\(\int \frac {(d+e x)^m (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx\) [252]

Optimal result
Mathematica [A] (warning: unable to verify)
Rubi [A] (warning: unable to verify)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 44, antiderivative size = 171 \[ \int \frac {(d+e x)^m (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {g (d+e x)^m \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{c e^2 (1+m)}+\frac {(b e g (1+2 m)-2 c (d g m+e f (1+m))) (d+e x)^m \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} \operatorname {Hypergeometric2F1}\left (1,1+m,\frac {3}{2},\frac {c d-b e-c e x}{2 c d-b e}\right )}{c e^2 (2 c d-b e) (1+m)} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.75 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x)^m (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {2 (d+e x)^m \sqrt {(d+e x) (-b e+c (d-e x))} \left (e (e f-d g)+\frac {e (b e g (1+2 m)-2 c (d g m+e f (1+m))) \left (\frac {c (d+e x)}{2 c d-b e}\right )^{-\frac {1}{2}-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {1}{2}-m,\frac {3}{2},\frac {-c d+b e+c e x}{-2 c d+b e}\right )}{c}\right )}{e^3 (-2 c d+b e) (1+2 m)} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.84 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.24, 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.

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

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1139

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

\(\Big \downarrow \) 1138

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

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

Input:

int((e*x+d)^m*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),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)^(1/2),x)
 

Fricas [F]

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

integrate((e*x+d)^m*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),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*e^2*x^2 + b*e^2*x - c*d^2 + b*d*e), x)
 

Sympy [F]

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

integrate((e*x+d)**m*(g*x+f)/(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(1/2),x 
)
 

Output:

Integral((d + e*x)**m*(f + g*x)/sqrt(-(d + e*x)*(b*e - c*d + c*e*x)), x)
 

Maxima [F]

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

integrate((e*x+d)^m*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algo 
rithm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*x+d)^m*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algo 
rithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(((f + g*x)*(d + e*x)^m)/(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2),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)^(1/2), x 
)
 

Reduce [F]

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

int((e*x+d)^m*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x)
 

Output:

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