\(\int \frac {(b x)^{5/2} (c+d x)^n}{(e+f x)^2} \, dx\) [570]

Optimal result

Integrand size = 22, antiderivative size = 61 \[ \int \frac {(b x)^{5/2} (c+d x)^n}{(e+f x)^2} \, dx=\frac {2 (b x)^{7/2} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \operatorname {AppellF1}\left (\frac {7}{2},-n,2,\frac {9}{2},-\frac {d x}{c},-\frac {f x}{e}\right )}{7 b e^2} \] Output:

2/7*(b*x)^(7/2)*(d*x+c)^n*AppellF1(7/2,-n,2,9/2,-d*x/c,-f*x/e)/b/e^2/((1+d 
*x/c)^n)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(133\) vs. \(2(61)=122\).

Time = 0.67 (sec) , antiderivative size = 133, normalized size of antiderivative = 2.18 \[ \int \frac {(b x)^{5/2} (c+d x)^n}{(e+f x)^2} \, dx=\frac {2 b^2 \sqrt {b x} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \left (9 e \operatorname {AppellF1}\left (\frac {1}{2},-n,1,\frac {3}{2},-\frac {d x}{c},-\frac {f x}{e}\right )-3 e \operatorname {AppellF1}\left (\frac {1}{2},-n,2,\frac {3}{2},-\frac {d x}{c},-\frac {f x}{e}\right )-6 e \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )+f x \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-n,\frac {5}{2},-\frac {d x}{c}\right )\right )}{3 f^3} \] Input:

Integrate[((b*x)^(5/2)*(c + d*x)^n)/(e + f*x)^2,x]
 

Output:

(2*b^2*Sqrt[b*x]*(c + d*x)^n*(9*e*AppellF1[1/2, -n, 1, 3/2, -((d*x)/c), -( 
(f*x)/e)] - 3*e*AppellF1[1/2, -n, 2, 3/2, -((d*x)/c), -((f*x)/e)] - 6*e*Hy 
pergeometric2F1[1/2, -n, 3/2, -((d*x)/c)] + f*x*Hypergeometric2F1[3/2, -n, 
 5/2, -((d*x)/c)]))/(3*f^3*(1 + (d*x)/c)^n)
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {148, 27, 395, 394}

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.

\(\displaystyle \int \frac {(b x)^{5/2} (c+d x)^n}{(e+f x)^2} \, dx\)

\(\Big \downarrow \) 148

\(\displaystyle \frac {2 \int \frac {b^5 x^3 (c+d x)^n}{(b e+b f x)^2}d\sqrt {b x}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle 2 b \int \frac {b^3 x^3 (c+d x)^n}{(b e+b f x)^2}d\sqrt {b x}\)

\(\Big \downarrow \) 395

\(\displaystyle 2 b (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \int \frac {b^3 x^3 \left (\frac {d x}{c}+1\right )^n}{(b e+b f x)^2}d\sqrt {b x}\)

\(\Big \downarrow \) 394

\(\displaystyle \frac {2 (b x)^{7/2} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {7}{2},-n,2,\frac {9}{2},-\frac {d x}{c},-\frac {f x}{e}\right )}{7 b e^2}\)

Input:

Int[((b*x)^(5/2)*(c + d*x)^n)/(e + f*x)^2,x]
 

Output:

(2*(b*x)^(7/2)*(c + d*x)^n*AppellF1[7/2, -n, 2, 9/2, -((d*x)/c), -((f*x)/e 
)])/(7*b*e^2*(1 + (d*x)/c)^n)
 

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 148

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), 
x_] :> With[{k = Denominator[m]}, Simp[k/b   Subst[Int[x^(k*(m + 1) - 1)*(c 
 + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, 
 d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
 

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ 
FracPart[p])   Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ 
[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 
1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

Maple [F]

\[\int \frac {\left (b x \right )^{\frac {5}{2}} \left (x d +c \right )^{n}}{\left (f x +e \right )^{2}}d x\]

Input:

int((b*x)^(5/2)*(d*x+c)^n/(f*x+e)^2,x)
 

Output:

int((b*x)^(5/2)*(d*x+c)^n/(f*x+e)^2,x)
 

Fricas [F]

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

integrate((b*x)^(5/2)*(d*x+c)^n/(f*x+e)^2,x, algorithm="fricas")
 

Output:

integral(sqrt(b*x)*(d*x + c)^n*b^2*x^2/(f^2*x^2 + 2*e*f*x + e^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(b x)^{5/2} (c+d x)^n}{(e+f x)^2} \, dx=\text {Timed out} \] Input:

integrate((b*x)**(5/2)*(d*x+c)**n/(f*x+e)**2,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((b*x)^(5/2)*(d*x+c)^n/(f*x+e)^2,x, algorithm="maxima")
 

Output:

integrate((b*x)^(5/2)*(d*x + c)^n/(f*x + e)^2, x)
 

Giac [F]

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

integrate((b*x)^(5/2)*(d*x+c)^n/(f*x+e)^2,x, algorithm="giac")
 

Output:

integrate((b*x)^(5/2)*(d*x + c)^n/(f*x + e)^2, x)
 

Mupad [F(-1)]

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

int(((b*x)^(5/2)*(c + d*x)^n)/(e + f*x)^2,x)
 

Output:

int(((b*x)^(5/2)*(c + d*x)^n)/(e + f*x)^2, x)
 

Reduce [F]

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

int((b*x)^(5/2)*(d*x+c)^n/(f*x+e)^2,x)
 

Output:

(sqrt(b)*b**2*(12*sqrt(x)*(c + d*x)**n*c**2*e*f*n + 4*sqrt(x)*(c + d*x)**n 
*c**2*f**2*n*x - 12*sqrt(x)*(c + d*x)**n*c*d*e**2*n - 30*sqrt(x)*(c + d*x) 
**n*c*d*e**2 - 8*sqrt(x)*(c + d*x)**n*c*d*e*f*n**2*x - 8*sqrt(x)*(c + d*x) 
**n*c*d*e*f*n*x - 10*sqrt(x)*(c + d*x)**n*c*d*e*f*x + 4*sqrt(x)*(c + d*x)* 
*n*c*d*f**2*n*x**2 + 2*sqrt(x)*(c + d*x)**n*c*d*f**2*x**2 + 8*sqrt(x)*(c + 
 d*x)**n*d**2*e**2*n**2*x + 24*sqrt(x)*(c + d*x)**n*d**2*e**2*n*x + 10*sqr 
t(x)*(c + d*x)**n*d**2*e**2*x - 8*sqrt(x)*(c + d*x)**n*d**2*e*f*n**2*x**2 
- 8*sqrt(x)*(c + d*x)**n*d**2*e*f*n*x**2 - 2*sqrt(x)*(c + d*x)**n*d**2*e*f 
*x**2 - 24*int((c + d*x)**n/(4*sqrt(x)*c**2*e**2*f*n**2 + 8*sqrt(x)*c**2*e 
**2*f*n + 3*sqrt(x)*c**2*e**2*f + 8*sqrt(x)*c**2*e*f**2*n**2*x + 16*sqrt(x 
)*c**2*e*f**2*n*x + 6*sqrt(x)*c**2*e*f**2*x + 4*sqrt(x)*c**2*f**3*n**2*x** 
2 + 8*sqrt(x)*c**2*f**3*n*x**2 + 3*sqrt(x)*c**2*f**3*x**2 - 8*sqrt(x)*c*d* 
e**3*n**3 - 20*sqrt(x)*c*d*e**3*n**2 - 14*sqrt(x)*c*d*e**3*n - 3*sqrt(x)*c 
*d*e**3 - 16*sqrt(x)*c*d*e**2*f*n**3*x - 36*sqrt(x)*c*d*e**2*f*n**2*x - 20 
*sqrt(x)*c*d*e**2*f*n*x - 3*sqrt(x)*c*d*e**2*f*x - 8*sqrt(x)*c*d*e*f**2*n* 
*3*x**2 - 12*sqrt(x)*c*d*e*f**2*n**2*x**2 + 2*sqrt(x)*c*d*e*f**2*n*x**2 + 
3*sqrt(x)*c*d*e*f**2*x**2 + 4*sqrt(x)*c*d*f**3*n**2*x**3 + 8*sqrt(x)*c*d*f 
**3*n*x**3 + 3*sqrt(x)*c*d*f**3*x**3 - 8*sqrt(x)*d**2*e**3*n**3*x - 20*sqr 
t(x)*d**2*e**3*n**2*x - 14*sqrt(x)*d**2*e**3*n*x - 3*sqrt(x)*d**2*e**3*x - 
 16*sqrt(x)*d**2*e**2*f*n**3*x**2 - 40*sqrt(x)*d**2*e**2*f*n**2*x**2 - ...