\(\int \frac {(e x)^m (A+B x)}{(a+b x+c x^2)^{5/2}} \, dx\) [238]

Optimal result

Integrand size = 25, antiderivative size = 281 \[ \int \frac {(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {A (e x)^{1+m} \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )^{5/2} \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )^{5/2} \operatorname {AppellF1}\left (1+m,\frac {5}{2},\frac {5}{2},2+m,-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e (1+m) \left (a+b x+c x^2\right )^{5/2}}+\frac {B (e x)^{2+m} \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )^{5/2} \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )^{5/2} \operatorname {AppellF1}\left (2+m,\frac {5}{2},\frac {5}{2},3+m,-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e^2 (2+m) \left (a+b x+c x^2\right )^{5/2}} \] Output:

A*(e*x)^(1+m)*(1+2*c*x/(b-(-4*a*c+b^2)^(1/2)))^(5/2)*(1+2*c*x/(b+(-4*a*c+b 
^2)^(1/2)))^(5/2)*AppellF1(1+m,5/2,5/2,2+m,-2*c*x/(b-(-4*a*c+b^2)^(1/2)),- 
2*c*x/(b+(-4*a*c+b^2)^(1/2)))/e/(1+m)/(c*x^2+b*x+a)^(5/2)+B*(e*x)^(2+m)*(1 
+2*c*x/(b-(-4*a*c+b^2)^(1/2)))^(5/2)*(1+2*c*x/(b+(-4*a*c+b^2)^(1/2)))^(5/2 
)*AppellF1(2+m,5/2,5/2,3+m,-2*c*x/(b-(-4*a*c+b^2)^(1/2)),-2*c*x/(b+(-4*a*c 
+b^2)^(1/2)))/e^2/(2+m)/(c*x^2+b*x+a)^(5/2)
 

Mathematica [A] (warning: unable to verify)

Time = 4.76 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.84 \[ \int \frac {(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {x (e x)^m \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}} \left (A (2+m) \operatorname {AppellF1}\left (1+m,\frac {5}{2},\frac {5}{2},2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+B (1+m) x \operatorname {AppellF1}\left (2+m,\frac {5}{2},\frac {5}{2},3+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )\right )}{a^2 (1+m) (2+m) \sqrt {a+x (b+c x)}} \] Input:

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

Output:

(x*(e*x)^m*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(b - Sqrt[b^2 - 4*a*c])]*S 
qrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(b + Sqrt[b^2 - 4*a*c])]*(A*(2 + m)*Ap 
pellF1[1 + m, 5/2, 5/2, 2 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/( 
-b + Sqrt[b^2 - 4*a*c])] + B*(1 + m)*x*AppellF1[2 + m, 5/2, 5/2, 3 + m, (- 
2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/(-b + Sqrt[b^2 - 4*a*c])]))/(a^2*( 
1 + m)*(2 + m)*Sqrt[a + x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1269, 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.

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

\(\Big \downarrow \) 1269

\(\displaystyle A \int \frac {(e x)^m}{\left (c x^2+b x+a\right )^{5/2}}dx+\frac {B \int \frac {(e x)^{m+1}}{\left (c x^2+b x+a\right )^{5/2}}dx}{e}\)

\(\Big \downarrow \) 1179

\(\displaystyle \frac {A \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{5/2} \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )^{5/2} \int \frac {(e x)^m}{\left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{5/2} \left (\frac {2 c x}{b+\sqrt {b^2-4 a c}}+1\right )^{5/2}}d(e x)}{e \left (a+b x+c x^2\right )^{5/2}}+\frac {B \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{5/2} \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )^{5/2} \int \frac {(e x)^{m+1}}{\left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{5/2} \left (\frac {2 c x}{b+\sqrt {b^2-4 a c}}+1\right )^{5/2}}d(e x)}{e^2 \left (a+b x+c x^2\right )^{5/2}}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {A (e x)^{m+1} \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{5/2} \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )^{5/2} \operatorname {AppellF1}\left (m+1,\frac {5}{2},\frac {5}{2},m+2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e (m+1) \left (a+b x+c x^2\right )^{5/2}}+\frac {B (e x)^{m+2} \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{5/2} \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )^{5/2} \operatorname {AppellF1}\left (m+2,\frac {5}{2},\frac {5}{2},m+3,-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e^2 (m+2) \left (a+b x+c x^2\right )^{5/2}}\)

Input:

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

Output:

(A*(e*x)^(1 + m)*(1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^(5/2)*(1 + (2*c*x)/ 
(b + Sqrt[b^2 - 4*a*c]))^(5/2)*AppellF1[1 + m, 5/2, 5/2, 2 + m, (-2*c*x)/( 
b - Sqrt[b^2 - 4*a*c]), (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(e*(1 + m)*(a + 
 b*x + c*x^2)^(5/2)) + (B*(e*x)^(2 + m)*(1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c 
]))^(5/2)*(1 + (2*c*x)/(b + Sqrt[b^2 - 4*a*c]))^(5/2)*AppellF1[2 + m, 5/2, 
 5/2, 3 + m, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x)/(b + Sqrt[b^2 - 4* 
a*c])])/(e^2*(2 + m)*(a + 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]
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

Maple [F]

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

Input:

int((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(5/2),x)
 

Output:

int((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(5/2),x)
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

integrate((e*x)**m*(B*x+A)/(c*x**2+b*x+a)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((B*x + A)*(e*x)^m/(c*x^2 + b*x + a)^(5/2), x)
 

Giac [F]

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

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

Output:

integrate((B*x + A)*(e*x)^m/(c*x^2 + b*x + a)^(5/2), x)
 

Mupad [F(-1)]

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

int(((e*x)^m*(A + B*x))/(a + b*x + c*x^2)^(5/2),x)
 

Output:

int(((e*x)^m*(A + B*x))/(a + b*x + c*x^2)^(5/2), x)
 

Reduce [F]

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

int((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(5/2),x)
 

Output:

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