\(\int \frac {(a+b x)^3}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx\) [1654]

Optimal result

Integrand size = 33, antiderivative size = 182 \[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\frac {3 (b c-a d)^2 (c+d x)^{2/3}}{8 d^4 \sqrt [3]{b c+a d+2 b d x}}-\frac {45 (b c-a d) (c+d x)^{2/3} (b c+a d+2 b d x)^{2/3}}{112 d^4}+\frac {3 b (c+d x)^{5/3} (b c+a d+2 b d x)^{2/3}}{28 d^4}-\frac {81 (b c-a d) (c+d x)^{2/3} (b c+a d+2 b d x)^{2/3} \operatorname {Hypergeometric2F1}\left (1,\frac {4}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )}{112 d^4} \] Output:

3/8*(-a*d+b*c)^2*(d*x+c)^(2/3)/d^4/(2*b*d*x+a*d+b*c)^(1/3)-45/112*(-a*d+b* 
c)*(d*x+c)^(2/3)*(2*b*d*x+a*d+b*c)^(2/3)/d^4+3/28*b*(d*x+c)^(5/3)*(2*b*d*x 
+a*d+b*c)^(2/3)/d^4-81/112*(-a*d+b*c)*(d*x+c)^(2/3)*(2*b*d*x+a*d+b*c)^(2/3 
)*hypergeom([1, 4/3],[5/3],2*b*(d*x+c)/(-a*d+b*c))/d^4
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.16 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.79 \[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\frac {3 (c+d x)^{2/3} \left (29 a^2 d^2+2 a b d (-12 c+17 d x)+b^2 \left (3 c^2-18 c d x+8 d^2 x^2\right )+27 (b c-a d)^2 \sqrt [3]{\frac {a d+b (c+2 d x)}{-b c+a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )\right )}{112 d^4 \sqrt [3]{a d+b (c+2 d x)}} \] Input:

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

Output:

(3*(c + d*x)^(2/3)*(29*a^2*d^2 + 2*a*b*d*(-12*c + 17*d*x) + b^2*(3*c^2 - 1 
8*c*d*x + 8*d^2*x^2) + 27*(b*c - a*d)^2*((a*d + b*(c + 2*d*x))/(-(b*c) + a 
*d))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, (2*b*(c + d*x))/(b*c - a*d)])) 
/(112*d^4*(a*d + b*(c + 2*d*x))^(1/3))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {111, 27, 163, 80, 79}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 111

\(\displaystyle \frac {3 \int -\frac {2 b (b c-a d) (a+b x) (b c+2 a d+3 b d x)}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}}dx}{14 b d^2}+\frac {3 (a+b x)^2 (c+d x)^{2/3}}{14 d^2 \sqrt [3]{a d+b c+2 b d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 (a+b x)^2 (c+d x)^{2/3}}{14 d^2 \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (b c-a d) \int \frac {(a+b x) (b c+2 a d+3 b d x)}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}}dx}{7 d^2}\)

\(\Big \downarrow \) 163

\(\displaystyle \frac {3 (a+b x)^2 (c+d x)^{2/3}}{14 d^2 \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (b c-a d) \left (-\frac {9 (b c-a d) \int \frac {1}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}}dx}{8 d}-\frac {3 (c+d x)^{2/3} (-7 a d+b c-6 b d x)}{16 d^2 \sqrt [3]{a d+b c+2 b d x}}\right )}{7 d^2}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {3 (a+b x)^2 (c+d x)^{2/3}}{14 d^2 \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (b c-a d) \left (-\frac {9 (b c-a d) \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} \int \frac {1}{\sqrt [3]{c+d x} \sqrt [3]{-\frac {b c+a d}{b c-a d}-\frac {2 b d x}{b c-a d}}}dx}{8 d \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (c+d x)^{2/3} (-7 a d+b c-6 b d x)}{16 d^2 \sqrt [3]{a d+b c+2 b d x}}\right )}{7 d^2}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {3 (a+b x)^2 (c+d x)^{2/3}}{14 d^2 \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (b c-a d) \left (-\frac {27 (c+d x)^{2/3} (b c-a d) \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )}{16 d^2 \sqrt [3]{a d+b c+2 b d x}}-\frac {3 (c+d x)^{2/3} (-7 a d+b c-6 b d x)}{16 d^2 \sqrt [3]{a d+b c+2 b d x}}\right )}{7 d^2}\)

Input:

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

Output:

(3*(a + b*x)^2*(c + d*x)^(2/3))/(14*d^2*(b*c + a*d + 2*b*d*x)^(1/3)) - (3* 
(b*c - a*d)*((-3*(c + d*x)^(2/3)*(b*c - 7*a*d - 6*b*d*x))/(16*d^2*(b*c + a 
*d + 2*b*d*x)^(1/3)) - (27*(b*c - a*d)*(c + d*x)^(2/3)*(-((b*c + a*d + 2*b 
*d*x)/(b*c - a*d)))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, (2*b*(c + d*x)) 
/(b*c - a*d)])/(16*d^2*(b*c + a*d + 2*b*d*x)^(1/3))))/(7*d^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(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*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 111

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

rule 163

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) 
)*((g_.) + (h_.)*(x_)), x_] :> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n 
+ 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(b*c - a*d)* 
(m + 1)*x)/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)))*(a + b*x)^(m + 1)*(c + 
d*x)^(n + 1), x] - Simp[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f 
*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c* 
d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2* 
d*(b*c - a*d)*(m + 1)*(m + n + 3))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((GeQ[m, -2] && LtQ[m, - 
1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]
 

Maple [F]

\[\int \frac {\left (b x +a \right )^{3}}{\left (x d +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {4}{3}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*(2*b*d*x + b*c + a*d)^( 
2/3)*(d*x + c)^(2/3)/(4*b^2*d^3*x^3 + b^2*c^3 + 2*a*b*c^2*d + a^2*c*d^2 + 
4*(2*b^2*c*d^2 + a*b*d^3)*x^2 + (5*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x), 
x)
 

Sympy [F]

\[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\int \frac {\left (a + b x\right )^{3}}{\sqrt [3]{c + d x} \left (a d + b c + 2 b d x\right )^{\frac {4}{3}}}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^3}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\int \frac {{\left (a+b\,x\right )}^3}{{\left (c+d\,x\right )}^{1/3}\,{\left (a\,d+b\,c+2\,b\,d\,x\right )}^{4/3}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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