\(\int \frac {A+B x}{(a+b x)^{5/2} (d+e x)^{5/2}} \, dx\) [228]

Optimal result

Integrand size = 24, antiderivative size = 192 \[ \int \frac {A+B x}{(a+b x)^{5/2} (d+e x)^{5/2}} \, dx=-\frac {2 (A b-a B)}{3 b (b d-a e) (a+b x)^{3/2} (d+e x)^{3/2}}-\frac {2 (b B d-2 A b e+a B e)}{b (b d-a e)^2 \sqrt {a+b x} (d+e x)^{3/2}}+\frac {8 e (2 A b e-B (b d+a e)) \sqrt {a+b x}}{3 b (b d-a e)^3 (d+e x)^{3/2}}+\frac {16 e (2 A b e-B (b d+a e)) \sqrt {a+b x}}{3 (b d-a e)^4 \sqrt {d+e x}} \] Output:

1/3*(-2*A*b+2*B*a)/b/(-a*e+b*d)/(b*x+a)^(3/2)/(e*x+d)^(3/2)-2*(-2*A*b*e+B* 
a*e+B*b*d)/b/(-a*e+b*d)^2/(b*x+a)^(1/2)/(e*x+d)^(3/2)+8/3*e*(2*A*b*e-B*(a* 
e+b*d))*(b*x+a)^(1/2)/b/(-a*e+b*d)^3/(e*x+d)^(3/2)+16/3*e*(2*A*b*e-B*(a*e+ 
b*d))*(b*x+a)^(1/2)/(-a*e+b*d)^4/(e*x+d)^(1/2)
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x}{(a+b x)^{5/2} (d+e x)^{5/2}} \, dx=-\frac {2 \left (-B d e^2 (a+b x)^3+A e^3 (a+b x)^3+6 b B d e (a+b x)^2 (d+e x)-9 A b e^2 (a+b x)^2 (d+e x)+3 a B e^2 (a+b x)^2 (d+e x)+3 b^2 B d (a+b x) (d+e x)^2-9 A b^2 e (a+b x) (d+e x)^2+6 a b B e (a+b x) (d+e x)^2+A b^3 (d+e x)^3-a b^2 B (d+e x)^3\right )}{3 (b d-a e)^4 (a+b x)^{3/2} (d+e x)^{3/2}} \] Input:

Integrate[(A + B*x)/((a + b*x)^(5/2)*(d + e*x)^(5/2)),x]
 

Output:

(-2*(-(B*d*e^2*(a + b*x)^3) + A*e^3*(a + b*x)^3 + 6*b*B*d*e*(a + b*x)^2*(d 
 + e*x) - 9*A*b*e^2*(a + b*x)^2*(d + e*x) + 3*a*B*e^2*(a + b*x)^2*(d + e*x 
) + 3*b^2*B*d*(a + b*x)*(d + e*x)^2 - 9*A*b^2*e*(a + b*x)*(d + e*x)^2 + 6* 
a*b*B*e*(a + b*x)*(d + e*x)^2 + A*b^3*(d + e*x)^3 - a*b^2*B*(d + e*x)^3))/ 
(3*(b*d - a*e)^4*(a + b*x)^(3/2)*(d + e*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {87, 55, 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.

\(\displaystyle \int \frac {A+B x}{(a+b x)^{5/2} (d+e x)^{5/2}} \, dx\)

\(\Big \downarrow \) 87

\(\displaystyle -\frac {(a B e-2 A b e+b B d) \int \frac {1}{(a+b x)^{5/2} (d+e x)^{3/2}}dx}{e (b d-a e)}-\frac {2 (B d-A e)}{3 e (a+b x)^{3/2} (d+e x)^{3/2} (b d-a e)}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {(a B e-2 A b e+b B d) \left (-\frac {4 e \int \frac {1}{(a+b x)^{3/2} (d+e x)^{3/2}}dx}{3 (b d-a e)}-\frac {2}{3 (a+b x)^{3/2} \sqrt {d+e x} (b d-a e)}\right )}{e (b d-a e)}-\frac {2 (B d-A e)}{3 e (a+b x)^{3/2} (d+e x)^{3/2} (b d-a e)}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {(a B e-2 A b e+b B d) \left (-\frac {4 e \left (-\frac {2 e \int \frac {1}{\sqrt {a+b x} (d+e x)^{3/2}}dx}{b d-a e}-\frac {2}{\sqrt {a+b x} \sqrt {d+e x} (b d-a e)}\right )}{3 (b d-a e)}-\frac {2}{3 (a+b x)^{3/2} \sqrt {d+e x} (b d-a e)}\right )}{e (b d-a e)}-\frac {2 (B d-A e)}{3 e (a+b x)^{3/2} (d+e x)^{3/2} (b d-a e)}\)

\(\Big \downarrow \) 48

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

Input:

Int[(A + B*x)/((a + b*x)^(5/2)*(d + e*x)^(5/2)),x]
 

Output:

(-2*(B*d - A*e))/(3*e*(b*d - a*e)*(a + b*x)^(3/2)*(d + e*x)^(3/2)) - ((b*B 
*d - 2*A*b*e + a*B*e)*(-2/(3*(b*d - a*e)*(a + b*x)^(3/2)*Sqrt[d + e*x]) - 
(4*e*(-2/((b*d - a*e)*Sqrt[a + b*x]*Sqrt[d + e*x]) - (4*e*Sqrt[a + b*x])/( 
(b*d - a*e)^2*Sqrt[d + e*x])))/(3*(b*d - a*e))))/(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 = 0.33 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.45

method	result	size
default	\(-\frac {2 \left (-16 A \,b^{3} e^{3} x^{3}+8 B a \,b^{2} e^{3} x^{3}+8 B \,b^{3} d \,e^{2} x^{3}-24 A a \,b^{2} e^{3} x^{2}-24 A \,b^{3} d \,e^{2} x^{2}+12 B \,a^{2} b \,e^{3} x^{2}+24 B a \,b^{2} d \,e^{2} x^{2}+12 B \,b^{3} d^{2} e \,x^{2}-6 A \,a^{2} b \,e^{3} x -36 A a \,b^{2} d \,e^{2} x -6 A \,b^{3} d^{2} e x +3 B \,a^{3} e^{3} x +21 B \,a^{2} b d \,e^{2} x +21 B a \,b^{2} d^{2} e x +3 b^{3} B \,d^{3} x +a^{3} A \,e^{3}-9 A \,a^{2} b d \,e^{2}-9 A a \,b^{2} d^{2} e +A \,b^{3} d^{3}+2 B \,a^{3} d \,e^{2}+12 B \,a^{2} b \,d^{2} e +2 B a \,b^{2} d^{3}\right )}{3 \left (a e -d b \right )^{4} \left (b x +a \right )^{\frac {3}{2}} \left (e x +d \right )^{\frac {3}{2}}}\)	\(279\)
gosper	\(-\frac {2 \left (-16 A \,b^{3} e^{3} x^{3}+8 B a \,b^{2} e^{3} x^{3}+8 B \,b^{3} d \,e^{2} x^{3}-24 A a \,b^{2} e^{3} x^{2}-24 A \,b^{3} d \,e^{2} x^{2}+12 B \,a^{2} b \,e^{3} x^{2}+24 B a \,b^{2} d \,e^{2} x^{2}+12 B \,b^{3} d^{2} e \,x^{2}-6 A \,a^{2} b \,e^{3} x -36 A a \,b^{2} d \,e^{2} x -6 A \,b^{3} d^{2} e x +3 B \,a^{3} e^{3} x +21 B \,a^{2} b d \,e^{2} x +21 B a \,b^{2} d^{2} e x +3 b^{3} B \,d^{3} x +a^{3} A \,e^{3}-9 A \,a^{2} b d \,e^{2}-9 A a \,b^{2} d^{2} e +A \,b^{3} d^{3}+2 B \,a^{3} d \,e^{2}+12 B \,a^{2} b \,d^{2} e +2 B a \,b^{2} d^{3}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right )^{\frac {3}{2}} \left (e^{4} a^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}\)	\(320\)
orering	\(-\frac {2 \left (-16 A \,b^{3} e^{3} x^{3}+8 B a \,b^{2} e^{3} x^{3}+8 B \,b^{3} d \,e^{2} x^{3}-24 A a \,b^{2} e^{3} x^{2}-24 A \,b^{3} d \,e^{2} x^{2}+12 B \,a^{2} b \,e^{3} x^{2}+24 B a \,b^{2} d \,e^{2} x^{2}+12 B \,b^{3} d^{2} e \,x^{2}-6 A \,a^{2} b \,e^{3} x -36 A a \,b^{2} d \,e^{2} x -6 A \,b^{3} d^{2} e x +3 B \,a^{3} e^{3} x +21 B \,a^{2} b d \,e^{2} x +21 B a \,b^{2} d^{2} e x +3 b^{3} B \,d^{3} x +a^{3} A \,e^{3}-9 A \,a^{2} b d \,e^{2}-9 A a \,b^{2} d^{2} e +A \,b^{3} d^{3}+2 B \,a^{3} d \,e^{2}+12 B \,a^{2} b \,d^{2} e +2 B a \,b^{2} d^{3}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right )^{\frac {3}{2}} \left (e^{4} a^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}\)	\(320\)

Input:

int((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-2/3*(-16*A*b^3*e^3*x^3+8*B*a*b^2*e^3*x^3+8*B*b^3*d*e^2*x^3-24*A*a*b^2*e^3 
*x^2-24*A*b^3*d*e^2*x^2+12*B*a^2*b*e^3*x^2+24*B*a*b^2*d*e^2*x^2+12*B*b^3*d 
^2*e*x^2-6*A*a^2*b*e^3*x-36*A*a*b^2*d*e^2*x-6*A*b^3*d^2*e*x+3*B*a^3*e^3*x+ 
21*B*a^2*b*d*e^2*x+21*B*a*b^2*d^2*e*x+3*B*b^3*d^3*x+A*a^3*e^3-9*A*a^2*b*d* 
e^2-9*A*a*b^2*d^2*e+A*b^3*d^3+2*B*a^3*d*e^2+12*B*a^2*b*d^2*e+2*B*a*b^2*d^3 
)/(a*e-b*d)^4/(b*x+a)^(3/2)/(e*x+d)^(3/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 565 vs. \(2 (170) = 340\).

Time = 3.15 (sec) , antiderivative size = 565, normalized size of antiderivative = 2.94 \[ \int \frac {A+B x}{(a+b x)^{5/2} (d+e x)^{5/2}} \, dx=-\frac {2 \, {\left (A a^{3} e^{3} + {\left (2 \, B a b^{2} + A b^{3}\right )} d^{3} + 3 \, {\left (4 \, B a^{2} b - 3 \, A a b^{2}\right )} d^{2} e + {\left (2 \, B a^{3} - 9 \, A a^{2} b\right )} d e^{2} + 8 \, {\left (B b^{3} d e^{2} + {\left (B a b^{2} - 2 \, A b^{3}\right )} e^{3}\right )} x^{3} + 12 \, {\left (B b^{3} d^{2} e + 2 \, {\left (B a b^{2} - A b^{3}\right )} d e^{2} + {\left (B a^{2} b - 2 \, A a b^{2}\right )} e^{3}\right )} x^{2} + 3 \, {\left (B b^{3} d^{3} + {\left (7 \, B a b^{2} - 2 \, A b^{3}\right )} d^{2} e + {\left (7 \, B a^{2} b - 12 \, A a b^{2}\right )} d e^{2} + {\left (B a^{3} - 2 \, A a^{2} b\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{3 \, {\left (a^{2} b^{4} d^{6} - 4 \, a^{3} b^{3} d^{5} e + 6 \, a^{4} b^{2} d^{4} e^{2} - 4 \, a^{5} b d^{3} e^{3} + a^{6} d^{2} e^{4} + {\left (b^{6} d^{4} e^{2} - 4 \, a b^{5} d^{3} e^{3} + 6 \, a^{2} b^{4} d^{2} e^{4} - 4 \, a^{3} b^{3} d e^{5} + a^{4} b^{2} e^{6}\right )} x^{4} + 2 \, {\left (b^{6} d^{5} e - 3 \, a b^{5} d^{4} e^{2} + 2 \, a^{2} b^{4} d^{3} e^{3} + 2 \, a^{3} b^{3} d^{2} e^{4} - 3 \, a^{4} b^{2} d e^{5} + a^{5} b e^{6}\right )} x^{3} + {\left (b^{6} d^{6} - 9 \, a^{2} b^{4} d^{4} e^{2} + 16 \, a^{3} b^{3} d^{3} e^{3} - 9 \, a^{4} b^{2} d^{2} e^{4} + a^{6} e^{6}\right )} x^{2} + 2 \, {\left (a b^{5} d^{6} - 3 \, a^{2} b^{4} d^{5} e + 2 \, a^{3} b^{3} d^{4} e^{2} + 2 \, a^{4} b^{2} d^{3} e^{3} - 3 \, a^{5} b d^{2} e^{4} + a^{6} d e^{5}\right )} x\right )}} \] Input:

integrate((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(5/2),x, algorithm="fricas")
 

Output:

-2/3*(A*a^3*e^3 + (2*B*a*b^2 + A*b^3)*d^3 + 3*(4*B*a^2*b - 3*A*a*b^2)*d^2* 
e + (2*B*a^3 - 9*A*a^2*b)*d*e^2 + 8*(B*b^3*d*e^2 + (B*a*b^2 - 2*A*b^3)*e^3 
)*x^3 + 12*(B*b^3*d^2*e + 2*(B*a*b^2 - A*b^3)*d*e^2 + (B*a^2*b - 2*A*a*b^2 
)*e^3)*x^2 + 3*(B*b^3*d^3 + (7*B*a*b^2 - 2*A*b^3)*d^2*e + (7*B*a^2*b - 12* 
A*a*b^2)*d*e^2 + (B*a^3 - 2*A*a^2*b)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d)/( 
a^2*b^4*d^6 - 4*a^3*b^3*d^5*e + 6*a^4*b^2*d^4*e^2 - 4*a^5*b*d^3*e^3 + a^6* 
d^2*e^4 + (b^6*d^4*e^2 - 4*a*b^5*d^3*e^3 + 6*a^2*b^4*d^2*e^4 - 4*a^3*b^3*d 
*e^5 + a^4*b^2*e^6)*x^4 + 2*(b^6*d^5*e - 3*a*b^5*d^4*e^2 + 2*a^2*b^4*d^3*e 
^3 + 2*a^3*b^3*d^2*e^4 - 3*a^4*b^2*d*e^5 + a^5*b*e^6)*x^3 + (b^6*d^6 - 9*a 
^2*b^4*d^4*e^2 + 16*a^3*b^3*d^3*e^3 - 9*a^4*b^2*d^2*e^4 + a^6*e^6)*x^2 + 2 
*(a*b^5*d^6 - 3*a^2*b^4*d^5*e + 2*a^3*b^3*d^4*e^2 + 2*a^4*b^2*d^3*e^3 - 3* 
a^5*b*d^2*e^4 + a^6*d*e^5)*x)
 

Sympy [F]

\[ \int \frac {A+B x}{(a+b x)^{5/2} (d+e x)^{5/2}} \, dx=\int \frac {A + B x}{\left (a + b x\right )^{\frac {5}{2}} \left (d + e x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((B*x+A)/(b*x+a)**(5/2)/(e*x+d)**(5/2),x)
 

Output:

Integral((A + B*x)/((a + b*x)**(5/2)*(d + e*x)**(5/2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B x}{(a+b x)^{5/2} (d+e x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(5/2),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*e-b*d>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. 1183 vs. \(2 (170) = 340\).

Time = 0.46 (sec) , antiderivative size = 1183, normalized size of antiderivative = 6.16 \[ \int \frac {A+B x}{(a+b x)^{5/2} (d+e x)^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(5/2),x, algorithm="giac")
 

Output:

-2/3*sqrt(b*x + a)*((5*B*b^7*d^4*e^3*abs(b) - 12*B*a*b^6*d^3*e^4*abs(b) - 
8*A*b^7*d^3*e^4*abs(b) + 6*B*a^2*b^5*d^2*e^5*abs(b) + 24*A*a*b^6*d^2*e^5*a 
bs(b) + 4*B*a^3*b^4*d*e^6*abs(b) - 24*A*a^2*b^5*d*e^6*abs(b) - 3*B*a^4*b^3 
*e^7*abs(b) + 8*A*a^3*b^4*e^7*abs(b))*(b*x + a)/(b^9*d^7*e - 7*a*b^8*d^6*e 
^2 + 21*a^2*b^7*d^5*e^3 - 35*a^3*b^6*d^4*e^4 + 35*a^4*b^5*d^3*e^5 - 21*a^5 
*b^4*d^2*e^6 + 7*a^6*b^3*d*e^7 - a^7*b^2*e^8) + 3*(2*B*b^8*d^5*e^2*abs(b) 
- 7*B*a*b^7*d^4*e^3*abs(b) - 3*A*b^8*d^4*e^3*abs(b) + 8*B*a^2*b^6*d^3*e^4* 
abs(b) + 12*A*a*b^7*d^3*e^4*abs(b) - 2*B*a^3*b^5*d^2*e^5*abs(b) - 18*A*a^2 
*b^6*d^2*e^5*abs(b) - 2*B*a^4*b^4*d*e^6*abs(b) + 12*A*a^3*b^5*d*e^6*abs(b) 
 + B*a^5*b^3*e^7*abs(b) - 3*A*a^4*b^4*e^7*abs(b))/(b^9*d^7*e - 7*a*b^8*d^6 
*e^2 + 21*a^2*b^7*d^5*e^3 - 35*a^3*b^6*d^4*e^4 + 35*a^4*b^5*d^3*e^5 - 21*a 
^5*b^4*d^2*e^6 + 7*a^6*b^3*d*e^7 - a^7*b^2*e^8))/(b^2*d + (b*x + a)*b*e - 
a*b*e)^(3/2) - 4/3*(3*sqrt(b*e)*B*b^7*d^3 - sqrt(b*e)*B*a*b^6*d^2*e - 8*sq 
rt(b*e)*A*b^7*d^2*e - 7*sqrt(b*e)*B*a^2*b^5*d*e^2 + 16*sqrt(b*e)*A*a*b^6*d 
*e^2 + 5*sqrt(b*e)*B*a^3*b^4*e^3 - 8*sqrt(b*e)*A*a^2*b^5*e^3 - 6*sqrt(b*e) 
*(sqrt(b*e)*sqrt(b*x + a) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*b^5*d 
^2 - 6*sqrt(b*e)*(sqrt(b*e)*sqrt(b*x + a) - sqrt(b^2*d + (b*x + a)*b*e - a 
*b*e))^2*B*a*b^4*d*e + 18*sqrt(b*e)*(sqrt(b*e)*sqrt(b*x + a) - sqrt(b^2*d 
+ (b*x + a)*b*e - a*b*e))^2*A*b^5*d*e + 12*sqrt(b*e)*(sqrt(b*e)*sqrt(b*x + 
 a) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*a^2*b^3*e^2 - 18*sqrt(b*...
 

Mupad [B] (verification not implemented)

Time = 1.87 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.58 \[ \int \frac {A+B x}{(a+b x)^{5/2} (d+e x)^{5/2}} \, dx=-\frac {\sqrt {d+e\,x}\,\left (\frac {16\,b\,x^3\,\left (B\,a\,e-2\,A\,b\,e+B\,b\,d\right )}{3\,{\left (a\,e-b\,d\right )}^4}+\frac {4\,B\,a^3\,d\,e^2+2\,A\,a^3\,e^3+24\,B\,a^2\,b\,d^2\,e-18\,A\,a^2\,b\,d\,e^2+4\,B\,a\,b^2\,d^3-18\,A\,a\,b^2\,d^2\,e+2\,A\,b^3\,d^3}{3\,b\,e^2\,{\left (a\,e-b\,d\right )}^4}+\frac {8\,x^2\,\left (a\,e+b\,d\right )\,\left (B\,a\,e-2\,A\,b\,e+B\,b\,d\right )}{e\,{\left (a\,e-b\,d\right )}^4}+\frac {2\,x\,\left (a^2\,e^2+6\,a\,b\,d\,e+b^2\,d^2\right )\,\left (B\,a\,e-2\,A\,b\,e+B\,b\,d\right )}{b\,e^2\,{\left (a\,e-b\,d\right )}^4}\right )}{x^3\,\sqrt {a+b\,x}+\frac {a\,d^2\,\sqrt {a+b\,x}}{b\,e^2}+\frac {x^2\,\left (a\,e+2\,b\,d\right )\,\sqrt {a+b\,x}}{b\,e}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {a+b\,x}}{b\,e^2}} \] Input:

int((A + B*x)/((a + b*x)^(5/2)*(d + e*x)^(5/2)),x)
 

Output:

-((d + e*x)^(1/2)*((16*b*x^3*(B*a*e - 2*A*b*e + B*b*d))/(3*(a*e - b*d)^4) 
+ (2*A*a^3*e^3 + 2*A*b^3*d^3 + 4*B*a*b^2*d^3 + 4*B*a^3*d*e^2 - 18*A*a*b^2* 
d^2*e - 18*A*a^2*b*d*e^2 + 24*B*a^2*b*d^2*e)/(3*b*e^2*(a*e - b*d)^4) + (8* 
x^2*(a*e + b*d)*(B*a*e - 2*A*b*e + B*b*d))/(e*(a*e - b*d)^4) + (2*x*(a^2*e 
^2 + b^2*d^2 + 6*a*b*d*e)*(B*a*e - 2*A*b*e + B*b*d))/(b*e^2*(a*e - b*d)^4) 
))/(x^3*(a + b*x)^(1/2) + (a*d^2*(a + b*x)^(1/2))/(b*e^2) + (x^2*(a*e + 2* 
b*d)*(a + b*x)^(1/2))/(b*e) + (d*x*(2*a*e + b*d)*(a + b*x)^(1/2))/(b*e^2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.50 \[ \int \frac {A+B x}{(a+b x)^{5/2} (d+e x)^{5/2}} \, dx=\frac {-\frac {16 \sqrt {e}\, \sqrt {b}\, \sqrt {b x +a}\, b \,d^{2}}{3}-\frac {32 \sqrt {e}\, \sqrt {b}\, \sqrt {b x +a}\, b d e x}{3}-\frac {16 \sqrt {e}\, \sqrt {b}\, \sqrt {b x +a}\, b \,e^{2} x^{2}}{3}-\frac {2 \sqrt {e x +d}\, a^{2} e^{2}}{3}+4 \sqrt {e x +d}\, a b d e +\frac {8 \sqrt {e x +d}\, a b \,e^{2} x}{3}+2 \sqrt {e x +d}\, b^{2} d^{2}+8 \sqrt {e x +d}\, b^{2} d e x +\frac {16 \sqrt {e x +d}\, b^{2} e^{2} x^{2}}{3}}{\sqrt {b x +a}\, \left (a^{3} e^{5} x^{2}-3 a^{2} b d \,e^{4} x^{2}+3 a \,b^{2} d^{2} e^{3} x^{2}-b^{3} d^{3} e^{2} x^{2}+2 a^{3} d \,e^{4} x -6 a^{2} b \,d^{2} e^{3} x +6 a \,b^{2} d^{3} e^{2} x -2 b^{3} d^{4} e x +a^{3} d^{2} e^{3}-3 a^{2} b \,d^{3} e^{2}+3 a \,b^{2} d^{4} e -b^{3} d^{5}\right )} \] Input:

int((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(5/2),x)
 

Output:

(2*( - 8*sqrt(e)*sqrt(b)*sqrt(a + b*x)*b*d**2 - 16*sqrt(e)*sqrt(b)*sqrt(a 
+ b*x)*b*d*e*x - 8*sqrt(e)*sqrt(b)*sqrt(a + b*x)*b*e**2*x**2 - sqrt(d + e* 
x)*a**2*e**2 + 6*sqrt(d + e*x)*a*b*d*e + 4*sqrt(d + e*x)*a*b*e**2*x + 3*sq 
rt(d + e*x)*b**2*d**2 + 12*sqrt(d + e*x)*b**2*d*e*x + 8*sqrt(d + e*x)*b**2 
*e**2*x**2))/(3*sqrt(a + b*x)*(a**3*d**2*e**3 + 2*a**3*d*e**4*x + a**3*e** 
5*x**2 - 3*a**2*b*d**3*e**2 - 6*a**2*b*d**2*e**3*x - 3*a**2*b*d*e**4*x**2 
+ 3*a*b**2*d**4*e + 6*a*b**2*d**3*e**2*x + 3*a*b**2*d**2*e**3*x**2 - b**3* 
d**5 - 2*b**3*d**4*e*x - b**3*d**3*e**2*x**2))