\(\int \frac {(a+b x) (g+h x)}{(c+d x)^2 \sqrt {e+f x}} \, dx\) [134]

Optimal result

Integrand size = 27, antiderivative size = 154 \[ \int \frac {(a+b x) (g+h x)}{(c+d x)^2 \sqrt {e+f x}} \, dx=\frac {2 b h \sqrt {e+f x}}{d^2 f}+\frac {(b c-a d) (d g-c h) \sqrt {e+f x}}{d^2 (d e-c f) (c+d x)}+\frac {\left (f \left (a d^2 g-b c^2 h\right )-(2 d e-c f) (b d g-2 b c h+a d h)\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{5/2} (d e-c f)^{3/2}} \] Output:

2*b*h*(f*x+e)^(1/2)/d^2/f+(-a*d+b*c)*(-c*h+d*g)*(f*x+e)^(1/2)/d^2/(-c*f+d* 
e)/(d*x+c)+(f*(a*d^2*g-b*c^2*h)-(-c*f+2*d*e)*(a*d*h-2*b*c*h+b*d*g))*arctan 
h(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^(1/2))/d^(5/2)/(-c*f+d*e)^(3/2)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b x) (g+h x)}{(c+d x)^2 \sqrt {e+f x}} \, dx=\frac {\sqrt {e+f x} \left (a d f (-d g+c h)+b \left (-3 c^2 f h+2 d^2 e h x+c d (f g+2 e h-2 f h x)\right )\right )}{d^2 f (d e-c f) (c+d x)}-\frac {\left (-a d (d f g-2 d e h+c f h)+b \left (2 d^2 e g+3 c^2 f h-c d (f g+4 e h)\right )\right ) \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{d^{5/2} (-d e+c f)^{3/2}} \] Input:

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

Output:

(Sqrt[e + f*x]*(a*d*f*(-(d*g) + c*h) + b*(-3*c^2*f*h + 2*d^2*e*h*x + c*d*( 
f*g + 2*e*h - 2*f*h*x))))/(d^2*f*(d*e - c*f)*(c + d*x)) - ((-(a*d*(d*f*g - 
 2*d*e*h + c*f*h)) + b*(2*d^2*e*g + 3*c^2*f*h - c*d*(f*g + 4*e*h)))*ArcTan 
[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c*f]])/(d^(5/2)*(-(d*e) + c*f)^(3/2 
))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {163, 73, 221}

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.

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

\(\Big \downarrow \) 163

\(\displaystyle -\frac {\left (a d (c f h-2 d e h+d f g)-b \left (3 c^2 f h-c d (4 e h+f g)+2 d^2 e g\right )\right ) \int \frac {1}{(c+d x) \sqrt {e+f x}}dx}{2 d^2 (d e-c f)}-\frac {\sqrt {e+f x} (a d f (d g-c h)+b c (3 c f h-d (2 e h+f g))-2 b d h x (d e-c f))}{d^2 f (c+d x) (d e-c f)}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\left (a d (c f h-2 d e h+d f g)-b \left (3 c^2 f h-c d (4 e h+f g)+2 d^2 e g\right )\right ) \int \frac {1}{c+\frac {d (e+f x)}{f}-\frac {d e}{f}}d\sqrt {e+f x}}{d^2 f (d e-c f)}-\frac {\sqrt {e+f x} (a d f (d g-c h)+b c (3 c f h-d (2 e h+f g))-2 b d h x (d e-c f))}{d^2 f (c+d x) (d e-c f)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right ) \left (a d (c f h-2 d e h+d f g)-b \left (3 c^2 f h-c d (4 e h+f g)+2 d^2 e g\right )\right )}{d^{5/2} (d e-c f)^{3/2}}-\frac {\sqrt {e+f x} (a d f (d g-c h)+b c (3 c f h-d (2 e h+f g))-2 b d h x (d e-c f))}{d^2 f (c+d x) (d e-c f)}\)

Input:

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

Output:

-((Sqrt[e + f*x]*(a*d*f*(d*g - c*h) + b*c*(3*c*f*h - d*(f*g + 2*e*h)) - 2* 
b*d*(d*e - c*f)*h*x))/(d^2*f*(d*e - c*f)*(c + d*x))) + ((a*d*(d*f*g - 2*d* 
e*h + c*f*h) - b*(2*d^2*e*g + 3*c^2*f*h - c*d*(f*g + 4*e*h)))*ArcTanh[(Sqr 
t[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d^(5/2)*(d*e - c*f)^(3/2))
 

Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 163

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) 
)*((g_.) + (h_.)*(x_)), x_] :> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n 
+ 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(b*c - a*d)* 
(m + 1)*x)/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)))*(a + b*x)^(m + 1)*(c + 
d*x)^(n + 1), x] - Simp[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f 
*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c* 
d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2* 
d*(b*c - a*d)*(m + 1)*(m + n + 3))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((GeQ[m, -2] && LtQ[m, - 
1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.18

method	result	size
risch	\(\frac {2 b h \sqrt {f x +e}}{d^{2} f}+\frac {-\frac {f \left (a c d h -a \,d^{2} g -b \,c^{2} h +b c d g \right ) \sqrt {f x +e}}{\left (c f -d e \right ) \left (\left (f x +e \right ) d +c f -d e \right )}+\frac {\left (a c d f h -2 a \,d^{2} e h +a \,d^{2} f g -3 b \,c^{2} f h +4 b c d e h +b c d f g -2 b \,d^{2} e g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}}}{d^{2}}\)	\(182\)
pseudoelliptic	\(\frac {-\left (\left (2 b e h x -a f g \right ) d^{2}+c \left (\left (\left (-2 b x +a \right ) h +b g \right ) f +2 e h b \right ) d -3 b \,c^{2} f h \right ) \sqrt {\left (c f -d e \right ) d}\, \sqrt {f x +e}+\left (\left (a f g -2 \left (a h +b g \right ) e \right ) d^{2}+c \left (\left (a h +b g \right ) f +4 e h b \right ) d -3 b \,c^{2} f h \right ) \left (x d +c \right ) f \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, f \,d^{2} \left (c f -d e \right ) \left (x d +c \right )}\)	\(182\)
derivativedivides	\(\frac {\frac {2 h b \sqrt {f x +e}}{d^{2}}-\frac {2 f \left (\frac {f \left (a c d h -a \,d^{2} g -b \,c^{2} h +b c d g \right ) \sqrt {f x +e}}{2 \left (c f -d e \right ) \left (\left (f x +e \right ) d +c f -d e \right )}-\frac {\left (a c d f h -2 a \,d^{2} e h +a \,d^{2} f g -3 b \,c^{2} f h +4 b c d e h +b c d f g -2 b \,d^{2} e g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{2 \left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}}\right )}{d^{2}}}{f}\)	\(186\)
default	\(\frac {\frac {2 h b \sqrt {f x +e}}{d^{2}}-\frac {2 f \left (\frac {f \left (a c d h -a \,d^{2} g -b \,c^{2} h +b c d g \right ) \sqrt {f x +e}}{2 \left (c f -d e \right ) \left (\left (f x +e \right ) d +c f -d e \right )}-\frac {\left (a c d f h -2 a \,d^{2} e h +a \,d^{2} f g -3 b \,c^{2} f h +4 b c d e h +b c d f g -2 b \,d^{2} e g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{2 \left (c f -d e \right ) \sqrt {\left (c f -d e \right ) d}}\right )}{d^{2}}}{f}\)	\(186\)

Input:

int((b*x+a)*(h*x+g)/(d*x+c)^2/(f*x+e)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2*b*h*(f*x+e)^(1/2)/d^2/f+1/d^2*(-f*(a*c*d*h-a*d^2*g-b*c^2*h+b*c*d*g)/(c*f 
-d*e)*(f*x+e)^(1/2)/((f*x+e)*d+c*f-d*e)+(a*c*d*f*h-2*a*d^2*e*h+a*d^2*f*g-3 
*b*c^2*f*h+4*b*c*d*e*h+b*c*d*f*g-2*b*d^2*e*g)/(c*f-d*e)/((c*f-d*e)*d)^(1/2 
)*arctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 396 vs. \(2 (141) = 282\).

Time = 0.11 (sec) , antiderivative size = 807, normalized size of antiderivative = 5.24 \[ \int \frac {(a+b x) (g+h x)}{(c+d x)^2 \sqrt {e+f x}} \, dx =\text {Too large to display} \] Input:

integrate((b*x+a)*(h*x+g)/(d*x+c)^2/(f*x+e)^(1/2),x, algorithm="fricas")
 

Output:

[1/2*(sqrt(d^2*e - c*d*f)*((2*b*c*d^2*e*f - (b*c^2*d + a*c*d^2)*f^2)*g - ( 
2*(2*b*c^2*d - a*c*d^2)*e*f - (3*b*c^3 - a*c^2*d)*f^2)*h + ((2*b*d^3*e*f - 
 (b*c*d^2 + a*d^3)*f^2)*g - (2*(2*b*c*d^2 - a*d^3)*e*f - (3*b*c^2*d - a*c* 
d^2)*f^2)*h)*x)*log((d*f*x + 2*d*e - c*f - 2*sqrt(d^2*e - c*d*f)*sqrt(f*x 
+ e))/(d*x + c)) + 2*(2*(b*d^4*e^2 - 2*b*c*d^3*e*f + b*c^2*d^2*f^2)*h*x + 
((b*c*d^3 - a*d^4)*e*f - (b*c^2*d^2 - a*c*d^3)*f^2)*g + (2*b*c*d^3*e^2 - ( 
5*b*c^2*d^2 - a*c*d^3)*e*f + (3*b*c^3*d - a*c^2*d^2)*f^2)*h)*sqrt(f*x + e) 
)/(c*d^5*e^2*f - 2*c^2*d^4*e*f^2 + c^3*d^3*f^3 + (d^6*e^2*f - 2*c*d^5*e*f^ 
2 + c^2*d^4*f^3)*x), (sqrt(-d^2*e + c*d*f)*((2*b*c*d^2*e*f - (b*c^2*d + a* 
c*d^2)*f^2)*g - (2*(2*b*c^2*d - a*c*d^2)*e*f - (3*b*c^3 - a*c^2*d)*f^2)*h 
+ ((2*b*d^3*e*f - (b*c*d^2 + a*d^3)*f^2)*g - (2*(2*b*c*d^2 - a*d^3)*e*f - 
(3*b*c^2*d - a*c*d^2)*f^2)*h)*x)*arctan(sqrt(-d^2*e + c*d*f)*sqrt(f*x + e) 
/(d*f*x + d*e)) + (2*(b*d^4*e^2 - 2*b*c*d^3*e*f + b*c^2*d^2*f^2)*h*x + ((b 
*c*d^3 - a*d^4)*e*f - (b*c^2*d^2 - a*c*d^3)*f^2)*g + (2*b*c*d^3*e^2 - (5*b 
*c^2*d^2 - a*c*d^3)*e*f + (3*b*c^3*d - a*c^2*d^2)*f^2)*h)*sqrt(f*x + e))/( 
c*d^5*e^2*f - 2*c^2*d^4*e*f^2 + c^3*d^3*f^3 + (d^6*e^2*f - 2*c*d^5*e*f^2 + 
 c^2*d^4*f^3)*x)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b x) (g+h x)}{(c+d x)^2 \sqrt {e+f x}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x+a)*(h*x+g)/(d*x+c)^2/(f*x+e)^(1/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(c*f-d*e>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.39 \[ \int \frac {(a+b x) (g+h x)}{(c+d x)^2 \sqrt {e+f x}} \, dx=\frac {{\left (2 \, b d^{2} e g - b c d f g - a d^{2} f g - 4 \, b c d e h + 2 \, a d^{2} e h + 3 \, b c^{2} f h - a c d f h\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{{\left (d^{3} e - c d^{2} f\right )} \sqrt {-d^{2} e + c d f}} + \frac {2 \, \sqrt {f x + e} b h}{d^{2} f} + \frac {\sqrt {f x + e} b c d f g - \sqrt {f x + e} a d^{2} f g - \sqrt {f x + e} b c^{2} f h + \sqrt {f x + e} a c d f h}{{\left (d^{3} e - c d^{2} f\right )} {\left ({\left (f x + e\right )} d - d e + c f\right )}} \] Input:

integrate((b*x+a)*(h*x+g)/(d*x+c)^2/(f*x+e)^(1/2),x, algorithm="giac")
 

Output:

(2*b*d^2*e*g - b*c*d*f*g - a*d^2*f*g - 4*b*c*d*e*h + 2*a*d^2*e*h + 3*b*c^2 
*f*h - a*c*d*f*h)*arctan(sqrt(f*x + e)*d/sqrt(-d^2*e + c*d*f))/((d^3*e - c 
*d^2*f)*sqrt(-d^2*e + c*d*f)) + 2*sqrt(f*x + e)*b*h/(d^2*f) + (sqrt(f*x + 
e)*b*c*d*f*g - sqrt(f*x + e)*a*d^2*f*g - sqrt(f*x + e)*b*c^2*f*h + sqrt(f* 
x + e)*a*c*d*f*h)/((d^3*e - c*d^2*f)*((f*x + e)*d - d*e + c*f))
 

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b x) (g+h x)}{(c+d x)^2 \sqrt {e+f x}} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}}{\sqrt {c\,f-d\,e}}\right )\,\left (a\,d^2\,f\,g-2\,a\,d^2\,e\,h-2\,b\,d^2\,e\,g-3\,b\,c^2\,f\,h+a\,c\,d\,f\,h+4\,b\,c\,d\,e\,h+b\,c\,d\,f\,g\right )}{d^{5/2}\,{\left (c\,f-d\,e\right )}^{3/2}}+\frac {\sqrt {e+f\,x}\,\left (a\,d^2\,f\,g+b\,c^2\,f\,h-a\,c\,d\,f\,h-b\,c\,d\,f\,g\right )}{\left (c\,f-d\,e\right )\,\left (d^3\,\left (e+f\,x\right )-d^3\,e+c\,d^2\,f\right )}+\frac {2\,b\,h\,\sqrt {e+f\,x}}{d^2\,f} \] Input:

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

Output:

(atan((d^(1/2)*(e + f*x)^(1/2))/(c*f - d*e)^(1/2))*(a*d^2*f*g - 2*a*d^2*e* 
h - 2*b*d^2*e*g - 3*b*c^2*f*h + a*c*d*f*h + 4*b*c*d*e*h + b*c*d*f*g))/(d^( 
5/2)*(c*f - d*e)^(3/2)) + ((e + f*x)^(1/2)*(a*d^2*f*g + b*c^2*f*h - a*c*d* 
f*h - b*c*d*f*g))/((c*f - d*e)*(d^3*(e + f*x) - d^3*e + c*d^2*f)) + (2*b*h 
*(e + f*x)^(1/2))/(d^2*f)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 900, normalized size of antiderivative = 5.84 \[ \int \frac {(a+b x) (g+h x)}{(c+d x)^2 \sqrt {e+f x}} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e))) 
*a*c**2*d*f**2*h - 2*sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt( 
d)*sqrt(c*f - d*e)))*a*c*d**2*e*f*h + sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e 
 + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a*c*d**2*f**2*g + sqrt(d)*sqrt(c*f - 
 d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a*c*d**2*f**2*h*x 
- 2*sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e 
)))*a*d**3*e*f*h*x + sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt( 
d)*sqrt(c*f - d*e)))*a*d**3*f**2*g*x - 3*sqrt(d)*sqrt(c*f - d*e)*atan((sqr 
t(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b*c**3*f**2*h + 4*sqrt(d)*sqrt(c* 
f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b*c**2*d*e*f*h 
+ sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)) 
)*b*c**2*d*f**2*g - 3*sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt 
(d)*sqrt(c*f - d*e)))*b*c**2*d*f**2*h*x - 2*sqrt(d)*sqrt(c*f - d*e)*atan(( 
sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b*c*d**2*e*f*g + 4*sqrt(d)*sqr 
t(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b*c*d**2*e* 
f*h*x + sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - 
 d*e)))*b*c*d**2*f**2*g*x - 2*sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)* 
d)/(sqrt(d)*sqrt(c*f - d*e)))*b*d**3*e*f*g*x - sqrt(e + f*x)*a*c**2*d**2*f 
**2*h + sqrt(e + f*x)*a*c*d**3*e*f*h + sqrt(e + f*x)*a*c*d**3*f**2*g - sqr 
t(e + f*x)*a*d**4*e*f*g + 3*sqrt(e + f*x)*b*c**3*d*f**2*h - 5*sqrt(e + ...