\(\int \frac {(d+e x)^m}{(b x+c x^2)^{5/2}} \, dx\) [248]

Optimal result

Integrand size = 21, antiderivative size = 81 \[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \sqrt {1+\frac {c x}{b}} (d+e x)^m \left (1+\frac {e x}{d}\right )^{-m} \operatorname {AppellF1}\left (-\frac {3}{2},\frac {5}{2},-m,-\frac {1}{2},-\frac {c x}{b},-\frac {e x}{d}\right )}{3 b^2 x \sqrt {b x+c x^2}} \] Output:

-2/3*(1+c*x/b)^(1/2)*(e*x+d)^m*AppellF1(-3/2,5/2,-m,-1/2,-c*x/b,-e*x/d)/b^ 
2/x/((1+e*x/d)^m)/(c*x^2+b*x)^(1/2)
 

Mathematica [A] (verified)

Time = 2.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.94 \[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 x \left (\frac {b+c x}{b}\right )^{5/2} (d+e x)^m \left (\frac {d+e x}{d}\right )^{-m} \operatorname {AppellF1}\left (-\frac {3}{2},\frac {5}{2},-m,-\frac {1}{2},-\frac {c x}{b},-\frac {e x}{d}\right )}{3 (x (b+c x))^{5/2}} \] Input:

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

Output:

(-2*x*((b + c*x)/b)^(5/2)*(d + e*x)^m*AppellF1[-3/2, 5/2, -m, -1/2, -((c*x 
)/b), -((e*x)/d)])/(3*(x*(b + c*x))^(5/2)*((d + e*x)/d)^m)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.30, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1179, 150}

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.

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

\(\Big \downarrow \) 1179

\(\displaystyle \frac {\left (-\frac {e x}{d}\right )^{5/2} \left (1-\frac {c (d+e x)}{c d-b e}\right )^{5/2} \int \frac {(d+e x)^m}{\left (1-\frac {d+e x}{d}\right )^{5/2} \left (1-\frac {c (d+e x)}{c d-b e}\right )^{5/2}}d(d+e x)}{e \left (b x+c x^2\right )^{5/2}}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {\left (-\frac {e x}{d}\right )^{5/2} (d+e x)^{m+1} \left (1-\frac {c (d+e x)}{c d-b e}\right )^{5/2} \operatorname {AppellF1}\left (m+1,\frac {5}{2},\frac {5}{2},m+2,\frac {d+e x}{d},\frac {c (d+e x)}{c d-b e}\right )}{e (m+1) \left (b x+c x^2\right )^{5/2}}\)

Input:

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

Output:

((-((e*x)/d))^(5/2)*(d + e*x)^(1 + m)*(1 - (c*(d + e*x))/(c*d - b*e))^(5/2 
)*AppellF1[1 + m, 5/2, 5/2, 2 + m, (d + e*x)/d, (c*(d + e*x))/(c*d - b*e)] 
)/(e*(1 + m)*(b*x + c*x^2)^(5/2))
 

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 1179

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(a + b*x + c*x^2)^p/(e*(1 - ( 
d + e*x)/(d - e*((b - q)/(2*c))))^p*(1 - (d + e*x)/(d - e*((b + q)/(2*c)))) 
^p)   Subst[Int[x^m*Simp[1 - x/(d - e*((b - q)/(2*c))), x]^p*Simp[1 - x/(d 
- e*((b + q)/(2*c))), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, b, c, d, e, m 
, p}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral((d + e*x)**m/(x*(b + c*x))**(5/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {\left (e x +d \right )^{m}}{\sqrt {x}\, \sqrt {c x +b}\, b^{2} x^{2}+2 \sqrt {x}\, \sqrt {c x +b}\, b c \,x^{3}+\sqrt {x}\, \sqrt {c x +b}\, c^{2} x^{4}}d x \] Input:

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

Output:

int((d + e*x)**m/(sqrt(x)*sqrt(b + c*x)*b**2*x**2 + 2*sqrt(x)*sqrt(b + c*x 
)*b*c*x**3 + sqrt(x)*sqrt(b + c*x)*c**2*x**4),x)