3.1576 \(\int \frac{\sqrt [3]{c+d x}}{\sqrt [3]{a+b x}} \, dx\)

Optimal. Leaf size=172 \[ -\frac{(b c-a d) \log (c+d x)}{6 b^{4/3} d^{2/3}}-\frac{(b c-a d) \log \left (\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{2 b^{4/3} d^{2/3}}-\frac{(b c-a d) \tan ^{-1}\left (\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt{3} \sqrt [3]{b} \sqrt [3]{c+d x}}+\frac{1}{\sqrt{3}}\right )}{\sqrt{3} b^{4/3} d^{2/3}}+\frac{(a+b x)^{2/3} \sqrt [3]{c+d x}}{b} \]

[Out]

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

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Rubi [A]  time = 0.0455278, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {50, 59} \[ -\frac{(b c-a d) \log (c+d x)}{6 b^{4/3} d^{2/3}}-\frac{(b c-a d) \log \left (\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{2 b^{4/3} d^{2/3}}-\frac{(b c-a d) \tan ^{-1}\left (\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt{3} \sqrt [3]{b} \sqrt [3]{c+d x}}+\frac{1}{\sqrt{3}}\right )}{\sqrt{3} b^{4/3} d^{2/3}}+\frac{(a+b x)^{2/3} \sqrt [3]{c+d x}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

Mathematica [C]  time = 0.0235564, size = 73, normalized size = 0.42 \[ \frac{3 (a+b x)^{2/3} \sqrt [3]{c+d x} \, _2F_1\left (-\frac{1}{3},\frac{2}{3};\frac{5}{3};\frac{d (a+b x)}{a d-b c}\right )}{2 b \sqrt [3]{\frac{b (c+d x)}{b c-a d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b*x)^(2/3)*(c + d*x)^(1/3)*Hypergeometric2F1[-1/3, 2/3, 5/3, (d*(a + b*x))/(-(b*c) + a*d)])/(2*b*((b*(
c + d*x))/(b*c - a*d))^(1/3))

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Maple [F]  time = 0.015, size = 0, normalized size = 0. \begin{align*} \int{\sqrt [3]{dx+c}{\frac{1}{\sqrt [3]{bx+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{\frac{1}{3}}}{{\left (b x + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.00917, size = 1547, normalized size = 8.99 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt [3]{c + d x}}{\sqrt [3]{a + b x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{\frac{1}{3}}}{{\left (b x + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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