\(\int \frac {(c+d x) (e+f x)^{3/2} (g+h x)}{(a+b x)^2} \, dx\) [65]

Optimal result

Integrand size = 27, antiderivative size = 247 \[ \int \frac {(c+d x) (e+f x)^{3/2} (g+h x)}{(a+b x)^2} \, dx=\frac {\left (7 a^2 d f h+b^2 (2 d e g+3 c f g+2 c e h)-a b (5 d f g+4 d e h+5 c f h)\right ) \sqrt {e+f x}}{b^4}+\frac {2 (b d g+b c h-2 a d h) (e+f x)^{3/2}}{3 b^3}-\frac {(b c-a d) (b g-a h) (e+f x)^{3/2}}{b^3 (a+b x)}+\frac {2 d h (e+f x)^{5/2}}{5 b^2 f}-\frac {\sqrt {b e-a f} \left (7 a^2 d f h+b^2 (2 d e g+3 c f g+2 c e h)-a b (5 d f g+4 d e h+5 c f h)\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {b e-a f}}\right )}{b^{9/2}} \] Output:

(7*a^2*d*f*h+b^2*(2*c*e*h+3*c*f*g+2*d*e*g)-a*b*(5*c*f*h+4*d*e*h+5*d*f*g))* 
(f*x+e)^(1/2)/b^4+2/3*(-2*a*d*h+b*c*h+b*d*g)*(f*x+e)^(3/2)/b^3-(-a*d+b*c)* 
(-a*h+b*g)*(f*x+e)^(3/2)/b^3/(b*x+a)+2/5*d*h*(f*x+e)^(5/2)/b^2/f-(-a*f+b*e 
)^(1/2)*(7*a^2*d*f*h+b^2*(2*c*e*h+3*c*f*g+2*d*e*g)-a*b*(5*c*f*h+4*d*e*h+5* 
d*f*g))*arctanh(b^(1/2)*(f*x+e)^(1/2)/(-a*f+b*e)^(1/2))/b^(9/2)
 

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.16 \[ \int \frac {(c+d x) (e+f x)^{3/2} (g+h x)}{(a+b x)^2} \, dx=\frac {\sqrt {e+f x} \left (105 a^3 d f^2 h-5 a^2 b f (15 c f h+d (15 f g+19 e h-14 f h x))+a b^2 \left (5 c f (9 f g+11 e h-10 f h x)+d \left (6 e^2 h+e f (55 g-68 h x)-2 f^2 x (25 g+7 h x)\right )\right )+b^3 \left (2 d x \left (3 e^2 h+f^2 x (5 g+3 h x)+e f (20 g+6 h x)\right )+5 c f (2 f x (3 g+h x)+e (-3 g+8 h x))\right )\right )}{15 b^4 f (a+b x)}-\frac {\sqrt {-b e+a f} \left (7 a^2 d f h+b^2 (2 d e g+3 c f g+2 c e h)-a b (5 d f g+4 d e h+5 c f h)\right ) \arctan \left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {-b e+a f}}\right )}{b^{9/2}} \] Input:

Integrate[((c + d*x)*(e + f*x)^(3/2)*(g + h*x))/(a + b*x)^2,x]
 

Output:

(Sqrt[e + f*x]*(105*a^3*d*f^2*h - 5*a^2*b*f*(15*c*f*h + d*(15*f*g + 19*e*h 
 - 14*f*h*x)) + a*b^2*(5*c*f*(9*f*g + 11*e*h - 10*f*h*x) + d*(6*e^2*h + e* 
f*(55*g - 68*h*x) - 2*f^2*x*(25*g + 7*h*x))) + b^3*(2*d*x*(3*e^2*h + f^2*x 
*(5*g + 3*h*x) + e*f*(20*g + 6*h*x)) + 5*c*f*(2*f*x*(3*g + h*x) + e*(-3*g 
+ 8*h*x)))))/(15*b^4*f*(a + b*x)) - (Sqrt[-(b*e) + a*f]*(7*a^2*d*f*h + b^2 
*(2*d*e*g + 3*c*f*g + 2*c*e*h) - a*b*(5*d*f*g + 4*d*e*h + 5*c*f*h))*ArcTan 
[(Sqrt[b]*Sqrt[e + f*x])/Sqrt[-(b*e) + a*f]])/b^(9/2)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {163, 60, 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.

\(\displaystyle \int \frac {(c+d x) (e+f x)^{3/2} (g+h x)}{(a+b x)^2} \, dx\)

\(\Big \downarrow \) 163

\(\displaystyle \frac {\left (7 a^2 d f h-a b (5 c f h+4 d e h+5 d f g)+b^2 (2 c e h+3 c f g+2 d e g)\right ) \int \frac {(e+f x)^{3/2}}{a+b x}dx}{2 b^2 (b e-a f)}-\frac {(e+f x)^{5/2} \left (7 a^2 d f h-a b (5 c f h+2 d e h+5 d f g)-2 b d h x (b e-a f)+5 b^2 c f g\right )}{5 b^2 f (a+b x) (b e-a f)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\left (7 a^2 d f h-a b (5 c f h+4 d e h+5 d f g)+b^2 (2 c e h+3 c f g+2 d e g)\right ) \left (\frac {(b e-a f) \int \frac {\sqrt {e+f x}}{a+b x}dx}{b}+\frac {2 (e+f x)^{3/2}}{3 b}\right )}{2 b^2 (b e-a f)}-\frac {(e+f x)^{5/2} \left (7 a^2 d f h-a b (5 c f h+2 d e h+5 d f g)-2 b d h x (b e-a f)+5 b^2 c f g\right )}{5 b^2 f (a+b x) (b e-a f)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\left (7 a^2 d f h-a b (5 c f h+4 d e h+5 d f g)+b^2 (2 c e h+3 c f g+2 d e g)\right ) \left (\frac {(b e-a f) \left (\frac {(b e-a f) \int \frac {1}{(a+b x) \sqrt {e+f x}}dx}{b}+\frac {2 \sqrt {e+f x}}{b}\right )}{b}+\frac {2 (e+f x)^{3/2}}{3 b}\right )}{2 b^2 (b e-a f)}-\frac {(e+f x)^{5/2} \left (7 a^2 d f h-a b (5 c f h+2 d e h+5 d f g)-2 b d h x (b e-a f)+5 b^2 c f g\right )}{5 b^2 f (a+b x) (b e-a f)}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\left (7 a^2 d f h-a b (5 c f h+4 d e h+5 d f g)+b^2 (2 c e h+3 c f g+2 d e g)\right ) \left (\frac {(b e-a f) \left (\frac {2 (b e-a f) \int \frac {1}{a+\frac {b (e+f x)}{f}-\frac {b e}{f}}d\sqrt {e+f x}}{b f}+\frac {2 \sqrt {e+f x}}{b}\right )}{b}+\frac {2 (e+f x)^{3/2}}{3 b}\right )}{2 b^2 (b e-a f)}-\frac {(e+f x)^{5/2} \left (7 a^2 d f h-a b (5 c f h+2 d e h+5 d f g)-2 b d h x (b e-a f)+5 b^2 c f g\right )}{5 b^2 f (a+b x) (b e-a f)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\left (\frac {(b e-a f) \left (\frac {2 \sqrt {e+f x}}{b}-\frac {2 \sqrt {b e-a f} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {b e-a f}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (e+f x)^{3/2}}{3 b}\right ) \left (7 a^2 d f h-a b (5 c f h+4 d e h+5 d f g)+b^2 (2 c e h+3 c f g+2 d e g)\right )}{2 b^2 (b e-a f)}-\frac {(e+f x)^{5/2} \left (7 a^2 d f h-a b (5 c f h+2 d e h+5 d f g)-2 b d h x (b e-a f)+5 b^2 c f g\right )}{5 b^2 f (a+b x) (b e-a f)}\)

Input:

Int[((c + d*x)*(e + f*x)^(3/2)*(g + h*x))/(a + b*x)^2,x]
 

Output:

-1/5*((e + f*x)^(5/2)*(5*b^2*c*f*g + 7*a^2*d*f*h - a*b*(5*d*f*g + 2*d*e*h 
+ 5*c*f*h) - 2*b*d*(b*e - a*f)*h*x))/(b^2*f*(b*e - a*f)*(a + b*x)) + ((7*a 
^2*d*f*h + b^2*(2*d*e*g + 3*c*f*g + 2*c*e*h) - a*b*(5*d*f*g + 4*d*e*h + 5* 
c*f*h))*((2*(e + f*x)^(3/2))/(3*b) + ((b*e - a*f)*((2*Sqrt[e + f*x])/b - ( 
2*Sqrt[b*e - a*f]*ArcTanh[(Sqrt[b]*Sqrt[e + f*x])/Sqrt[b*e - a*f]])/b^(3/2 
)))/b))/(2*b^2*(b*e - a*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 = 0.49 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.18

method	result	size
pseudoelliptic	\(-\frac {7 \left (\left (b x +a \right ) \left (\frac {\left (3 c f g +2 e \left (c h +d g \right )\right ) b^{2}}{7}-\frac {5 a \left (f \left (c h +d g \right )+\frac {4 d e h}{5}\right ) b}{7}+a^{2} d f h \right ) f \left (a f -b e \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )-\left (\frac {\left (2 x \left (\frac {x \left (\frac {3 h x}{5}+g \right ) d}{3}+c \left (\frac {h x}{3}+g \right )\right ) f^{2}-\left (4 \left (-\frac {1}{5} h \,x^{2}-\frac {2}{3} g x \right ) d +c \left (-\frac {8 h x}{3}+g \right )\right ) e f +\frac {2 d \,e^{2} h x}{5}\right ) b^{3}}{7}+\frac {11 a \left (\frac {\left (-10 x \left (\frac {7 h x}{25}+g \right ) d +9 \left (-\frac {10 h x}{9}+g \right ) c \right ) f^{2}}{11}+\left (\left (-\frac {68 h x}{55}+g \right ) d +c h \right ) e f +\frac {6 d \,e^{2} h}{55}\right ) b^{2}}{21}-\frac {5 a^{2} \left (\left (\left (-\frac {14 h x}{15}+g \right ) d +c h \right ) f +\frac {19 d e h}{15}\right ) f b}{7}+a^{3} d \,f^{2} h \right ) \sqrt {\left (a f -b e \right ) b}\, \sqrt {f x +e}\right )}{\sqrt {\left (a f -b e \right ) b}\, b^{4} f \left (b x +a \right )}\)	\(292\)
risch	\(\frac {2 \left (3 d h \,x^{2} b^{2} f^{2}-10 a b d \,f^{2} h x +5 b^{2} c \,f^{2} h x +6 b^{2} d e f h x +5 b^{2} d \,f^{2} g x +45 a^{2} d \,f^{2} h -30 a b c \,f^{2} h -40 a b d e f h -30 a b d \,f^{2} g +20 b^{2} c e f h +15 b^{2} c \,f^{2} g +3 b^{2} d \,e^{2} h +20 b^{2} d e f g \right ) \sqrt {f x +e}}{15 f \,b^{4}}-\frac {\left (2 a f -2 b e \right ) \left (\frac {\left (-\frac {1}{2} a^{2} d f h +\frac {1}{2} a b c f h +\frac {1}{2} a b d f g -\frac {1}{2} b^{2} c f g \right ) \sqrt {f x +e}}{\left (f x +e \right ) b +a f -b e}+\frac {\left (7 a^{2} d f h -5 a b c f h -4 a b d e h -5 a b d f g +2 b^{2} c e h +3 b^{2} c f g +2 b^{2} d e g \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{2 \sqrt {\left (a f -b e \right ) b}}\right )}{b^{4}}\)	\(308\)
derivativedivides	\(\frac {\frac {2 \left (\frac {d h \left (f x +e \right )^{\frac {5}{2}} b^{2}}{5}-\frac {2 a b d f h \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b^{2} c f h \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b^{2} d f g \left (f x +e \right )^{\frac {3}{2}}}{3}+3 a^{2} d \,f^{2} h \sqrt {f x +e}-2 a b c \,f^{2} h \sqrt {f x +e}-2 a b d e f h \sqrt {f x +e}-2 a b d \,f^{2} g \sqrt {f x +e}+b^{2} c e f h \sqrt {f x +e}+b^{2} c \,f^{2} g \sqrt {f x +e}+b^{2} d e f g \sqrt {f x +e}\right )}{b^{4}}-\frac {2 f \left (\frac {\left (-\frac {1}{2} a^{3} d \,f^{2} h +\frac {1}{2} a^{2} b c \,f^{2} h +\frac {1}{2} a^{2} b d e f h +\frac {1}{2} a^{2} b d \,f^{2} g -\frac {1}{2} a \,b^{2} c e f h -\frac {1}{2} a \,b^{2} c \,f^{2} g -\frac {1}{2} a \,b^{2} d e f g +\frac {1}{2} b^{3} c e f g \right ) \sqrt {f x +e}}{\left (f x +e \right ) b +a f -b e}+\frac {\left (7 a^{3} d \,f^{2} h -5 a^{2} b c \,f^{2} h -11 a^{2} b d e f h -5 a^{2} b d \,f^{2} g +7 a \,b^{2} c e f h +3 a \,b^{2} c \,f^{2} g +4 a \,b^{2} d \,e^{2} h +7 a \,b^{2} d e f g -2 b^{3} c \,e^{2} h -3 b^{3} c e f g -2 b^{3} d \,e^{2} g \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{2 \sqrt {\left (a f -b e \right ) b}}\right )}{b^{4}}}{f}\)	\(445\)
default	\(\frac {\frac {2 \left (\frac {d h \left (f x +e \right )^{\frac {5}{2}} b^{2}}{5}-\frac {2 a b d f h \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b^{2} c f h \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b^{2} d f g \left (f x +e \right )^{\frac {3}{2}}}{3}+3 a^{2} d \,f^{2} h \sqrt {f x +e}-2 a b c \,f^{2} h \sqrt {f x +e}-2 a b d e f h \sqrt {f x +e}-2 a b d \,f^{2} g \sqrt {f x +e}+b^{2} c e f h \sqrt {f x +e}+b^{2} c \,f^{2} g \sqrt {f x +e}+b^{2} d e f g \sqrt {f x +e}\right )}{b^{4}}-\frac {2 f \left (\frac {\left (-\frac {1}{2} a^{3} d \,f^{2} h +\frac {1}{2} a^{2} b c \,f^{2} h +\frac {1}{2} a^{2} b d e f h +\frac {1}{2} a^{2} b d \,f^{2} g -\frac {1}{2} a \,b^{2} c e f h -\frac {1}{2} a \,b^{2} c \,f^{2} g -\frac {1}{2} a \,b^{2} d e f g +\frac {1}{2} b^{3} c e f g \right ) \sqrt {f x +e}}{\left (f x +e \right ) b +a f -b e}+\frac {\left (7 a^{3} d \,f^{2} h -5 a^{2} b c \,f^{2} h -11 a^{2} b d e f h -5 a^{2} b d \,f^{2} g +7 a \,b^{2} c e f h +3 a \,b^{2} c \,f^{2} g +4 a \,b^{2} d \,e^{2} h +7 a \,b^{2} d e f g -2 b^{3} c \,e^{2} h -3 b^{3} c e f g -2 b^{3} d \,e^{2} g \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{2 \sqrt {\left (a f -b e \right ) b}}\right )}{b^{4}}}{f}\)	\(445\)

Input:

int((d*x+c)*(f*x+e)^(3/2)*(h*x+g)/(b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

-7*((b*x+a)*(1/7*(3*c*f*g+2*e*(c*h+d*g))*b^2-5/7*a*(f*(c*h+d*g)+4/5*d*e*h) 
*b+a^2*d*f*h)*f*(a*f-b*e)*arctan(b*(f*x+e)^(1/2)/((a*f-b*e)*b)^(1/2))-(1/7 
*(2*x*(1/3*x*(3/5*h*x+g)*d+c*(1/3*h*x+g))*f^2-(4*(-1/5*h*x^2-2/3*g*x)*d+c* 
(-8/3*h*x+g))*e*f+2/5*d*e^2*h*x)*b^3+11/21*a*(1/11*(-10*x*(7/25*h*x+g)*d+9 
*(-10/9*h*x+g)*c)*f^2+((-68/55*h*x+g)*d+c*h)*e*f+6/55*d*e^2*h)*b^2-5/7*a^2 
*(((-14/15*h*x+g)*d+c*h)*f+19/15*d*e*h)*f*b+a^3*d*f^2*h)*((a*f-b*e)*b)^(1/ 
2)*(f*x+e)^(1/2))/((a*f-b*e)*b)^(1/2)/b^4/f/(b*x+a)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 459 vs. \(2 (225) = 450\).

Time = 0.14 (sec) , antiderivative size = 928, normalized size of antiderivative = 3.76 \[ \int \frac {(c+d x) (e+f x)^{3/2} (g+h x)}{(a+b x)^2} \, dx =\text {Too large to display} \] Input:

integrate((d*x+c)*(f*x+e)^(3/2)*(h*x+g)/(b*x+a)^2,x, algorithm="fricas")
 

Output:

[1/30*(15*((2*a*b^2*d*e*f + (3*a*b^2*c - 5*a^2*b*d)*f^2)*g + (2*(a*b^2*c - 
 2*a^2*b*d)*e*f - (5*a^2*b*c - 7*a^3*d)*f^2)*h + ((2*b^3*d*e*f + (3*b^3*c 
- 5*a*b^2*d)*f^2)*g + (2*(b^3*c - 2*a*b^2*d)*e*f - (5*a*b^2*c - 7*a^2*b*d) 
*f^2)*h)*x)*sqrt((b*e - a*f)/b)*log((b*f*x + 2*b*e - a*f - 2*sqrt(f*x + e) 
*b*sqrt((b*e - a*f)/b))/(b*x + a)) + 2*(6*b^3*d*f^2*h*x^3 + 2*(5*b^3*d*f^2 
*g + (6*b^3*d*e*f + (5*b^3*c - 7*a*b^2*d)*f^2)*h)*x^2 - 5*((3*b^3*c - 11*a 
*b^2*d)*e*f - 3*(3*a*b^2*c - 5*a^2*b*d)*f^2)*g + (6*a*b^2*d*e^2 + 5*(11*a* 
b^2*c - 19*a^2*b*d)*e*f - 15*(5*a^2*b*c - 7*a^3*d)*f^2)*h + 2*(5*(4*b^3*d* 
e*f + (3*b^3*c - 5*a*b^2*d)*f^2)*g + (3*b^3*d*e^2 + 2*(10*b^3*c - 17*a*b^2 
*d)*e*f - 5*(5*a*b^2*c - 7*a^2*b*d)*f^2)*h)*x)*sqrt(f*x + e))/(b^5*f*x + a 
*b^4*f), -1/15*(15*((2*a*b^2*d*e*f + (3*a*b^2*c - 5*a^2*b*d)*f^2)*g + (2*( 
a*b^2*c - 2*a^2*b*d)*e*f - (5*a^2*b*c - 7*a^3*d)*f^2)*h + ((2*b^3*d*e*f + 
(3*b^3*c - 5*a*b^2*d)*f^2)*g + (2*(b^3*c - 2*a*b^2*d)*e*f - (5*a*b^2*c - 7 
*a^2*b*d)*f^2)*h)*x)*sqrt(-(b*e - a*f)/b)*arctan(-sqrt(f*x + e)*b*sqrt(-(b 
*e - a*f)/b)/(b*e - a*f)) - (6*b^3*d*f^2*h*x^3 + 2*(5*b^3*d*f^2*g + (6*b^3 
*d*e*f + (5*b^3*c - 7*a*b^2*d)*f^2)*h)*x^2 - 5*((3*b^3*c - 11*a*b^2*d)*e*f 
 - 3*(3*a*b^2*c - 5*a^2*b*d)*f^2)*g + (6*a*b^2*d*e^2 + 5*(11*a*b^2*c - 19* 
a^2*b*d)*e*f - 15*(5*a^2*b*c - 7*a^3*d)*f^2)*h + 2*(5*(4*b^3*d*e*f + (3*b^ 
3*c - 5*a*b^2*d)*f^2)*g + (3*b^3*d*e^2 + 2*(10*b^3*c - 17*a*b^2*d)*e*f - 5 
*(5*a*b^2*c - 7*a^2*b*d)*f^2)*h)*x)*sqrt(f*x + e))/(b^5*f*x + a*b^4*f)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x) (e+f x)^{3/2} (g+h x)}{(a+b x)^2} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)*(f*x+e)**(3/2)*(h*x+g)/(b*x+a)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x) (e+f x)^{3/2} (g+h x)}{(a+b x)^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x+c)*(f*x+e)^(3/2)*(h*x+g)/(b*x+a)^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(a*f-b*e>0)', see `assume?` for m 
ore detail
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (225) = 450\).

Time = 0.14 (sec) , antiderivative size = 518, normalized size of antiderivative = 2.10 \[ \int \frac {(c+d x) (e+f x)^{3/2} (g+h x)}{(a+b x)^2} \, dx=\frac {{\left (2 \, b^{3} d e^{2} g + 3 \, b^{3} c e f g - 7 \, a b^{2} d e f g - 3 \, a b^{2} c f^{2} g + 5 \, a^{2} b d f^{2} g + 2 \, b^{3} c e^{2} h - 4 \, a b^{2} d e^{2} h - 7 \, a b^{2} c e f h + 11 \, a^{2} b d e f h + 5 \, a^{2} b c f^{2} h - 7 \, a^{3} d f^{2} h\right )} \arctan \left (\frac {\sqrt {f x + e} b}{\sqrt {-b^{2} e + a b f}}\right )}{\sqrt {-b^{2} e + a b f} b^{4}} - \frac {\sqrt {f x + e} b^{3} c e f g - \sqrt {f x + e} a b^{2} d e f g - \sqrt {f x + e} a b^{2} c f^{2} g + \sqrt {f x + e} a^{2} b d f^{2} g - \sqrt {f x + e} a b^{2} c e f h + \sqrt {f x + e} a^{2} b d e f h + \sqrt {f x + e} a^{2} b c f^{2} h - \sqrt {f x + e} a^{3} d f^{2} h}{{\left ({\left (f x + e\right )} b - b e + a f\right )} b^{4}} + \frac {2 \, {\left (5 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{8} d f^{5} g + 15 \, \sqrt {f x + e} b^{8} d e f^{5} g + 15 \, \sqrt {f x + e} b^{8} c f^{6} g - 30 \, \sqrt {f x + e} a b^{7} d f^{6} g + 3 \, {\left (f x + e\right )}^{\frac {5}{2}} b^{8} d f^{4} h + 5 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{8} c f^{5} h - 10 \, {\left (f x + e\right )}^{\frac {3}{2}} a b^{7} d f^{5} h + 15 \, \sqrt {f x + e} b^{8} c e f^{5} h - 30 \, \sqrt {f x + e} a b^{7} d e f^{5} h - 30 \, \sqrt {f x + e} a b^{7} c f^{6} h + 45 \, \sqrt {f x + e} a^{2} b^{6} d f^{6} h\right )}}{15 \, b^{10} f^{5}} \] Input:

integrate((d*x+c)*(f*x+e)^(3/2)*(h*x+g)/(b*x+a)^2,x, algorithm="giac")
 

Output:

(2*b^3*d*e^2*g + 3*b^3*c*e*f*g - 7*a*b^2*d*e*f*g - 3*a*b^2*c*f^2*g + 5*a^2 
*b*d*f^2*g + 2*b^3*c*e^2*h - 4*a*b^2*d*e^2*h - 7*a*b^2*c*e*f*h + 11*a^2*b* 
d*e*f*h + 5*a^2*b*c*f^2*h - 7*a^3*d*f^2*h)*arctan(sqrt(f*x + e)*b/sqrt(-b^ 
2*e + a*b*f))/(sqrt(-b^2*e + a*b*f)*b^4) - (sqrt(f*x + e)*b^3*c*e*f*g - sq 
rt(f*x + e)*a*b^2*d*e*f*g - sqrt(f*x + e)*a*b^2*c*f^2*g + sqrt(f*x + e)*a^ 
2*b*d*f^2*g - sqrt(f*x + e)*a*b^2*c*e*f*h + sqrt(f*x + e)*a^2*b*d*e*f*h + 
sqrt(f*x + e)*a^2*b*c*f^2*h - sqrt(f*x + e)*a^3*d*f^2*h)/(((f*x + e)*b - b 
*e + a*f)*b^4) + 2/15*(5*(f*x + e)^(3/2)*b^8*d*f^5*g + 15*sqrt(f*x + e)*b^ 
8*d*e*f^5*g + 15*sqrt(f*x + e)*b^8*c*f^6*g - 30*sqrt(f*x + e)*a*b^7*d*f^6* 
g + 3*(f*x + e)^(5/2)*b^8*d*f^4*h + 5*(f*x + e)^(3/2)*b^8*c*f^5*h - 10*(f* 
x + e)^(3/2)*a*b^7*d*f^5*h + 15*sqrt(f*x + e)*b^8*c*e*f^5*h - 30*sqrt(f*x 
+ e)*a*b^7*d*e*f^5*h - 30*sqrt(f*x + e)*a*b^7*c*f^6*h + 45*sqrt(f*x + e)*a 
^2*b^6*d*f^6*h)/(b^10*f^5)
 

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.55 \[ \int \frac {(c+d x) (e+f x)^{3/2} (g+h x)}{(a+b x)^2} \, dx={\left (e+f\,x\right )}^{3/2}\,\left (\frac {2\,c\,f\,h-4\,d\,e\,h+2\,d\,f\,g}{3\,b^2\,f}-\frac {4\,d\,h\,\left (a\,f-b\,e\right )}{3\,b^3\,f}\right )-\sqrt {e+f\,x}\,\left (\frac {2\,\left (a\,f-b\,e\right )\,\left (\frac {2\,c\,f\,h-4\,d\,e\,h+2\,d\,f\,g}{b^2\,f}-\frac {4\,d\,h\,\left (a\,f-b\,e\right )}{b^3\,f}\right )}{b}+\frac {2\,\left (c\,f-d\,e\right )\,\left (e\,h-f\,g\right )}{b^2\,f}+\frac {2\,d\,h\,{\left (a\,f-b\,e\right )}^2}{b^4\,f}\right )+\frac {\sqrt {e+f\,x}\,\left (a^3\,d\,f^2\,h+a\,b^2\,c\,f^2\,g-a^2\,b\,c\,f^2\,h-a^2\,b\,d\,f^2\,g-b^3\,c\,e\,f\,g+a\,b^2\,c\,e\,f\,h+a\,b^2\,d\,e\,f\,g-a^2\,b\,d\,e\,f\,h\right )}{b^5\,\left (e+f\,x\right )-b^5\,e+a\,b^4\,f}+\frac {2\,d\,h\,{\left (e+f\,x\right )}^{5/2}}{5\,b^2\,f}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e+f\,x}\,1{}\mathrm {i}}{\sqrt {b\,e-a\,f}}\right )\,\sqrt {b\,e-a\,f}\,\left (2\,b^2\,c\,e\,h+3\,b^2\,c\,f\,g+2\,b^2\,d\,e\,g+7\,a^2\,d\,f\,h-5\,a\,b\,c\,f\,h-4\,a\,b\,d\,e\,h-5\,a\,b\,d\,f\,g\right )\,1{}\mathrm {i}}{b^{9/2}} \] Input:

int(((e + f*x)^(3/2)*(g + h*x)*(c + d*x))/(a + b*x)^2,x)
 

Output:

(e + f*x)^(3/2)*((2*c*f*h - 4*d*e*h + 2*d*f*g)/(3*b^2*f) - (4*d*h*(a*f - b 
*e))/(3*b^3*f)) - (e + f*x)^(1/2)*((2*(a*f - b*e)*((2*c*f*h - 4*d*e*h + 2* 
d*f*g)/(b^2*f) - (4*d*h*(a*f - b*e))/(b^3*f)))/b + (2*(c*f - d*e)*(e*h - f 
*g))/(b^2*f) + (2*d*h*(a*f - b*e)^2)/(b^4*f)) + ((e + f*x)^(1/2)*(a^3*d*f^ 
2*h + a*b^2*c*f^2*g - a^2*b*c*f^2*h - a^2*b*d*f^2*g - b^3*c*e*f*g + a*b^2* 
c*e*f*h + a*b^2*d*e*f*g - a^2*b*d*e*f*h))/(b^5*(e + f*x) - b^5*e + a*b^4*f 
) + (atan((b^(1/2)*(e + f*x)^(1/2)*1i)/(b*e - a*f)^(1/2))*(b*e - a*f)^(1/2 
)*(2*b^2*c*e*h + 3*b^2*c*f*g + 2*b^2*d*e*g + 7*a^2*d*f*h - 5*a*b*c*f*h - 4 
*a*b*d*e*h - 5*a*b*d*f*g)*1i)/b^(9/2) + (2*d*h*(e + f*x)^(5/2))/(5*b^2*f)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1045, normalized size of antiderivative = 4.23 \[ \int \frac {(c+d x) (e+f x)^{3/2} (g+h x)}{(a+b x)^2} \, dx =\text {Too large to display} \] Input:

int((d*x+c)*(f*x+e)^(3/2)*(h*x+g)/(b*x+a)^2,x)
 

Output:

( - 105*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - 
 b*e)))*a**3*d*f**2*h + 75*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/ 
(sqrt(b)*sqrt(a*f - b*e)))*a**2*b*c*f**2*h + 60*sqrt(b)*sqrt(a*f - b*e)*at 
an((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a**2*b*d*e*f*h + 75*sqrt(b 
)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a**2*b 
*d*f**2*g - 105*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sq 
rt(a*f - b*e)))*a**2*b*d*f**2*h*x - 30*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt( 
e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*c*e*f*h - 45*sqrt(b)*sqrt(a* 
f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*c*f**2*g 
 + 75*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b 
*e)))*a*b**2*c*f**2*h*x - 30*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b 
)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*d*e*f*g + 60*sqrt(b)*sqrt(a*f - b*e)*a 
tan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*d*e*f*h*x + 75*sqr 
t(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b 
**2*d*f**2*g*x - 30*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b 
)*sqrt(a*f - b*e)))*b**3*c*e*f*h*x - 45*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt 
(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*b**3*c*f**2*g*x - 30*sqrt(b)*sqrt( 
a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*b**3*d*e*f*g* 
x + 105*sqrt(e + f*x)*a**3*b*d*f**2*h - 75*sqrt(e + f*x)*a**2*b**2*c*f**2* 
h - 95*sqrt(e + f*x)*a**2*b**2*d*e*f*h - 75*sqrt(e + f*x)*a**2*b**2*d*f...