\(\int \frac {(d+e x)^m (f+g x)}{(c d^2-b d e-b e^2 x-c e^2 x^2)^{5/2}} \, dx\) [254]

Optimal result

Integrand size = 44, antiderivative size = 180 \[ \int \frac {(d+e x)^m (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx=\frac {g (d+e x)^m}{c e^2 (3-m) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {(b e g (3-2 m)-2 c (e f (3-m)-d g m)) (d+e x)^m \operatorname {Hypergeometric2F1}\left (1,-3+m,-\frac {1}{2},\frac {c d-b e-c e x}{2 c d-b e}\right )}{3 c e^2 (2 c d-b e) (3-m) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}} \] Output:

g*(e*x+d)^m/c/e^2/(3-m)/(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(3/2)-1/3*(b*e*g* 
(3-2*m)-2*c*(e*f*(3-m)-d*g*m))*(e*x+d)^m*hypergeom([1, -3+m],[-1/2],(-c*e* 
x-b*e+c*d)/(-b*e+2*c*d))/c/e^2/(-b*e+2*c*d)/(3-m)/(d*(-b*e+c*d)-b*e^2*x-c* 
e^2*x^2)^(3/2)
 

Mathematica [A] (warning: unable to verify)

Time = 1.11 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.03 \[ \int \frac {(d+e x)^m (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx=\frac {2 (d+e x)^m \left (-\frac {e (-2 c d+b e)^2 (c e f+c d g-b e g)}{c}-e (b e g (3-2 m)+2 c (e f (-3+m)+d g m)) (d+e x) \left (\frac {c (d+e x)}{2 c d-b e}\right )^{\frac {1}{2}-m} (-c d+b e+c e x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {5}{2}-m,\frac {1}{2},\frac {-c d+b e+c e x}{-2 c d+b e}\right )\right )}{3 e^3 (-2 c d+b e)^3 ((d+e x) (-b e+c (d-e x)))^{3/2}} \] Input:

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

Output:

(2*(d + e*x)^m*(-((e*(-2*c*d + b*e)^2*(c*e*f + c*d*g - b*e*g))/c) - e*(b*e 
*g*(3 - 2*m) + 2*c*(e*f*(-3 + m) + d*g*m))*(d + e*x)*((c*(d + e*x))/(2*c*d 
 - b*e))^(1/2 - m)*(-(c*d) + b*e + c*e*x)*Hypergeometric2F1[-1/2, 5/2 - m, 
 1/2, (-(c*d) + b*e + c*e*x)/(-2*c*d + b*e)]))/(3*e^3*(-2*c*d + b*e)^3*((d 
 + e*x)*(-(b*e) + c*(d - e*x)))^(3/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.53 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.27, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1221, 1139, 1138, 80, 79}

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.

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

\(\Big \downarrow \) 1221

\(\displaystyle \frac {g (d+e x)^m}{c e^2 (3-m) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {(b e g (3-2 m)-2 c (e f (3-m)-d g m)) \int \frac {(d+e x)^m}{\left (-c x^2 e^2-b x e^2+d (c d-b e)\right )^{5/2}}dx}{2 c e (3-m)}\)

\(\Big \downarrow \) 1139

\(\displaystyle \frac {g (d+e x)^m}{c e^2 (3-m) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {(d+e x)^m \left (\frac {e x}{d}+1\right )^{-m} (b e g (3-2 m)-2 c (e f (3-m)-d g m)) \int \frac {\left (\frac {e x}{d}+1\right )^m}{\left (-c x^2 e^2-b x e^2+d (c d-b e)\right )^{5/2}}dx}{2 c e (3-m)}\)

\(\Big \downarrow \) 1138

\(\displaystyle \frac {g (d+e x)^m}{c e^2 (3-m) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {(d+e x)^m \left (\frac {e x}{d}+1\right )^{\frac {1}{2}-m} \sqrt {d (c d-b e)-c d e x} (b e g (3-2 m)-2 c (e f (3-m)-d g m)) \int \frac {\left (\frac {e x}{d}+1\right )^{m-\frac {5}{2}}}{(d (c d-b e)-c d e x)^{5/2}}dx}{2 c e (3-m) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {g (d+e x)^m}{c e^2 (3-m) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {c d^2 (d+e x)^m \sqrt {d (c d-b e)-c d e x} \left (\frac {c (d+e x)}{2 c d-b e}\right )^{\frac {1}{2}-m} (b e g (3-2 m)-2 c (e f (3-m)-d g m)) \int \frac {\left (\frac {c d}{2 c d-b e}+\frac {c e x}{2 c d-b e}\right )^{m-\frac {5}{2}}}{(d (c d-b e)-c d e x)^{5/2}}dx}{2 e (3-m) (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {g (d+e x)^m}{c e^2 (3-m) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}-\frac {d (d+e x)^m \left (\frac {c (d+e x)}{2 c d-b e}\right )^{\frac {1}{2}-m} (b e g (3-2 m)-2 c (e f (3-m)-d g m)) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{2}-m,-\frac {1}{2},\frac {c d-b e-c e x}{2 c d-b e}\right )}{3 e^2 (3-m) (2 c d-b e)^2 (d (c d-b e)-c d e x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\)

Input:

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

Output:

(g*(d + e*x)^m)/(c*e^2*(3 - m)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2) 
) - (d*(b*e*g*(3 - 2*m) - 2*c*(e*f*(3 - m) - d*g*m))*(d + e*x)^m*((c*(d + 
e*x))/(2*c*d - b*e))^(1/2 - m)*Hypergeometric2F1[-3/2, 5/2 - m, -1/2, (c*d 
 - b*e - c*e*x)/(2*c*d - b*e)])/(3*e^2*(2*c*d - b*e)^2*(3 - m)*(d*(c*d - b 
*e) - c*d*e*x)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])
 

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 1138

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[d^m*((a + b*x + c*x^2)^FracPart[p]/((1 + e*(x/d))^FracPart[p] 
*(a/d + (c*x)/e)^FracPart[p]))   Int[(1 + e*(x/d))^(m + p)*(a/d + (c/e)*x)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
&& (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (IntegerQ[3*p] || Integer 
Q[4*p]))
 

rule 1139

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

rule 1221

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((e*x+d)^m*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x, algo 
rithm="fricas")
 

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^m (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

integrate((e*x+d)^m*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x, algo 
rithm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*x+d)^m*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x, algo 
rithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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