\(\int \frac {\sqrt [3]{c+d x}}{(a+b x)^{5/2}} \, dx\) [419]

Optimal result

Integrand size = 19, antiderivative size = 417 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{5/2}} \, dx=-\frac {2 \sqrt [3]{c+d x}}{3 b (a+b x)^{3/2}}-\frac {4 d \sqrt [3]{c+d x}}{9 b (b c-a d) \sqrt {a+b x}}+\frac {4 \sqrt {2-\sqrt {3}} d \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),-7+4 \sqrt {3}\right )}{9 \sqrt [4]{3} b^{4/3} (b c-a d) \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \] Output:

-2/3*(d*x+c)^(1/3)/b/(b*x+a)^(3/2)-4/9*d*(d*x+c)^(1/3)/b/(-a*d+b*c)/(b*x+a 
)^(1/2)+4/27*(1/2*6^(1/2)-1/2*2^(1/2))*d*((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c) 
^(1/3))*(((-a*d+b*c)^(2/3)+b^(1/3)*(-a*d+b*c)^(1/3)*(d*x+c)^(1/3)+b^(2/3)* 
(d*x+c)^(2/3))/((1-3^(1/2))*(-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))^2)^(1/ 
2)*EllipticF(((1+3^(1/2))*(-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))/((1-3^(1 
/2))*(-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)),2*I-I*3^(1/2))*3^(3/4)/b^(4/3 
)/(-a*d+b*c)/(b*x+a)^(1/2)/(-(-a*d+b*c)^(1/3)*((-a*d+b*c)^(1/3)-b^(1/3)*(d 
*x+c)^(1/3))/((1-3^(1/2))*(-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.18 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{5/2}} \, dx=-\frac {2 \sqrt [3]{c+d x} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{3},-\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b (a+b x)^{3/2} \sqrt [3]{\frac {b (c+d x)}{b c-a d}}} \] Input:

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

Output:

(-2*(c + d*x)^(1/3)*Hypergeometric2F1[-3/2, -1/3, -1/2, (d*(a + b*x))/(-(b 
*c) + a*d)])/(3*b*(a + b*x)^(3/2)*((b*(c + d*x))/(b*c - a*d))^(1/3))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {57, 61, 73, 760}

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.

\(\displaystyle \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{5/2}} \, dx\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {2 d \int \frac {1}{(a+b x)^{3/2} (c+d x)^{2/3}}dx}{9 b}-\frac {2 \sqrt [3]{c+d x}}{3 b (a+b x)^{3/2}}\)

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 760

\(\displaystyle \frac {2 d \left (\frac {2 \sqrt {2-\sqrt {3}} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt [3]{b} (b c-a d) \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}-\frac {2 \sqrt [3]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{9 b}-\frac {2 \sqrt [3]{c+d x}}{3 b (a+b x)^{3/2}}\)

Input:

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

Output:

(-2*(c + d*x)^(1/3))/(3*b*(a + b*x)^(3/2)) + (2*d*((-2*(c + d*x)^(1/3))/(( 
b*c - a*d)*Sqrt[a + b*x]) + (2*Sqrt[2 - Sqrt[3]]*((b*c - a*d)^(1/3) - b^(1 
/3)*(c + d*x)^(1/3))*Sqrt[((b*c - a*d)^(2/3) + b^(1/3)*(b*c - a*d)^(1/3)*( 
c + d*x)^(1/3) + b^(2/3)*(c + d*x)^(2/3))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) 
 - b^(1/3)*(c + d*x)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*(b*c - a*d) 
^(1/3) - b^(1/3)*(c + d*x)^(1/3))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/ 
3)*(c + d*x)^(1/3))], -7 + 4*Sqrt[3]])/(3^(1/4)*b^(1/3)*(b*c - a*d)*Sqrt[- 
(((b*c - a*d)^(1/3)*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3)))/((1 - S 
qrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2)]*Sqrt[a - (b*c)/d 
+ (b*(c + d*x))/d])))/(9*b)
 

Defintions of rubi rules used

rule 57

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

rule 61

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

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 760

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- 
s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) 
*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x 
] && NegQ[a]
 

Maple [F]

\[\int \frac {\left (x d +c \right )^{\frac {1}{3}}}{\left (b x +a \right )^{\frac {5}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{5/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x+c)^(1/3)/(b*x+a)^(5/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{5/2}} \, dx=\int \frac {\sqrt [3]{c + d x}}{\left (a + b x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((d*x+c)**(1/3)/(b*x+a)**(5/2),x)
 

Output:

Integral((c + d*x)**(1/3)/(a + b*x)**(5/2), x)
 

Maxima [F]

\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{5/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x+c)^(1/3)/(b*x+a)^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{5/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x+c)^(1/3)/(b*x+a)^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{5/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{1/3}}{{\left (a+b\,x\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*(3*(c + d*x)**(1/3)*sqrt(a + b*x)*c + 2*int(((c + d*x)**(1/3)*sqrt(a + 
b*x)*x)/(2*a**4*c*d + 2*a**4*d**2*x - 9*a**3*b*c**2 - 3*a**3*b*c*d*x + 6*a 
**3*b*d**2*x**2 - 27*a**2*b**2*c**2*x - 21*a**2*b**2*c*d*x**2 + 6*a**2*b** 
2*d**2*x**3 - 27*a*b**3*c**2*x**2 - 25*a*b**3*c*d*x**3 + 2*a*b**3*d**2*x** 
4 - 9*b**4*c**2*x**3 - 9*b**4*c*d*x**4),x)*a**4*d**3 - 11*int(((c + d*x)** 
(1/3)*sqrt(a + b*x)*x)/(2*a**4*c*d + 2*a**4*d**2*x - 9*a**3*b*c**2 - 3*a** 
3*b*c*d*x + 6*a**3*b*d**2*x**2 - 27*a**2*b**2*c**2*x - 21*a**2*b**2*c*d*x* 
*2 + 6*a**2*b**2*d**2*x**3 - 27*a*b**3*c**2*x**2 - 25*a*b**3*c*d*x**3 + 2* 
a*b**3*d**2*x**4 - 9*b**4*c**2*x**3 - 9*b**4*c*d*x**4),x)*a**3*b*c*d**2 + 
4*int(((c + d*x)**(1/3)*sqrt(a + b*x)*x)/(2*a**4*c*d + 2*a**4*d**2*x - 9*a 
**3*b*c**2 - 3*a**3*b*c*d*x + 6*a**3*b*d**2*x**2 - 27*a**2*b**2*c**2*x - 2 
1*a**2*b**2*c*d*x**2 + 6*a**2*b**2*d**2*x**3 - 27*a*b**3*c**2*x**2 - 25*a* 
b**3*c*d*x**3 + 2*a*b**3*d**2*x**4 - 9*b**4*c**2*x**3 - 9*b**4*c*d*x**4),x 
)*a**3*b*d**3*x + 9*int(((c + d*x)**(1/3)*sqrt(a + b*x)*x)/(2*a**4*c*d + 2 
*a**4*d**2*x - 9*a**3*b*c**2 - 3*a**3*b*c*d*x + 6*a**3*b*d**2*x**2 - 27*a* 
*2*b**2*c**2*x - 21*a**2*b**2*c*d*x**2 + 6*a**2*b**2*d**2*x**3 - 27*a*b**3 
*c**2*x**2 - 25*a*b**3*c*d*x**3 + 2*a*b**3*d**2*x**4 - 9*b**4*c**2*x**3 - 
9*b**4*c*d*x**4),x)*a**2*b**2*c**2*d - 22*int(((c + d*x)**(1/3)*sqrt(a + b 
*x)*x)/(2*a**4*c*d + 2*a**4*d**2*x - 9*a**3*b*c**2 - 3*a**3*b*c*d*x + 6*a* 
*3*b*d**2*x**2 - 27*a**2*b**2*c**2*x - 21*a**2*b**2*c*d*x**2 + 6*a**2*b...