3.31.42 \(\int \frac {(a+b x)^{4/3}}{(c+d x)^{4/3} (e+f x)^3} \, dx\) [3042]

Optimal. Leaf size=434 \[ \frac {3 d (a+b x)^{7/3}}{(b c-a d) (d e-c f) \sqrt [3]{c+d x} (e+f x)^2}-\frac {(6 b d e+b c f-7 a d f) (a+b x)^{4/3} (c+d x)^{2/3}}{2 (b c-a d) (d e-c f)^2 (e+f x)^2}+\frac {2 (6 b d e+b c f-7 a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 (d e-c f)^3 (e+f x)}+\frac {2 (b c-a d) (6 b d e+b c f-7 a d f) \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d e-c f} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} (b e-a f)^{2/3} (d e-c f)^{10/3}}-\frac {(b c-a d) (6 b d e+b c f-7 a d f) \log (e+f x)}{9 (b e-a f)^{2/3} (d e-c f)^{10/3}}+\frac {(b c-a d) (6 b d e+b c f-7 a d f) \log \left (-\sqrt [3]{a+b x}+\frac {\sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt [3]{d e-c f}}\right )}{3 (b e-a f)^{2/3} (d e-c f)^{10/3}} \]

[Out]

3*d*(b*x+a)^(7/3)/(-a*d+b*c)/(-c*f+d*e)/(d*x+c)^(1/3)/(f*x+e)^2-1/2*(-7*a*d*f+b*c*f+6*b*d*e)*(b*x+a)^(4/3)*(d*
x+c)^(2/3)/(-a*d+b*c)/(-c*f+d*e)^2/(f*x+e)^2+2/3*(-7*a*d*f+b*c*f+6*b*d*e)*(b*x+a)^(1/3)*(d*x+c)^(2/3)/(-c*f+d*
e)^3/(f*x+e)-1/9*(-a*d+b*c)*(-7*a*d*f+b*c*f+6*b*d*e)*ln(f*x+e)/(-a*f+b*e)^(2/3)/(-c*f+d*e)^(10/3)+1/3*(-a*d+b*
c)*(-7*a*d*f+b*c*f+6*b*d*e)*ln(-(b*x+a)^(1/3)+(-a*f+b*e)^(1/3)*(d*x+c)^(1/3)/(-c*f+d*e)^(1/3))/(-a*f+b*e)^(2/3
)/(-c*f+d*e)^(10/3)+2/9*(-a*d+b*c)*(-7*a*d*f+b*c*f+6*b*d*e)*arctan(1/3*3^(1/2)+2/3*(-a*f+b*e)^(1/3)*(d*x+c)^(1
/3)/(-c*f+d*e)^(1/3)/(b*x+a)^(1/3)*3^(1/2))/(-a*f+b*e)^(2/3)/(-c*f+d*e)^(10/3)*3^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 434, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {98, 96, 93} \begin {gather*} \frac {2 (b c-a d) (-7 a d f+b c f+6 b d e) \text {ArcTan}\left (\frac {2 \sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt {3} \sqrt [3]{a+b x} \sqrt [3]{d e-c f}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3} (b e-a f)^{2/3} (d e-c f)^{10/3}}+\frac {3 d (a+b x)^{7/3}}{\sqrt [3]{c+d x} (e+f x)^2 (b c-a d) (d e-c f)}-\frac {(a+b x)^{4/3} (c+d x)^{2/3} (-7 a d f+b c f+6 b d e)}{2 (e+f x)^2 (b c-a d) (d e-c f)^2}+\frac {2 \sqrt [3]{a+b x} (c+d x)^{2/3} (-7 a d f+b c f+6 b d e)}{3 (e+f x) (d e-c f)^3}-\frac {(b c-a d) \log (e+f x) (-7 a d f+b c f+6 b d e)}{9 (b e-a f)^{2/3} (d e-c f)^{10/3}}+\frac {(b c-a d) (-7 a d f+b c f+6 b d e) \log \left (\frac {\sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt [3]{d e-c f}}-\sqrt [3]{a+b x}\right )}{3 (b e-a f)^{2/3} (d e-c f)^{10/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d*(a + b*x)^(7/3))/((b*c - a*d)*(d*e - c*f)*(c + d*x)^(1/3)*(e + f*x)^2) - ((6*b*d*e + b*c*f - 7*a*d*f)*(a
+ b*x)^(4/3)*(c + d*x)^(2/3))/(2*(b*c - a*d)*(d*e - c*f)^2*(e + f*x)^2) + (2*(6*b*d*e + b*c*f - 7*a*d*f)*(a +
b*x)^(1/3)*(c + d*x)^(2/3))/(3*(d*e - c*f)^3*(e + f*x)) + (2*(b*c - a*d)*(6*b*d*e + b*c*f - 7*a*d*f)*ArcTan[1/
Sqrt[3] + (2*(b*e - a*f)^(1/3)*(c + d*x)^(1/3))/(Sqrt[3]*(d*e - c*f)^(1/3)*(a + b*x)^(1/3))])/(3*Sqrt[3]*(b*e
- a*f)^(2/3)*(d*e - c*f)^(10/3)) - ((b*c - a*d)*(6*b*d*e + b*c*f - 7*a*d*f)*Log[e + f*x])/(9*(b*e - a*f)^(2/3)
*(d*e - c*f)^(10/3)) + ((b*c - a*d)*(6*b*d*e + b*c*f - 7*a*d*f)*Log[-(a + b*x)^(1/3) + ((b*e - a*f)^(1/3)*(c +
d*x)^(1/3))/(d*e - c*f)^(1/3)])/(3*(b*e - a*f)^(2/3)*(d*e - c*f)^(10/3))

Rule 93

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

Rule 96

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

Rule 98

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)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rubi steps

\begin {align*} \int \frac {(a+b x)^{4/3}}{(c+d x)^{4/3} (e+f x)^3} \, dx &=\frac {3 d (a+b x)^{7/3}}{(b c-a d) (d e-c f) \sqrt [3]{c+d x} (e+f x)^2}-\frac {(6 b d e+b c f-7 a d f) \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x} (e+f x)^3} \, dx}{(b c-a d) (d e-c f)}\\ &=\frac {3 d (a+b x)^{7/3}}{(b c-a d) (d e-c f) \sqrt [3]{c+d x} (e+f x)^2}-\frac {(6 b d e+b c f-7 a d f) (a+b x)^{4/3} (c+d x)^{2/3}}{2 (b c-a d) (d e-c f)^2 (e+f x)^2}+\frac {(2 (6 b d e+b c f-7 a d f)) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^2} \, dx}{3 (d e-c f)^2}\\ &=\frac {3 d (a+b x)^{7/3}}{(b c-a d) (d e-c f) \sqrt [3]{c+d x} (e+f x)^2}-\frac {(6 b d e+b c f-7 a d f) (a+b x)^{4/3} (c+d x)^{2/3}}{2 (b c-a d) (d e-c f)^2 (e+f x)^2}+\frac {2 (6 b d e+b c f-7 a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 (d e-c f)^3 (e+f x)}-\frac {(2 (b c-a d) (6 b d e+b c f-7 a d f)) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)} \, dx}{9 (d e-c f)^3}\\ &=\frac {3 d (a+b x)^{7/3}}{(b c-a d) (d e-c f) \sqrt [3]{c+d x} (e+f x)^2}-\frac {(6 b d e+b c f-7 a d f) (a+b x)^{4/3} (c+d x)^{2/3}}{2 (b c-a d) (d e-c f)^2 (e+f x)^2}+\frac {2 (6 b d e+b c f-7 a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 (d e-c f)^3 (e+f x)}+\frac {2 (b c-a d) (6 b d e+b c f-7 a d f) \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d e-c f} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} (b e-a f)^{2/3} (d e-c f)^{10/3}}-\frac {(b c-a d) (6 b d e+b c f-7 a d f) \log (e+f x)}{9 (b e-a f)^{2/3} (d e-c f)^{10/3}}+\frac {(b c-a d) (6 b d e+b c f-7 a d f) \log \left (-\sqrt [3]{a+b x}+\frac {\sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt [3]{d e-c f}}\right )}{3 (b e-a f)^{2/3} (d e-c f)^{10/3}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 10.30, size = 214, normalized size = 0.49 \begin {gather*} \frac {-6 d (a+b x)^{7/3}+\frac {(6 b d e+b c f-7 a d f) \sqrt [3]{a+b x} \left (3 (b e-a f) (d e-c f) (a+b x) (c+d x)-4 (b c-a d) (e+f x) \left ((b e-a f) (c+d x)-(b c-a d) (e+f x) \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )\right )\right )}{3 (b e-a f) (d e-c f)^2}}{2 (b c-a d) (-d e+c f) \sqrt [3]{c+d x} (e+f x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{\frac {4}{3}}}{\left (d x +c \right )^{\frac {4}{3}} \left (f x +e \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 3287 vs. \(2 (402) = 804\).
time = 3.98, size = 6733, normalized size = 15.51 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/18*(6*sqrt(1/3)*((a*b^2*c^3*d - 8*a^2*b*c^2*d^2 + 7*a^3*c*d^3)*f^5*x^3 + (a*b^2*c^4 - 8*a^2*b*c^3*d + 7*a^
3*c^2*d^2)*f^5*x^2 + 6*(b^3*c^2*d^2 - a*b^2*c*d^3 + (b^3*c*d^3 - a*b^2*d^4)*x)*e^5 + (12*(b^3*c*d^3 - a*b^2*d^
4)*f*x^2 + (7*b^3*c^2*d^2 - 20*a*b^2*c*d^3 + 13*a^2*b*d^4)*f*x - (5*b^3*c^3*d + 8*a*b^2*c^2*d^2 - 13*a^2*b*c*d
^3)*f)*e^4 + (6*(b^3*c*d^3 - a*b^2*d^4)*f^2*x^3 - 2*(2*b^3*c^2*d^2 + 11*a*b^2*c*d^3 - 13*a^2*b*d^4)*f^2*x^2 -
(11*b^3*c^3*d + 3*a*b^2*c^2*d^2 - 21*a^2*b*c*d^3 + 7*a^3*d^4)*f^2*x - (b^3*c^4 - 13*a*b^2*c^3*d + 5*a^2*b*c^2*
d^2 + 7*a^3*c*d^3)*f^2)*e^3 - ((5*b^3*c^2*d^2 + 8*a*b^2*c*d^3 - 13*a^2*b*d^4)*f^3*x^3 + (7*b^3*c^3*d - 18*a*b^
2*c^2*d^2 - 3*a^2*b*c*d^3 + 14*a^3*d^4)*f^3*x^2 + (2*b^3*c^4 - 27*a*b^2*c^3*d + 18*a^2*b*c^2*d^2 + 7*a^3*c*d^3
)*f^3*x - (a*b^2*c^4 - 8*a^2*b*c^3*d + 7*a^3*c^2*d^2)*f^3)*e^2 - ((b^3*c^3*d - 13*a*b^2*c^2*d^2 + 5*a^2*b*c*d^
3 + 7*a^3*d^4)*f^4*x^3 + (b^3*c^4 - 15*a*b^2*c^3*d + 21*a^2*b*c^2*d^2 - 7*a^3*c*d^3)*f^4*x^2 - 2*(a*b^2*c^4 -
8*a^2*b*c^3*d + 7*a^3*c^2*d^2)*f^4*x)*e)*sqrt(-(a^2*c*f^3 - b^2*d*e^3 - (2*a*b*c + a^2*d)*f^2*e + (b^2*c + 2*a
*b*d)*f*e^2)^(1/3)/(c*f - d*e))*log((3*a^2*c*f^2 + (2*a*b*c + a^2*d)*f^2*x - 3*(a^2*c*f^3 - b^2*d*e^3 - (2*a*b
*c + a^2*d)*f^2*e + (b^2*c + 2*a*b*d)*f*e^2)^(1/3)*(a*f - b*e)*(b*x + a)^(1/3)*(d*x + c)^(2/3) + 3*sqrt(1/3)*(
2*(a*c*f^2 + b*d*e^2 - (b*c + a*d)*f*e)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (a^2*c*f^3 - b^2*d*e^3 - (2*a*b*c +
a^2*d)*f^2*e + (b^2*c + 2*a*b*d)*f*e^2)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) - (a^2*c*f^3 - b^2*d*e^3 - (2*a*
b*c + a^2*d)*f^2*e + (b^2*c + 2*a*b*d)*f*e^2)^(1/3)*(a*d*f*x + a*c*f - (b*d*x + b*c)*e))*sqrt(-(a^2*c*f^3 - b^
2*d*e^3 - (2*a*b*c + a^2*d)*f^2*e + (b^2*c + 2*a*b*d)*f*e^2)^(1/3)/(c*f - d*e)) + (3*b^2*d*x + b^2*c + 2*a*b*d
)*e^2 - 2*((b^2*c + 2*a*b*d)*f*x + (2*a*b*c + a^2*d)*f)*e)/(f*x + e)) + 2*((b^2*c^2*d - 8*a*b*c*d^2 + 7*a^2*d^
3)*f^3*x^3 + (b^2*c^3 - 8*a*b*c^2*d + 7*a^2*c*d^2)*f^3*x^2 + 6*(b^2*c^2*d - a*b*c*d^2 + (b^2*c*d^2 - a*b*d^3)*
x)*e^3 + (12*(b^2*c*d^2 - a*b*d^3)*f*x^2 + (13*b^2*c^2*d - 20*a*b*c*d^2 + 7*a^2*d^3)*f*x + (b^2*c^3 - 8*a*b*c^
2*d + 7*a^2*c*d^2)*f)*e^2 + 2*(3*(b^2*c*d^2 - a*b*d^3)*f^2*x^3 + (4*b^2*c^2*d - 11*a*b*c*d^2 + 7*a^2*d^3)*f^2*
x^2 + (b^2*c^3 - 8*a*b*c^2*d + 7*a^2*c*d^2)*f^2*x)*e)*(a^2*c*f^3 - b^2*d*e^3 - (2*a*b*c + a^2*d)*f^2*e + (b^2*
c + 2*a*b*d)*f*e^2)^(2/3)*log(((a*c*f^2 + b*d*e^2 - (b*c + a*d)*f*e)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (a^2*c*
f^3 - b^2*d*e^3 - (2*a*b*c + a^2*d)*f^2*e + (b^2*c + 2*a*b*d)*f*e^2)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) + (
a^2*c*f^3 - b^2*d*e^3 - (2*a*b*c + a^2*d)*f^2*e + (b^2*c + 2*a*b*d)*f*e^2)^(1/3)*(a*d*f*x + a*c*f - (b*d*x + b
*c)*e))/(d*x + c)) - 4*((b^2*c^2*d - 8*a*b*c*d^2 + 7*a^2*d^3)*f^3*x^3 + (b^2*c^3 - 8*a*b*c^2*d + 7*a^2*c*d^2)*
f^3*x^2 + 6*(b^2*c^2*d - a*b*c*d^2 + (b^2*c*d^2 - a*b*d^3)*x)*e^3 + (12*(b^2*c*d^2 - a*b*d^3)*f*x^2 + (13*b^2*
c^2*d - 20*a*b*c*d^2 + 7*a^2*d^3)*f*x + (b^2*c^3 - 8*a*b*c^2*d + 7*a^2*c*d^2)*f)*e^2 + 2*(3*(b^2*c*d^2 - a*b*d
^3)*f^2*x^3 + (4*b^2*c^2*d - 11*a*b*c*d^2 + 7*a^2*d^3)*f^2*x^2 + (b^2*c^3 - 8*a*b*c^2*d + 7*a^2*c*d^2)*f^2*x)*
e)*(a^2*c*f^3 - b^2*d*e^3 - (2*a*b*c + a^2*d)*f^2*e + (b^2*c + 2*a*b*d)*f*e^2)^(2/3)*log(((a*c*f^2 + b*d*e^2 -
(b*c + a*d)*f*e)*(b*x + a)^(1/3)*(d*x + c)^(2/3) - (a^2*c*f^3 - b^2*d*e^3 - (2*a*b*c + a^2*d)*f^2*e + (b^2*c
+ 2*a*b*d)*f*e^2)^(2/3)*(d*x + c))/(d*x + c)) + 3*(3*a^3*c^3*f^5 + (25*a^2*b*c^2*d - 28*a^3*c*d^2)*f^5*x^2 + 7
*(a^2*b*c^3 - a^3*c^2*d)*f^5*x - 6*(b^3*d^3*x + 4*b^3*c*d^2 - 3*a*b^2*d^3)*e^5 - (3*b^3*d^3*f*x^2 + (37*b^3*c*
d^2 - 61*a*b^2*d^3)*f*x - (20*b^3*c^2*d + 43*a*b^2*c*d^2 - 36*a^2*b*d^3)*f)*e^4 - 2*((11*b^3*c*d^2 - 17*a*b^2*
d^3)*f^2*x^2 - 2*(9*b^3*c^2*d + 8*a*b^2*c*d^2 - 26*a^2*b*d^3)*f^2*x - (2*b^3*c^3 - 28*a*b^2*c^2*d - 7*a^2*b*c*
d^2 + 9*a^3*d^3)*f^2)*e^3 + ((25*b^3*c^2*d + 16*a*b^2*c*d^2 - 59*a^2*b*d^3)*f^3*x^2 + (7*b^3*c^3 - 79*a*b^2*c^
2*d + 47*a^2*b*c*d^2 + 49*a^3*d^3)*f^3*x - (5*a*b^2*c^3 - 52*a^2*b*c^2*d + 5*a^3*c*d^2)*f^3)*e^2 - 2*((25*a*b^
2*c^2*d - 17*a^2*b*c*d^2 - 14*a^3*d^3)*f^4*x^2 + (7*a*b^2*c^3 - 25*a^2*b*c^2*d + 21*a^3*c*d^2)*f^4*x + (a^2*b*
c^3 + 8*a^3*c^2*d)*f^4)*e)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(a^2*c^4*d*f^8*x^3 + a^2*c^5*f^8*x^2 + (b^2*d^5*x
+ b^2*c*d^4)*e^8 + 2*(b^2*d^5*f*x^2 - (b^2*c*d^4 + a*b*d^5)*f*x - (2*b^2*c^2*d^3 + a*b*c*d^4)*f)*e^7 + (b^2*d^
5*f^2*x^3 - (7*b^2*c*d^4 + 4*a*b*d^5)*f^2*x^2 - (2*b^2*c^2*d^3 - 4*a*b*c*d^4 - a^2*d^5)*f^2*x + (6*b^2*c^3*d^2
+ 8*a*b*c^2*d^3 + a^2*c*d^4)*f^2)*e^6 - 2*((2*b^2*c*d^4 + a*b*d^5)*f^3*x^3 - (4*b^2*c^2*d^3 + 7*a*b*c*d^4 + a
^2*d^5)*f^3*x^2 - (4*b^2*c^3*d^2 + 2*a*b*c^2*d^3 - a^2*c*d^4)*f^3*x + 2*(b^2*c^4*d + 3*a*b*c^3*d^2 + a^2*c^2*d
^3)*f^3)*e^5 + ((6*b^2*c^2*d^3 + 8*a*b*c*d^4 + a^2*d^5)*f^4*x^3 - (2*b^2*c^3*d^2 + 16*a*b*c^2*d^3 + 7*a^2*c*d^
4)*f^4*x^2 - (7*b^2*c^4*d + 16*a*b*c^3*d^2 + 2*a^2*c^2*d^3)*f^4*x + (b^2*c^5 + 8*a*b*c^4*d + 6*a^2*c^3*d^2)*f^
4)*e^4 - 2*(2*(b^2*c^3*d^2 + 3*a*b*c^2*d^3 + a^2*c*d^4)*f^5*x^3 + (b^2*c^4*d - 2*a*b*c^3*d^2 - 4*a^2*c^2*d^3)*
f^5*x^2 - (b^2*c^5 + 7*a*b*c^4*d + 4*a^2*c^3*d^...

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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