3.3020 \(\int \frac {1}{\sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)^2} \, dx\)
Optimal. Leaf size=293 \[ -\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x}}{(e+f x) (b e-a f) (d e-c f)}+\frac {\log (e+f x) (-2 a d f-b c f+3 b d e)}{6 (b e-a f)^{4/3} (d e-c f)^{5/3}}-\frac {(-2 a d f-b c f+3 b d e) \log \left (\frac {\sqrt [3]{a+b x} \sqrt [3]{d e-c f}}{\sqrt [3]{b e-a f}}-\sqrt [3]{c+d x}\right )}{2 (b e-a f)^{4/3} (d e-c f)^{5/3}}-\frac {(-2 a d f-b c f+3 b d e) \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x} \sqrt [3]{d e-c f}}{\sqrt {3} \sqrt [3]{c+d x} \sqrt [3]{b e-a f}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} (b e-a f)^{4/3} (d e-c f)^{5/3}} \]
[Out]
-f*(b*x+a)^(2/3)*(d*x+c)^(1/3)/(-a*f+b*e)/(-c*f+d*e)/(f*x+e)+1/6*(-2*a*d*f-b*c*f+3*b*d*e)*ln(f*x+e)/(-a*f+b*e)
^(4/3)/(-c*f+d*e)^(5/3)-1/2*(-2*a*d*f-b*c*f+3*b*d*e)*ln((-c*f+d*e)^(1/3)*(b*x+a)^(1/3)/(-a*f+b*e)^(1/3)-(d*x+c
)^(1/3))/(-a*f+b*e)^(4/3)/(-c*f+d*e)^(5/3)-1/3*(-2*a*d*f-b*c*f+3*b*d*e)*arctan(1/3*3^(1/2)+2/3*(-c*f+d*e)^(1/3
)*(b*x+a)^(1/3)/(-a*f+b*e)^(1/3)/(d*x+c)^(1/3)*3^(1/2))/(-a*f+b*e)^(4/3)/(-c*f+d*e)^(5/3)*3^(1/2)
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Rubi [A] time = 0.17, antiderivative size = 293, normalized size of antiderivative = 1.00,
number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used =
{96, 91} \[ -\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x}}{(e+f x) (b e-a f) (d e-c f)}+\frac {\log (e+f x) (-2 a d f-b c f+3 b d e)}{6 (b e-a f)^{4/3} (d e-c f)^{5/3}}-\frac {(-2 a d f-b c f+3 b d e) \log \left (\frac {\sqrt [3]{a+b x} \sqrt [3]{d e-c f}}{\sqrt [3]{b e-a f}}-\sqrt [3]{c+d x}\right )}{2 (b e-a f)^{4/3} (d e-c f)^{5/3}}-\frac {(-2 a d f-b c f+3 b d e) \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x} \sqrt [3]{d e-c f}}{\sqrt {3} \sqrt [3]{c+d x} \sqrt [3]{b e-a f}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} (b e-a f)^{4/3} (d e-c f)^{5/3}} \]
Antiderivative was successfully verified.
[In]
Int[1/((a + b*x)^(1/3)*(c + d*x)^(2/3)*(e + f*x)^2),x]
[Out]
-((f*(a + b*x)^(2/3)*(c + d*x)^(1/3))/((b*e - a*f)*(d*e - c*f)*(e + f*x))) - ((3*b*d*e - b*c*f - 2*a*d*f)*ArcT
an[1/Sqrt[3] + (2*(d*e - c*f)^(1/3)*(a + b*x)^(1/3))/(Sqrt[3]*(b*e - a*f)^(1/3)*(c + d*x)^(1/3))])/(Sqrt[3]*(b
*e - a*f)^(4/3)*(d*e - c*f)^(5/3)) + ((3*b*d*e - b*c*f - 2*a*d*f)*Log[e + f*x])/(6*(b*e - a*f)^(4/3)*(d*e - c*
f)^(5/3)) - ((3*b*d*e - b*c*f - 2*a*d*f)*Log[((d*e - c*f)^(1/3)*(a + b*x)^(1/3))/(b*e - a*f)^(1/3) - (c + d*x)
^(1/3)])/(2*(b*e - a*f)^(4/3)*(d*e - c*f)^(5/3))
Rule 91
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[(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 {1}{\sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)^2} \, dx &=-\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x}}{(b e-a f) (d e-c f) (e+f x)}+\frac {(3 b d e-b c f-2 a d f) \int \frac {1}{\sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)} \, dx}{3 (b e-a f) (d e-c f)}\\ &=-\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x}}{(b e-a f) (d e-c f) (e+f x)}-\frac {(3 b d e-b c f-2 a d f) \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{d e-c f} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b e-a f} \sqrt [3]{c+d x}}\right )}{\sqrt {3} (b e-a f)^{4/3} (d e-c f)^{5/3}}+\frac {(3 b d e-b c f-2 a d f) \log (e+f x)}{6 (b e-a f)^{4/3} (d e-c f)^{5/3}}-\frac {(3 b d e-b c f-2 a d f) \log \left (\frac {\sqrt [3]{d e-c f} \sqrt [3]{a+b x}}{\sqrt [3]{b e-a f}}-\sqrt [3]{c+d x}\right )}{2 (b e-a f)^{4/3} (d e-c f)^{5/3}}\\ \end {align*}
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Mathematica [C] time = 0.05, size = 124, normalized size = 0.42 \[ \frac {(a+b x)^{2/3} \left (\frac {(-2 a d f-b c f+3 b d e) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{b e-a f}-\frac {2 f (c+d x)}{e+f x}\right )}{2 (c+d x)^{2/3} (b e-a f) (d e-c f)} \]
Antiderivative was successfully verified.
[In]
Integrate[1/((a + b*x)^(1/3)*(c + d*x)^(2/3)*(e + f*x)^2),x]
[Out]
((a + b*x)^(2/3)*((-2*f*(c + d*x))/(e + f*x) + ((3*b*d*e - b*c*f - 2*a*d*f)*Hypergeometric2F1[2/3, 1, 5/3, ((d
*e - c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))])/(b*e - a*f)))/(2*(b*e - a*f)*(d*e - c*f)*(c + d*x)^(2/3))
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fricas [B] time = 2.30, size = 2859, normalized size = 9.76 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*x+a)^(1/3)/(d*x+c)^(2/3)/(f*x+e)^2,x, algorithm="fricas")
[Out]
[-1/6*(3*sqrt(1/3)*(3*b^2*d^2*e^4 - (4*b^2*c*d + 5*a*b*d^2)*e^3*f + (b^2*c^2 + 6*a*b*c*d + 2*a^2*d^2)*e^2*f^2
- (a*b*c^2 + 2*a^2*c*d)*e*f^3 + (3*b^2*d^2*e^3*f - (4*b^2*c*d + 5*a*b*d^2)*e^2*f^2 + (b^2*c^2 + 6*a*b*c*d + 2*
a^2*d^2)*e*f^3 - (a*b*c^2 + 2*a^2*c*d)*f^4)*x)*sqrt((-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2
+ 2*a*c*d)*e*f^2)^(1/3)/(b*e - a*f))*log((3*a*c^2*f^2 + (2*b*c*d + a*d^2)*e^2 - 2*(b*c^2 + 2*a*c*d)*e*f + 3*(
-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^(1/3)*(d*e - c*f)*(b*x + a)^(2/3)*
(d*x + c)^(1/3) + (3*b*d^2*e^2 - 2*(2*b*c*d + a*d^2)*e*f + (b*c^2 + 2*a*c*d)*f^2)*x - 3*sqrt(1/3)*(2*(b*d*e^2
+ a*c*f^2 - (b*c + a*d)*e*f)*(b*x + a)^(1/3)*(d*x + c)^(2/3) - (-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2
*f - (b*c^2 + 2*a*c*d)*e*f^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d
^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^(1/3)*(a*d*e - a*c*f + (b*d*e - b*c*f)*x))*sqrt((-b*d^2*e^3 + a*c^2*f^3 +
(2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^(1/3)/(b*e - a*f)))/(f*x + e)) + 2*(-b*d^2*e^3 + a*c^2*f^3
+ (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^(2/3)*(3*b*d*e^2 - (b*c + 2*a*d)*e*f + (3*b*d*e*f - (b*c
+ 2*a*d)*f^2)*x)*log(((b*d*e^2 + a*c*f^2 - (b*c + a*d)*e*f)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b*d^2*e^3 + a
*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^(2/3)*(b*x + a))/(b*x + a)) - (-b*d^2*e^3 + a*c^
2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^(2/3)*(3*b*d*e^2 - (b*c + 2*a*d)*e*f + (3*b*d*e*f -
(b*c + 2*a*d)*f^2)*x)*log(((b*d*e^2 + a*c*f^2 - (b*c + a*d)*e*f)*(b*x + a)^(1/3)*(d*x + c)^(2/3) + (-b*d^2*e^
3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b
*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^(1/3)*(a*d*e - a*c*f + (b*d*e - b*c*
f)*x))/(b*x + a)) + 6*(b*d^2*e^3*f - a*c^2*f^4 - (2*b*c*d + a*d^2)*e^2*f^2 + (b*c^2 + 2*a*c*d)*e*f^3)*(b*x + a
)^(2/3)*(d*x + c)^(1/3))/(b^2*d^3*e^6 - a^2*c^3*e*f^5 - (3*b^2*c*d^2 + 2*a*b*d^3)*e^5*f + (3*b^2*c^2*d + 6*a*b
*c*d^2 + a^2*d^3)*e^4*f^2 - (b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*e^3*f^3 + (2*a*b*c^3 + 3*a^2*c^2*d)*e^2*f^4
+ (b^2*d^3*e^5*f - a^2*c^3*f^6 - (3*b^2*c*d^2 + 2*a*b*d^3)*e^4*f^2 + (3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*e^3
*f^3 - (b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*e^2*f^4 + (2*a*b*c^3 + 3*a^2*c^2*d)*e*f^5)*x), 1/6*(6*sqrt(1/3)*(
3*b^2*d^2*e^4 - (4*b^2*c*d + 5*a*b*d^2)*e^3*f + (b^2*c^2 + 6*a*b*c*d + 2*a^2*d^2)*e^2*f^2 - (a*b*c^2 + 2*a^2*c
*d)*e*f^3 + (3*b^2*d^2*e^3*f - (4*b^2*c*d + 5*a*b*d^2)*e^2*f^2 + (b^2*c^2 + 6*a*b*c*d + 2*a^2*d^2)*e*f^3 - (a*
b*c^2 + 2*a^2*c*d)*f^4)*x)*sqrt(-(-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^
(1/3)/(b*e - a*f))*arctan(sqrt(1/3)*(2*(-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e
*f^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a
*c*d)*e*f^2)^(1/3)*(a*d*e - a*c*f + (b*d*e - b*c*f)*x))*sqrt(-(-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*
f - (b*c^2 + 2*a*c*d)*e*f^2)^(1/3)/(b*e - a*f))/(a*d^2*e^2 - 2*a*c*d*e*f + a*c^2*f^2 + (b*d^2*e^2 - 2*b*c*d*e*
f + b*c^2*f^2)*x)) - 2*(-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^(2/3)*(3*b
*d*e^2 - (b*c + 2*a*d)*e*f + (3*b*d*e*f - (b*c + 2*a*d)*f^2)*x)*log(((b*d*e^2 + a*c*f^2 - (b*c + a*d)*e*f)*(b*
x + a)^(2/3)*(d*x + c)^(1/3) - (-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^(2
/3)*(b*x + a))/(b*x + a)) + (-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^(2/3)
*(3*b*d*e^2 - (b*c + 2*a*d)*e*f + (3*b*d*e*f - (b*c + 2*a*d)*f^2)*x)*log(((b*d*e^2 + a*c*f^2 - (b*c + a*d)*e*f
)*(b*x + a)^(1/3)*(d*x + c)^(2/3) + (-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^
2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*
d)*e*f^2)^(1/3)*(a*d*e - a*c*f + (b*d*e - b*c*f)*x))/(b*x + a)) - 6*(b*d^2*e^3*f - a*c^2*f^4 - (2*b*c*d + a*d^
2)*e^2*f^2 + (b*c^2 + 2*a*c*d)*e*f^3)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b^2*d^3*e^6 - a^2*c^3*e*f^5 - (3*b^2*c
*d^2 + 2*a*b*d^3)*e^5*f + (3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*e^4*f^2 - (b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2
)*e^3*f^3 + (2*a*b*c^3 + 3*a^2*c^2*d)*e^2*f^4 + (b^2*d^3*e^5*f - a^2*c^3*f^6 - (3*b^2*c*d^2 + 2*a*b*d^3)*e^4*f
^2 + (3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*e^3*f^3 - (b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*e^2*f^4 + (2*a*b*c^
3 + 3*a^2*c^2*d)*e*f^5)*x)]
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} {\left (f x + e\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*x+a)^(1/3)/(d*x+c)^(2/3)/(f*x+e)^2,x, algorithm="giac")
[Out]
integrate(1/((b*x + a)^(1/3)*(d*x + c)^(2/3)*(f*x + e)^2), x)
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maple [F] time = 0.23, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b x +a \right )^{\frac {1}{3}} \left (d x +c \right )^{\frac {2}{3}} \left (f x +e \right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(b*x+a)^(1/3)/(d*x+c)^(2/3)/(f*x+e)^2,x)
[Out]
int(1/(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 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} {\left (f x + e\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*x+a)^(1/3)/(d*x+c)^(2/3)/(f*x+e)^2,x, algorithm="maxima")
[Out]
integrate(1/((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 \[ \int \frac {1}{{\left (e+f\,x\right )}^2\,{\left (a+b\,x\right )}^{1/3}\,{\left (c+d\,x\right )}^{2/3}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/((e + f*x)^2*(a + b*x)^(1/3)*(c + d*x)^(2/3)),x)
[Out]
int(1/((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 \[ \int \frac {1}{\sqrt [3]{a + b x} \left (c + d x\right )^{\frac {2}{3}} \left (e + f x\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*x+a)**(1/3)/(d*x+c)**(2/3)/(f*x+e)**2,x)
[Out]
Integral(1/((a + b*x)**(1/3)*(c + d*x)**(2/3)*(e + f*x)**2), x)
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