\(\int \frac {(a+b x)^{5/2} (A+B x)}{(d+e x)^{3/2}} \, dx\) [197]

Optimal result

Integrand size = 24, antiderivative size = 218 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {2 (B d-A e) (a+b x)^{5/2}}{e^2 \sqrt {d+e x}}+\frac {5 (b d-a e) (7 b B d-6 A b e-a B e) \sqrt {a+b x} \sqrt {d+e x}}{8 e^4}-\frac {5 (7 b B d-6 A b e-a B e) (a+b x)^{3/2} \sqrt {d+e x}}{12 e^3}+\frac {B (a+b x)^{5/2} \sqrt {d+e x}}{3 e^2}-\frac {5 (b d-a e)^2 (7 b B d-6 A b e-a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 \sqrt {b} e^{9/2}} \] Output:

2*(-A*e+B*d)*(b*x+a)^(5/2)/e^2/(e*x+d)^(1/2)+5/8*(-a*e+b*d)*(-6*A*b*e-B*a* 
e+7*B*b*d)*(b*x+a)^(1/2)*(e*x+d)^(1/2)/e^4-5/12*(-6*A*b*e-B*a*e+7*B*b*d)*( 
b*x+a)^(3/2)*(e*x+d)^(1/2)/e^3+1/3*B*(b*x+a)^(5/2)*(e*x+d)^(1/2)/e^2-5/8*( 
-a*e+b*d)^2*(-6*A*b*e-B*a*e+7*B*b*d)*arctanh(e^(1/2)*(b*x+a)^(1/2)/b^(1/2) 
/(e*x+d)^(1/2))/b^(1/2)/e^(9/2)
 

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {\sqrt {a+b x} \left (3 a^2 e^2 (27 B d-16 A e+11 B e x)+2 a b e \left (3 A e (25 d+9 e x)+B \left (-95 d^2-34 d e x+13 e^2 x^2\right )\right )+b^2 \left (6 A e \left (-15 d^2-5 d e x+2 e^2 x^2\right )+B \left (105 d^3+35 d^2 e x-14 d e^2 x^2+8 e^3 x^3\right )\right )\right )}{24 e^4 \sqrt {d+e x}}+\frac {5 (b d-a e)^2 (-7 b B d+6 A b e+a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 \sqrt {b} e^{9/2}} \] Input:

Integrate[((a + b*x)^(5/2)*(A + B*x))/(d + e*x)^(3/2),x]
 

Output:

(Sqrt[a + b*x]*(3*a^2*e^2*(27*B*d - 16*A*e + 11*B*e*x) + 2*a*b*e*(3*A*e*(2 
5*d + 9*e*x) + B*(-95*d^2 - 34*d*e*x + 13*e^2*x^2)) + b^2*(6*A*e*(-15*d^2 
- 5*d*e*x + 2*e^2*x^2) + B*(105*d^3 + 35*d^2*e*x - 14*d*e^2*x^2 + 8*e^3*x^ 
3))))/(24*e^4*Sqrt[d + e*x]) + (5*(b*d - a*e)^2*(-7*b*B*d + 6*A*b*e + a*B* 
e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*Sqrt[b]*e^ 
(9/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {87, 60, 60, 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.

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {(-a B e-6 A b e+7 b B d) \int \frac {(a+b x)^{5/2}}{\sqrt {d+e x}}dx}{e (b d-a e)}-\frac {2 (a+b x)^{7/2} (B d-A e)}{e \sqrt {d+e x} (b d-a e)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(-a B e-6 A b e+7 b B d) \left (\frac {(a+b x)^{5/2} \sqrt {d+e x}}{3 e}-\frac {5 (b d-a e) \int \frac {(a+b x)^{3/2}}{\sqrt {d+e x}}dx}{6 e}\right )}{e (b d-a e)}-\frac {2 (a+b x)^{7/2} (B d-A e)}{e \sqrt {d+e x} (b d-a e)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(-a B e-6 A b e+7 b B d) \left (\frac {(a+b x)^{5/2} \sqrt {d+e x}}{3 e}-\frac {5 (b d-a e) \left (\frac {(a+b x)^{3/2} \sqrt {d+e x}}{2 e}-\frac {3 (b d-a e) \int \frac {\sqrt {a+b x}}{\sqrt {d+e x}}dx}{4 e}\right )}{6 e}\right )}{e (b d-a e)}-\frac {2 (a+b x)^{7/2} (B d-A e)}{e \sqrt {d+e x} (b d-a e)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(-a B e-6 A b e+7 b B d) \left (\frac {(a+b x)^{5/2} \sqrt {d+e x}}{3 e}-\frac {5 (b d-a e) \left (\frac {(a+b x)^{3/2} \sqrt {d+e x}}{2 e}-\frac {3 (b d-a e) \left (\frac {\sqrt {a+b x} \sqrt {d+e x}}{e}-\frac {(b d-a e) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}}dx}{2 e}\right )}{4 e}\right )}{6 e}\right )}{e (b d-a e)}-\frac {2 (a+b x)^{7/2} (B d-A e)}{e \sqrt {d+e x} (b d-a e)}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {(-a B e-6 A b e+7 b B d) \left (\frac {(a+b x)^{5/2} \sqrt {d+e x}}{3 e}-\frac {5 (b d-a e) \left (\frac {(a+b x)^{3/2} \sqrt {d+e x}}{2 e}-\frac {3 (b d-a e) \left (\frac {\sqrt {a+b x} \sqrt {d+e x}}{e}-\frac {(b d-a e) \int \frac {1}{b-\frac {e (a+b x)}{d+e x}}d\frac {\sqrt {a+b x}}{\sqrt {d+e x}}}{e}\right )}{4 e}\right )}{6 e}\right )}{e (b d-a e)}-\frac {2 (a+b x)^{7/2} (B d-A e)}{e \sqrt {d+e x} (b d-a e)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(-a B e-6 A b e+7 b B d) \left (\frac {(a+b x)^{5/2} \sqrt {d+e x}}{3 e}-\frac {5 (b d-a e) \left (\frac {(a+b x)^{3/2} \sqrt {d+e x}}{2 e}-\frac {3 (b d-a e) \left (\frac {\sqrt {a+b x} \sqrt {d+e x}}{e}-\frac {(b d-a e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} e^{3/2}}\right )}{4 e}\right )}{6 e}\right )}{e (b d-a e)}-\frac {2 (a+b x)^{7/2} (B d-A e)}{e \sqrt {d+e x} (b d-a e)}\)

Input:

Int[((a + b*x)^(5/2)*(A + B*x))/(d + e*x)^(3/2),x]
 

Output:

(-2*(B*d - A*e)*(a + b*x)^(7/2))/(e*(b*d - a*e)*Sqrt[d + e*x]) + ((7*b*B*d 
 - 6*A*b*e - a*B*e)*(((a + b*x)^(5/2)*Sqrt[d + e*x])/(3*e) - (5*(b*d - a*e 
)*(((a + b*x)^(3/2)*Sqrt[d + e*x])/(2*e) - (3*(b*d - a*e)*((Sqrt[a + b*x]* 
Sqrt[d + e*x])/e - ((b*d - a*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*S 
qrt[d + e*x])])/(Sqrt[b]*e^(3/2))))/(4*e)))/(6*e)))/(e*(b*d - a*e))
 

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 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]))))
 

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. \(1183\) vs. \(2(182)=364\).

Time = 0.27 (sec) , antiderivative size = 1184, normalized size of antiderivative = 5.43

method	result	size
default	\(\text {Expression too large to display}\)	\(1184\)

Input:

int((b*x+a)^(5/2)*(B*x+A)/(e*x+d)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/48*(b*x+a)^(1/2)*(-96*A*a^2*e^3*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+210* 
B*b^2*d^3*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)-136*B*a*b*d*e^2*x*((e*x+d)*( 
b*x+a))^(1/2)*(b*e)^(1/2)+15*B*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*( 
b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*a^3*e^4*x+24*A*b^2*e^3*x^2*((e*x+d)*(b*x+ 
a))^(1/2)*(b*e)^(1/2)-105*B*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e 
)^(1/2)+a*e+d*b)/(b*e)^(1/2))*b^3*d^3*e*x+225*B*ln(1/2*(2*b*e*x+2*((e*x+d) 
*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*a*b^2*d^3*e+66*B*a^2*e^3 
*x*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+162*B*a^2*d*e^2*((e*x+d)*(b*x+a))^( 
1/2)*(b*e)^(1/2)+225*B*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/ 
2)+a*e+d*b)/(b*e)^(1/2))*a*b^2*d^2*e^2*x+52*B*a*b*e^3*x^2*((e*x+d)*(b*x+a) 
)^(1/2)*(b*e)^(1/2)-28*B*b^2*d*e^2*x^2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2) 
+108*A*a*b*e^3*x*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+70*B*b^2*d^2*e*x*((e* 
x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)-60*A*b^2*d*e^2*x*((e*x+d)*(b*x+a))^(1/2)*( 
b*e)^(1/2)+300*A*a*b*d*e^2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)-180*A*ln(1/ 
2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*a*b 
^2*d*e^3*x-135*B*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e 
+d*b)/(b*e)^(1/2))*a^2*b*d*e^3*x-105*B*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a)) 
^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*b^3*d^4-380*B*a*b*d^2*e*((e*x+d)* 
(b*x+a))^(1/2)*(b*e)^(1/2)+90*A*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)* 
(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*b^3*d^3*e+15*B*ln(1/2*(2*b*e*x+2*((e*...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (182) = 364\).

Time = 0.75 (sec) , antiderivative size = 858, normalized size of antiderivative = 3.94 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{(d+e x)^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate((b*x+a)^(5/2)*(B*x+A)/(e*x+d)^(3/2),x, algorithm="fricas")
 

Output:

[-1/96*(15*(7*B*b^3*d^4 - 3*(5*B*a*b^2 + 2*A*b^3)*d^3*e + 3*(3*B*a^2*b + 4 
*A*a*b^2)*d^2*e^2 - (B*a^3 + 6*A*a^2*b)*d*e^3 + (7*B*b^3*d^3*e - 3*(5*B*a* 
b^2 + 2*A*b^3)*d^2*e^2 + 3*(3*B*a^2*b + 4*A*a*b^2)*d*e^3 - (B*a^3 + 6*A*a^ 
2*b)*e^4)*x)*sqrt(b*e)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 
 4*(2*b*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x + a)*sqrt(e*x + d) + 8*(b^2*d* 
e + a*b*e^2)*x) - 4*(8*B*b^3*e^4*x^3 + 105*B*b^3*d^3*e - 48*A*a^2*b*e^4 - 
10*(19*B*a*b^2 + 9*A*b^3)*d^2*e^2 + 3*(27*B*a^2*b + 50*A*a*b^2)*d*e^3 - 2* 
(7*B*b^3*d*e^3 - (13*B*a*b^2 + 6*A*b^3)*e^4)*x^2 + (35*B*b^3*d^2*e^2 - 2*( 
34*B*a*b^2 + 15*A*b^3)*d*e^3 + 3*(11*B*a^2*b + 18*A*a*b^2)*e^4)*x)*sqrt(b* 
x + a)*sqrt(e*x + d))/(b*e^6*x + b*d*e^5), 1/48*(15*(7*B*b^3*d^4 - 3*(5*B* 
a*b^2 + 2*A*b^3)*d^3*e + 3*(3*B*a^2*b + 4*A*a*b^2)*d^2*e^2 - (B*a^3 + 6*A* 
a^2*b)*d*e^3 + (7*B*b^3*d^3*e - 3*(5*B*a*b^2 + 2*A*b^3)*d^2*e^2 + 3*(3*B*a 
^2*b + 4*A*a*b^2)*d*e^3 - (B*a^3 + 6*A*a^2*b)*e^4)*x)*sqrt(-b*e)*arctan(1/ 
2*(2*b*e*x + b*d + a*e)*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e^2*x^ 
2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)) + 2*(8*B*b^3*e^4*x^3 + 105*B*b^3*d^3 
*e - 48*A*a^2*b*e^4 - 10*(19*B*a*b^2 + 9*A*b^3)*d^2*e^2 + 3*(27*B*a^2*b + 
50*A*a*b^2)*d*e^3 - 2*(7*B*b^3*d*e^3 - (13*B*a*b^2 + 6*A*b^3)*e^4)*x^2 + ( 
35*B*b^3*d^2*e^2 - 2*(34*B*a*b^2 + 15*A*b^3)*d*e^3 + 3*(11*B*a^2*b + 18*A* 
a*b^2)*e^4)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b*e^6*x + b*d*e^5)]
 

Sympy [F]

\[ \int \frac {(a+b x)^{5/2} (A+B x)}{(d+e x)^{3/2}} \, dx=\int \frac {\left (A + B x\right ) \left (a + b x\right )^{\frac {5}{2}}}{\left (d + e x\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((b*x+a)**(5/2)*(B*x+A)/(e*x+d)**(3/2),x)
 

Output:

Integral((A + B*x)*(a + b*x)**(5/2)/(d + e*x)**(3/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b x)^{5/2} (A+B x)}{(d+e x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x+a)^(5/2)*(B*x+A)/(e*x+d)^(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*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. 419 vs. \(2 (182) = 364\).

Time = 0.20 (sec) , antiderivative size = 419, normalized size of antiderivative = 1.92 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {{\left ({\left (b x + a\right )} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} B {\left | b \right |}}{b e} - \frac {7 \, B b^{2} d e^{5} {\left | b \right |} - B a b e^{6} {\left | b \right |} - 6 \, A b^{2} e^{6} {\left | b \right |}}{b^{2} e^{7}}\right )} + \frac {5 \, {\left (7 \, B b^{3} d^{2} e^{4} {\left | b \right |} - 8 \, B a b^{2} d e^{5} {\left | b \right |} - 6 \, A b^{3} d e^{5} {\left | b \right |} + B a^{2} b e^{6} {\left | b \right |} + 6 \, A a b^{2} e^{6} {\left | b \right |}\right )}}{b^{2} e^{7}}\right )} + \frac {15 \, {\left (7 \, B b^{4} d^{3} e^{3} {\left | b \right |} - 15 \, B a b^{3} d^{2} e^{4} {\left | b \right |} - 6 \, A b^{4} d^{2} e^{4} {\left | b \right |} + 9 \, B a^{2} b^{2} d e^{5} {\left | b \right |} + 12 \, A a b^{3} d e^{5} {\left | b \right |} - B a^{3} b e^{6} {\left | b \right |} - 6 \, A a^{2} b^{2} e^{6} {\left | b \right |}\right )}}{b^{2} e^{7}}\right )} \sqrt {b x + a}}{24 \, \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}} + \frac {5 \, {\left (7 \, B b^{3} d^{3} {\left | b \right |} - 15 \, B a b^{2} d^{2} e {\left | b \right |} - 6 \, A b^{3} d^{2} e {\left | b \right |} + 9 \, B a^{2} b d e^{2} {\left | b \right |} + 12 \, A a b^{2} d e^{2} {\left | b \right |} - B a^{3} e^{3} {\left | b \right |} - 6 \, A a^{2} b e^{3} {\left | b \right |}\right )} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{8 \, \sqrt {b e} b e^{4}} \] Input:

integrate((b*x+a)^(5/2)*(B*x+A)/(e*x+d)^(3/2),x, algorithm="giac")
 

Output:

1/24*((b*x + a)*(2*(b*x + a)*(4*(b*x + a)*B*abs(b)/(b*e) - (7*B*b^2*d*e^5* 
abs(b) - B*a*b*e^6*abs(b) - 6*A*b^2*e^6*abs(b))/(b^2*e^7)) + 5*(7*B*b^3*d^ 
2*e^4*abs(b) - 8*B*a*b^2*d*e^5*abs(b) - 6*A*b^3*d*e^5*abs(b) + B*a^2*b*e^6 
*abs(b) + 6*A*a*b^2*e^6*abs(b))/(b^2*e^7)) + 15*(7*B*b^4*d^3*e^3*abs(b) - 
15*B*a*b^3*d^2*e^4*abs(b) - 6*A*b^4*d^2*e^4*abs(b) + 9*B*a^2*b^2*d*e^5*abs 
(b) + 12*A*a*b^3*d*e^5*abs(b) - B*a^3*b*e^6*abs(b) - 6*A*a^2*b^2*e^6*abs(b 
))/(b^2*e^7))*sqrt(b*x + a)/sqrt(b^2*d + (b*x + a)*b*e - a*b*e) + 5/8*(7*B 
*b^3*d^3*abs(b) - 15*B*a*b^2*d^2*e*abs(b) - 6*A*b^3*d^2*e*abs(b) + 9*B*a^2 
*b*d*e^2*abs(b) + 12*A*a*b^2*d*e^2*abs(b) - B*a^3*e^3*abs(b) - 6*A*a^2*b*e 
^3*abs(b))*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt(b^2*d + (b*x + a)*b*e - 
 a*b*e)))/(sqrt(b*e)*b*e^4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{5/2} (A+B x)}{(d+e x)^{3/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{5/2}}{{\left (d+e\,x\right )}^{3/2}} \,d x \] Input:

int(((A + B*x)*(a + b*x)^(5/2))/(d + e*x)^(3/2),x)
 

Output:

int(((A + B*x)*(a + b*x)^(5/2))/(d + e*x)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 724, normalized size of antiderivative = 3.32 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{(d+e x)^{3/2}} \, dx =\text {Too large to display} \] Input:

int((b*x+a)^(5/2)*(B*x+A)/(e*x+d)^(3/2),x)
 

Output:

( - 384*sqrt(d + e*x)*sqrt(a + b*x)*a**3*e**4 + 1848*sqrt(d + e*x)*sqrt(a 
+ b*x)*a**2*b*d*e**3 + 696*sqrt(d + e*x)*sqrt(a + b*x)*a**2*b*e**4*x - 224 
0*sqrt(d + e*x)*sqrt(a + b*x)*a*b**2*d**2*e**2 - 784*sqrt(d + e*x)*sqrt(a 
+ b*x)*a*b**2*d*e**3*x + 304*sqrt(d + e*x)*sqrt(a + b*x)*a*b**2*e**4*x**2 
+ 840*sqrt(d + e*x)*sqrt(a + b*x)*b**3*d**3*e + 280*sqrt(d + e*x)*sqrt(a + 
 b*x)*b**3*d**2*e**2*x - 112*sqrt(d + e*x)*sqrt(a + b*x)*b**3*d*e**3*x**2 
+ 64*sqrt(d + e*x)*sqrt(a + b*x)*b**3*e**4*x**3 + 840*sqrt(e)*sqrt(b)*log( 
(sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*a**3*d*e* 
*3 + 840*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x 
))/sqrt(a*e - b*d))*a**3*e**4*x - 2520*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a 
 + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*a**2*b*d**2*e**2 - 2520* 
sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a 
*e - b*d))*a**2*b*d*e**3*x + 2520*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b* 
x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*a*b**2*d**3*e + 2520*sqrt(e)* 
sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d 
))*a*b**2*d**2*e**2*x - 840*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + s 
qrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*b**3*d**4 - 840*sqrt(e)*sqrt(b)*log 
((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*b**3*d** 
3*e*x - 525*sqrt(e)*sqrt(b)*a**3*d*e**3 - 525*sqrt(e)*sqrt(b)*a**3*e**4*x 
+ 1575*sqrt(e)*sqrt(b)*a**2*b*d**2*e**2 + 1575*sqrt(e)*sqrt(b)*a**2*b*d...