\(\int \frac {\sqrt [4]{c+d x}}{(a+b x)^{5/2}} \, dx\) [435]

Optimal result

Integrand size = 19, antiderivative size = 145 \[ \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{5/2}} \, dx=-\frac {2 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/2}}-\frac {d \sqrt [4]{c+d x}}{3 b (b c-a d) \sqrt {a+b x}}-\frac {d \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 b^{5/4} (b c-a d)^{3/4} \sqrt {a+b x}} \] Output:

-2/3*(d*x+c)^(1/4)/b/(b*x+a)^(3/2)-1/3*d*(d*x+c)^(1/4)/b/(-a*d+b*c)/(b*x+a 
)^(1/2)-1/3*d*(-d*(b*x+a)/(-a*d+b*c))^(1/2)*EllipticF(b^(1/4)*(d*x+c)^(1/4 
)/(-a*d+b*c)^(1/4),I)/b^(5/4)/(-a*d+b*c)^(3/4)/(b*x+a)^(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 = 0.50 \[ \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{5/2}} \, dx=-\frac {2 \sqrt [4]{c+d x} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{4},-\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b (a+b x)^{3/2} \sqrt [4]{\frac {b (c+d x)}{b c-a d}}} \] Input:

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

Output:

(-2*(c + d*x)^(1/4)*Hypergeometric2F1[-3/2, -1/4, -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))^(1/4))
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {57, 61, 73, 765, 762}

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.

\(\displaystyle \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{5/2}} \, dx\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {d \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/4}}dx}{6 b}-\frac {2 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/2}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {d \left (-\frac {d \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}}dx}{2 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 b}-\frac {2 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {d \left (-\frac {2 \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [4]{c+d x}}{b c-a d}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 b}-\frac {2 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/2}}\)

\(\Big \downarrow \) 765

\(\displaystyle \frac {d \left (-\frac {2 \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [4]{c+d x}}{(b c-a d) \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 b}-\frac {2 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {d \left (-\frac {2 \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{\sqrt [4]{b} (b c-a d)^{3/4} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 b}-\frac {2 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/2}}\)

Input:

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

Output:

(-2*(c + d*x)^(1/4))/(3*b*(a + b*x)^(3/2)) + (d*((-2*(c + d*x)^(1/4))/((b* 
c - a*d)*Sqrt[a + b*x]) - (2*Sqrt[1 - (b*(c + d*x))/(b*c - a*d)]*EllipticF 
[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(b^(1/4)*(b*c - 
 a*d)^(3/4)*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])))/(6*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 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 765

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt 
[a + b*x^4]   Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ 
[b/a] &&  !GtQ[a, 0]
 

Maple [F]

\[\int \frac {\left (x d +c \right )^{\frac {1}{4}}}{\left (b x +a \right )^{\frac {5}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{5/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{4}}}{{\left (b x + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x+c)^(1/4)/(b*x+a)^(5/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{5/2}} \, dx=\int \frac {\sqrt [4]{c + d x}}{\left (a + b x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((d*x+c)**(1/4)/(b*x+a)**(5/2),x)
 

Output:

Integral((c + d*x)**(1/4)/(a + b*x)**(5/2), x)
 

Maxima [F]

\[ \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{5/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{4}}}{{\left (b x + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x+c)^(1/4)/(b*x+a)^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{5/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{4}}}{{\left (b x + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x+c)^(1/4)/(b*x+a)^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{5/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{1/4}}{{\left (a+b\,x\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(4*(c + d*x)**(1/4)*sqrt(a + b*x)*c + int(((c + d*x)**(1/4)*sqrt(a + b*x)* 
x)/(a**4*c*d + a**4*d**2*x - 6*a**3*b*c**2 - 3*a**3*b*c*d*x + 3*a**3*b*d** 
2*x**2 - 18*a**2*b**2*c**2*x - 15*a**2*b**2*c*d*x**2 + 3*a**2*b**2*d**2*x* 
*3 - 18*a*b**3*c**2*x**2 - 17*a*b**3*c*d*x**3 + a*b**3*d**2*x**4 - 6*b**4* 
c**2*x**3 - 6*b**4*c*d*x**4),x)*a**4*d**3 - 7*int(((c + d*x)**(1/4)*sqrt(a 
 + b*x)*x)/(a**4*c*d + a**4*d**2*x - 6*a**3*b*c**2 - 3*a**3*b*c*d*x + 3*a* 
*3*b*d**2*x**2 - 18*a**2*b**2*c**2*x - 15*a**2*b**2*c*d*x**2 + 3*a**2*b**2 
*d**2*x**3 - 18*a*b**3*c**2*x**2 - 17*a*b**3*c*d*x**3 + a*b**3*d**2*x**4 - 
 6*b**4*c**2*x**3 - 6*b**4*c*d*x**4),x)*a**3*b*c*d**2 + 2*int(((c + d*x)** 
(1/4)*sqrt(a + b*x)*x)/(a**4*c*d + a**4*d**2*x - 6*a**3*b*c**2 - 3*a**3*b* 
c*d*x + 3*a**3*b*d**2*x**2 - 18*a**2*b**2*c**2*x - 15*a**2*b**2*c*d*x**2 + 
 3*a**2*b**2*d**2*x**3 - 18*a*b**3*c**2*x**2 - 17*a*b**3*c*d*x**3 + a*b**3 
*d**2*x**4 - 6*b**4*c**2*x**3 - 6*b**4*c*d*x**4),x)*a**3*b*d**3*x + 6*int( 
((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**4*c*d + a**4*d**2*x - 6*a**3*b*c**2 
 - 3*a**3*b*c*d*x + 3*a**3*b*d**2*x**2 - 18*a**2*b**2*c**2*x - 15*a**2*b** 
2*c*d*x**2 + 3*a**2*b**2*d**2*x**3 - 18*a*b**3*c**2*x**2 - 17*a*b**3*c*d*x 
**3 + a*b**3*d**2*x**4 - 6*b**4*c**2*x**3 - 6*b**4*c*d*x**4),x)*a**2*b**2* 
c**2*d - 14*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**4*c*d + a**4*d**2*x 
 - 6*a**3*b*c**2 - 3*a**3*b*c*d*x + 3*a**3*b*d**2*x**2 - 18*a**2*b**2*c**2 
*x - 15*a**2*b**2*c*d*x**2 + 3*a**2*b**2*d**2*x**3 - 18*a*b**3*c**2*x**...