\(\int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{7/2}} \, dx\) [202]

Optimal result

Integrand size = 29, antiderivative size = 206 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{7/2}} \, dx=-\frac {2 f h \sqrt {c+d x}}{b^3 \sqrt {a+b x}}-\frac {2 (b e-a f) (b g-a h) (c+d x)^{3/2}}{5 b^2 (b c-a d) (a+b x)^{5/2}}-\frac {2 \left (8 a^2 d f h-b^2 (2 d e g-5 c (f g+e h))-a b (10 c f h+3 d (f g+e h))\right ) (c+d x)^{3/2}}{15 b^2 (b c-a d)^2 (a+b x)^{3/2}}+\frac {2 \sqrt {d} f h \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{7/2}} \] Output:

-2*f*h*(d*x+c)^(1/2)/b^3/(b*x+a)^(1/2)-2/5*(-a*f+b*e)*(-a*h+b*g)*(d*x+c)^( 
3/2)/b^2/(-a*d+b*c)/(b*x+a)^(5/2)-2/15*(8*a^2*d*f*h-b^2*(2*d*e*g-5*c*(e*h+ 
f*g))-a*b*(10*c*f*h+3*d*(e*h+f*g)))*(d*x+c)^(3/2)/b^2/(-a*d+b*c)^2/(b*x+a) 
^(3/2)+2*d^(1/2)*f*h*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/ 
b^(7/2)
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{7/2}} \, dx=-\frac {2 \sqrt {c+d x} \left (15 a^4 d^2 f h+5 a^3 b d f h (-5 c+7 d x)+a^2 b^2 f h \left (8 c^2-59 c d x+23 d^2 x^2\right )+b^4 \left (-2 d^2 e g x^2+c d x (5 f g x+e (g+5 h x))+c^2 (5 f x (g+3 h x)+e (3 g+5 h x))\right )+a b^3 \left (-d^2 x (5 e g+3 f g x+3 e h x)+2 c^2 (e h+f (g+10 h x))-c d (e (5 g+h x)+f x (g+40 h x))\right )\right )}{15 b^3 (b c-a d)^2 (a+b x)^{5/2}}+\frac {2 \sqrt {d} f h \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{7/2}} \] Input:

Integrate[(Sqrt[c + d*x]*(e + f*x)*(g + h*x))/(a + b*x)^(7/2),x]
 

Output:

(-2*Sqrt[c + d*x]*(15*a^4*d^2*f*h + 5*a^3*b*d*f*h*(-5*c + 7*d*x) + a^2*b^2 
*f*h*(8*c^2 - 59*c*d*x + 23*d^2*x^2) + b^4*(-2*d^2*e*g*x^2 + c*d*x*(5*f*g* 
x + e*(g + 5*h*x)) + c^2*(5*f*x*(g + 3*h*x) + e*(3*g + 5*h*x))) + a*b^3*(- 
(d^2*x*(5*e*g + 3*f*g*x + 3*e*h*x)) + 2*c^2*(e*h + f*(g + 10*h*x)) - c*d*( 
e*(5*g + h*x) + f*x*(g + 40*h*x)))))/(15*b^3*(b*c - a*d)^2*(a + b*x)^(5/2) 
) + (2*Sqrt[d]*f*h*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x]) 
])/b^(7/2)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {162, 57, 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.

\(\displaystyle \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{7/2}} \, dx\)

\(\Big \downarrow \) 162

\(\displaystyle \frac {f h \int \frac {\sqrt {c+d x}}{(a+b x)^{3/2}}dx}{b^2}-\frac {2 (c+d x)^{3/2} \left (5 a^3 d f h+b x \left (8 a^2 d f h-a b (10 c f h+3 d (e h+f g))-b^2 (2 d e g-5 c (e h+f g))\right )-7 a^2 b c f h-a b^2 (5 d e g-2 c (e h+f g))+3 b^3 c e g\right )}{15 b^2 (a+b x)^{5/2} (b c-a d)^2}\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {f h \left (\frac {d \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{b}-\frac {2 \sqrt {c+d x}}{b \sqrt {a+b x}}\right )}{b^2}-\frac {2 (c+d x)^{3/2} \left (5 a^3 d f h+b x \left (8 a^2 d f h-a b (10 c f h+3 d (e h+f g))-b^2 (2 d e g-5 c (e h+f g))\right )-7 a^2 b c f h-a b^2 (5 d e g-2 c (e h+f g))+3 b^3 c e g\right )}{15 b^2 (a+b x)^{5/2} (b c-a d)^2}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {f h \left (\frac {2 d \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}-\frac {2 \sqrt {c+d x}}{b \sqrt {a+b x}}\right )}{b^2}-\frac {2 (c+d x)^{3/2} \left (5 a^3 d f h+b x \left (8 a^2 d f h-a b (10 c f h+3 d (e h+f g))-b^2 (2 d e g-5 c (e h+f g))\right )-7 a^2 b c f h-a b^2 (5 d e g-2 c (e h+f g))+3 b^3 c e g\right )}{15 b^2 (a+b x)^{5/2} (b c-a d)^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {f h \left (\frac {2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}}-\frac {2 \sqrt {c+d x}}{b \sqrt {a+b x}}\right )}{b^2}-\frac {2 (c+d x)^{3/2} \left (5 a^3 d f h+b x \left (8 a^2 d f h-a b (10 c f h+3 d (e h+f g))-b^2 (2 d e g-5 c (e h+f g))\right )-7 a^2 b c f h-a b^2 (5 d e g-2 c (e h+f g))+3 b^3 c e g\right )}{15 b^2 (a+b x)^{5/2} (b c-a d)^2}\)

Input:

Int[(Sqrt[c + d*x]*(e + f*x)*(g + h*x))/(a + b*x)^(7/2),x]
 

Output:

(-2*(c + d*x)^(3/2)*(3*b^3*c*e*g - 7*a^2*b*c*f*h + 5*a^3*d*f*h - a*b^2*(5* 
d*e*g - 2*c*(f*g + e*h)) + b*(8*a^2*d*f*h - b^2*(2*d*e*g - 5*c*(f*g + e*h) 
) - a*b*(10*c*f*h + 3*d*(f*g + e*h)))*x))/(15*b^2*(b*c - a*d)^2*(a + b*x)^ 
(5/2)) + (f*h*((-2*Sqrt[c + d*x])/(b*Sqrt[a + b*x]) + (2*Sqrt[d]*ArcTanh[( 
Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/b^(3/2)))/b^2
 

Defintions of rubi rules used

rule 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[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 162

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) 
)*((g_.) + (h_.)*(x_)), x_] :> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) 
 - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g + e*h) + d*e 
*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + 
e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)/(b^2*(b 
*c - a*d)^2*(m + 1)*(m + 2)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] + Sim 
p[(f*(h/b^2) - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d 
*(f*g + e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/( 
b^2*(b*c - a*d)^2*(m + 1)*(m + 2)))   Int[(a + b*x)^(m + 2)*(c + d*x)^n, x] 
, x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + 
 n + 3, 0] &&  !LtQ[n, -2]))
 

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. \(1416\) vs. \(2(178)=356\).

Time = 0.33 (sec) , antiderivative size = 1417, normalized size of antiderivative = 6.88

method	result	size
default	\(\text {Expression too large to display}\)	\(1417\)

Input:

int((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

1/15*(-16*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*a^2*b^2*c^2*f*h+4*(d*b)^(1/2 
)*((b*x+a)*(d*x+c))^(1/2)*b^4*d^2*e*g*x^2+45*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^4*b*d^3*f*h*x-70*(d*b)^(1 
/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*b*d^2*f*h*x+10*(d*b)^(1/2)*((b*x+a)*(d*x+c 
))^(1/2)*a*b^3*d^2*e*g*x+50*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*b*c*d* 
f*h+6*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^3*d^2*e*h*x^2+6*(d*b)^(1/2)* 
((b*x+a)*(d*x+c))^(1/2)*a*b^3*d^2*f*g*x^2-10*(d*b)^(1/2)*((b*x+a)*(d*x+c)) 
^(1/2)*b^4*c*d*e*h*x^2-10*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^4*c*d*f*g* 
x^2-90*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^3*b^2*c*d^2*f*h*x+45*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^3*c^2*d*f*h*x-90*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^3*c*d^2*f 
*h*x^2+45*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^4*c^2*d*f*h*x^2-46*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2 
*b^2*d^2*f*h*x^2+80*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^3*c*d*f*h*x^2+ 
118*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b^2*c*d*f*h*x+2*(d*b)^(1/2)*(( 
b*x+a)*(d*x+c))^(1/2)*a*b^3*c*d*e*h*x-40*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/ 
2)*a*b^3*c^2*f*h*x+10*a*b^3*c*d*e*g*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-2* 
b^4*c*d*e*g*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-30*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^4*b*c*d^2*f*h+...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 621 vs. \(2 (178) = 356\).

Time = 29.93 (sec) , antiderivative size = 1267, normalized size of antiderivative = 6.15 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{7/2}} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(7/2),x, algorithm="fricas 
")
 

Output:

[1/30*(15*((b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*f*h*x^3 + 3*(a*b^4*c^2 - 
2*a^2*b^3*c*d + a^3*b^2*d^2)*f*h*x^2 + 3*(a^2*b^3*c^2 - 2*a^3*b^2*c*d + a^ 
4*b*d^2)*f*h*x + (a^3*b^2*c^2 - 2*a^4*b*c*d + a^5*d^2)*f*h)*sqrt(d/b)*log( 
8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b^2*d*x + b^2*c + a*b 
*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) + 4*( 
((2*b^4*d^2*e - (5*b^4*c*d - 3*a*b^3*d^2)*f)*g - ((5*b^4*c*d - 3*a*b^3*d^2 
)*e + (15*b^4*c^2 - 40*a*b^3*c*d + 23*a^2*b^2*d^2)*f)*h)*x^2 - (2*a*b^3*c^ 
2*f + (3*b^4*c^2 - 5*a*b^3*c*d)*e)*g - (2*a*b^3*c^2*e + (8*a^2*b^2*c^2 - 2 
5*a^3*b*c*d + 15*a^4*d^2)*f)*h - (((b^4*c*d - 5*a*b^3*d^2)*e + (5*b^4*c^2 
- a*b^3*c*d)*f)*g + ((5*b^4*c^2 - a*b^3*c*d)*e + (20*a*b^3*c^2 - 59*a^2*b^ 
2*c*d + 35*a^3*b*d^2)*f)*h)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*b^5*c^2 - 
 2*a^4*b^4*c*d + a^5*b^3*d^2 + (b^8*c^2 - 2*a*b^7*c*d + a^2*b^6*d^2)*x^3 + 
 3*(a*b^7*c^2 - 2*a^2*b^6*c*d + a^3*b^5*d^2)*x^2 + 3*(a^2*b^6*c^2 - 2*a^3* 
b^5*c*d + a^4*b^4*d^2)*x), -1/15*(15*((b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2 
)*f*h*x^3 + 3*(a*b^4*c^2 - 2*a^2*b^3*c*d + a^3*b^2*d^2)*f*h*x^2 + 3*(a^2*b 
^3*c^2 - 2*a^3*b^2*c*d + a^4*b*d^2)*f*h*x + (a^3*b^2*c^2 - 2*a^4*b*c*d + a 
^5*d^2)*f*h)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqr 
t(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) - 2*(((2*b^ 
4*d^2*e - (5*b^4*c*d - 3*a*b^3*d^2)*f)*g - ((5*b^4*c*d - 3*a*b^3*d^2)*e + 
(15*b^4*c^2 - 40*a*b^3*c*d + 23*a^2*b^2*d^2)*f)*h)*x^2 - (2*a*b^3*c^2*f...
 

Sympy [F]

\[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{7/2}} \, dx=\int \frac {\sqrt {c + d x} \left (e + f x\right ) \left (g + h x\right )}{\left (a + b x\right )^{\frac {7}{2}}}\, dx \] Input:

integrate((d*x+c)**(1/2)*(f*x+e)*(h*x+g)/(b*x+a)**(7/2),x)
 

Output:

Integral(sqrt(c + d*x)*(e + f*x)*(g + h*x)/(a + b*x)**(7/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{7/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(7/2),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. 2589 vs. \(2 (178) = 356\).

Time = 0.51 (sec) , antiderivative size = 2589, normalized size of antiderivative = 12.57 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{7/2}} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(7/2),x, algorithm="giac")
 

Output:

-sqrt(b*d)*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)/b^5 + 4/15*(2*sqrt(b*d)*b^9*c^3*d^2*e*g*abs(b) - 6*sqrt 
(b*d)*a*b^8*c^2*d^3*e*g*abs(b) + 6*sqrt(b*d)*a^2*b^7*c*d^4*e*g*abs(b) - 2* 
sqrt(b*d)*a^3*b^6*d^5*e*g*abs(b) - 5*sqrt(b*d)*b^9*c^4*d*f*g*abs(b) + 18*s 
qrt(b*d)*a*b^8*c^3*d^2*f*g*abs(b) - 24*sqrt(b*d)*a^2*b^7*c^2*d^3*f*g*abs(b 
) + 14*sqrt(b*d)*a^3*b^6*c*d^4*f*g*abs(b) - 3*sqrt(b*d)*a^4*b^5*d^5*f*g*ab 
s(b) - 5*sqrt(b*d)*b^9*c^4*d*e*h*abs(b) + 18*sqrt(b*d)*a*b^8*c^3*d^2*e*h*a 
bs(b) - 24*sqrt(b*d)*a^2*b^7*c^2*d^3*e*h*abs(b) + 14*sqrt(b*d)*a^3*b^6*c*d 
^4*e*h*abs(b) - 3*sqrt(b*d)*a^4*b^5*d^5*e*h*abs(b) - 15*sqrt(b*d)*b^9*c^5* 
f*h*abs(b) + 85*sqrt(b*d)*a*b^8*c^4*d*f*h*abs(b) - 188*sqrt(b*d)*a^2*b^7*c 
^3*d^2*f*h*abs(b) + 204*sqrt(b*d)*a^3*b^6*c^2*d^3*f*h*abs(b) - 109*sqrt(b* 
d)*a^4*b^5*c*d^4*f*h*abs(b) + 23*sqrt(b*d)*a^5*b^4*d^5*f*h*abs(b) - 10*sqr 
t(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b 
^7*c^2*d^2*e*g*abs(b) + 20*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c 
 + (b*x + a)*b*d - a*b*d))^2*a*b^6*c*d^3*e*g*abs(b) - 10*sqrt(b*d)*(sqrt(b 
*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^5*d^4*e*g 
*abs(b) + 10*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b 
*d - a*b*d))^2*b^7*c^3*d*f*g*abs(b) - 20*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a 
) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^6*c^2*d^2*f*g*abs(b) + 10*s 
qrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{7/2}} \, dx=\int \frac {\left (e+f\,x\right )\,\left (g+h\,x\right )\,\sqrt {c+d\,x}}{{\left (a+b\,x\right )}^{7/2}} \,d x \] Input:

int(((e + f*x)*(g + h*x)*(c + d*x)^(1/2))/(a + b*x)^(7/2),x)
 

Output:

int(((e + f*x)*(g + h*x)*(c + d*x)^(1/2))/(a + b*x)^(7/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 1567, normalized size of antiderivative = 7.61 \[ \int \frac {\sqrt {c+d x} (e+f x) (g+h x)}{(a+b x)^{7/2}} \, dx =\text {Too large to display} \] Input:

int((d*x+c)^(1/2)*(f*x+e)*(h*x+g)/(b*x+a)^(7/2),x)
 

Output:

(2*(15*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**4*d**2*f*h - 30*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**3*b*c*d*f*h + 30*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**3*b*d**2*f*h*x + 15*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**2*b**2*c**2*f*h - 60*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**2 
*b**2*c*d*f*h*x + 15*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**2*b**2*d**2*f*h*x**2 + 30 
*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**3*c**2*f*h*x - 30*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**3*c*d*f*h*x**2 + 15*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**4*c**2*f*h*x**2 + 5*s 
qrt(d)*sqrt(b)*sqrt(a + b*x)*a**4*d**2*f*h - 16*sqrt(d)*sqrt(b)*sqrt(a + b 
*x)*a**3*b*c*d*f*h + 3*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**3*b*d**2*e*h + 3*s 
qrt(d)*sqrt(b)*sqrt(a + b*x)*a**3*b*d**2*f*g + 10*sqrt(d)*sqrt(b)*sqrt(a + 
 b*x)*a**3*b*d**2*f*h*x + 9*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**2*b**2*c**2*f 
*h - sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**2*b**2*c*d*e*h - sqrt(d)*sqrt(b)*...