3.3034 \(\int \frac {1}{(a+b x)^2 \sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx\)

Optimal. Leaf size=114 \[ -\frac {3 d (c+d x)^{2/3} \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} F_1\left (\frac {2}{3};\frac {4}{3},2;\frac {5}{3};\frac {2 b (c+d x)}{b c-a d},\frac {b (c+d x)}{b c-a d}\right )}{2 (b c-a d)^3 \sqrt [3]{a d+b c+2 b d x}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 114, 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 = {137, 136} \[ -\frac {3 d (c+d x)^{2/3} \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} F_1\left (\frac {2}{3};\frac {4}{3},2;\frac {5}{3};\frac {2 b (c+d x)}{b c-a d},\frac {b (c+d x)}{b c-a d}\right )}{2 (b c-a d)^3 \sqrt [3]{a d+b c+2 b d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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]

Rubi steps

\begin {align*} \int \frac {1}{(a+b x)^2 \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)^2 \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 (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},2;\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)^3 \sqrt [3]{b c+a d+2 b d x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 1.12, size = 289, normalized size = 2.54 \[ \frac {(c+d x)^{2/3} \left (\frac {d \left (-7\ 2^{2/3} (b c-a d)^2 \sqrt [3]{\frac {a d+b c+2 b d x}{b c+b d x}} F_1\left (\frac {5}{3};\frac {1}{3},1;\frac {8}{3};\frac {b c-a d}{2 b c+2 b d x},\frac {b c-a d}{b c+b d x}\right )+40\ 2^{2/3} b (c+d x) (b c-a d) \sqrt [3]{\frac {a d+b c+2 b d x}{b c+b d x}} F_1\left (\frac {2}{3};\frac {1}{3},1;\frac {5}{3};\frac {b c-a d}{2 b c+2 b d x},\frac {b c-a d}{b c+b d x}\right )+70 b (c+d x) (a d+b (c+2 d x))\right )}{b^2 (c+d x)^2}-\frac {10 (13 a d+b (c+14 d x))}{a+b x}\right )}{10 (b c-a d)^3 \sqrt [3]{a d+b (c+2 d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((c + d*x)^(2/3)*((-10*(13*a*d + b*(c + 14*d*x)))/(a + b*x) + (d*(70*b*(c + d*x)*(a*d + b*(c + 2*d*x)) + 40*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)] - 7*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)^3*(a*d + b*(c + 2*d*x))^(1/3))

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b x +a \right )^{2} \left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________