3.3033 \(\int \frac{1}{(a+b x) \sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx\)

Optimal. Leaf size=113 \[ \frac{3 (c+d x)^{2/3} \sqrt [3]{-\frac{a d+b c+2 b d x}{b c-a d}} F_1\left (\frac{2}{3};\frac{4}{3},1;\frac{5}{3};\frac{2 b (c+d x)}{b c-a d},\frac{b (c+d x)}{b c-a d}\right )}{2 (b c-a d)^2 \sqrt [3]{a d+b c+2 b d x}} \]

[Out]

(3*(c + d*x)^(2/3)*(-((b*c + a*d + 2*b*d*x)/(b*c - a*d)))^(1/3)*AppellF1[2/3, 4/3, 1, 5/3, (2*b*(c + d*x))/(b*
c - a*d), (b*(c + d*x))/(b*c - a*d)])/(2*(b*c - a*d)^2*(b*c + a*d + 2*b*d*x)^(1/3))

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Rubi [A]  time = 0.0601176, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {137, 136} \[ \frac{3 (c+d x)^{2/3} \sqrt [3]{-\frac{a d+b c+2 b d x}{b c-a d}} F_1\left (\frac{2}{3};\frac{4}{3},1;\frac{5}{3};\frac{2 b (c+d x)}{b c-a d},\frac{b (c+d x)}{b c-a d}\right )}{2 (b c-a d)^2 \sqrt [3]{a d+b c+2 b d x}} \]

Antiderivative was successfully verified.

[In]

Int[1/((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)*(-((b*c + a*d + 2*b*d*x)/(b*c - a*d)))^(1/3)*AppellF1[2/3, 4/3, 1, 5/3, (2*b*(c + d*x))/(b*
c - a*d), (b*(c + d*x))/(b*c - a*d)])/(2*(b*c - a*d)^2*(b*c + a*d + 2*b*d*x)^(1/3))

Rule 137

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rubi steps

\begin{align*} \int \frac{1}{(a+b x) \sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx &=\frac{\left (d \sqrt [3]{\frac{d (b c+a d+2 b d x)}{-2 b c d+d (b c+a d)}}\right ) \int \frac{1}{(a+b x) \sqrt [3]{c+d x} \left (\frac{d (b c+a d)}{-2 b c d+d (b c+a d)}+\frac{2 b d^2 x}{-2 b c d+d (b c+a d)}\right )^{4/3}} \, dx}{(-2 b c d+d (b c+a d)) \sqrt [3]{b c+a d+2 b d x}}\\ &=\frac{3 (c+d x)^{2/3} \sqrt [3]{-\frac{b c+a d+2 b d x}{b c-a d}} F_1\left (\frac{2}{3};\frac{4}{3},1;\frac{5}{3};\frac{2 b (c+d x)}{b c-a d},\frac{b (c+d x)}{b c-a d}\right )}{2 (b c-a d)^2 \sqrt [3]{b c+a d+2 b d x}}\\ \end{align*}

Mathematica [A]  time = 0.770157, size = 145, normalized size = 1.28 \[ \frac{3 \left (2 b (c+d x)-\frac{(a d+b (c+2 d x)) F_1\left (\frac{2}{3};-\frac{2}{3},1;\frac{5}{3};\frac{b c-a d}{2 b c+2 b d x},\frac{b c-a d}{b c+b d x}\right )}{\left (\frac{a d+b c+2 b d x}{2 b c+2 b d x}\right )^{2/3}}\right )}{2 b^2 (c+d x)^{4/3} (b c-a d) \sqrt [3]{a d+b (c+2 d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3*(2*b*(c + d*x) - ((a*d + b*(c + 2*d*x))*AppellF1[2/3, -2/3, 1, 5/3, (b*c - a*d)/(2*b*c + 2*b*d*x), (b*c - a
*d)/(b*c + b*d*x)])/((b*c + a*d + 2*b*d*x)/(2*b*c + 2*b*d*x))^(2/3)))/(2*b^2*(b*c - a*d)*(c + d*x)^(4/3)*(a*d
+ b*(c + 2*d*x))^(1/3))

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{bx+a}{\frac{1}{\sqrt [3]{dx+c}}} \left ( 2\,bdx+ad+bc \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

Integral(1/((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{1}{{\left (2 \, b d x + b c + a d\right )}^{\frac{4}{3}}{\left (b x + a\right )}{\left (d x + c\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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