3.16.77 \(\int \frac {\sqrt [3]{c+d x}}{(a+b x)^{4/3}} \, dx\) [1577]

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

[Out]

-3*(d*x+c)^(1/3)/b/(b*x+a)^(1/3)-1/2*d^(1/3)*ln(d*x+c)/b^(4/3)-3/2*d^(1/3)*ln(-1+d^(1/3)*(b*x+a)^(1/3)/b^(1/3)
/(d*x+c)^(1/3))/b^(4/3)-d^(1/3)*arctan(1/3*3^(1/2)+2/3*d^(1/3)*(b*x+a)^(1/3)/b^(1/3)/(d*x+c)^(1/3)*3^(1/2))*3^
(1/2)/b^(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 {\sqrt {3} \sqrt [3]{d} \text {ArcTan}\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 )}{b^{4/3}}-\frac {3 \sqrt [3]{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}}-\frac {3 \sqrt [3]{c+d x}}{b \sqrt [3]{a+b x}}-\frac {\sqrt [3]{d} \log (c+d x)}{2 b^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (d x +c \right )^{\frac {1}{3}}}{\left (b x +a \right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((d*x + c)^(1/3)/(b*x + a)^(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 = 0.88, size = 233, normalized size = 1.56 \begin {gather*} -\frac {2 \, \sqrt {3} {\left (b x + a\right )} \left (-\frac {d}{b}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b \left (-\frac {d}{b}\right )^{\frac {2}{3}} + \sqrt {3} {\left (b d x + a d\right )}}{3 \, {\left (b d x + a d\right )}}\right ) + {\left (b x + a\right )} \left (-\frac {d}{b}\right )^{\frac {1}{3}} \log \left (\frac {{\left (b x + a\right )} \left (-\frac {d}{b}\right )^{\frac {2}{3}} - {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} \left (-\frac {d}{b}\right )^{\frac {1}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{b x + a}\right ) - 2 \, {\left (b x + a\right )} \left (-\frac {d}{b}\right )^{\frac {1}{3}} \log \left (\frac {{\left (b x + a\right )} \left (-\frac {d}{b}\right )^{\frac {1}{3}} + {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{b x + a}\right ) + 6 \, {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{2 \, {\left (b^{2} x + a b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^{1/3}}{{\left (a+b\,x\right )}^{4/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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