3.3031 \(\int \frac{a+b x}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx\)

Optimal. Leaf size=32 \[ \frac{3 (c+d x)^{2/3}}{2 d^2 \sqrt [3]{a d+b c+2 b d x}} \]

[Out]

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

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Rubi [A]  time = 0.0048773, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.032, Rules used = {74} \[ \frac{3 (c+d x)^{2/3}}{2 d^2 \sqrt [3]{a d+b c+2 b d x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/((c + d*x)^(1/3)*(b*c + a*d + 2*b*d*x)^(4/3)),x]

[Out]

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

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rubi steps

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

Mathematica [A]  time = 0.0255556, size = 32, normalized size = 1. \[ \frac{3 (c+d x)^{2/3}}{2 d^2 \sqrt [3]{a d+b (c+2 d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/((c + d*x)^(1/3)*(b*c + a*d + 2*b*d*x)^(4/3)),x]

[Out]

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

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Maple [A]  time = 0.005, size = 27, normalized size = 0.8 \begin{align*}{\frac{3}{2\,{d}^{2}} \left ( dx+c \right ) ^{{\frac{2}{3}}}{\frac{1}{\sqrt [3]{2\,bdx+ad+bc}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x)

[Out]

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

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Maxima [A]  time = 2.53377, size = 35, normalized size = 1.09 \begin{align*} \frac{3 \,{\left (d x + c\right )}^{\frac{2}{3}}}{2 \,{\left (2 \, b d x + b c + a d\right )}^{\frac{1}{3}} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 3.38088, size = 107, normalized size = 3.34 \begin{align*} \frac{3 \,{\left (2 \, b d x + b c + a d\right )}^{\frac{2}{3}}{\left (d x + c\right )}^{\frac{2}{3}}}{2 \,{\left (2 \, b d^{3} x + b c d^{2} + a d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x, algorithm="fricas")

[Out]

3/2*(2*b*d*x + b*c + a*d)^(2/3)*(d*x + c)^(2/3)/(2*b*d^3*x + b*c*d^2 + a*d^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b x}{\sqrt [3]{c + d x} \left (a d + b c + 2 b d x\right )^{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)**(1/3)/(2*b*d*x+a*d+b*c)**(4/3),x)

[Out]

Integral((a + b*x)/((c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**(4/3)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b x + a}{{\left (2 \, b d x + b c + a d\right )}^{\frac{4}{3}}{\left (d x + c\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x, algorithm="giac")

[Out]

integrate((b*x + a)/((2*b*d*x + b*c + a*d)^(4/3)*(d*x + c)^(1/3)), x)