\(\int \frac {(c+d x)^{5/4}}{(a+b x)^{9/2}} \, dx\) [450]

Optimal result

Integrand size = 19, antiderivative size = 213 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/2}} \, dx=-\frac {d \sqrt [4]{c+d x}}{7 b^2 (a+b x)^{5/2}}-\frac {d^2 \sqrt [4]{c+d x}}{42 b^2 (b c-a d) (a+b x)^{3/2}}+\frac {5 d^3 \sqrt [4]{c+d x}}{84 b^2 (b c-a d)^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{7 b (a+b x)^{7/2}}+\frac {5 d^3 \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{84 b^{9/4} (b c-a d)^{7/4} \sqrt {a+b x}} \] Output:

-1/7*d*(d*x+c)^(1/4)/b^2/(b*x+a)^(5/2)-1/42*d^2*(d*x+c)^(1/4)/b^2/(-a*d+b* 
c)/(b*x+a)^(3/2)+5/84*d^3*(d*x+c)^(1/4)/b^2/(-a*d+b*c)^2/(b*x+a)^(1/2)-2/7 
*(d*x+c)^(5/4)/b/(b*x+a)^(7/2)+5/84*d^3*(-d*(b*x+a)/(-a*d+b*c))^(1/2)*Elli 
pticF(b^(1/4)*(d*x+c)^(1/4)/(-a*d+b*c)^(1/4),I)/b^(9/4)/(-a*d+b*c)^(7/4)/( 
b*x+a)^(1/2)
 

Mathematica [C] (verified)

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

Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.34 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/2}} \, dx=-\frac {2 (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},-\frac {5}{4},-\frac {5}{2},\frac {d (a+b x)}{-b c+a d}\right )}{7 b (a+b x)^{7/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \] Input:

Integrate[(c + d*x)^(5/4)/(a + b*x)^(9/2),x]
 

Output:

(-2*(c + d*x)^(5/4)*Hypergeometric2F1[-7/2, -5/4, -5/2, (d*(a + b*x))/(-(b 
*c) + a*d)])/(7*b*(a + b*x)^(7/2)*((b*(c + d*x))/(b*c - a*d))^(5/4))
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {57, 57, 61, 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.

\(\displaystyle \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/2}} \, dx\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {5 d \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{7/2}}dx}{14 b}-\frac {2 (c+d x)^{5/4}}{7 b (a+b x)^{7/2}}\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {5 d \left (\frac {d \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/4}}dx}{10 b}-\frac {2 \sqrt [4]{c+d x}}{5 b (a+b x)^{5/2}}\right )}{14 b}-\frac {2 (c+d x)^{5/4}}{7 b (a+b x)^{7/2}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {5 d \left (\frac {d \left (-\frac {5 d \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/4}}dx}{6 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\right )}{10 b}-\frac {2 \sqrt [4]{c+d x}}{5 b (a+b x)^{5/2}}\right )}{14 b}-\frac {2 (c+d x)^{5/4}}{7 b (a+b x)^{7/2}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {5 d \left (\frac {d \left (-\frac {5 d \left (-\frac {d \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}}dx}{2 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\right )}{10 b}-\frac {2 \sqrt [4]{c+d x}}{5 b (a+b x)^{5/2}}\right )}{14 b}-\frac {2 (c+d x)^{5/4}}{7 b (a+b x)^{7/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {5 d \left (\frac {d \left (-\frac {5 d \left (-\frac {2 \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [4]{c+d x}}{b c-a d}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\right )}{10 b}-\frac {2 \sqrt [4]{c+d x}}{5 b (a+b x)^{5/2}}\right )}{14 b}-\frac {2 (c+d x)^{5/4}}{7 b (a+b x)^{7/2}}\)

\(\Big \downarrow \) 765

\(\displaystyle \frac {5 d \left (\frac {d \left (-\frac {5 d \left (-\frac {2 \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [4]{c+d x}}{(b c-a d) \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\right )}{10 b}-\frac {2 \sqrt [4]{c+d x}}{5 b (a+b x)^{5/2}}\right )}{14 b}-\frac {2 (c+d x)^{5/4}}{7 b (a+b x)^{7/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {5 d \left (\frac {d \left (-\frac {5 d \left (-\frac {2 \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{\sqrt [4]{b} (b c-a d)^{3/4} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\right )}{10 b}-\frac {2 \sqrt [4]{c+d x}}{5 b (a+b x)^{5/2}}\right )}{14 b}-\frac {2 (c+d x)^{5/4}}{7 b (a+b x)^{7/2}}\)

Input:

Int[(c + d*x)^(5/4)/(a + b*x)^(9/2),x]
 

Output:

(-2*(c + d*x)^(5/4))/(7*b*(a + b*x)^(7/2)) + (5*d*((-2*(c + d*x)^(1/4))/(5 
*b*(a + b*x)^(5/2)) + (d*((-2*(c + d*x)^(1/4))/(3*(b*c - a*d)*(a + b*x)^(3 
/2)) - (5*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*c - a*d))))/(10*b)))/(14*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]

\[\int \frac {\left (x d +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {9}{2}}}d x\]

Input:

int((d*x+c)^(5/4)/(b*x+a)^(9/2),x)
 

Output:

int((d*x+c)^(5/4)/(b*x+a)^(9/2),x)
 

Fricas [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {9}{2}}} \,d x } \] Input:

integrate((d*x+c)^(5/4)/(b*x+a)^(9/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b*x + a)*(d*x + c)^(5/4)/(b^5*x^5 + 5*a*b^4*x^4 + 10*a^2*b^3 
*x^3 + 10*a^3*b^2*x^2 + 5*a^4*b*x + a^5), x)
 

Sympy [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\left (a + b x\right )^{\frac {9}{2}}}\, dx \] Input:

integrate((d*x+c)**(5/4)/(b*x+a)**(9/2),x)
 

Output:

Integral((c + d*x)**(5/4)/(a + b*x)**(9/2), x)
 

Maxima [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {9}{2}}} \,d x } \] Input:

integrate((d*x+c)^(5/4)/(b*x+a)^(9/2),x, algorithm="maxima")
 

Output:

integrate((d*x + c)^(5/4)/(b*x + a)^(9/2), x)
 

Giac [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {9}{2}}} \,d x } \] Input:

integrate((d*x+c)^(5/4)/(b*x+a)^(9/2),x, algorithm="giac")
 

Output:

integrate((d*x + c)^(5/4)/(b*x + a)^(9/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{9/2}} \,d x \] Input:

int((c + d*x)^(5/4)/(a + b*x)^(9/2),x)
 

Output:

int((c + d*x)^(5/4)/(a + b*x)^(9/2), x)
 

Reduce [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/2}} \, dx=\text {too large to display} \] Input:

int((d*x+c)^(5/4)/(b*x+a)^(9/2),x)
 

Output:

(16*(c + d*x)**(1/4)*sqrt(a + b*x)*a*c*d - 4*(c + d*x)**(1/4)*sqrt(a + b*x 
)*a*d**2*x + 36*(c + d*x)**(1/4)*sqrt(a + b*x)*b*c**2 + 56*(c + d*x)**(1/4 
)*sqrt(a + b*x)*b*c*d*x + 5*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**6*c 
*d + a**6*d**2*x - 14*a**5*b*c**2 - 9*a**5*b*c*d*x + 5*a**5*b*d**2*x**2 - 
70*a**4*b**2*c**2*x - 60*a**4*b**2*c*d*x**2 + 10*a**4*b**2*d**2*x**3 - 140 
*a**3*b**3*c**2*x**2 - 130*a**3*b**3*c*d*x**3 + 10*a**3*b**3*d**2*x**4 - 1 
40*a**2*b**4*c**2*x**3 - 135*a**2*b**4*c*d*x**4 + 5*a**2*b**4*d**2*x**5 - 
70*a*b**5*c**2*x**4 - 69*a*b**5*c*d*x**5 + a*b**5*d**2*x**6 - 14*b**6*c**2 
*x**5 - 14*b**6*c*d*x**6),x)*a**7*d**4 - 80*int(((c + d*x)**(1/4)*sqrt(a + 
 b*x)*x)/(a**6*c*d + a**6*d**2*x - 14*a**5*b*c**2 - 9*a**5*b*c*d*x + 5*a** 
5*b*d**2*x**2 - 70*a**4*b**2*c**2*x - 60*a**4*b**2*c*d*x**2 + 10*a**4*b**2 
*d**2*x**3 - 140*a**3*b**3*c**2*x**2 - 130*a**3*b**3*c*d*x**3 + 10*a**3*b* 
*3*d**2*x**4 - 140*a**2*b**4*c**2*x**3 - 135*a**2*b**4*c*d*x**4 + 5*a**2*b 
**4*d**2*x**5 - 70*a*b**5*c**2*x**4 - 69*a*b**5*c*d*x**5 + a*b**5*d**2*x** 
6 - 14*b**6*c**2*x**5 - 14*b**6*c*d*x**6),x)*a**6*b*c*d**3 + 20*int(((c + 
d*x)**(1/4)*sqrt(a + b*x)*x)/(a**6*c*d + a**6*d**2*x - 14*a**5*b*c**2 - 9* 
a**5*b*c*d*x + 5*a**5*b*d**2*x**2 - 70*a**4*b**2*c**2*x - 60*a**4*b**2*c*d 
*x**2 + 10*a**4*b**2*d**2*x**3 - 140*a**3*b**3*c**2*x**2 - 130*a**3*b**3*c 
*d*x**3 + 10*a**3*b**3*d**2*x**4 - 140*a**2*b**4*c**2*x**3 - 135*a**2*b**4 
*c*d*x**4 + 5*a**2*b**4*d**2*x**5 - 70*a*b**5*c**2*x**4 - 69*a*b**5*c*d...