\(\int \frac {(a+b x) \sqrt {e+f x} (g+h x)}{(c+d x)^2} \, dx\) [44]

Optimal result

Integrand size = 27, antiderivative size = 182 \[ \int \frac {(a+b x) \sqrt {e+f x} (g+h x)}{(c+d x)^2} \, dx=\frac {2 (b d g-2 b c h+a d h) \sqrt {e+f x}}{d^3}+\frac {(b c-a d) (d g-c h) \sqrt {e+f x}}{d^3 (c+d x)}+\frac {2 b h (e+f x)^{3/2}}{3 d^2 f}-\frac {\left (a d (d f g+2 d e h-3 c f h)+b \left (2 d^2 e g+5 c^2 f h-c d (3 f g+4 e h)\right )\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2} \sqrt {d e-c f}} \] Output:

2*(a*d*h-2*b*c*h+b*d*g)*(f*x+e)^(1/2)/d^3+(-a*d+b*c)*(-c*h+d*g)*(f*x+e)^(1 
/2)/d^3/(d*x+c)+2/3*b*h*(f*x+e)^(3/2)/d^2/f-(a*d*(-3*c*f*h+2*d*e*h+d*f*g)+ 
b*(2*d^2*e*g+5*c^2*f*h-c*d*(4*e*h+3*f*g)))*arctanh(d^(1/2)*(f*x+e)^(1/2)/( 
-c*f+d*e)^(1/2))/d^(7/2)/(-c*f+d*e)^(1/2)
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.03 \[ \int \frac {(a+b x) \sqrt {e+f x} (g+h x)}{(c+d x)^2} \, dx=\frac {\sqrt {e+f x} \left (3 a d f (-d g+3 c h+2 d h x)+b \left (-15 c^2 f h+c d (9 f g+2 e h-10 f h x)+2 d^2 x (3 f g+e h+f h x)\right )\right )}{3 d^3 f (c+d x)}+\frac {\left (a d (d f g+2 d e h-3 c f h)+b \left (2 d^2 e g+5 c^2 f h-c d (3 f g+4 e h)\right )\right ) \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{d^{7/2} \sqrt {-d e+c f}} \] Input:

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

Output:

(Sqrt[e + f*x]*(3*a*d*f*(-(d*g) + 3*c*h + 2*d*h*x) + b*(-15*c^2*f*h + c*d* 
(9*f*g + 2*e*h - 10*f*h*x) + 2*d^2*x*(3*f*g + e*h + f*h*x))))/(3*d^3*f*(c 
+ d*x)) + ((a*d*(d*f*g + 2*d*e*h - 3*c*f*h) + b*(2*d^2*e*g + 5*c^2*f*h - c 
*d*(3*f*g + 4*e*h)))*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c*f]])/( 
d^(7/2)*Sqrt[-(d*e) + c*f])
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.17, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {163, 60, 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.

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

\(\Big \downarrow \) 163

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {\left (a d (-3 c f h+2 d e h+d f g)+b \left (5 c^2 f h-c d (4 e h+3 f g)+2 d^2 e g\right )\right ) \left (\frac {(d e-c f) \int \frac {1}{(c+d x) \sqrt {e+f x}}dx}{d}+\frac {2 \sqrt {e+f x}}{d}\right )}{2 d^2 (d e-c f)}-\frac {(e+f x)^{3/2} (3 a d f (d g-c h)-b c (-5 c f h+2 d e h+3 d f g)-2 b d h x (d e-c f))}{3 d^2 f (c+d x) (d e-c f)}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\left (a d (-3 c f h+2 d e h+d f g)+b \left (5 c^2 f h-c d (4 e h+3 f g)+2 d^2 e g\right )\right ) \left (\frac {2 (d e-c f) \int \frac {1}{c+\frac {d (e+f x)}{f}-\frac {d e}{f}}d\sqrt {e+f x}}{d f}+\frac {2 \sqrt {e+f x}}{d}\right )}{2 d^2 (d e-c f)}-\frac {(e+f x)^{3/2} (3 a d f (d g-c h)-b c (-5 c f h+2 d e h+3 d f g)-2 b d h x (d e-c f))}{3 d^2 f (c+d x) (d e-c f)}\)

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 60

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

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 = 193, normalized size of antiderivative = 1.06

method	result	size
pseudoelliptic	\(\frac {-3 \left (\frac {\left (-a f g -2 \left (a h +b g \right ) e \right ) d^{2}}{3}+\left (\left (a h +b g \right ) f +\frac {4 e h b}{3}\right ) c d -\frac {5 b \,c^{2} f h}{3}\right ) \left (x d +c \right ) f \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )+3 \sqrt {\left (c f -d e \right ) d}\, \left (\frac {\left (\left (2 x \left (\frac {h x}{3}+g \right ) b -a \left (-2 h x +g \right )\right ) f +\frac {2 b e h x}{3}\right ) d^{2}}{3}+c \left (\left (\left (-\frac {10 h x}{9}+g \right ) b +a h \right ) f +\frac {2 e h b}{9}\right ) d -\frac {5 b \,c^{2} f h}{3}\right ) \sqrt {f x +e}}{d^{3} f \left (x d +c \right ) \sqrt {\left (c f -d e \right ) d}}\)	\(193\)
risch	\(\frac {2 \left (h x d b f +3 a d f h -6 b c f h +b d e h +3 b d f g \right ) \sqrt {f x +e}}{3 f \,d^{3}}-\frac {\frac {2 \left (-\frac {1}{2} a c d f h +\frac {1}{2} a \,d^{2} f g +\frac {1}{2} b \,c^{2} f h -\frac {1}{2} b c d f g \right ) \sqrt {f x +e}}{\left (f x +e \right ) d +c f -d e}+\frac {\left (3 a c d f h -2 a \,d^{2} e h -a \,d^{2} f g -5 b \,c^{2} f h +4 b c d e h +3 b c d f g -2 b \,d^{2} e g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}}}{d^{3}}\)	\(199\)
derivativedivides	\(\frac {\frac {2 \left (\frac {d h \left (f x +e \right )^{\frac {3}{2}} b}{3}+a d f h \sqrt {f x +e}-2 b c f h \sqrt {f x +e}+b d f g \sqrt {f x +e}\right )}{d^{3}}-\frac {2 f \left (\frac {\left (-\frac {1}{2} a c d f h +\frac {1}{2} a \,d^{2} f g +\frac {1}{2} b \,c^{2} f h -\frac {1}{2} b c d f g \right ) \sqrt {f x +e}}{\left (f x +e \right ) d +c f -d e}+\frac {\left (3 a c d f h -2 a \,d^{2} e h -a \,d^{2} f g -5 b \,c^{2} f h +4 b c d e h +3 b c d f g -2 b \,d^{2} e g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{2 \sqrt {\left (c f -d e \right ) d}}\right )}{d^{3}}}{f}\)	\(214\)
default	\(\frac {\frac {2 \left (\frac {d h \left (f x +e \right )^{\frac {3}{2}} b}{3}+a d f h \sqrt {f x +e}-2 b c f h \sqrt {f x +e}+b d f g \sqrt {f x +e}\right )}{d^{3}}-\frac {2 f \left (\frac {\left (-\frac {1}{2} a c d f h +\frac {1}{2} a \,d^{2} f g +\frac {1}{2} b \,c^{2} f h -\frac {1}{2} b c d f g \right ) \sqrt {f x +e}}{\left (f x +e \right ) d +c f -d e}+\frac {\left (3 a c d f h -2 a \,d^{2} e h -a \,d^{2} f g -5 b \,c^{2} f h +4 b c d e h +3 b c d f g -2 b \,d^{2} e g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{2 \sqrt {\left (c f -d e \right ) d}}\right )}{d^{3}}}{f}\)	\(214\)

Input:

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

Output:

3/((c*f-d*e)*d)^(1/2)*(-(1/3*(-a*f*g-2*(a*h+b*g)*e)*d^2+((a*h+b*g)*f+4/3*e 
*h*b)*c*d-5/3*b*c^2*f*h)*(d*x+c)*f*arctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1 
/2))+((c*f-d*e)*d)^(1/2)*(1/3*((2*x*(1/3*h*x+g)*b-a*(-2*h*x+g))*f+2/3*b*e* 
h*x)*d^2+c*(((-10/9*h*x+g)*b+a*h)*f+2/9*e*h*b)*d-5/3*b*c^2*f*h)*(f*x+e)^(1 
/2))/d^3/f/(d*x+c)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 446 vs. \(2 (164) = 328\).

Time = 0.15 (sec) , antiderivative size = 906, normalized size of antiderivative = 4.98 \[ \int \frac {(a+b x) \sqrt {e+f x} (g+h x)}{(c+d x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/6*(3*sqrt(d^2*e - c*d*f)*((2*b*c*d^2*e*f - (3*b*c^2*d - a*c*d^2)*f^2)* 
g - (2*(2*b*c^2*d - a*c*d^2)*e*f - (5*b*c^3 - 3*a*c^2*d)*f^2)*h + ((2*b*d^ 
3*e*f - (3*b*c*d^2 - a*d^3)*f^2)*g - (2*(2*b*c*d^2 - a*d^3)*e*f - (5*b*c^2 
*d - 3*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*f - b*c*d^3*f^2)*h*x^2 + 3*((3 
*b*c*d^3 - a*d^4)*e*f - (3*b*c^2*d^2 - a*c*d^3)*f^2)*g + (2*b*c*d^3*e^2 - 
(17*b*c^2*d^2 - 9*a*c*d^3)*e*f + 3*(5*b*c^3*d - 3*a*c^2*d^2)*f^2)*h + 2*(3 
*(b*d^4*e*f - b*c*d^3*f^2)*g + (b*d^4*e^2 - 3*(2*b*c*d^3 - a*d^4)*e*f + (5 
*b*c^2*d^2 - 3*a*c*d^3)*f^2)*h)*x)*sqrt(f*x + e))/(c*d^5*e*f - c^2*d^4*f^2 
 + (d^6*e*f - c*d^5*f^2)*x), 1/3*(3*sqrt(-d^2*e + c*d*f)*((2*b*c*d^2*e*f - 
 (3*b*c^2*d - a*c*d^2)*f^2)*g - (2*(2*b*c^2*d - a*c*d^2)*e*f - (5*b*c^3 - 
3*a*c^2*d)*f^2)*h + ((2*b*d^3*e*f - (3*b*c*d^2 - a*d^3)*f^2)*g - (2*(2*b*c 
*d^2 - a*d^3)*e*f - (5*b*c^2*d - 3*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*f - b*c*d^3*f^2)*h*x^2 
 + 3*((3*b*c*d^3 - a*d^4)*e*f - (3*b*c^2*d^2 - a*c*d^3)*f^2)*g + (2*b*c*d^ 
3*e^2 - (17*b*c^2*d^2 - 9*a*c*d^3)*e*f + 3*(5*b*c^3*d - 3*a*c^2*d^2)*f^2)* 
h + 2*(3*(b*d^4*e*f - b*c*d^3*f^2)*g + (b*d^4*e^2 - 3*(2*b*c*d^3 - a*d^4)* 
e*f + (5*b*c^2*d^2 - 3*a*c*d^3)*f^2)*h)*x)*sqrt(f*x + e))/(c*d^5*e*f - c^2 
*d^4*f^2 + (d^6*e*f - c*d^5*f^2)*x)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((b*x+a)*(f*x+e)^(1/2)*(h*x+g)/(d*x+c)^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 = 249, normalized size of antiderivative = 1.37 \[ \int \frac {(a+b x) \sqrt {e+f x} (g+h x)}{(c+d x)^2} \, dx=\frac {{\left (2 \, b d^{2} e g - 3 \, b c d f g + a d^{2} f g - 4 \, b c d e h + 2 \, a d^{2} e h + 5 \, b c^{2} f h - 3 \, a c d f h\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{\sqrt {-d^{2} e + c d f} d^{3}} + \frac {\sqrt {f x + e} b c d f g - \sqrt {f x + e} a d^{2} f g - \sqrt {f x + e} b c^{2} f h + \sqrt {f x + e} a c d f h}{{\left ({\left (f x + e\right )} d - d e + c f\right )} d^{3}} + \frac {2 \, {\left (3 \, \sqrt {f x + e} b d^{4} f^{3} g + {\left (f x + e\right )}^{\frac {3}{2}} b d^{4} f^{2} h - 6 \, \sqrt {f x + e} b c d^{3} f^{3} h + 3 \, \sqrt {f x + e} a d^{4} f^{3} h\right )}}{3 \, d^{6} f^{3}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b x) \sqrt {e+f x} (g+h x)}{(c+d x)^2} \, dx=\sqrt {e+f\,x}\,\left (\frac {2\,a\,f\,h-4\,b\,e\,h+2\,b\,f\,g}{d^2\,f}-\frac {4\,b\,h\,\left (c\,f-d\,e\right )}{d^3\,f}\right )-\frac {\sqrt {e+f\,x}\,\left (a\,d^2\,f\,g+b\,c^2\,f\,h-a\,c\,d\,f\,h-b\,c\,d\,f\,g\right )}{d^4\,\left (e+f\,x\right )-d^4\,e+c\,d^3\,f}+\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}}{\sqrt {c\,f-d\,e}}\right )\,\left (2\,a\,d^2\,e\,h+a\,d^2\,f\,g+2\,b\,d^2\,e\,g+5\,b\,c^2\,f\,h-3\,a\,c\,d\,f\,h-4\,b\,c\,d\,e\,h-3\,b\,c\,d\,f\,g\right )}{d^{7/2}\,\sqrt {c\,f-d\,e}}+\frac {2\,b\,h\,{\left (e+f\,x\right )}^{3/2}}{3\,d^2\,f} \] Input:

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

Output:

(e + f*x)^(1/2)*((2*a*f*h - 4*b*e*h + 2*b*f*g)/(d^2*f) - (4*b*h*(c*f - d*e 
))/(d^3*f)) - ((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))/(d^4*(e + f*x) - d^4*e + c*d^3*f) + (atan((d^(1/2)*(e + f*x)^(1/2))/ 
(c*f - d*e)^(1/2))*(2*a*d^2*e*h + a*d^2*f*g + 2*b*d^2*e*g + 5*b*c^2*f*h - 
3*a*c*d*f*h - 4*b*c*d*e*h - 3*b*c*d*f*g))/(d^(7/2)*(c*f - d*e)^(1/2)) + (2 
*b*h*(e + f*x)^(3/2))/(3*d^2*f)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 989, normalized size of antiderivative = 5.43 \[ \int \frac {(a+b x) \sqrt {e+f x} (g+h x)}{(c+d x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 9*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 + 6*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 + 3*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 - 9*sqrt(d)*s 
qrt(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 + 6*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 + 3*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 + 15*sqrt(d)*sqrt(c*f - d* 
e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b*c**3*f**2*h - 12*sq 
rt(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 - 9*sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*s 
qrt(c*f - d*e)))*b*c**2*d*f**2*g + 15*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 + 6*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 
 - 12*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*h*x - 9*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 + 6*sqrt(d)*sqrt(c*f - d*e)*a 
tan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b*d**3*e*f*g*x + 9*sqrt(e 
 + f*x)*a*c**2*d**2*f**2*h - 9*sqrt(e + f*x)*a*c*d**3*e*f*h - 3*sqrt(e + f 
*x)*a*c*d**3*f**2*g + 6*sqrt(e + f*x)*a*c*d**3*f**2*h*x + 3*sqrt(e + f*...