3.3035 \(\int \frac{1}{(a+b x)^3 \sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx\)

Optimal. Leaf size=116 \[ \frac{3 d^2 (c+d x)^{2/3} \sqrt [3]{-\frac{a d+b c+2 b d x}{b c-a d}} F_1\left (\frac{2}{3};\frac{4}{3},3;\frac{5}{3};\frac{2 b (c+d x)}{b c-a d},\frac{b (c+d x)}{b c-a d}\right )}{2 (b c-a d)^4 \sqrt [3]{a d+b c+2 b d x}} \]

[Out]

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

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Rubi [A]  time = 0.0509298, antiderivative size = 116, 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} \[ \frac{3 d^2 (c+d x)^{2/3} \sqrt [3]{-\frac{a d+b c+2 b d x}{b c-a d}} F_1\left (\frac{2}{3};\frac{4}{3},3;\frac{5}{3};\frac{2 b (c+d x)}{b c-a d},\frac{b (c+d x)}{b c-a d}\right )}{2 (b c-a d)^4 \sqrt [3]{a d+b c+2 b d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d^2*(c + d*x)^(2/3)*(-((b*c + a*d + 2*b*d*x)/(b*c - a*d)))^(1/3)*AppellF1[2/3, 4/3, 3, 5/3, (2*b*(c + d*x))
/(b*c - a*d), (b*(c + d*x))/(b*c - a*d)])/(2*(b*c - a*d)^4*(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)^3 \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)^3 \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 d^2 (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},3;\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)^4 \sqrt [3]{b c+a d+2 b d x}}\\ \end{align*}

Mathematica [B]  time = 1.185, size = 324, normalized size = 2.79 \[ \frac{(c+d x)^{2/3} \left (\frac{5 \left (57 a^2 d^2+2 a b d (4 c+61 d x)+b^2 \left (-c^2+6 c d x+64 d^2 x^2\right )\right )}{(a+b x)^2}-\frac{d^2 \left (-16\ 2^{2/3} (b c-a d)^2 \sqrt [3]{\frac{a d+b c+2 b d x}{b c+b d x}} F_1\left (\frac{5}{3};\frac{1}{3},1;\frac{8}{3};\frac{b c-a d}{2 b c+2 b d x},\frac{b c-a d}{b c+b d x}\right )+95\ 2^{2/3} b (c+d x) (b c-a d) \sqrt [3]{\frac{a d+b c+2 b d x}{b c+b d x}} F_1\left (\frac{2}{3};\frac{1}{3},1;\frac{5}{3};\frac{b c-a d}{2 b c+2 b d x},\frac{b c-a d}{b c+b d x}\right )+160 b (c+d x) (a d+b (c+2 d x))\right )}{b^2 (c+d x)^2}\right )}{10 (b c-a d)^4 \sqrt [3]{a d+b (c+2 d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((c + d*x)^(2/3)*((5*(57*a^2*d^2 + 2*a*b*d*(4*c + 61*d*x) + b^2*(-c^2 + 6*c*d*x + 64*d^2*x^2)))/(a + b*x)^2 -
(d^2*(160*b*(c + d*x)*(a*d + b*(c + 2*d*x)) + 95*2^(2/3)*b*(b*c - a*d)*(c + d*x)*((b*c + a*d + 2*b*d*x)/(b*c +
b*d*x))^(1/3)*AppellF1[2/3, 1/3, 1, 5/3, (b*c - a*d)/(2*b*c + 2*b*d*x), (b*c - a*d)/(b*c + b*d*x)] - 16*2^(2/
3)*(b*c - a*d)^2*((b*c + a*d + 2*b*d*x)/(b*c + b*d*x))^(1/3)*AppellF1[5/3, 1/3, 1, 8/3, (b*c - a*d)/(2*b*c + 2
*b*d*x), (b*c - a*d)/(b*c + b*d*x)]))/(b^2*(c + d*x)^2)))/(10*(b*c - a*d)^4*(a*d + b*(c + 2*d*x))^(1/3))

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( bx+a \right ) ^{3}}{\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)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x)

[Out]

int(1/(b*x+a)^3/(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 )}^{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(1/(b*x+a)^3/(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)^3*(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)^3/(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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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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 )}^{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(1/(b*x+a)^3/(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)^3*(d*x + c)^(1/3)), x)