\(\int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{3/2}} \, dx\) [200]

Optimal result

Integrand size = 29, antiderivative size = 223 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{3/2}} \, dx=-\frac {2 (b e-a f) (b g-a h) \sqrt {c+d x}}{b^3 \sqrt {a+b x}}+\frac {(b c f h-9 a d f h+4 b d (f g+e h)) \sqrt {a+b x} \sqrt {c+d x}}{4 b^3 d}+\frac {f h (a+b x)^{3/2} \sqrt {c+d x}}{2 b^3}+\frac {\left (15 a^2 d^2 f h-6 a b d (c f h+2 d (f g+e h))+b^2 \left (8 d^2 e g-c^2 f h+4 c d (f g+e h)\right )\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{7/2} d^{3/2}} \] Output:

-2*(-a*f+b*e)*(-a*h+b*g)*(d*x+c)^(1/2)/b^3/(b*x+a)^(1/2)+1/4*(b*c*f*h-9*a* 
d*f*h+4*b*d*(e*h+f*g))*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^3/d+1/2*f*h*(b*x+a)^( 
3/2)*(d*x+c)^(1/2)/b^3+1/4*(15*a^2*d^2*f*h-6*a*b*d*(c*f*h+2*d*(e*h+f*g))+b 
^2*(8*d^2*e*g-c^2*f*h+4*c*d*(e*h+f*g)))*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1 
/2)/(d*x+c)^(1/2))/b^(7/2)/d^(3/2)
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {c+d x} \left (-15 a^2 d f h+a b (c f h+d (12 f g+12 e h-5 f h x))+b^2 \left (c f h x+d \left (-8 e g+4 f g x+4 e h x+2 f h x^2\right )\right )\right )}{4 b^3 d \sqrt {a+b x}}+\frac {\left (15 a^2 d^2 f h-6 a b d (c f h+2 d (f g+e h))+b^2 \left (8 d^2 e g-c^2 f h+4 c d (f g+e h)\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{4 b^{7/2} d^{3/2}} \] Input:

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

Output:

(Sqrt[c + d*x]*(-15*a^2*d*f*h + a*b*(c*f*h + d*(12*f*g + 12*e*h - 5*f*h*x) 
) + b^2*(c*f*h*x + d*(-8*e*g + 4*f*g*x + 4*e*h*x + 2*f*h*x^2))))/(4*b^3*d* 
Sqrt[a + b*x]) + ((15*a^2*d^2*f*h - 6*a*b*d*(c*f*h + 2*d*(f*g + e*h)) + b^ 
2*(8*d^2*e*g - c^2*f*h + 4*c*d*(f*g + e*h)))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x 
])/(Sqrt[d]*Sqrt[a + b*x])])/(4*b^(7/2)*d^(3/2))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {163, 60, 66, 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.

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

\(\Big \downarrow \) 163

\(\displaystyle \frac {\left (15 a^2 d^2 f h-6 a b d (c f h+2 d (e h+f g))+b^2 \left (c^2 (-f) h+4 c d (e h+f g)+8 d^2 e g\right )\right ) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}}dx}{4 b^2 d (b c-a d)}-\frac {(c+d x)^{3/2} \left (5 a^2 d f h-a b (c f h+4 d (e h+f g))-b f h x (b c-a d)+4 b^2 d e g\right )}{2 b^2 d \sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\left (15 a^2 d^2 f h-6 a b d (c f h+2 d (e h+f g))+b^2 \left (c^2 (-f) h+4 c d (e h+f g)+8 d^2 e g\right )\right ) \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b^2 d (b c-a d)}-\frac {(c+d x)^{3/2} \left (5 a^2 d f h-a b (c f h+4 d (e h+f g))-b f h x (b c-a d)+4 b^2 d e g\right )}{2 b^2 d \sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\left (15 a^2 d^2 f h-6 a b d (c f h+2 d (e h+f g))+b^2 \left (c^2 (-f) h+4 c d (e h+f g)+8 d^2 e g\right )\right ) \left (\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b^2 d (b c-a d)}-\frac {(c+d x)^{3/2} \left (5 a^2 d f h-a b (c f h+4 d (e h+f g))-b f h x (b c-a d)+4 b^2 d e g\right )}{2 b^2 d \sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right ) \left (15 a^2 d^2 f h-6 a b d (c f h+2 d (e h+f g))+b^2 \left (c^2 (-f) h+4 c d (e h+f g)+8 d^2 e g\right )\right )}{4 b^2 d (b c-a d)}-\frac {(c+d x)^{3/2} \left (5 a^2 d f h-a b (c f h+4 d (e h+f g))-b f h x (b c-a d)+4 b^2 d e g\right )}{2 b^2 d \sqrt {a+b x} (b c-a d)}\)

Input:

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

Output:

-1/2*((c + d*x)^(3/2)*(4*b^2*d*e*g + 5*a^2*d*f*h - a*b*(c*f*h + 4*d*(f*g + 
 e*h)) - b*(b*c - a*d)*f*h*x))/(b^2*d*(b*c - a*d)*Sqrt[a + b*x]) + ((15*a^ 
2*d^2*f*h - 6*a*b*d*(c*f*h + 2*d*(f*g + e*h)) + b^2*(8*d^2*e*g - c^2*f*h + 
 4*c*d*(f*g + e*h)))*((Sqrt[a + b*x]*Sqrt[c + d*x])/b + ((b*c - a*d)*ArcTa 
nh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(b^(3/2)*Sqrt[d])))/( 
4*b^2*d*(b*c - a*d))
 

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 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1117\) vs. \(2(193)=386\).

Time = 0.33 (sec) , antiderivative size = 1118, normalized size of antiderivative = 5.01

method	result	size
default	\(\text {Expression too large to display}\)	\(1118\)

Input:

int((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/8*(d*x+c)^(1/2)*(-10*a*b*d*f*h*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-6*l 
n(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2)) 
*a*b^2*c*d*f*h*x+8*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a 
*d+b*c)/(d*b)^(1/2))*b^3*d^2*e*g*x-ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/ 
2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*b^3*c^2*f*h*x+24*a*b*d*f*g*((b*x+a)*( 
d*x+c))^(1/2)*(d*b)^(1/2)-30*a^2*d*f*h*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2) 
-16*b^2*d*e*g*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+15*ln(1/2*(2*b*d*x+2*((b 
*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b*d^2*f*h*x-12* 
ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2) 
)*a*b^2*d^2*e*h*x-12*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2) 
+a*d+b*c)/(d*b)^(1/2))*a*b^2*d^2*f*g*x+4*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c 
))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*b^3*c*d*e*h*x+4*ln(1/2*(2*b*d*x 
+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*b^3*c*d*f*g*x 
-6*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1 
/2))*a^2*b*c*d*f*h+4*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2) 
+a*d+b*c)/(d*b)^(1/2))*a*b^2*c*d*e*h+8*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c)) 
^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^2*d^2*e*g-12*ln(1/2*(2*b*d*x+ 
2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b*d^2*e*h- 
12*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1 
/2))*a^2*b*d^2*f*g-ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (193) = 386\).

Time = 3.84 (sec) , antiderivative size = 802, normalized size of antiderivative = 3.60 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(3/2),x, algorithm="fricas 
")
 

Output:

[1/16*(sqrt(b*d)*(4*(2*a*b^2*d^2*e + (a*b^2*c*d - 3*a^2*b*d^2)*f)*g + (4*( 
a*b^2*c*d - 3*a^2*b*d^2)*e - (a*b^2*c^2 + 6*a^2*b*c*d - 15*a^3*d^2)*f)*h + 
 (4*(2*b^3*d^2*e + (b^3*c*d - 3*a*b^2*d^2)*f)*g + (4*(b^3*c*d - 3*a*b^2*d^ 
2)*e - (b^3*c^2 + 6*a*b^2*c*d - 15*a^2*b*d^2)*f)*h)*x)*log(8*b^2*d^2*x^2 + 
 b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b* 
x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(2*b^3*d^2*f*h*x^2 - 4 
*(2*b^3*d^2*e - 3*a*b^2*d^2*f)*g + (12*a*b^2*d^2*e + (a*b^2*c*d - 15*a^2*b 
*d^2)*f)*h + (4*b^3*d^2*f*g + (4*b^3*d^2*e + (b^3*c*d - 5*a*b^2*d^2)*f)*h) 
*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^5*d^2*x + a*b^4*d^2), -1/8*(sqrt(-b*d) 
*(4*(2*a*b^2*d^2*e + (a*b^2*c*d - 3*a^2*b*d^2)*f)*g + (4*(a*b^2*c*d - 3*a^ 
2*b*d^2)*e - (a*b^2*c^2 + 6*a^2*b*c*d - 15*a^3*d^2)*f)*h + (4*(2*b^3*d^2*e 
 + (b^3*c*d - 3*a*b^2*d^2)*f)*g + (4*(b^3*c*d - 3*a*b^2*d^2)*e - (b^3*c^2 
+ 6*a*b^2*c*d - 15*a^2*b*d^2)*f)*h)*x)*arctan(1/2*(2*b*d*x + b*c + a*d)*sq 
rt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a 
*b*d^2)*x)) - 2*(2*b^3*d^2*f*h*x^2 - 4*(2*b^3*d^2*e - 3*a*b^2*d^2*f)*g + ( 
12*a*b^2*d^2*e + (a*b^2*c*d - 15*a^2*b*d^2)*f)*h + (4*b^3*d^2*f*g + (4*b^3 
*d^2*e + (b^3*c*d - 5*a*b^2*d^2)*f)*h)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^ 
5*d^2*x + a*b^4*d^2)]
 

Sympy [F]

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

integrate((d*x+c)**(1/2)*(f*x+e)*(h*x+g)/(b*x+a)**(3/2),x)
 

Output:

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

Maxima [F(-2)]

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

integrate((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(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(a*d-b*c>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. 404 vs. \(2 (193) = 386\).

Time = 0.29 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.81 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{3/2}} \, dx=\frac {1}{4} \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} f h {\left | b \right |}}{b^{5}} + \frac {4 \, b^{10} d^{2} f g {\left | b \right |} + 4 \, b^{10} d^{2} e h {\left | b \right |} + b^{10} c d f h {\left | b \right |} - 9 \, a b^{9} d^{2} f h {\left | b \right |}}{b^{14} d^{2}}\right )} - \frac {4 \, {\left (b^{3} c d e g {\left | b \right |} - a b^{2} d^{2} e g {\left | b \right |} - a b^{2} c d f g {\left | b \right |} + a^{2} b d^{2} f g {\left | b \right |} - a b^{2} c d e h {\left | b \right |} + a^{2} b d^{2} e h {\left | b \right |} + a^{2} b c d f h {\left | b \right |} - a^{3} d^{2} f h {\left | b \right |}\right )}}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} \sqrt {b d} b^{3}} - \frac {{\left (8 \, b^{2} d^{2} e g {\left | b \right |} + 4 \, b^{2} c d f g {\left | b \right |} - 12 \, a b d^{2} f g {\left | b \right |} + 4 \, b^{2} c d e h {\left | b \right |} - 12 \, a b d^{2} e h {\left | b \right |} - b^{2} c^{2} f h {\left | b \right |} - 6 \, a b c d f h {\left | b \right |} + 15 \, a^{2} d^{2} f h {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{8 \, \sqrt {b d} b^{4} d} \] Input:

integrate((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(3/2),x, algorithm="giac")
 

Output:

1/4*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*f*h*abs 
(b)/b^5 + (4*b^10*d^2*f*g*abs(b) + 4*b^10*d^2*e*h*abs(b) + b^10*c*d*f*h*ab 
s(b) - 9*a*b^9*d^2*f*h*abs(b))/(b^14*d^2)) - 4*(b^3*c*d*e*g*abs(b) - a*b^2 
*d^2*e*g*abs(b) - a*b^2*c*d*f*g*abs(b) + a^2*b*d^2*f*g*abs(b) - a*b^2*c*d* 
e*h*abs(b) + a^2*b*d^2*e*h*abs(b) + a^2*b*c*d*f*h*abs(b) - a^3*d^2*f*h*abs 
(b))/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b 
*d - a*b*d))^2)*sqrt(b*d)*b^3) - 1/8*(8*b^2*d^2*e*g*abs(b) + 4*b^2*c*d*f*g 
*abs(b) - 12*a*b*d^2*f*g*abs(b) + 4*b^2*c*d*e*h*abs(b) - 12*a*b*d^2*e*h*ab 
s(b) - b^2*c^2*f*h*abs(b) - 6*a*b*c*d*f*h*abs(b) + 15*a^2*d^2*f*h*abs(b))* 
log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sq 
rt(b*d)*b^4*d)
 

Mupad [F(-1)]

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

int(((e + f*x)*(g + h*x)*(c + d*x)^(1/2))/(a + b*x)^(3/2),x)
 

Output:

int(((e + f*x)*(g + h*x)*(c + d*x)^(1/2))/(a + b*x)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 729, normalized size of antiderivative = 3.27 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{3/2}} \, dx =\text {Too large to display} \] Input:

int((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(3/2),x)
 

Output:

(15*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqr 
t(c + d*x))/sqrt(a*d - b*c))*a**2*d**2*f*h - 6*sqrt(d)*sqrt(b)*sqrt(a + b* 
x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a* 
b*c*d*f*h - 12*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + 
sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b*d**2*e*h - 12*sqrt(d)*sqrt(b)* 
sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d 
 - b*c))*a*b*d**2*f*g - sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a 
+ b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**2*c**2*f*h + 4*sqrt(d) 
*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x)) 
/sqrt(a*d - b*c))*b**2*c*d*e*h + 4*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt 
(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**2*c*d*f*g + 
 8*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt 
(c + d*x))/sqrt(a*d - b*c))*b**2*d**2*e*g - 10*sqrt(d)*sqrt(b)*sqrt(a + b* 
x)*a**2*d**2*f*h + 2*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b*c*d*f*h + 9*sqrt(d) 
*sqrt(b)*sqrt(a + b*x)*a*b*d**2*e*h + 9*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b* 
d**2*f*g - sqrt(d)*sqrt(b)*sqrt(a + b*x)*b**2*c*d*e*h - sqrt(d)*sqrt(b)*sq 
rt(a + b*x)*b**2*c*d*f*g - 8*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b**2*d**2*e*g - 
 15*sqrt(c + d*x)*a**2*b*d**2*f*h + sqrt(c + d*x)*a*b**2*c*d*f*h + 12*sqrt 
(c + d*x)*a*b**2*d**2*e*h + 12*sqrt(c + d*x)*a*b**2*d**2*f*g - 5*sqrt(c + 
d*x)*a*b**2*d**2*f*h*x + sqrt(c + d*x)*b**3*c*d*f*h*x - 8*sqrt(c + d*x)...