3.31.33 \(\int \frac {1}{(a+b x) \sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx\) [3033]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 113, normalized size of antiderivative = 1.00, 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 = {142, 141} \begin {gather*} \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}} \end {gather*}

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 141

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)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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])

Rule 142

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]

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*}

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Mathematica [A]
time = 20.66, size = 145, normalized size = 1.28 \begin {gather*} \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 {b c+a d+2 b d x}{2 b c+2 b d x}\right )^{2/3}}\right )}{2 b^2 (b c-a d) (c+d x)^{4/3} \sqrt [3]{a d+b (c+2 d x)}} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b x +a \right ) \left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {4}{3}}}\, dx\]

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 = 0.00, 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(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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, 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(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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\left (a+b\,x\right )\,{\left (c+d\,x\right )}^{1/3}\,{\left (a\,d+b\,c+2\,b\,d\,x\right )}^{4/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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