\(\int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{9/2}} \, dx\) [179]

Optimal result

Integrand size = 24, antiderivative size = 147 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{9/2}} \, dx=-\frac {2 (B d-A e) (a+b x)^{3/2}}{7 e (b d-a e) (d+e x)^{7/2}}+\frac {2 (3 b B d+4 A b e-7 a B e) (a+b x)^{3/2}}{35 e (b d-a e)^2 (d+e x)^{5/2}}+\frac {4 b (3 b B d+4 A b e-7 a B e) (a+b x)^{3/2}}{105 e (b d-a e)^3 (d+e x)^{3/2}} \] Output:

-2/7*(-A*e+B*d)*(b*x+a)^(3/2)/e/(-a*e+b*d)/(e*x+d)^(7/2)+2/35*(4*A*b*e-7*B 
*a*e+3*B*b*d)*(b*x+a)^(3/2)/e/(-a*e+b*d)^2/(e*x+d)^(5/2)+4/105*b*(4*A*b*e- 
7*B*a*e+3*B*b*d)*(b*x+a)^(3/2)/e/(-a*e+b*d)^3/(e*x+d)^(3/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{9/2}} \, dx=\frac {2 (a+b x)^{3/2} \left (35 A b^2-35 a b B-\frac {15 B d e (a+b x)^2}{(d+e x)^2}+\frac {15 A e^2 (a+b x)^2}{(d+e x)^2}+\frac {21 b B d (a+b x)}{d+e x}-\frac {42 A b e (a+b x)}{d+e x}+\frac {21 a B e (a+b x)}{d+e x}\right )}{105 (b d-a e)^3 (d+e x)^{3/2}} \] Input:

Integrate[(Sqrt[a + b*x]*(A + B*x))/(d + e*x)^(9/2),x]
 

Output:

(2*(a + b*x)^(3/2)*(35*A*b^2 - 35*a*b*B - (15*B*d*e*(a + b*x)^2)/(d + e*x) 
^2 + (15*A*e^2*(a + b*x)^2)/(d + e*x)^2 + (21*b*B*d*(a + b*x))/(d + e*x) - 
 (42*A*b*e*(a + b*x))/(d + e*x) + (21*a*B*e*(a + b*x))/(d + e*x)))/(105*(b 
*d - a*e)^3*(d + e*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {87, 55, 48}

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.

\(\displaystyle \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{9/2}} \, dx\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {(-7 a B e+4 A b e+3 b B d) \int \frac {\sqrt {a+b x}}{(d+e x)^{7/2}}dx}{7 e (b d-a e)}-\frac {2 (a+b x)^{3/2} (B d-A e)}{7 e (d+e x)^{7/2} (b d-a e)}\)

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 48

\(\displaystyle \frac {\left (\frac {4 b (a+b x)^{3/2}}{15 (d+e x)^{3/2} (b d-a e)^2}+\frac {2 (a+b x)^{3/2}}{5 (d+e x)^{5/2} (b d-a e)}\right ) (-7 a B e+4 A b e+3 b B d)}{7 e (b d-a e)}-\frac {2 (a+b x)^{3/2} (B d-A e)}{7 e (d+e x)^{7/2} (b d-a e)}\)

Input:

Int[(Sqrt[a + b*x]*(A + B*x))/(d + e*x)^(9/2),x]
 

Output:

(-2*(B*d - A*e)*(a + b*x)^(3/2))/(7*e*(b*d - a*e)*(d + e*x)^(7/2)) + ((3*b 
*B*d + 4*A*b*e - 7*a*B*e)*((2*(a + b*x)^(3/2))/(5*(b*d - a*e)*(d + e*x)^(5 
/2)) + (4*b*(a + b*x)^(3/2))/(15*(b*d - a*e)^2*(d + e*x)^(3/2))))/(7*e*(b* 
d - a*e))
 

Defintions of rubi rules used

rule 48

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] /; FreeQ[{ 
a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
 

rule 55

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*(S 
implify[m + n + 2]/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^Simplify[m + 1]*( 
c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 
 2], 0] && NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ 
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] ||  !SumSimp 
lerQ[n, 1])
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.01

method	result	size
default	\(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (8 A \,b^{2} e^{2} x^{2}-14 B a b \,e^{2} x^{2}+6 B \,b^{2} d e \,x^{2}-12 A a b \,e^{2} x +28 A \,b^{2} d e x +21 B \,a^{2} e^{2} x -58 B a b d e x +21 b^{2} B \,d^{2} x +15 a^{2} A \,e^{2}-42 A a b d e +35 A \,b^{2} d^{2}+6 B \,a^{2} d e -14 B a b \,d^{2}\right )}{105 \left (e x +d \right )^{\frac {7}{2}} \left (a e -d b \right )^{3}}\)	\(149\)
gosper	\(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (8 A \,b^{2} e^{2} x^{2}-14 B a b \,e^{2} x^{2}+6 B \,b^{2} d e \,x^{2}-12 A a b \,e^{2} x +28 A \,b^{2} d e x +21 B \,a^{2} e^{2} x -58 B a b d e x +21 b^{2} B \,d^{2} x +15 a^{2} A \,e^{2}-42 A a b d e +35 A \,b^{2} d^{2}+6 B \,a^{2} d e -14 B a b \,d^{2}\right )}{105 \left (e x +d \right )^{\frac {7}{2}} \left (a^{3} e^{3}-3 a^{2} b \,e^{2} d +3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}\)	\(177\)
orering	\(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (8 A \,b^{2} e^{2} x^{2}-14 B a b \,e^{2} x^{2}+6 B \,b^{2} d e \,x^{2}-12 A a b \,e^{2} x +28 A \,b^{2} d e x +21 B \,a^{2} e^{2} x -58 B a b d e x +21 b^{2} B \,d^{2} x +15 a^{2} A \,e^{2}-42 A a b d e +35 A \,b^{2} d^{2}+6 B \,a^{2} d e -14 B a b \,d^{2}\right )}{105 \left (e x +d \right )^{\frac {7}{2}} \left (a^{3} e^{3}-3 a^{2} b \,e^{2} d +3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}\)	\(177\)

Input:

int((b*x+a)^(1/2)*(B*x+A)/(e*x+d)^(9/2),x,method=_RETURNVERBOSE)
 

Output:

-2/105*(b*x+a)^(3/2)/(e*x+d)^(7/2)*(8*A*b^2*e^2*x^2-14*B*a*b*e^2*x^2+6*B*b 
^2*d*e*x^2-12*A*a*b*e^2*x+28*A*b^2*d*e*x+21*B*a^2*e^2*x-58*B*a*b*d*e*x+21* 
B*b^2*d^2*x+15*A*a^2*e^2-42*A*a*b*d*e+35*A*b^2*d^2+6*B*a^2*d*e-14*B*a*b*d^ 
2)/(a*e-b*d)^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 440 vs. \(2 (129) = 258\).

Time = 6.10 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.99 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{9/2}} \, dx=\frac {2 \, {\left (15 \, A a^{3} e^{2} + 2 \, {\left (3 \, B b^{3} d e - {\left (7 \, B a b^{2} - 4 \, A b^{3}\right )} e^{2}\right )} x^{3} - 7 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} d^{2} + 6 \, {\left (B a^{3} - 7 \, A a^{2} b\right )} d e + {\left (21 \, B b^{3} d^{2} - 4 \, {\left (13 \, B a b^{2} - 7 \, A b^{3}\right )} d e + {\left (7 \, B a^{2} b - 4 \, A a b^{2}\right )} e^{2}\right )} x^{2} + {\left (7 \, {\left (B a b^{2} + 5 \, A b^{3}\right )} d^{2} - 2 \, {\left (26 \, B a^{2} b + 7 \, A a b^{2}\right )} d e + 3 \, {\left (7 \, B a^{3} + A a^{2} b\right )} e^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{105 \, {\left (b^{3} d^{7} - 3 \, a b^{2} d^{6} e + 3 \, a^{2} b d^{5} e^{2} - a^{3} d^{4} e^{3} + {\left (b^{3} d^{3} e^{4} - 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} b d e^{6} - a^{3} e^{7}\right )} x^{4} + 4 \, {\left (b^{3} d^{4} e^{3} - 3 \, a b^{2} d^{3} e^{4} + 3 \, a^{2} b d^{2} e^{5} - a^{3} d e^{6}\right )} x^{3} + 6 \, {\left (b^{3} d^{5} e^{2} - 3 \, a b^{2} d^{4} e^{3} + 3 \, a^{2} b d^{3} e^{4} - a^{3} d^{2} e^{5}\right )} x^{2} + 4 \, {\left (b^{3} d^{6} e - 3 \, a b^{2} d^{5} e^{2} + 3 \, a^{2} b d^{4} e^{3} - a^{3} d^{3} e^{4}\right )} x\right )}} \] Input:

integrate((b*x+a)^(1/2)*(B*x+A)/(e*x+d)^(9/2),x, algorithm="fricas")
 

Output:

2/105*(15*A*a^3*e^2 + 2*(3*B*b^3*d*e - (7*B*a*b^2 - 4*A*b^3)*e^2)*x^3 - 7* 
(2*B*a^2*b - 5*A*a*b^2)*d^2 + 6*(B*a^3 - 7*A*a^2*b)*d*e + (21*B*b^3*d^2 - 
4*(13*B*a*b^2 - 7*A*b^3)*d*e + (7*B*a^2*b - 4*A*a*b^2)*e^2)*x^2 + (7*(B*a* 
b^2 + 5*A*b^3)*d^2 - 2*(26*B*a^2*b + 7*A*a*b^2)*d*e + 3*(7*B*a^3 + A*a^2*b 
)*e^2)*x)*sqrt(b*x + a)*sqrt(e*x + d)/(b^3*d^7 - 3*a*b^2*d^6*e + 3*a^2*b*d 
^5*e^2 - a^3*d^4*e^3 + (b^3*d^3*e^4 - 3*a*b^2*d^2*e^5 + 3*a^2*b*d*e^6 - a^ 
3*e^7)*x^4 + 4*(b^3*d^4*e^3 - 3*a*b^2*d^3*e^4 + 3*a^2*b*d^2*e^5 - a^3*d*e^ 
6)*x^3 + 6*(b^3*d^5*e^2 - 3*a*b^2*d^4*e^3 + 3*a^2*b*d^3*e^4 - a^3*d^2*e^5) 
*x^2 + 4*(b^3*d^6*e - 3*a*b^2*d^5*e^2 + 3*a^2*b*d^4*e^3 - a^3*d^3*e^4)*x)
 

Sympy [F]

\[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{9/2}} \, dx=\int \frac {\left (A + B x\right ) \sqrt {a + b x}}{\left (d + e x\right )^{\frac {9}{2}}}\, dx \] Input:

integrate((b*x+a)**(1/2)*(B*x+A)/(e*x+d)**(9/2),x)
 

Output:

Integral((A + B*x)*sqrt(a + b*x)/(d + e*x)**(9/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{9/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x+a)^(1/2)*(B*x+A)/(e*x+d)^(9/2),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(e*(a*e-b*d)>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (129) = 258\).

Time = 0.26 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.58 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{9/2}} \, dx=\frac {2 \, {\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (3 \, B b^{8} d e^{4} {\left | b \right |} - 7 \, B a b^{7} e^{5} {\left | b \right |} + 4 \, A b^{8} e^{5} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{5} d^{3} e^{3} - 3 \, a b^{4} d^{2} e^{4} + 3 \, a^{2} b^{3} d e^{5} - a^{3} b^{2} e^{6}} + \frac {7 \, {\left (3 \, B b^{9} d^{2} e^{3} {\left | b \right |} - 10 \, B a b^{8} d e^{4} {\left | b \right |} + 4 \, A b^{9} d e^{4} {\left | b \right |} + 7 \, B a^{2} b^{7} e^{5} {\left | b \right |} - 4 \, A a b^{8} e^{5} {\left | b \right |}\right )}}{b^{5} d^{3} e^{3} - 3 \, a b^{4} d^{2} e^{4} + 3 \, a^{2} b^{3} d e^{5} - a^{3} b^{2} e^{6}}\right )} - \frac {35 \, {\left (B a b^{9} d^{2} e^{3} {\left | b \right |} - A b^{10} d^{2} e^{3} {\left | b \right |} - 2 \, B a^{2} b^{8} d e^{4} {\left | b \right |} + 2 \, A a b^{9} d e^{4} {\left | b \right |} + B a^{3} b^{7} e^{5} {\left | b \right |} - A a^{2} b^{8} e^{5} {\left | b \right |}\right )}}{b^{5} d^{3} e^{3} - 3 \, a b^{4} d^{2} e^{4} + 3 \, a^{2} b^{3} d e^{5} - a^{3} b^{2} e^{6}}\right )} {\left (b x + a\right )}^{\frac {3}{2}}}{105 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {7}{2}}} \] Input:

integrate((b*x+a)^(1/2)*(B*x+A)/(e*x+d)^(9/2),x, algorithm="giac")
 

Output:

2/105*((b*x + a)*(2*(3*B*b^8*d*e^4*abs(b) - 7*B*a*b^7*e^5*abs(b) + 4*A*b^8 
*e^5*abs(b))*(b*x + a)/(b^5*d^3*e^3 - 3*a*b^4*d^2*e^4 + 3*a^2*b^3*d*e^5 - 
a^3*b^2*e^6) + 7*(3*B*b^9*d^2*e^3*abs(b) - 10*B*a*b^8*d*e^4*abs(b) + 4*A*b 
^9*d*e^4*abs(b) + 7*B*a^2*b^7*e^5*abs(b) - 4*A*a*b^8*e^5*abs(b))/(b^5*d^3* 
e^3 - 3*a*b^4*d^2*e^4 + 3*a^2*b^3*d*e^5 - a^3*b^2*e^6)) - 35*(B*a*b^9*d^2* 
e^3*abs(b) - A*b^10*d^2*e^3*abs(b) - 2*B*a^2*b^8*d*e^4*abs(b) + 2*A*a*b^9* 
d*e^4*abs(b) + B*a^3*b^7*e^5*abs(b) - A*a^2*b^8*e^5*abs(b))/(b^5*d^3*e^3 - 
 3*a*b^4*d^2*e^4 + 3*a^2*b^3*d*e^5 - a^3*b^2*e^6))*(b*x + a)^(3/2)/(b^2*d 
+ (b*x + a)*b*e - a*b*e)^(7/2)
 

Mupad [B] (verification not implemented)

Time = 1.52 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.01 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{9/2}} \, dx=-\frac {\sqrt {d+e\,x}\,\left (\frac {\sqrt {a+b\,x}\,\left (12\,B\,a^3\,d\,e+30\,A\,a^3\,e^2-28\,B\,a^2\,b\,d^2-84\,A\,a^2\,b\,d\,e+70\,A\,a\,b^2\,d^2\right )}{105\,e^4\,{\left (a\,e-b\,d\right )}^3}+\frac {x\,\sqrt {a+b\,x}\,\left (42\,B\,a^3\,e^2-104\,B\,a^2\,b\,d\,e+6\,A\,a^2\,b\,e^2+14\,B\,a\,b^2\,d^2-28\,A\,a\,b^2\,d\,e+70\,A\,b^3\,d^2\right )}{105\,e^4\,{\left (a\,e-b\,d\right )}^3}+\frac {4\,b^2\,x^3\,\sqrt {a+b\,x}\,\left (4\,A\,b\,e-7\,B\,a\,e+3\,B\,b\,d\right )}{105\,e^3\,{\left (a\,e-b\,d\right )}^3}-\frac {2\,b\,x^2\,\left (a\,e-7\,b\,d\right )\,\sqrt {a+b\,x}\,\left (4\,A\,b\,e-7\,B\,a\,e+3\,B\,b\,d\right )}{105\,e^4\,{\left (a\,e-b\,d\right )}^3}\right )}{x^4+\frac {d^4}{e^4}+\frac {4\,d\,x^3}{e}+\frac {4\,d^3\,x}{e^3}+\frac {6\,d^2\,x^2}{e^2}} \] Input:

int(((A + B*x)*(a + b*x)^(1/2))/(d + e*x)^(9/2),x)
 

Output:

-((d + e*x)^(1/2)*(((a + b*x)^(1/2)*(30*A*a^3*e^2 + 12*B*a^3*d*e + 70*A*a* 
b^2*d^2 - 28*B*a^2*b*d^2 - 84*A*a^2*b*d*e))/(105*e^4*(a*e - b*d)^3) + (x*( 
a + b*x)^(1/2)*(70*A*b^3*d^2 + 42*B*a^3*e^2 + 6*A*a^2*b*e^2 + 14*B*a*b^2*d 
^2 - 28*A*a*b^2*d*e - 104*B*a^2*b*d*e))/(105*e^4*(a*e - b*d)^3) + (4*b^2*x 
^3*(a + b*x)^(1/2)*(4*A*b*e - 7*B*a*e + 3*B*b*d))/(105*e^3*(a*e - b*d)^3) 
- (2*b*x^2*(a*e - 7*b*d)*(a + b*x)^(1/2)*(4*A*b*e - 7*B*a*e + 3*B*b*d))/(1 
05*e^4*(a*e - b*d)^3)))/(x^4 + d^4/e^4 + (4*d*x^3)/e + (4*d^3*x)/e^3 + (6* 
d^2*x^2)/e^2)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.82 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{9/2}} \, dx=\frac {-\frac {2 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{3} e^{4}}{7}+\frac {2 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{2} b d \,e^{3}}{5}-\frac {16 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{2} b \,e^{4} x}{35}+\frac {4 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,b^{2} d \,e^{3} x}{5}-\frac {2 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,b^{2} e^{4} x^{2}}{35}+\frac {2 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{3} d \,e^{3} x^{2}}{5}+\frac {4 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{3} e^{4} x^{3}}{35}-\frac {4 \sqrt {e}\, \sqrt {b}\, b^{3} d^{4}}{35}-\frac {16 \sqrt {e}\, \sqrt {b}\, b^{3} d^{3} e x}{35}-\frac {24 \sqrt {e}\, \sqrt {b}\, b^{3} d^{2} e^{2} x^{2}}{35}-\frac {16 \sqrt {e}\, \sqrt {b}\, b^{3} d \,e^{3} x^{3}}{35}-\frac {4 \sqrt {e}\, \sqrt {b}\, b^{3} e^{4} x^{4}}{35}}{e^{3} \left (a^{2} e^{6} x^{4}-2 a b d \,e^{5} x^{4}+b^{2} d^{2} e^{4} x^{4}+4 a^{2} d \,e^{5} x^{3}-8 a b \,d^{2} e^{4} x^{3}+4 b^{2} d^{3} e^{3} x^{3}+6 a^{2} d^{2} e^{4} x^{2}-12 a b \,d^{3} e^{3} x^{2}+6 b^{2} d^{4} e^{2} x^{2}+4 a^{2} d^{3} e^{3} x -8 a b \,d^{4} e^{2} x +4 b^{2} d^{5} e x +a^{2} d^{4} e^{2}-2 a b \,d^{5} e +b^{2} d^{6}\right )} \] Input:

int((b*x+a)^(1/2)*(B*x+A)/(e*x+d)^(9/2),x)
 

Output:

(2*( - 5*sqrt(d + e*x)*sqrt(a + b*x)*a**3*e**4 + 7*sqrt(d + e*x)*sqrt(a + 
b*x)*a**2*b*d*e**3 - 8*sqrt(d + e*x)*sqrt(a + b*x)*a**2*b*e**4*x + 14*sqrt 
(d + e*x)*sqrt(a + b*x)*a*b**2*d*e**3*x - sqrt(d + e*x)*sqrt(a + b*x)*a*b* 
*2*e**4*x**2 + 7*sqrt(d + e*x)*sqrt(a + b*x)*b**3*d*e**3*x**2 + 2*sqrt(d + 
 e*x)*sqrt(a + b*x)*b**3*e**4*x**3 - 2*sqrt(e)*sqrt(b)*b**3*d**4 - 8*sqrt( 
e)*sqrt(b)*b**3*d**3*e*x - 12*sqrt(e)*sqrt(b)*b**3*d**2*e**2*x**2 - 8*sqrt 
(e)*sqrt(b)*b**3*d*e**3*x**3 - 2*sqrt(e)*sqrt(b)*b**3*e**4*x**4))/(35*e**3 
*(a**2*d**4*e**2 + 4*a**2*d**3*e**3*x + 6*a**2*d**2*e**4*x**2 + 4*a**2*d*e 
**5*x**3 + a**2*e**6*x**4 - 2*a*b*d**5*e - 8*a*b*d**4*e**2*x - 12*a*b*d**3 
*e**3*x**2 - 8*a*b*d**2*e**4*x**3 - 2*a*b*d*e**5*x**4 + b**2*d**6 + 4*b**2 
*d**5*e*x + 6*b**2*d**4*e**2*x**2 + 4*b**2*d**3*e**3*x**3 + b**2*d**2*e**4 
*x**4))