3.17.17 \(\int \frac {\sqrt [3]{a+b x}}{(c+d x)^{4/3}} \, dx\) [1617]

Optimal. Leaf size=149 \[ -\frac {3 \sqrt [3]{a+b x}}{d \sqrt [3]{c+d x}}-\frac {\sqrt {3} \sqrt [3]{b} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{d^{4/3}}-\frac {\sqrt [3]{b} \log (a+b x)}{2 d^{4/3}}-\frac {3 \sqrt [3]{b} \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{2 d^{4/3}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 149, 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 = {49, 61} \begin {gather*} -\frac {3 \sqrt [3]{b} \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{2 d^{4/3}}-\frac {\sqrt {3} \sqrt [3]{b} \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 )}{d^{4/3}}-\frac {3 \sqrt [3]{a+b x}}{d \sqrt [3]{c+d x}}-\frac {\sqrt [3]{b} \log (a+b x)}{2 d^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(1/3)/(c + d*x)^(4/3),x]

[Out]

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

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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}}{(c+d x)^{4/3}} \, dx &=-\frac {3 \sqrt [3]{a+b x}}{d \sqrt [3]{c+d x}}+\frac {b \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}} \, dx}{d}\\ &=-\frac {3 \sqrt [3]{a+b x}}{d \sqrt [3]{c+d x}}-\frac {\sqrt {3} \sqrt [3]{b} \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 )}{d^{4/3}}-\frac {\sqrt [3]{b} \log (a+b x)}{2 d^{4/3}}-\frac {3 \sqrt [3]{b} \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{2 d^{4/3}}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 191, normalized size = 1.28 \begin {gather*} \frac {-\frac {6 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}+2 \sqrt {3} \sqrt [3]{b} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}}{\sqrt {3}}\right )-2 \sqrt [3]{b} \log \left (\sqrt [3]{b}-\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}\right )+\sqrt [3]{b} \log \left (b^{2/3}+\frac {d^{2/3} (a+b x)^{2/3}}{(c+d x)^{2/3}}+\frac {\sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}\right )}{2 d^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(1/3)/(c + d*x)^(4/3),x]

[Out]

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

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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded while calling a Python object} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[(a + b*x)^(1/3)/(c + d*x)^(4/3),x]')

[Out]

cought exception: maximum recursion depth exceeded while calling a Python object

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/3)/(d*x+c)^(4/3),x)

[Out]

int((b*x+a)^(1/3)/(d*x+c)^(4/3),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)^(4/3),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(1/3)/(d*x + c)^(4/3), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (109) = 218\).
time = 0.30, size = 233, normalized size = 1.56 \begin {gather*} -\frac {2 \, \sqrt {3} {\left (d x + c\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} d \left (-\frac {b}{d}\right )^{\frac {2}{3}} + \sqrt {3} {\left (b d x + b c\right )}}{3 \, {\left (b d x + b c\right )}}\right ) + {\left (d x + c\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} \log \left (\frac {{\left (d x + c\right )} \left (-\frac {b}{d}\right )^{\frac {2}{3}} - {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} \left (-\frac {b}{d}\right )^{\frac {1}{3}} + {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{d x + c}\right ) - 2 \, {\left (d x + c\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} \log \left (\frac {{\left (d x + c\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{d x + c}\right ) + 6 \, {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{2 \, {\left (d^{2} x + c d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/3)/(d*x+c)^(4/3),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{a + b x}}{\left (c + d x\right )^{\frac {4}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/3)/(d*x+c)**(4/3),x)

[Out]

Integral((a + b*x)**(1/3)/(c + d*x)**(4/3), x)

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Giac [F] N/A
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)^(4/3),x)

[Out]

Could not integrate

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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 )}^{4/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(1/3)/(c + d*x)^(4/3),x)

[Out]

int((a + b*x)^(1/3)/(c + d*x)^(4/3), x)

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