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

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

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Rubi [A]  time = 0.03, 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 = {47, 59} \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 47

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 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]{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 [C]  time = 0.04, size = 73, normalized size = 0.49 \begin {gather*} \frac {3 (a+b x)^{4/3} \left (\frac {b (c+d x)}{b c-a d}\right )^{4/3} \, _2F_1\left (\frac {4}{3},\frac {4}{3};\frac {7}{3};\frac {d (a+b x)}{a d-b c}\right )}{4 b (c+d x)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b*x)^(4/3)*((b*(c + d*x))/(b*c - a*d))^(4/3)*Hypergeometric2F1[4/3, 4/3, 7/3, (d*(a + b*x))/(-(b*c) +
a*d)])/(4*b*(c + d*x)^(4/3))

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IntegrateAlgebraic [A]  time = 0.16, size = 200, normalized size = 1.34 \begin {gather*} \frac {\sqrt [3]{b} \log \left (\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}}+b^{2/3}\right )}{2 d^{4/3}}-\frac {\sqrt [3]{b} \log \left (\sqrt [3]{b}-\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}\right )}{d^{4/3}}+\frac {\sqrt {3} \sqrt [3]{b} \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 )}{d^{4/3}}-\frac {3 \sqrt [3]{a+b x}}{d \sqrt [3]{c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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*d^(1/3)*(a + b*x)^(1/3))/(Sq
rt[3]*b^(1/3)*(c + d*x)^(1/3))])/d^(4/3) - (b^(1/3)*Log[b^(1/3) - (d^(1/3)*(a + b*x)^(1/3))/(c + d*x)^(1/3)])/
d^(4/3) + (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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fricas [B]  time = 1.46, 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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x + a\right )}^{\frac {1}{3}}}{{\left (d x + c\right )}^{\frac {4}{3}}}\,{d x} \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="giac")

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b x +a \right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {4}{3}}}\, dx \end {gather*}

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*} \int \frac {{\left (b x + a\right )}^{\frac {1}{3}}}{{\left (d x + c\right )}^{\frac {4}{3}}}\,{d x} \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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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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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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