3.16.88 \(\int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx\) [1588]

Optimal. Leaf size=171 \[ \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{d}+\frac {(b c-a d) \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{\sqrt {3} b^{2/3} d^{4/3}}+\frac {(b c-a d) \log (a+b x)}{6 b^{2/3} d^{4/3}}+\frac {(b c-a d) \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{2 b^{2/3} d^{4/3}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 171, normalized size of antiderivative = 1.00, 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 = {52, 61} \begin {gather*} \frac {(b c-a d) \text {ArcTan}\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^{2/3} d^{4/3}}+\frac {(b c-a d) \log (a+b x)}{6 b^{2/3} d^{4/3}}+\frac {(b c-a d) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{2 b^{2/3} d^{4/3}}+\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 52

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 61

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, Simp[(-Sqrt
[3])*(q/d)*ArcTan[2*q*((a + b*x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3))) + 1/Sqrt[3]], x] + (-Simp[3*(q/(2*d))*Log[q*
((a + b*x)^(1/3)/(c + d*x)^(1/3)) - 1], x] - Simp[(q/(2*d))*Log[c + d*x], x])] /; FreeQ[{a, b, c, d}, x] && Ne
Q[b*c - a*d, 0] && PosQ[d/b]

Rubi steps

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

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Mathematica [A]
time = 10.02, size = 278, normalized size = 1.63 \begin {gather*} \frac {(a+b x)^{4/3} \left (6 b^{2/3} \sqrt [3]{d (a+b x)} (c+d x)^{2/3}+2 \sqrt {3} (b c-a d) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}{2 \sqrt [3]{d (a+b x)}+\sqrt [3]{b} \sqrt [3]{c+d x}}\right )+2 (b c-a d) \log \left (\sqrt [3]{d (a+b x)}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )-b c \log \left ((d (a+b x))^{2/3}+\sqrt [3]{b} \sqrt [3]{d (a+b x)} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}\right )+a d \log \left ((d (a+b x))^{2/3}+\sqrt [3]{b} \sqrt [3]{d (a+b x)} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}\right )\right )}{6 b^{2/3} (d (a+b x))^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (132) = 264\).
time = 1.35, size = 618, normalized size = 3.61 \begin {gather*} \left [\frac {6 \, {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b^{2} d - 3 \, \sqrt {\frac {1}{3}} {\left (b^{2} c d - a b d^{2}\right )} \sqrt {\frac {\left (-b^{2} d\right )^{\frac {1}{3}}}{d}} \log \left (3 \, b^{2} d x + b^{2} c + 2 \, a b d + 3 \, \left (-b^{2} d\right )^{\frac {1}{3}} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d - \left (-b^{2} d\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} + \left (-b^{2} d\right )^{\frac {1}{3}} {\left (b d x + b c\right )}\right )} \sqrt {\frac {\left (-b^{2} d\right )^{\frac {1}{3}}}{d}}\right ) - \left (-b^{2} d\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (\frac {{\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d + \left (-b^{2} d\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} - \left (-b^{2} d\right )^{\frac {1}{3}} {\left (b d x + b c\right )}}{d x + c}\right ) + 2 \, \left (-b^{2} d\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (\frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d - \left (-b^{2} d\right )^{\frac {2}{3}} {\left (d x + c\right )}}{d x + c}\right )}{6 \, b^{2} d^{2}}, \frac {6 \, {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b^{2} d - 6 \, \sqrt {\frac {1}{3}} {\left (b^{2} c d - a b d^{2}\right )} \sqrt {-\frac {\left (-b^{2} d\right )^{\frac {1}{3}}}{d}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (-b^{2} d\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} - \left (-b^{2} d\right )^{\frac {1}{3}} {\left (b d x + b c\right )}\right )} \sqrt {-\frac {\left (-b^{2} d\right )^{\frac {1}{3}}}{d}}}{b^{2} d x + b^{2} c}\right ) - \left (-b^{2} d\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (\frac {{\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d + \left (-b^{2} d\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} - \left (-b^{2} d\right )^{\frac {1}{3}} {\left (b d x + b c\right )}}{d x + c}\right ) + 2 \, \left (-b^{2} d\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (\frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d - \left (-b^{2} d\right )^{\frac {2}{3}} {\left (d x + c\right )}}{d x + c}\right )}{6 \, b^{2} d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{a + b x}}{\sqrt [3]{c + d x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{1/3}}{{\left (c+d\,x\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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