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

Optimal. Leaf size=200 \[ -\frac {\log (c+d x) (-a d f-2 b c f+3 b d e)}{6 b^{4/3} d^{5/3}}-\frac {(-a d f-2 b c f+3 b d e) \log \left (\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{2 b^{4/3} d^{5/3}}-\frac {(-a d f-2 b c f+3 b d e) \tan ^{-1}\left (\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} b^{4/3} d^{5/3}}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x}}{b d} \]

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Rubi [A]  time = 0.11, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {80, 59} \begin {gather*} -\frac {\log (c+d x) (-a d f-2 b c f+3 b d e)}{6 b^{4/3} d^{5/3}}-\frac {(-a d f-2 b c f+3 b d e) \log \left (\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{2 b^{4/3} d^{5/3}}-\frac {(-a d f-2 b c f+3 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 )}{\sqrt {3} b^{4/3} d^{5/3}}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x}}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(f*(a + b*x)^(2/3)*(c + d*x)^(1/3))/(b*d) - ((3*b*d*e - 2*b*c*f - 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))])/(Sqrt[3]*b^(4/3)*d^(5/3)) - ((3*b*d*e - 2*b*c*f - a*d*f)*Log[c +
d*x])/(6*b^(4/3)*d^(5/3)) - ((3*b*d*e - 2*b*c*f - a*d*f)*Log[-1 + (d^(1/3)*(a + b*x)^(1/3))/(b^(1/3)*(c + d*x
)^(1/3))])/(2*b^(4/3)*d^(5/3))

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 \frac {e+f x}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx &=\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x}}{b d}+\frac {\left (b d e-\left (\frac {2 b c}{3}+\frac {a d}{3}\right ) f\right ) \int \frac {1}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx}{b d}\\ &=\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x}}{b d}-\frac {(3 b d e-2 b c f-a d f) \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{\sqrt {3} b^{4/3} d^{5/3}}-\frac {(3 b d e-2 b c f-a d f) \log (c+d x)}{6 b^{4/3} d^{5/3}}-\frac {(3 b d e-2 b c f-a d f) \log \left (-1+\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{2 b^{4/3} d^{5/3}}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 103, normalized size = 0.52 \begin {gather*} \frac {(a+b x)^{2/3} \left (\left (\frac {b (c+d x)}{b c-a d}\right )^{2/3} (-a d f-2 b c f+3 b d e) \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {d (a+b x)}{a d-b c}\right )+2 b f (c+d x)\right )}{2 b^2 d (c+d x)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(2/3)*(2*b*f*(c + d*x) + (3*b*d*e - 2*b*c*f - a*d*f)*((b*(c + d*x))/(b*c - a*d))^(2/3)*Hypergeometr
ic2F1[2/3, 2/3, 5/3, (d*(a + b*x))/(-(b*c) + a*d)]))/(2*b^2*d*(c + d*x)^(2/3))

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IntegrateAlgebraic [A]  time = 0.52, size = 281, normalized size = 1.40 \begin {gather*} \frac {(a d f+2 b c f-3 b d e) \log \left (\sqrt [3]{d}-\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{a+b x}}\right )}{3 b^{4/3} d^{5/3}}+\frac {(-a d f-2 b c f+3 b d e) \log \left (\frac {b^{2/3} (c+d x)^{2/3}}{(a+b x)^{2/3}}+\frac {\sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{c+d x}}{\sqrt [3]{a+b x}}+d^{2/3}\right )}{6 b^{4/3} d^{5/3}}+\frac {(-a d f-2 b c f+3 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 )}{\sqrt {3} b^{4/3} d^{5/3}}-\frac {f \sqrt [3]{c+d x} (a d-b c)}{b d \sqrt [3]{a+b x} \left (\frac {b (c+d x)}{a+b x}-d\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((-(b*c) + a*d)*f*(c + d*x)^(1/3))/(b*d*(a + b*x)^(1/3)*(-d + (b*(c + d*x))/(a + b*x)))) + ((3*b*d*e - 2*b*c
*f - 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))])/(Sqrt[3]*b^(4/3
)*d^(5/3)) + ((-3*b*d*e + 2*b*c*f + a*d*f)*Log[d^(1/3) - (b^(1/3)*(c + d*x)^(1/3))/(a + b*x)^(1/3)])/(3*b^(4/3
)*d^(5/3)) + ((3*b*d*e - 2*b*c*f - a*d*f)*Log[d^(2/3) + (b^(1/3)*d^(1/3)*(c + d*x)^(1/3))/(a + b*x)^(1/3) + (b
^(2/3)*(c + d*x)^(2/3))/(a + b*x)^(2/3)])/(6*b^(4/3)*d^(5/3))

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fricas [A]  time = 1.48, size = 660, normalized size = 3.30 \begin {gather*} \left [\frac {6 \, {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d^{2} f - 3 \, \sqrt {\frac {1}{3}} {\left (3 \, b^{2} d^{2} e - {\left (2 \, b^{2} c d + a b d^{2}\right )} f\right )} \sqrt {-\frac {\left (b d^{2}\right )^{\frac {1}{3}}}{b}} \log \left (-3 \, b d^{2} x - 2 \, b c d - a d^{2} + 3 \, \left (b d^{2}\right )^{\frac {1}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} d + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d - \left (b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} - \left (b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}\right )} \sqrt {-\frac {\left (b d^{2}\right )^{\frac {1}{3}}}{b}}\right ) - 2 \, \left (b d^{2}\right )^{\frac {2}{3}} {\left (3 \, b d e - {\left (2 \, b c + a d\right )} f\right )} \log \left (\frac {{\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d - \left (b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}}{b x + a}\right ) + \left (b d^{2}\right )^{\frac {2}{3}} {\left (3 \, b d e - {\left (2 \, b c + a d\right )} f\right )} \log \left (\frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d + \left (b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} + \left (b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}}{b x + a}\right )}{6 \, b^{2} d^{3}}, \frac {6 \, {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d^{2} f + 6 \, \sqrt {\frac {1}{3}} {\left (3 \, b^{2} d^{2} e - {\left (2 \, b^{2} c d + a b d^{2}\right )} f\right )} \sqrt {\frac {\left (b d^{2}\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} + \left (b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}\right )} \sqrt {\frac {\left (b d^{2}\right )^{\frac {1}{3}}}{b}}}{b d^{2} x + a d^{2}}\right ) - 2 \, \left (b d^{2}\right )^{\frac {2}{3}} {\left (3 \, b d e - {\left (2 \, b c + a d\right )} f\right )} \log \left (\frac {{\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d - \left (b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}}{b x + a}\right ) + \left (b d^{2}\right )^{\frac {2}{3}} {\left (3 \, b d e - {\left (2 \, b c + a d\right )} f\right )} \log \left (\frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d + \left (b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} + \left (b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}}{b x + a}\right )}{6 \, b^{2} d^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(6*(b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d^2*f - 3*sqrt(1/3)*(3*b^2*d^2*e - (2*b^2*c*d + a*b*d^2)*f)*sqrt(-(b
*d^2)^(1/3)/b)*log(-3*b*d^2*x - 2*b*c*d - a*d^2 + 3*(b*d^2)^(1/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3)*d + 3*sqrt(1
/3)*(2*(b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d - (b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (b*d^2)^(1/3)*(b*
d*x + a*d))*sqrt(-(b*d^2)^(1/3)/b)) - 2*(b*d^2)^(2/3)*(3*b*d*e - (2*b*c + a*d)*f)*log(((b*x + a)^(2/3)*(d*x +
c)^(1/3)*b*d - (b*d^2)^(2/3)*(b*x + a))/(b*x + a)) + (b*d^2)^(2/3)*(3*b*d*e - (2*b*c + a*d)*f)*log(((b*x + a)^
(1/3)*(d*x + c)^(2/3)*b*d + (b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (b*d^2)^(1/3)*(b*d*x + a*d))/(b*x
+ a)))/(b^2*d^3), 1/6*(6*(b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d^2*f + 6*sqrt(1/3)*(3*b^2*d^2*e - (2*b^2*c*d + a*b
*d^2)*f)*sqrt((b*d^2)^(1/3)/b)*arctan(sqrt(1/3)*(2*(b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (b*d^2)^(1/
3)*(b*d*x + a*d))*sqrt((b*d^2)^(1/3)/b)/(b*d^2*x + a*d^2)) - 2*(b*d^2)^(2/3)*(3*b*d*e - (2*b*c + a*d)*f)*log((
(b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (b*d^2)^(2/3)*(b*x + a))/(b*x + a)) + (b*d^2)^(2/3)*(3*b*d*e - (2*b*c +
a*d)*f)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d + (b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (b*d^2)^(1/
3)*(b*d*x + a*d))/(b*x + a)))/(b^2*d^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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