\(\int \frac {(a+b x) (g+h x)}{(c+d x)^3 (e+f x)^{3/2}} \, dx\) [188]

Optimal result

Integrand size = 27, antiderivative size = 280 \[ \int \frac {(a+b x) (g+h x)}{(c+d x)^3 (e+f x)^{3/2}} \, dx=-\frac {2 (b e-a f) (f g-e h)}{(d e-c f)^3 \sqrt {e+f x}}+\frac {(b c-a d) (d g-c h) \sqrt {e+f x}}{2 d (d e-c f)^2 (c+d x)^2}+\frac {\left (a d (7 d f g-4 d e h-3 c f h)-b \left (4 d^2 e g+c^2 f h+c d (3 f g-8 e h)\right )\right ) \sqrt {e+f x}}{4 d (d e-c f)^3 (c+d x)}-\frac {\left (3 a d f (5 d f g-4 d e h-c f h)-b \left (c^2 f^2 h+c d f (3 f g-8 e h)+4 d^2 e (3 f g-2 e h)\right )\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{4 d^{3/2} (d e-c f)^{7/2}} \] Output:

-2*(-a*f+b*e)*(-e*h+f*g)/(-c*f+d*e)^3/(f*x+e)^(1/2)+1/2*(-a*d+b*c)*(-c*h+d 
*g)*(f*x+e)^(1/2)/d/(-c*f+d*e)^2/(d*x+c)^2+1/4*(a*d*(-3*c*f*h-4*d*e*h+7*d* 
f*g)-b*(4*d^2*e*g+c^2*f*h+c*d*(-8*e*h+3*f*g)))*(f*x+e)^(1/2)/d/(-c*f+d*e)^ 
3/(d*x+c)-1/4*(3*a*d*f*(-c*f*h-4*d*e*h+5*d*f*g)-b*(c^2*f^2*h+c*d*f*(-8*e*h 
+3*f*g)+4*d^2*e*(-2*e*h+3*f*g)))*arctanh(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^ 
(1/2))/d^(3/2)/(-c*f+d*e)^(7/2)
 

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.25 \[ \int \frac {(a+b x) (g+h x)}{(c+d x)^3 (e+f x)^{3/2}} \, dx=\frac {\frac {\sqrt {d} \left (b \left (c^3 f h (e+f x)+4 d^3 e x (-e g-3 f g x+2 e h x)+c^2 d \left (14 e^2 h-f^2 x (5 g+h x)+e f (-13 g+5 h x)\right )+c d^2 \left (-3 f^2 g x^2-2 e^2 (g-12 h x)+e f x (-21 g+8 h x)\right )\right )-a d \left (c^2 f (-8 f g+13 e h+5 f h x)+d^2 \left (-15 f^2 g x^2+2 e^2 (g+2 h x)+e f x (-5 g+12 h x)\right )+c d \left (2 e^2 h+f^2 x (-25 g+3 h x)+e f (-9 g+21 h x)\right )\right )\right )}{(d e-c f)^3 (c+d x)^2 \sqrt {e+f x}}-\frac {\left (-3 a d f (-5 d f g+4 d e h+c f h)+b \left (-c^2 f^2 h+4 d^2 e (-3 f g+2 e h)+c d f (-3 f g+8 e h)\right )\right ) \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{(-d e+c f)^{7/2}}}{4 d^{3/2}} \] Input:

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

Output:

((Sqrt[d]*(b*(c^3*f*h*(e + f*x) + 4*d^3*e*x*(-(e*g) - 3*f*g*x + 2*e*h*x) + 
 c^2*d*(14*e^2*h - f^2*x*(5*g + h*x) + e*f*(-13*g + 5*h*x)) + c*d^2*(-3*f^ 
2*g*x^2 - 2*e^2*(g - 12*h*x) + e*f*x*(-21*g + 8*h*x))) - a*d*(c^2*f*(-8*f* 
g + 13*e*h + 5*f*h*x) + d^2*(-15*f^2*g*x^2 + 2*e^2*(g + 2*h*x) + e*f*x*(-5 
*g + 12*h*x)) + c*d*(2*e^2*h + f^2*x*(-25*g + 3*h*x) + e*f*(-9*g + 21*h*x) 
))))/((d*e - c*f)^3*(c + d*x)^2*Sqrt[e + f*x]) - ((-3*a*d*f*(-5*d*f*g + 4* 
d*e*h + c*f*h) + b*(-(c^2*f^2*h) + 4*d^2*e*(-3*f*g + 2*e*h) + c*d*f*(-3*f* 
g + 8*e*h)))*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c*f]])/(-(d*e) + 
 c*f)^(7/2))/(4*d^(3/2))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {161, 52, 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.

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

\(\Big \downarrow \) 161

\(\displaystyle -\frac {\left (3 a d f (-c f h-4 d e h+5 d f g)-b \left (c^2 f^2 h+c d f (3 f g-8 e h)+4 d^2 e (3 f g-2 e h)\right )\right ) \int \frac {1}{(c+d x)^2 \sqrt {e+f x}}dx}{4 d f (d e-c f)^2}-\frac {x \left (a d f (-c f h-4 d e h+5 d f g)-b \left (-c^2 f^2 h+c d f^2 g+4 d^2 e (f g-e h)\right )\right )+a d f (-5 c e h+4 c f g+d e g)-b c e (-c f h-4 d e h+5 d f g)}{2 d f (c+d x)^2 \sqrt {e+f x} (d e-c f)^2}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {\left (3 a d f (-c f h-4 d e h+5 d f g)-b \left (c^2 f^2 h+c d f (3 f g-8 e h)+4 d^2 e (3 f g-2 e h)\right )\right ) \left (-\frac {f \int \frac {1}{(c+d x) \sqrt {e+f x}}dx}{2 (d e-c f)}-\frac {\sqrt {e+f x}}{(c+d x) (d e-c f)}\right )}{4 d f (d e-c f)^2}-\frac {x \left (a d f (-c f h-4 d e h+5 d f g)-b \left (-c^2 f^2 h+c d f^2 g+4 d^2 e (f g-e h)\right )\right )+a d f (-5 c e h+4 c f g+d e g)-b c e (-c f h-4 d e h+5 d f g)}{2 d f (c+d x)^2 \sqrt {e+f x} (d e-c f)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\left (3 a d f (-c f h-4 d e h+5 d f g)-b \left (c^2 f^2 h+c d f (3 f g-8 e h)+4 d^2 e (3 f g-2 e h)\right )\right ) \left (-\frac {\int \frac {1}{c+\frac {d (e+f x)}{f}-\frac {d e}{f}}d\sqrt {e+f x}}{d e-c f}-\frac {\sqrt {e+f x}}{(c+d x) (d e-c f)}\right )}{4 d f (d e-c f)^2}-\frac {x \left (a d f (-c f h-4 d e h+5 d f g)-b \left (-c^2 f^2 h+c d f^2 g+4 d^2 e (f g-e h)\right )\right )+a d f (-5 c e h+4 c f g+d e g)-b c e (-c f h-4 d e h+5 d f g)}{2 d f (c+d x)^2 \sqrt {e+f x} (d e-c f)^2}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\left (\frac {f \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{\sqrt {d} (d e-c f)^{3/2}}-\frac {\sqrt {e+f x}}{(c+d x) (d e-c f)}\right ) \left (3 a d f (-c f h-4 d e h+5 d f g)-b \left (c^2 f^2 h+c d f (3 f g-8 e h)+4 d^2 e (3 f g-2 e h)\right )\right )}{4 d f (d e-c f)^2}-\frac {x \left (a d f (-c f h-4 d e h+5 d f g)-b \left (-c^2 f^2 h+c d f^2 g+4 d^2 e (f g-e h)\right )\right )+a d f (-5 c e h+4 c f g+d e g)-b c e (-c f h-4 d e h+5 d f g)}{2 d f (c+d x)^2 \sqrt {e+f x} (d e-c f)^2}\)

Input:

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

Output:

-1/2*(a*d*f*(d*e*g + 4*c*f*g - 5*c*e*h) - b*c*e*(5*d*f*g - 4*d*e*h - c*f*h 
) + (a*d*f*(5*d*f*g - 4*d*e*h - c*f*h) - b*(c*d*f^2*g - c^2*f^2*h + 4*d^2* 
e*(f*g - e*h)))*x)/(d*f*(d*e - c*f)^2*(c + d*x)^2*Sqrt[e + f*x]) - ((3*a*d 
*f*(5*d*f*g - 4*d*e*h - c*f*h) - b*(c^2*f^2*h + c*d*f*(3*f*g - 8*e*h) + 4* 
d^2*e*(3*f*g - 2*e*h)))*(-(Sqrt[e + f*x]/((d*e - c*f)*(c + d*x))) + (f*Arc 
Tanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(Sqrt[d]*(d*e - c*f)^(3/2)) 
))/(4*d*f*(d*e - c*f)^2)
 

Defintions of rubi rules used

rule 52

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

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 161

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_)) 
*((g_.) + (h_.)*(x_)), x_] :> Simp[((b^2*c*d*e*g*(n + 1) + a^2*c*d*f*h*(n + 
 1) + a*b*(d^2*e*g*(m + 1) + c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(m + n + 2)) 
 + (a^2*d^2*f*h*(n + 1) - a*b*d^2*(f*g + e*h)*(n + 1) + b^2*(c^2*f*h*(m + 1 
) - c*d*(f*g + e*h)*(m + 1) + d^2*e*g*(m + n + 2)))*x)/(b*d*(b*c - a*d)^2*( 
m + 1)*(n + 1)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] - Simp[(a^2*d^2*f* 
h*(2 + 3*n + 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*(2 + 3*m + m^2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + 
 d^2*e*g*(6 + m^2 + 5*n + n^2 + m*(2*n + 5))))/(b*d*(b*c - a*d)^2*(m + 1)*( 
n + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x], x] /; FreeQ[{a, b, c 
, d, e, f, g, h}, x] && LtQ[m, -1] && LtQ[n, -1]
 

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 = 0.58 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.28

method	result	size
pseudoelliptic	\(\frac {\frac {3 \left (x d +c \right )^{2} \sqrt {f x +e}\, \left (\left (-5 g \,f^{2} a +4 e \left (a h +b g \right ) f -\frac {8 b \,e^{2} h}{3}\right ) d^{2}+c \left (\left (a h +b g \right ) f -\frac {8 e h b}{3}\right ) f d +\frac {b \,c^{2} f^{2} h}{3}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{4}+\frac {13 \sqrt {\left (c f -d e \right ) d}\, \left (\frac {\left (-15 a \,f^{2} g \,x^{2}-5 x \left (\frac {12 \left (-a h -b g \right ) x}{5}+g a \right ) e f +2 \left (-4 b h \,x^{2}+2 \left (a h +b g \right ) x +g a \right ) e^{2}\right ) d^{3}}{13}+\frac {2 c \left (\frac {\left (3 \left (a h +b g \right ) x^{2}-25 a g x \right ) f^{2}}{2}-\frac {9 \left (\frac {8 b h \,x^{2}}{9}+\frac {7 \left (-a h -b g \right ) x}{3}+g a \right ) e f}{2}+e^{2} \left (-12 b h x +a h +b g \right )\right ) d^{2}}{13}+\left (\frac {\left (b h \,x^{2}+5 \left (a h +b g \right ) x -8 g a \right ) f^{2}}{13}+e \left (a h +b g -\frac {5}{13} b h x \right ) f -\frac {14 b \,e^{2} h}{13}\right ) c^{2} d -\frac {b \,c^{3} f h \left (f x +e \right )}{13}\right )}{4}}{\left (x d +c \right )^{2} \left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}\, d \sqrt {f x +e}}\)	\(358\)
derivativedivides	\(\frac {\frac {2 \left (\left (\frac {3}{8} a c d \,f^{2} h +\frac {1}{2} a \,d^{2} e f h -\frac {7}{8} a \,d^{2} f^{2} g +\frac {1}{8} b \,c^{2} f^{2} h -b c d e f h +\frac {3}{8} b c d \,f^{2} g +\frac {1}{2} b \,d^{2} e f g \right ) \left (f x +e \right )^{\frac {3}{2}}+\frac {f \left (5 a \,c^{2} d \,f^{2} h -a c \,d^{2} e f h -9 a c \,d^{2} f^{2} g -4 a \,d^{3} e^{2} h +9 a \,d^{3} e f g -b \,c^{3} f^{2} h -7 b \,c^{2} d e f h +5 b \,c^{2} d \,f^{2} g +8 b c \,d^{2} e^{2} h -b c \,d^{2} e f g -4 b \,d^{3} e^{2} g \right ) \sqrt {f x +e}}{8 d}\right )}{\left (\left (f x +e \right ) d +c f -d e \right )^{2}}+\frac {\left (3 a c d \,f^{2} h +12 a \,d^{2} e f h -15 a \,d^{2} f^{2} g +b \,c^{2} f^{2} h -8 b c d e f h +3 b c d \,f^{2} g -8 b \,e^{2} h \,d^{2}+12 b \,d^{2} e f g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{4 d \sqrt {\left (c f -d e \right ) d}}}{\left (c f -d e \right )^{3}}-\frac {2 \left (-a f h e +g \,f^{2} a +b \,e^{2} h -b f g e \right )}{\left (c f -d e \right )^{3} \sqrt {f x +e}}\)	\(391\)
default	\(\frac {\frac {2 \left (\left (\frac {3}{8} a c d \,f^{2} h +\frac {1}{2} a \,d^{2} e f h -\frac {7}{8} a \,d^{2} f^{2} g +\frac {1}{8} b \,c^{2} f^{2} h -b c d e f h +\frac {3}{8} b c d \,f^{2} g +\frac {1}{2} b \,d^{2} e f g \right ) \left (f x +e \right )^{\frac {3}{2}}+\frac {f \left (5 a \,c^{2} d \,f^{2} h -a c \,d^{2} e f h -9 a c \,d^{2} f^{2} g -4 a \,d^{3} e^{2} h +9 a \,d^{3} e f g -b \,c^{3} f^{2} h -7 b \,c^{2} d e f h +5 b \,c^{2} d \,f^{2} g +8 b c \,d^{2} e^{2} h -b c \,d^{2} e f g -4 b \,d^{3} e^{2} g \right ) \sqrt {f x +e}}{8 d}\right )}{\left (\left (f x +e \right ) d +c f -d e \right )^{2}}+\frac {\left (3 a c d \,f^{2} h +12 a \,d^{2} e f h -15 a \,d^{2} f^{2} g +b \,c^{2} f^{2} h -8 b c d e f h +3 b c d \,f^{2} g -8 b \,e^{2} h \,d^{2}+12 b \,d^{2} e f g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{4 d \sqrt {\left (c f -d e \right ) d}}}{\left (c f -d e \right )^{3}}-\frac {2 \left (-a f h e +g \,f^{2} a +b \,e^{2} h -b f g e \right )}{\left (c f -d e \right )^{3} \sqrt {f x +e}}\)	\(391\)

Input:

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

Output:

3/4*((d*x+c)^2*(f*x+e)^(1/2)*((-5*g*f^2*a+4*e*(a*h+b*g)*f-8/3*b*e^2*h)*d^2 
+c*((a*h+b*g)*f-8/3*e*h*b)*f*d+1/3*b*c^2*f^2*h)*arctan(d*(f*x+e)^(1/2)/((c 
*f-d*e)*d)^(1/2))+13/3*((c*f-d*e)*d)^(1/2)*(1/13*(-15*a*f^2*g*x^2-5*x*(12/ 
5*(-a*h-b*g)*x+g*a)*e*f+2*(-4*b*h*x^2+2*(a*h+b*g)*x+g*a)*e^2)*d^3+2/13*c*( 
1/2*(3*(a*h+b*g)*x^2-25*a*g*x)*f^2-9/2*(8/9*b*h*x^2+7/3*(-a*h-b*g)*x+g*a)* 
e*f+e^2*(-12*b*h*x+a*h+b*g))*d^2+(1/13*(b*h*x^2+5*(a*h+b*g)*x-8*g*a)*f^2+e 
*(a*h+b*g-5/13*b*h*x)*f-14/13*b*e^2*h)*c^2*d-1/13*b*c^3*f*h*(f*x+e)))/(f*x 
+e)^(1/2)/((c*f-d*e)*d)^(1/2)/(d*x+c)^2/(c*f-d*e)^3/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1241 vs. \(2 (258) = 516\).

Time = 0.24 (sec) , antiderivative size = 2496, normalized size of antiderivative = 8.91 \[ \int \frac {(a+b x) (g+h x)}{(c+d x)^3 (e+f x)^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/8*(((3*(4*b*d^4*e*f^2 + (b*c*d^3 - 5*a*d^4)*f^3)*g - (8*b*d^4*e^2*f + 
4*(2*b*c*d^3 - 3*a*d^4)*e*f^2 - (b*c^2*d^2 + 3*a*c*d^3)*f^3)*h)*x^3 + (3*( 
4*b*d^4*e^2*f + (9*b*c*d^3 - 5*a*d^4)*e*f^2 + 2*(b*c^2*d^2 - 5*a*c*d^3)*f^ 
3)*g - (8*b*d^4*e^3 + 12*(2*b*c*d^3 - a*d^4)*e^2*f + 3*(5*b*c^2*d^2 - 9*a* 
c*d^3)*e*f^2 - 2*(b*c^3*d + 3*a*c^2*d^2)*f^3)*h)*x^2 + 3*(4*b*c^2*d^2*e^2* 
f + (b*c^3*d - 5*a*c^2*d^2)*e*f^2)*g - (8*b*c^2*d^2*e^3 + 4*(2*b*c^3*d - 3 
*a*c^2*d^2)*e^2*f - (b*c^4 + 3*a*c^3*d)*e*f^2)*h + (3*(8*b*c*d^3*e^2*f + 2 
*(3*b*c^2*d^2 - 5*a*c*d^3)*e*f^2 + (b*c^3*d - 5*a*c^2*d^2)*f^3)*g - (16*b* 
c*d^3*e^3 + 24*(b*c^2*d^2 - a*c*d^3)*e^2*f + 6*(b*c^3*d - 3*a*c^2*d^2)*e*f 
^2 - (b*c^4 + 3*a*c^3*d)*f^3)*h)*x)*sqrt(d^2*e - c*d*f)*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*((3*(4*b*d^5* 
e^2*f - (3*b*c*d^4 + 5*a*d^5)*e*f^2 - (b*c^2*d^3 - 5*a*c*d^4)*f^3)*g - (8* 
b*d^5*e^3 - 12*a*d^5*e^2*f - 9*(b*c^2*d^3 - a*c*d^4)*e*f^2 + (b*c^3*d^2 + 
3*a*c^2*d^3)*f^3)*h)*x^2 + (8*a*c^3*d^2*f^3 + 2*(b*c*d^4 + a*d^5)*e^3 + 11 
*(b*c^2*d^3 - a*c*d^4)*e^2*f - (13*b*c^3*d^2 - a*c^2*d^3)*e*f^2)*g - (2*(7 
*b*c^2*d^3 - a*c*d^4)*e^3 - (13*b*c^3*d^2 + 11*a*c^2*d^3)*e^2*f - (b*c^4*d 
 - 13*a*c^3*d^2)*e*f^2)*h + ((4*b*d^5*e^3 + (17*b*c*d^4 - 5*a*d^5)*e^2*f - 
 4*(4*b*c^2*d^3 + 5*a*c*d^4)*e*f^2 - 5*(b*c^3*d^2 - 5*a*c^2*d^3)*f^3)*g - 
(4*(6*b*c*d^4 - a*d^5)*e^3 - (19*b*c^2*d^3 + 17*a*c*d^4)*e^2*f - 4*(b*c^3* 
d^2 - 4*a*c^2*d^3)*e*f^2 - (b*c^4*d - 5*a*c^3*d^2)*f^3)*h)*x)*sqrt(f*x ...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((b*x+a)*(h*x+g)/(d*x+c)^3/(f*x+e)^(3/2),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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 607 vs. \(2 (258) = 516\).

Time = 0.15 (sec) , antiderivative size = 607, normalized size of antiderivative = 2.17 \[ \int \frac {(a+b x) (g+h x)}{(c+d x)^3 (e+f x)^{3/2}} \, dx=-\frac {{\left (12 \, b d^{2} e f g + 3 \, b c d f^{2} g - 15 \, a d^{2} f^{2} g - 8 \, b d^{2} e^{2} h - 8 \, b c d e f h + 12 \, a d^{2} e f h + b c^{2} f^{2} h + 3 \, a c d f^{2} h\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{4 \, {\left (d^{4} e^{3} - 3 \, c d^{3} e^{2} f + 3 \, c^{2} d^{2} e f^{2} - c^{3} d f^{3}\right )} \sqrt {-d^{2} e + c d f}} - \frac {2 \, {\left (b e f g - a f^{2} g - b e^{2} h + a e f h\right )}}{{\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \sqrt {f x + e}} - \frac {4 \, {\left (f x + e\right )}^{\frac {3}{2}} b d^{3} e f g - 4 \, \sqrt {f x + e} b d^{3} e^{2} f g + 3 \, {\left (f x + e\right )}^{\frac {3}{2}} b c d^{2} f^{2} g - 7 \, {\left (f x + e\right )}^{\frac {3}{2}} a d^{3} f^{2} g - \sqrt {f x + e} b c d^{2} e f^{2} g + 9 \, \sqrt {f x + e} a d^{3} e f^{2} g + 5 \, \sqrt {f x + e} b c^{2} d f^{3} g - 9 \, \sqrt {f x + e} a c d^{2} f^{3} g - 8 \, {\left (f x + e\right )}^{\frac {3}{2}} b c d^{2} e f h + 4 \, {\left (f x + e\right )}^{\frac {3}{2}} a d^{3} e f h + 8 \, \sqrt {f x + e} b c d^{2} e^{2} f h - 4 \, \sqrt {f x + e} a d^{3} e^{2} f h + {\left (f x + e\right )}^{\frac {3}{2}} b c^{2} d f^{2} h + 3 \, {\left (f x + e\right )}^{\frac {3}{2}} a c d^{2} f^{2} h - 7 \, \sqrt {f x + e} b c^{2} d e f^{2} h - \sqrt {f x + e} a c d^{2} e f^{2} h - \sqrt {f x + e} b c^{3} f^{3} h + 5 \, \sqrt {f x + e} a c^{2} d f^{3} h}{4 \, {\left (d^{4} e^{3} - 3 \, c d^{3} e^{2} f + 3 \, c^{2} d^{2} e f^{2} - c^{3} d f^{3}\right )} {\left ({\left (f x + e\right )} d - d e + c f\right )}^{2}} \] Input:

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

Output:

-1/4*(12*b*d^2*e*f*g + 3*b*c*d*f^2*g - 15*a*d^2*f^2*g - 8*b*d^2*e^2*h - 8* 
b*c*d*e*f*h + 12*a*d^2*e*f*h + b*c^2*f^2*h + 3*a*c*d*f^2*h)*arctan(sqrt(f* 
x + e)*d/sqrt(-d^2*e + c*d*f))/((d^4*e^3 - 3*c*d^3*e^2*f + 3*c^2*d^2*e*f^2 
 - c^3*d*f^3)*sqrt(-d^2*e + c*d*f)) - 2*(b*e*f*g - a*f^2*g - b*e^2*h + a*e 
*f*h)/((d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*sqrt(f*x + e)) 
- 1/4*(4*(f*x + e)^(3/2)*b*d^3*e*f*g - 4*sqrt(f*x + e)*b*d^3*e^2*f*g + 3*( 
f*x + e)^(3/2)*b*c*d^2*f^2*g - 7*(f*x + e)^(3/2)*a*d^3*f^2*g - sqrt(f*x + 
e)*b*c*d^2*e*f^2*g + 9*sqrt(f*x + e)*a*d^3*e*f^2*g + 5*sqrt(f*x + e)*b*c^2 
*d*f^3*g - 9*sqrt(f*x + e)*a*c*d^2*f^3*g - 8*(f*x + e)^(3/2)*b*c*d^2*e*f*h 
 + 4*(f*x + e)^(3/2)*a*d^3*e*f*h + 8*sqrt(f*x + e)*b*c*d^2*e^2*f*h - 4*sqr 
t(f*x + e)*a*d^3*e^2*f*h + (f*x + e)^(3/2)*b*c^2*d*f^2*h + 3*(f*x + e)^(3/ 
2)*a*c*d^2*f^2*h - 7*sqrt(f*x + e)*b*c^2*d*e*f^2*h - sqrt(f*x + e)*a*c*d^2 
*e*f^2*h - sqrt(f*x + e)*b*c^3*f^3*h + 5*sqrt(f*x + e)*a*c^2*d*f^3*h)/((d^ 
4*e^3 - 3*c*d^3*e^2*f + 3*c^2*d^2*e*f^2 - c^3*d*f^3)*((f*x + e)*d - d*e + 
c*f)^2)
 

Mupad [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 442, normalized size of antiderivative = 1.58 \[ \int \frac {(a+b x) (g+h x)}{(c+d x)^3 (e+f x)^{3/2}} \, dx=-\frac {\frac {2\,\left (a\,f^2\,g+b\,e^2\,h-a\,e\,f\,h-b\,e\,f\,g\right )}{c\,f-d\,e}-\frac {{\left (e+f\,x\right )}^2\,\left (b\,c^2\,f^2\,h-15\,a\,d^2\,f^2\,g-8\,b\,d^2\,e^2\,h+3\,a\,c\,d\,f^2\,h+3\,b\,c\,d\,f^2\,g+12\,a\,d^2\,e\,f\,h+12\,b\,d^2\,e\,f\,g-8\,b\,c\,d\,e\,f\,h\right )}{4\,{\left (c\,f-d\,e\right )}^3}+\frac {\left (e+f\,x\right )\,\left (25\,a\,d^2\,f^2\,g+b\,c^2\,f^2\,h+16\,b\,d^2\,e^2\,h-5\,a\,c\,d\,f^2\,h-5\,b\,c\,d\,f^2\,g-20\,a\,d^2\,e\,f\,h-20\,b\,d^2\,e\,f\,g+8\,b\,c\,d\,e\,f\,h\right )}{4\,d\,{\left (c\,f-d\,e\right )}^2}}{d^2\,{\left (e+f\,x\right )}^{5/2}-{\left (e+f\,x\right )}^{3/2}\,\left (2\,d^2\,e-2\,c\,d\,f\right )+\sqrt {e+f\,x}\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2\right )}-\frac {\mathrm {atan}\left (\frac {\sqrt {e+f\,x}\,\left (-c^3\,d\,f^3+3\,c^2\,d^2\,e\,f^2-3\,c\,d^3\,e^2\,f+d^4\,e^3\right )}{\sqrt {d}\,{\left (c\,f-d\,e\right )}^{7/2}}\right )\,\left (b\,c^2\,f^2\,h-15\,a\,d^2\,f^2\,g-8\,b\,d^2\,e^2\,h+3\,a\,c\,d\,f^2\,h+3\,b\,c\,d\,f^2\,g+12\,a\,d^2\,e\,f\,h+12\,b\,d^2\,e\,f\,g-8\,b\,c\,d\,e\,f\,h\right )}{4\,d^{3/2}\,{\left (c\,f-d\,e\right )}^{7/2}} \] Input:

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

Output:

- ((2*(a*f^2*g + b*e^2*h - a*e*f*h - b*e*f*g))/(c*f - d*e) - ((e + f*x)^2* 
(b*c^2*f^2*h - 15*a*d^2*f^2*g - 8*b*d^2*e^2*h + 3*a*c*d*f^2*h + 3*b*c*d*f^ 
2*g + 12*a*d^2*e*f*h + 12*b*d^2*e*f*g - 8*b*c*d*e*f*h))/(4*(c*f - d*e)^3) 
+ ((e + f*x)*(25*a*d^2*f^2*g + b*c^2*f^2*h + 16*b*d^2*e^2*h - 5*a*c*d*f^2* 
h - 5*b*c*d*f^2*g - 20*a*d^2*e*f*h - 20*b*d^2*e*f*g + 8*b*c*d*e*f*h))/(4*d 
*(c*f - d*e)^2))/(d^2*(e + f*x)^(5/2) - (e + f*x)^(3/2)*(2*d^2*e - 2*c*d*f 
) + (e + f*x)^(1/2)*(c^2*f^2 + d^2*e^2 - 2*c*d*e*f)) - (atan(((e + f*x)^(1 
/2)*(d^4*e^3 - c^3*d*f^3 + 3*c^2*d^2*e*f^2 - 3*c*d^3*e^2*f))/(d^(1/2)*(c*f 
 - d*e)^(7/2)))*(b*c^2*f^2*h - 15*a*d^2*f^2*g - 8*b*d^2*e^2*h + 3*a*c*d*f^ 
2*h + 3*b*c*d*f^2*g + 12*a*d^2*e*f*h + 12*b*d^2*e*f*g - 8*b*c*d*e*f*h))/(4 
*d^(3/2)*(c*f - d*e)^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 2007, normalized size of antiderivative = 7.17 \[ \int \frac {(a+b x) (g+h x)}{(c+d x)^3 (e+f x)^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(3*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*s 
qrt(c*f - d*e)))*a*c**3*d*f**2*h + 12*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e 
)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a*c**2*d**2*e*f*h - 15 
*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqr 
t(c*f - d*e)))*a*c**2*d**2*f**2*g + 6*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e 
)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a*c**2*d**2*f**2*h*x + 
 24*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)* 
sqrt(c*f - d*e)))*a*c*d**3*e*f*h*x - 30*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d 
*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a*c*d**3*f**2*g*x + 
3*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sq 
rt(c*f - d*e)))*a*c*d**3*f**2*h*x**2 + 12*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - 
 d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a*d**4*e*f*h*x**2 
- 15*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d) 
*sqrt(c*f - d*e)))*a*d**4*f**2*g*x**2 + sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d 
*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b*c**4*f**2*h - 8*sq 
rt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c 
*f - d*e)))*b*c**3*d*e*f*h + 3*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan( 
(sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b*c**3*d*f**2*g + 2*sqrt(d)*s 
qrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d* 
e)))*b*c**3*d*f**2*h*x - 8*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((...