\(\int \frac {(A+B x) (d+e x)^{5/2}}{(a+b x)^3} \, dx\) [165]

Optimal result

Integrand size = 22, antiderivative size = 196 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{(a+b x)^3} \, dx=\frac {e (8 b B d+5 A b e-13 a B e) \sqrt {d+e x}}{2 b^4}-\frac {(b d-a e) (4 b B d+5 A b e-9 a B e) \sqrt {d+e x}}{4 b^4 (a+b x)}+\frac {2 B e (d+e x)^{3/2}}{3 b^3}-\frac {(A b-a B) (d+e x)^{5/2}}{2 b^2 (a+b x)^2}-\frac {5 e \sqrt {b d-a e} (4 b B d+3 A b e-7 a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{9/2}} \] Output:

1/2*e*(5*A*b*e-13*B*a*e+8*B*b*d)*(e*x+d)^(1/2)/b^4-1/4*(-a*e+b*d)*(5*A*b*e 
-9*B*a*e+4*B*b*d)*(e*x+d)^(1/2)/b^4/(b*x+a)+2/3*B*e*(e*x+d)^(3/2)/b^3-1/2* 
(A*b-B*a)*(e*x+d)^(5/2)/b^2/(b*x+a)^2-5/4*e*(-a*e+b*d)^(1/2)*(3*A*b*e-7*B* 
a*e+4*B*b*d)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(9/2)
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{(a+b x)^3} \, dx=-\frac {\sqrt {d+e x} \left (3 A b \left (-15 a^2 e^2+5 a b e (d-5 e x)+b^2 \left (2 d^2+9 d e x-8 e^2 x^2\right )\right )+B \left (105 a^3 e^2+5 a^2 b e (-19 d+35 e x)-4 b^3 x \left (-3 d^2+14 d e x+2 e^2 x^2\right )+a b^2 \left (6 d^2-163 d e x+56 e^2 x^2\right )\right )\right )}{12 b^4 (a+b x)^2}-\frac {5 e \sqrt {-b d+a e} (4 b B d+3 A b e-7 a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{4 b^{9/2}} \] Input:

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

Output:

-1/12*(Sqrt[d + e*x]*(3*A*b*(-15*a^2*e^2 + 5*a*b*e*(d - 5*e*x) + b^2*(2*d^ 
2 + 9*d*e*x - 8*e^2*x^2)) + B*(105*a^3*e^2 + 5*a^2*b*e*(-19*d + 35*e*x) - 
4*b^3*x*(-3*d^2 + 14*d*e*x + 2*e^2*x^2) + a*b^2*(6*d^2 - 163*d*e*x + 56*e^ 
2*x^2))))/(b^4*(a + b*x)^2) - (5*e*Sqrt[-(b*d) + a*e]*(4*b*B*d + 3*A*b*e - 
 7*a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(4*b^(9/2))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {87, 51, 60, 60, 73, 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.

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-1/2*((A*b - a*B)*(d + e*x)^(7/2))/(b*(b*d - a*e)*(a + b*x)^2) + ((4*b*B*d 
 + 3*A*b*e - 7*a*B*e)*(-((d + e*x)^(5/2)/(b*(a + b*x))) + (5*e*((2*(d + e* 
x)^(3/2))/(3*b) + ((b*d - a*e)*((2*Sqrt[d + e*x])/b - (2*Sqrt[b*d - a*e]*A 
rcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(3/2)))/b))/(2*b)))/(4* 
b*(b*d - a*e))
 

Defintions of rubi rules used

rule 51

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, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

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 [A] (verified)

Time = 0.59 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.03

method	result	size
risch	\(\frac {2 e \left (e b B x +3 A b e -9 B a e +7 B b d \right ) \sqrt {e x +d}}{3 b^{4}}-\frac {\left (2 a e -2 d b \right ) e \left (\frac {\left (-\frac {9}{8} A \,b^{2} e +\frac {13}{8} B a b e -\frac {1}{2} b^{2} B d \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} A a b \,e^{2}+\frac {7}{8} A \,b^{2} d e +\frac {11}{8} B \,a^{2} e^{2}-\frac {15}{8} B a b d e +\frac {1}{2} b^{2} B \,d^{2}\right ) \sqrt {e x +d}}{\left (\left (e x +d \right ) b +a e -d b \right )^{2}}+\frac {5 \left (3 A b e -7 B a e +4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{8 \sqrt {\left (a e -d b \right ) b}}\right )}{b^{4}}\)	\(202\)
pseudoelliptic	\(\frac {-\frac {15 \left (\left (A e +\frac {4 B d}{3}\right ) b -\frac {7 B a e}{3}\right ) \left (b x +a \right )^{2} \left (a e -d b \right ) e \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{4}+\frac {15 \sqrt {e x +d}\, \sqrt {\left (a e -d b \right ) b}\, \left (\left (\frac {8 \left (\frac {B x}{3}+A \right ) x^{2} e^{2}}{15}-\frac {3 \left (-\frac {56 B x}{27}+A \right ) x d e}{5}-\frac {2 d^{2} \left (2 B x +A \right )}{15}\right ) b^{3}-\frac {a \left (\left (\frac {56}{15} x^{2} B -5 x A \right ) e^{2}+d \left (-\frac {163 B x}{15}+A \right ) e +\frac {2 B \,d^{2}}{5}\right ) b^{2}}{3}+a^{2} e \left (\left (A -\frac {35 B x}{9}\right ) e +\frac {19 B d}{9}\right ) b -\frac {7 B \,a^{3} e^{2}}{3}\right )}{4}}{\sqrt {\left (a e -d b \right ) b}\, b^{4} \left (b x +a \right )^{2}}\)	\(209\)
derivativedivides	\(2 e \left (\frac {\frac {b B \left (e x +d \right )^{\frac {3}{2}}}{3}+A b e \sqrt {e x +d}-3 B a e \sqrt {e x +d}+2 B b d \sqrt {e x +d}}{b^{4}}-\frac {\frac {\left (-\frac {9}{8} b^{2} e^{2} a A +\frac {9}{8} A \,b^{3} d e +\frac {13}{8} B \,a^{2} b \,e^{2}-\frac {17}{8} B a \,b^{2} d e +\frac {1}{2} b^{3} B \,d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} A \,a^{2} b \,e^{3}+\frac {7}{4} A a \,b^{2} d \,e^{2}-\frac {7}{8} A \,b^{3} d^{2} e +\frac {11}{8} B \,a^{3} e^{3}-\frac {13}{4} B \,a^{2} b d \,e^{2}+\frac {19}{8} B a \,b^{2} d^{2} e -\frac {1}{2} b^{3} B \,d^{3}\right ) \sqrt {e x +d}}{\left (\left (e x +d \right ) b +a e -d b \right )^{2}}+\frac {5 \left (3 A a b \,e^{2}-3 A \,b^{2} d e -7 B \,a^{2} e^{2}+11 B a b d e -4 b^{2} B \,d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{8 \sqrt {\left (a e -d b \right ) b}}}{b^{4}}\right )\)	\(294\)
default	\(2 e \left (\frac {\frac {b B \left (e x +d \right )^{\frac {3}{2}}}{3}+A b e \sqrt {e x +d}-3 B a e \sqrt {e x +d}+2 B b d \sqrt {e x +d}}{b^{4}}-\frac {\frac {\left (-\frac {9}{8} b^{2} e^{2} a A +\frac {9}{8} A \,b^{3} d e +\frac {13}{8} B \,a^{2} b \,e^{2}-\frac {17}{8} B a \,b^{2} d e +\frac {1}{2} b^{3} B \,d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} A \,a^{2} b \,e^{3}+\frac {7}{4} A a \,b^{2} d \,e^{2}-\frac {7}{8} A \,b^{3} d^{2} e +\frac {11}{8} B \,a^{3} e^{3}-\frac {13}{4} B \,a^{2} b d \,e^{2}+\frac {19}{8} B a \,b^{2} d^{2} e -\frac {1}{2} b^{3} B \,d^{3}\right ) \sqrt {e x +d}}{\left (\left (e x +d \right ) b +a e -d b \right )^{2}}+\frac {5 \left (3 A a b \,e^{2}-3 A \,b^{2} d e -7 B \,a^{2} e^{2}+11 B a b d e -4 b^{2} B \,d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{8 \sqrt {\left (a e -d b \right ) b}}}{b^{4}}\right )\)	\(294\)

Input:

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

Output:

2/3*e*(B*b*e*x+3*A*b*e-9*B*a*e+7*B*b*d)*(e*x+d)^(1/2)/b^4-1/b^4*(2*a*e-2*b 
*d)*e*(((-9/8*A*b^2*e+13/8*B*a*b*e-1/2*b^2*B*d)*(e*x+d)^(3/2)+(-7/8*A*a*b* 
e^2+7/8*A*b^2*d*e+11/8*B*a^2*e^2-15/8*B*a*b*d*e+1/2*b^2*B*d^2)*(e*x+d)^(1/ 
2))/((e*x+d)*b+a*e-d*b)^2+5/8*(3*A*b*e-7*B*a*e+4*B*b*d)/((a*e-b*d)*b)^(1/2 
)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 680, normalized size of antiderivative = 3.47 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{(a+b x)^3} \, dx=\left [\frac {15 \, {\left (4 \, B a^{2} b d e - {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (8 \, B b^{3} e^{2} x^{3} - 6 \, {\left (B a b^{2} + A b^{3}\right )} d^{2} + 5 \, {\left (19 \, B a^{2} b - 3 \, A a b^{2}\right )} d e - 15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + 8 \, {\left (7 \, B b^{3} d e - {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} e^{2}\right )} x^{2} - {\left (12 \, B b^{3} d^{2} - {\left (163 \, B a b^{2} - 27 \, A b^{3}\right )} d e + 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, -\frac {15 \, {\left (4 \, B a^{2} b d e - {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (8 \, B b^{3} e^{2} x^{3} - 6 \, {\left (B a b^{2} + A b^{3}\right )} d^{2} + 5 \, {\left (19 \, B a^{2} b - 3 \, A a b^{2}\right )} d e - 15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + 8 \, {\left (7 \, B b^{3} d e - {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} e^{2}\right )} x^{2} - {\left (12 \, B b^{3} d^{2} - {\left (163 \, B a b^{2} - 27 \, A b^{3}\right )} d e + 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{12 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \] Input:

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

Output:

[1/24*(15*(4*B*a^2*b*d*e - (7*B*a^3 - 3*A*a^2*b)*e^2 + (4*B*b^3*d*e - (7*B 
*a*b^2 - 3*A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*d*e - (7*B*a^2*b - 3*A*a*b^2)*e^ 
2)*x)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*b*sqr 
t((b*d - a*e)/b))/(b*x + a)) + 2*(8*B*b^3*e^2*x^3 - 6*(B*a*b^2 + A*b^3)*d^ 
2 + 5*(19*B*a^2*b - 3*A*a*b^2)*d*e - 15*(7*B*a^3 - 3*A*a^2*b)*e^2 + 8*(7*B 
*b^3*d*e - (7*B*a*b^2 - 3*A*b^3)*e^2)*x^2 - (12*B*b^3*d^2 - (163*B*a*b^2 - 
 27*A*b^3)*d*e + 25*(7*B*a^2*b - 3*A*a*b^2)*e^2)*x)*sqrt(e*x + d))/(b^6*x^ 
2 + 2*a*b^5*x + a^2*b^4), -1/12*(15*(4*B*a^2*b*d*e - (7*B*a^3 - 3*A*a^2*b) 
*e^2 + (4*B*b^3*d*e - (7*B*a*b^2 - 3*A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*d*e - 
(7*B*a^2*b - 3*A*a*b^2)*e^2)*x)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d) 
*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (8*B*b^3*e^2*x^3 - 6*(B*a*b^2 + A*b 
^3)*d^2 + 5*(19*B*a^2*b - 3*A*a*b^2)*d*e - 15*(7*B*a^3 - 3*A*a^2*b)*e^2 + 
8*(7*B*b^3*d*e - (7*B*a*b^2 - 3*A*b^3)*e^2)*x^2 - (12*B*b^3*d^2 - (163*B*a 
*b^2 - 27*A*b^3)*d*e + 25*(7*B*a^2*b - 3*A*a*b^2)*e^2)*x)*sqrt(e*x + d))/( 
b^6*x^2 + 2*a*b^5*x + a^2*b^4)]
                                                                                    
                                                                                    
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((B*x+A)*(e*x+d)^(5/2)/(b*x+a)^3,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. 390 vs. \(2 (168) = 336\).

Time = 0.13 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.99 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{(a+b x)^3} \, dx=\frac {5 \, {\left (4 \, B b^{2} d^{2} e - 11 \, B a b d e^{2} + 3 \, A b^{2} d e^{2} + 7 \, B a^{2} e^{3} - 3 \, A a b e^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, \sqrt {-b^{2} d + a b e} b^{4}} - \frac {4 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{3} d^{2} e - 4 \, \sqrt {e x + d} B b^{3} d^{3} e - 17 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b^{2} d e^{2} + 9 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{3} d e^{2} + 19 \, \sqrt {e x + d} B a b^{2} d^{2} e^{2} - 7 \, \sqrt {e x + d} A b^{3} d^{2} e^{2} + 13 \, {\left (e x + d\right )}^{\frac {3}{2}} B a^{2} b e^{3} - 9 \, {\left (e x + d\right )}^{\frac {3}{2}} A a b^{2} e^{3} - 26 \, \sqrt {e x + d} B a^{2} b d e^{3} + 14 \, \sqrt {e x + d} A a b^{2} d e^{3} + 11 \, \sqrt {e x + d} B a^{3} e^{4} - 7 \, \sqrt {e x + d} A a^{2} b e^{4}}{4 \, {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2} b^{4}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} B b^{6} e + 6 \, \sqrt {e x + d} B b^{6} d e - 9 \, \sqrt {e x + d} B a b^{5} e^{2} + 3 \, \sqrt {e x + d} A b^{6} e^{2}\right )}}{3 \, b^{9}} \] Input:

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

Output:

5/4*(4*B*b^2*d^2*e - 11*B*a*b*d*e^2 + 3*A*b^2*d*e^2 + 7*B*a^2*e^3 - 3*A*a* 
b*e^3)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)* 
b^4) - 1/4*(4*(e*x + d)^(3/2)*B*b^3*d^2*e - 4*sqrt(e*x + d)*B*b^3*d^3*e - 
17*(e*x + d)^(3/2)*B*a*b^2*d*e^2 + 9*(e*x + d)^(3/2)*A*b^3*d*e^2 + 19*sqrt 
(e*x + d)*B*a*b^2*d^2*e^2 - 7*sqrt(e*x + d)*A*b^3*d^2*e^2 + 13*(e*x + d)^( 
3/2)*B*a^2*b*e^3 - 9*(e*x + d)^(3/2)*A*a*b^2*e^3 - 26*sqrt(e*x + d)*B*a^2* 
b*d*e^3 + 14*sqrt(e*x + d)*A*a*b^2*d*e^3 + 11*sqrt(e*x + d)*B*a^3*e^4 - 7* 
sqrt(e*x + d)*A*a^2*b*e^4)/(((e*x + d)*b - b*d + a*e)^2*b^4) + 2/3*((e*x + 
 d)^(3/2)*B*b^6*e + 6*sqrt(e*x + d)*B*b^6*d*e - 9*sqrt(e*x + d)*B*a*b^5*e^ 
2 + 3*sqrt(e*x + d)*A*b^6*e^2)/b^9
 

Mupad [B] (verification not implemented)

Time = 1.09 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.66 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{(a+b x)^3} \, dx=\left (\frac {2\,A\,e^2-2\,B\,d\,e}{b^3}+\frac {2\,B\,e\,\left (3\,b^3\,d-3\,a\,b^2\,e\right )}{b^6}\right )\,\sqrt {d+e\,x}-\frac {{\left (d+e\,x\right )}^{3/2}\,\left (\frac {13\,B\,a^2\,b\,e^3}{4}-\frac {17\,B\,a\,b^2\,d\,e^2}{4}-\frac {9\,A\,a\,b^2\,e^3}{4}+B\,b^3\,d^2\,e+\frac {9\,A\,b^3\,d\,e^2}{4}\right )-\sqrt {d+e\,x}\,\left (-\frac {11\,B\,a^3\,e^4}{4}+\frac {13\,B\,a^2\,b\,d\,e^3}{2}+\frac {7\,A\,a^2\,b\,e^4}{4}-\frac {19\,B\,a\,b^2\,d^2\,e^2}{4}-\frac {7\,A\,a\,b^2\,d\,e^3}{2}+B\,b^3\,d^3\,e+\frac {7\,A\,b^3\,d^2\,e^2}{4}\right )}{b^6\,{\left (d+e\,x\right )}^2-\left (2\,b^6\,d-2\,a\,b^5\,e\right )\,\left (d+e\,x\right )+b^6\,d^2+a^2\,b^4\,e^2-2\,a\,b^5\,d\,e}+\frac {2\,B\,e\,{\left (d+e\,x\right )}^{3/2}}{3\,b^3}+\frac {e\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,1{}\mathrm {i}}{\sqrt {b\,d-a\,e}}\right )\,\sqrt {b\,d-a\,e}\,\left (3\,A\,b\,e-7\,B\,a\,e+4\,B\,b\,d\right )\,5{}\mathrm {i}}{4\,b^{9/2}} \] Input:

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

Output:

((2*A*e^2 - 2*B*d*e)/b^3 + (2*B*e*(3*b^3*d - 3*a*b^2*e))/b^6)*(d + e*x)^(1 
/2) - ((d + e*x)^(3/2)*((13*B*a^2*b*e^3)/4 - (9*A*a*b^2*e^3)/4 + (9*A*b^3* 
d*e^2)/4 + B*b^3*d^2*e - (17*B*a*b^2*d*e^2)/4) - (d + e*x)^(1/2)*((7*A*a^2 
*b*e^4)/4 - (11*B*a^3*e^4)/4 + B*b^3*d^3*e + (7*A*b^3*d^2*e^2)/4 - (19*B*a 
*b^2*d^2*e^2)/4 - (7*A*a*b^2*d*e^3)/2 + (13*B*a^2*b*d*e^3)/2))/(b^6*(d + e 
*x)^2 - (2*b^6*d - 2*a*b^5*e)*(d + e*x) + b^6*d^2 + a^2*b^4*e^2 - 2*a*b^5* 
d*e) + (2*B*e*(d + e*x)^(3/2))/(3*b^3) + (e*atan((b^(1/2)*(d + e*x)^(1/2)* 
1i)/(b*d - a*e)^(1/2))*(b*d - a*e)^(1/2)*(3*A*b*e - 7*B*a*e + 4*B*b*d)*5i) 
/(4*b^(9/2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.39 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{(a+b x)^3} \, dx=\frac {15 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{2} e^{2}-15 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a b d e +15 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a b \,e^{2} x -15 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) b^{2} d e x -15 \sqrt {e x +d}\, a^{2} b \,e^{2}+20 \sqrt {e x +d}\, a \,b^{2} d e -10 \sqrt {e x +d}\, a \,b^{2} e^{2} x -3 \sqrt {e x +d}\, b^{3} d^{2}+14 \sqrt {e x +d}\, b^{3} d e x +2 \sqrt {e x +d}\, b^{3} e^{2} x^{2}}{3 b^{4} \left (b x +a \right )} \] Input:

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

Output:

(15*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d 
)))*a**2*e**2 - 15*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b) 
*sqrt(a*e - b*d)))*a*b*d*e + 15*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x 
)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a*b*e**2*x - 15*sqrt(b)*sqrt(a*e - b*d)*at 
an((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*b**2*d*e*x - 15*sqrt(d + e 
*x)*a**2*b*e**2 + 20*sqrt(d + e*x)*a*b**2*d*e - 10*sqrt(d + e*x)*a*b**2*e* 
*2*x - 3*sqrt(d + e*x)*b**3*d**2 + 14*sqrt(d + e*x)*b**3*d*e*x + 2*sqrt(d 
+ e*x)*b**3*e**2*x**2)/(3*b**4*(a + b*x))