d*e*f + 35*c^2*f^2))*Log[a + b*x])/(81*b^(5/3)*d^(13/3)) - (2*(b*c - a*d)*(a^2*
d^2*f^2 - a*b*d*f*(9*d*e - 7*c*f) - b^2*(27*d^2*e^2 - 63*c*d*e*f + 35*c^2*f^2))*Log[-1 + (b^(1/3)*(c + d*x)^(1
/3))/(d^(1/3)*(a + b*x)^(1/3))])/(27*b^(5/3)*d^(13/3))

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rubi steps

\begin {align*} \int \frac {(a+b x)^{4/3} (e+f x)^2}{(c+d x)^{4/3}} \, dx &=\frac {3 (d e-c f)^2 (a+b x)^{7/3}}{d^2 (b c-a d) \sqrt [3]{c+d x}}-\frac {3 \int \frac {(a+b x)^{4/3} \left (\frac {1}{3} \left (a d f (2 d e-c f)+b \left (6 d^2 e^2-14 c d e f+7 c^2 f^2\right )\right )-\frac {1}{3} d (b c-a d) f^2 x\right )}{\sqrt [3]{c+d x}} \, dx}{d^2 (b c-a d)}\\ &=\frac {3 (d e-c f)^2 (a+b x)^{7/3}}{d^2 (b c-a d) \sqrt [3]{c+d x}}+\frac {f^2 (a+b x)^{7/3} (c+d x)^{2/3}}{3 b d^2}+\frac {\left (2 \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right )\right ) \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx}{9 b d^2 (b c-a d)}\\ &=\frac {3 (d e-c f)^2 (a+b x)^{7/3}}{d^2 (b c-a d) \sqrt [3]{c+d x}}+\frac {\left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right ) (a+b x)^{4/3} (c+d x)^{2/3}}{9 b d^3 (b c-a d)}+\frac {f^2 (a+b x)^{7/3} (c+d x)^{2/3}}{3 b d^2}-\frac {\left (4 \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right )\right ) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx}{27 b d^3}\\ &=\frac {3 (d e-c f)^2 (a+b x)^{7/3}}{d^2 (b c-a d) \sqrt [3]{c+d x}}-\frac {4 \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{27 b d^4}+\frac {\left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right ) (a+b x)^{4/3} (c+d x)^{2/3}}{9 b d^3 (b c-a d)}+\frac {f^2 (a+b x)^{7/3} (c+d x)^{2/3}}{3 b d^2}+\frac {\left (4 (b c-a d) \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right )\right ) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}} \, dx}{81 b d^4}\\ &=\frac {3 (d e-c f)^2 (a+b x)^{7/3}}{d^2 (b c-a d) \sqrt [3]{c+d x}}-\frac {4 \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{27 b d^4}+\frac {\left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right ) (a+b x)^{4/3} (c+d x)^{2/3}}{9 b d^3 (b c-a d)}+\frac {f^2 (a+b x)^{7/3} (c+d x)^{2/3}}{3 b d^2}-\frac {4 (b c-a d) \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 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 )}{27 \sqrt {3} b^{5/3} d^{13/3}}-\frac {2 (b c-a d) \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right ) \log (a+b x)}{81 b^{5/3} d^{13/3}}-\frac {2 (b c-a d) \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 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 )}{27 b^{5/3} d^{13/3}}\\ \end {align*}

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Mathematica [C]  time = 0.23, size = 175, normalized size = 0.31 \begin {gather*} \frac {(a+b x)^{7/3} \left (\frac {2 \sqrt [3]{\frac {b (c+d x)}{b c-a d}} \left (-a^2 d^2 f^2+a b d f (9 d e-7 c f)+b^2 \left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right ) \, _2F_1\left (\frac {1}{3},\frac {7}{3};\frac {10}{3};\frac {d (a+b x)}{a d-b c}\right )}{b^2}-\frac {7 f^2 (c+d x) (b c-a d)}{b}-63 (d e-c f)^2\right )}{21 d^2 \sqrt [3]{c+d x} (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(4/3)*(e + f*x)^2)/(c + d*x)^(4/3),x]

[Out]

((a + b*x)^(7/3)*(-63*(d*e - c*f)^2 - (7*(b*c - a*d)*f^2*(c + d*x))/b + (2*(-(a^2*d^2*f^2) + a*b*d*f*(9*d*e -
7*c*f) + b^2*(27*d^2*e^2 - 63*c*d*e*f + 35*c^2*f^2))*((b*(c + d*x))/(b*c - a*d))^(1/3)*Hypergeometric2F1[1/3,
7/3, 10/3, (d*(a + b*x))/(-(b*c) + a*d)])/b^2))/(21*d^2*(-(b*c) + a*d)*(c + d*x)^(1/3))

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IntegrateAlgebraic [F]  time = 62.76, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x)^{4/3} (e+f x)^2}{(c+d x)^{4/3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((a + b*x)^(4/3)*(e + f*x)^2)/(c + d*x)^(4/3),x]

[Out]

Defer[IntegrateAlgebraic][((a + b*x)^(4/3)*(e + f*x)^2)/(c + d*x)^(4/3), x]

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fricas [B]  time = 2.04, size = 2267, normalized size = 4.03

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(4/3)*(f*x+e)^2/(d*x+c)^(4/3),x, algorithm="fricas")

[Out]

[1/81*(6*sqrt(1/3)*(27*(b^4*c^2*d^3 - a*b^3*c*d^4)*e^2 - 9*(7*b^4*c^3*d^2 - 8*a*b^3*c^2*d^3 + a^2*b^2*c*d^4)*e
*f + (35*b^4*c^4*d - 42*a*b^3*c^3*d^2 + 6*a^2*b^2*c^2*d^3 + a^3*b*c*d^4)*f^2 + (27*(b^4*c*d^4 - a*b^3*d^5)*e^2
- 9*(7*b^4*c^2*d^3 - 8*a*b^3*c*d^4 + a^2*b^2*d^5)*e*f + (35*b^4*c^3*d^2 - 42*a*b^3*c^2*d^3 + 6*a^2*b^2*c*d^4
+ a^3*b*d^5)*f^2)*x)*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^3*c^2*d^2 - a*b^2*c*d
^3)*e^2 - 9*(7*b^3*c^3*d - 8*a*b^2*c^2*d^2 + a^2*b*c*d^3)*e*f + (35*b^3*c^4 - 42*a*b^2*c^3*d + 6*a^2*b*c^2*d^2
+ a^3*c*d^3)*f^2 + (27*(b^3*c*d^3 - a*b^2*d^4)*e^2 - 9*(7*b^3*c^2*d^2 - 8*a*b^2*c*d^3 + a^2*b*d^4)*e*f + (35*
b^3*c^3*d - 42*a*b^2*c^2*d^2 + 6*a^2*b*c*d^3 + a^3*d^4)*f^2)*x)*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^
3*c^2*d^2 - a*b^2*c*d^3)*e^2 - 9*(7*b^3*c^3*d - 8*a*b^2*c^2*d^2 + a^2*b*c*d^3)*e*f + (35*b^3*c^4 - 42*a*b^2*c^
3*d + 6*a^2*b*c^2*d^2 + a^3*c*d^3)*f^2 + (27*(b^3*c*d^3 - a*b^2*d^4)*e^2 - 9*(7*b^3*c^2*d^2 - 8*a*b^2*c*d^3 +
a^2*b*d^4)*e*f + (35*b^3*c^3*d - 42*a*b^2*c^2*d^2 + 6*a^2*b*c*d^3 + a^3*d^4)*f^2)*x)*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*(9*b^4*d^4*f^2*x^3 + 27*(4*b^4*c*d^3 - 3*a*b^3*d^4)*
e^2 - 9*(28*b^4*c^2*d^2 - 25*a*b^3*c*d^3)*e*f + (140*b^4*c^3*d - 133*a*b^3*c^2*d^2 + 2*a^2*b^2*c*d^3)*f^2 + 3*
(9*b^4*d^4*e*f - 5*(b^4*c*d^3 - a*b^3*d^4)*f^2)*x^2 + (27*b^4*d^4*e^2 - 63*(b^4*c*d^3 - a*b^3*d^4)*e*f + (35*b
^4*c^2*d^2 - 37*a*b^3*c*d^3 + 2*a^2*b^2*d^4)*f^2)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^3*d^6*x + b^3*c*d^5),
-1/81*(12*sqrt(1/3)*(27*(b^4*c^2*d^3 - a*b^3*c*d^4)*e^2 - 9*(7*b^4*c^3*d^2 - 8*a*b^3*c^2*d^3 + a^2*b^2*c*d^4)
*e*f + (35*b^4*c^4*d - 42*a*b^3*c^3*d^2 + 6*a^2*b^2*c^2*d^3 + a^3*b*c*d^4)*f^2 + (27*(b^4*c*d^4 - a*b^3*d^5)*e
^2 - 9*(7*b^4*c^2*d^3 - 8*a*b^3*c*d^4 + a^2*b^2*d^5)*e*f + (35*b^4*c^3*d^2 - 42*a*b^3*c^2*d^3 + 6*a^2*b^2*c*d^
4 + a^3*b*d^5)*f^2)*x)*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^3*c^2*d^2 -
a*b^2*c*d^3)*e^2 - 9*(7*b^3*c^3*d - 8*a*b^2*c^2*d^2 + a^2*b*c*d^3)*e*f + (35*b^3*c^4 - 42*a*b^2*c^3*d + 6*a^2*
b*c^2*d^2 + a^3*c*d^3)*f^2 + (27*(b^3*c*d^3 - a*b^2*d^4)*e^2 - 9*(7*b^3*c^2*d^2 - 8*a*b^2*c*d^3 + a^2*b*d^4)*e
*f + (35*b^3*c^3*d - 42*a*b^2*c^2*d^2 + 6*a^2*b*c*d^3 + a^3*d^4)*f^2)*x)*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^3*c^2*d^2 - a*b^2*c*d^3)*e^2 - 9*(7*b^3*c^3*d - 8*a*b^2*c^2*d^2 + a^2*b*c*d^3)*e*f + (35*b^3*c^4 - 42
*a*b^2*c^3*d + 6*a^2*b*c^2*d^2 + a^3*c*d^3)*f^2 + (27*(b^3*c*d^3 - a*b^2*d^4)*e^2 - 9*(7*b^3*c^2*d^2 - 8*a*b^2
*c*d^3 + a^2*b*d^4)*e*f + (35*b^3*c^3*d - 42*a*b^2*c^2*d^2 + 6*a^2*b*c*d^3 + a^3*d^4)*f^2)*x)*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*(9*b^4*d^4*f^2*x^3 + 27*(4*b^4*c*d^3 - 3*a*
b^3*d^4)*e^2 - 9*(28*b^4*c^2*d^2 - 25*a*b^3*c*d^3)*e*f + (140*b^4*c^3*d - 133*a*b^3*c^2*d^2 + 2*a^2*b^2*c*d^3)
*f^2 + 3*(9*b^4*d^4*e*f - 5*(b^4*c*d^3 - a*b^3*d^4)*f^2)*x^2 + (27*b^4*d^4*e^2 - 63*(b^4*c*d^3 - a*b^3*d^4)*e*
f + (35*b^4*c^2*d^2 - 37*a*b^3*c*d^3 + 2*a^2*b^2*d^4)*f^2)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^3*d^6*x + b^
3*c*d^5)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x + a\right )}^{\frac {4}{3}} {\left (f x + e\right )}^{2}}{{\left (d x + c\right )}^{\frac {4}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(4/3)*(f*x+e)^2/(d*x+c)^(4/3),x, algorithm="giac")

[Out]

integrate((b*x + a)^(4/3)*(f*x + e)^2/(d*x + c)^(4/3), x)

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b x +a \right )^{\frac {4}{3}} \left (f x +e \right )^{2}}{\left (d x +c \right )^{\frac {4}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(4/3)*(f*x+e)^2/(d*x+c)^(4/3),x)

[Out]

int((b*x+a)^(4/3)*(f*x+e)^2/(d*x+c)^(4/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x + a\right )}^{\frac {4}{3}} {\left (f x + e\right )}^{2}}{{\left (d x + c\right )}^{\frac {4}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(4/3)*(f*x+e)^2/(d*x+c)^(4/3),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(4/3)*(f*x + e)^2/(d*x + c)^(4/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e+f\,x\right )}^2\,{\left (a+b\,x\right )}^{4/3}}{{\left (c+d\,x\right )}^{4/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e + f*x)^2*(a + b*x)^(4/3))/(c + d*x)^(4/3),x)

[Out]

int(((e + f*x)^2*(a + b*x)^(4/3))/(c + d*x)^(4/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right )^{\frac {4}{3}} \left (e + f x\right )^{2}}{\left (c + d x\right )^{\frac {4}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(4/3)*(f*x+e)**2/(d*x+c)**(4/3),x)

[Out]

Integral((a + b*x)**(4/3)*(e + f*x)**2/(c + d*x)**(4/3), x)

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