3.3009 \(\int \frac {\sqrt [3]{a+b x} (e+f x)}{\sqrt [3]{c+d x}} \, dx\)

Optimal. Leaf size=273 \[ \frac {(b c-a d) \log (a+b x) (-a d f-2 b c f+3 b d e)}{18 b^{5/3} d^{7/3}}+\frac {(b c-a d) (-a d f-2 b c f+3 b d e) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{6 b^{5/3} d^{7/3}}+\frac {(b c-a d) (-a d f-2 b c f+3 b d e) \tan ^{-1}\left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3} d^{7/3}}+\frac {\sqrt [3]{a+b x} (c+d x)^{2/3} (-a d f-2 b c f+3 b d e)}{3 b d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 b d} \]

[Out]

1/3*(-a*d*f-2*b*c*f+3*b*d*e)*(b*x+a)^(1/3)*(d*x+c)^(2/3)/b/d^2+1/2*f*(b*x+a)^(4/3)*(d*x+c)^(2/3)/b/d+1/18*(-a*
d+b*c)*(-a*d*f-2*b*c*f+3*b*d*e)*ln(b*x+a)/b^(5/3)/d^(7/3)+1/6*(-a*d+b*c)*(-a*d*f-2*b*c*f+3*b*d*e)*ln(-1+b^(1/3
)*(d*x+c)^(1/3)/d^(1/3)/(b*x+a)^(1/3))/b^(5/3)/d^(7/3)+1/9*(-a*d+b*c)*(-a*d*f-2*b*c*f+3*b*d*e)*arctan(1/3*3^(1
/2)+2/3*b^(1/3)*(d*x+c)^(1/3)/d^(1/3)/(b*x+a)^(1/3)*3^(1/2))/b^(5/3)/d^(7/3)*3^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {80, 50, 59} \[ \frac {(b c-a d) \log (a+b x) (-a d f-2 b c f+3 b d e)}{18 b^{5/3} d^{7/3}}+\frac {(b c-a d) (-a d f-2 b c f+3 b d e) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{6 b^{5/3} d^{7/3}}+\frac {(b c-a d) (-a d f-2 b c f+3 b d e) \tan ^{-1}\left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3} d^{7/3}}+\frac {\sqrt [3]{a+b x} (c+d x)^{2/3} (-a d f-2 b c f+3 b d e)}{3 b d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b*d*e - 2*b*c*f - a*d*f)*(a + b*x)^(1/3)*(c + d*x)^(2/3))/(3*b*d^2) + (f*(a + b*x)^(4/3)*(c + d*x)^(2/3))/
(2*b*d) + ((b*c - a*d)*(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))])/(3*Sqrt[3]*b^(5/3)*d^(7/3)) + ((b*c - a*d)*(3*b*d*e - 2*b*c*f - a*d*f)*Log[a + b*x])/(
18*b^(5/3)*d^(7/3)) + ((b*c - a*d)*(3*b*d*e - 2*b*c*f - a*d*f)*Log[-1 + (b^(1/3)*(c + d*x)^(1/3))/(d^(1/3)*(a
+ b*x)^(1/3))])/(6*b^(5/3)*d^(7/3))

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]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{a+b x} (e+f x)}{\sqrt [3]{c+d x}} \, dx &=\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 b d}+\frac {\left (2 b d e-\left (\frac {4 b c}{3}+\frac {2 a d}{3}\right ) f\right ) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx}{2 b d}\\ &=\frac {(3 b d e-2 b c f-a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 b d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 b d}-\frac {((b c-a d) (3 b d e-2 b c f-a d f)) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}} \, dx}{9 b d^2}\\ &=\frac {(3 b d e-2 b c f-a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 b d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 b d}+\frac {(b c-a d) (3 b d e-2 b c f-a d f) \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 )}{3 \sqrt {3} b^{5/3} d^{7/3}}+\frac {(b c-a d) (3 b d e-2 b c f-a d f) \log (a+b x)}{18 b^{5/3} d^{7/3}}+\frac {(b c-a d) (3 b d e-2 b c f-a d f) \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{6 b^{5/3} d^{7/3}}\\ \end {align*}

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Mathematica [C]  time = 0.17, size = 103, normalized size = 0.38 \[ \frac {(a+b x)^{4/3} \left (\sqrt [3]{\frac {b (c+d x)}{b c-a d}} (-a d f-2 b c f+3 b d e) \, _2F_1\left (\frac {1}{3},\frac {4}{3};\frac {7}{3};\frac {d (a+b x)}{a d-b c}\right )+2 b f (c+d x)\right )}{4 b^2 d \sqrt [3]{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(4/3)*(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))^(1/3)*Hypergeometr
ic2F1[1/3, 4/3, 7/3, (d*(a + b*x))/(-(b*c) + a*d)]))/(4*b^2*d*(c + d*x)^(1/3))

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fricas [A]  time = 1.13, size = 892, normalized size = 3.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/18*(3*sqrt(1/3)*(3*(b^3*c*d^2 - a*b^2*d^3)*e - (2*b^3*c^2*d - a*b^2*c*d^2 - a^2*b*d^3)*f)*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)) - (b^2*d)^(2/3)*(3*(b^2*c*d - a*b*d^2)*e - (2*b^2*c^2 - a*b*c*d - a^2*d^2)*f)*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)) + 2*(b^2*d)^(2/3)*(3*(b^2*c*d - a*b*d^2)*e - (2*b^2*c^2 - a*b*c*d - a^2*d^2)*f)*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*(3*b^3*d^2*f*x + 6*b^3*d^2*e - (4*b^3*c*d
- a*b^2*d^2)*f)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^3*d^3), -1/18*(6*sqrt(1/3)*(3*(b^3*c*d^2 - a*b^2*d^3)*e -
(2*b^3*c^2*d - a*b^2*c*d^2 - a^2*b*d^3)*f)*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)) + (b^2*d)^(2/3)*(
3*(b^2*c*d - a*b*d^2)*e - (2*b^2*c^2 - a*b*c*d - a^2*d^2)*f)*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)) - 2*(b^2*d)^(2/3)*(3*(b^2*c*
d - a*b*d^2)*e - (2*b^2*c^2 - a*b*c*d - a^2*d^2)*f)*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*(3*b^3*d^2*f*x + 6*b^3*d^2*e - (4*b^3*c*d - a*b^2*d^2)*f)*(b*x + a)^(1/3)*(d*x + c)^(
2/3))/(b^3*d^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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