\(\int \frac {x^{5/2} (A+B x)}{(a+b x)^{5/2}} \, dx\) [336]

Optimal result

Integrand size = 20, antiderivative size = 152 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{5/2}} \, dx=-\frac {2 (A b-a B) x^{5/2}}{3 b^2 (a+b x)^{3/2}}-\frac {2 (5 A b-8 a B) x^{3/2}}{3 b^3 \sqrt {a+b x}}+\frac {5 (4 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{4 b^4}+\frac {B x^{3/2} \sqrt {a+b x}}{2 b^3}-\frac {5 a (4 A b-7 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{9/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.80 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {\sqrt {x} \left (-105 a^3 B+a b^2 x (80 A-21 B x)+20 a^2 b (3 A-7 B x)+6 b^3 x^2 (2 A+B x)\right )}{12 b^4 (a+b x)^{3/2}}+\frac {5 a (-4 A b+7 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{2 b^{9/2}} \] Input:

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

Output:

(Sqrt[x]*(-105*a^3*B + a*b^2*x*(80*A - 21*B*x) + 20*a^2*b*(3*A - 7*B*x) + 
6*b^3*x^2*(2*A + B*x)))/(12*b^4*(a + b*x)^(3/2)) + (5*a*(-4*A*b + 7*a*B)*A 
rcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(2*b^(9/2))
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {87, 57, 60, 60, 65, 219}

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.

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 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 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(258\) vs. \(2(118)=236\).

Time = 0.18 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.70

method	result	size
risch	\(\frac {\left (2 b B x +4 A b -11 B a \right ) \sqrt {x}\, \sqrt {b x +a}}{4 b^{4}}-\frac {a \left (20 A \sqrt {b}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )-\frac {35 B a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{\sqrt {b}}-\frac {16 \left (3 A b -4 B a \right ) \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{b \left (x +\frac {a}{b}\right )}+\frac {8 a^{2} \left (A b -B a \right ) \left (\frac {2 \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{3 a \left (x +\frac {a}{b}\right )^{2}}+\frac {4 b \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{3 a^{2} \left (x +\frac {a}{b}\right )}\right )}{b^{2}}\right ) \sqrt {x \left (b x +a \right )}}{8 b^{4} \sqrt {x}\, \sqrt {b x +a}}\)	\(259\)
default	\(-\frac {\left (-12 B \,b^{\frac {7}{2}} x^{3} \sqrt {x \left (b x +a \right )}+60 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a \,b^{3} x^{2}-24 A \,b^{\frac {7}{2}} x^{2} \sqrt {x \left (b x +a \right )}-105 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b^{2} x^{2}+42 B a \,b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+120 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b^{2} x -160 A \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} a x -210 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3} b x +280 B \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{2} x +60 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3} b -120 A \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{2}-105 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{4}+210 B \sqrt {x \left (b x +a \right )}\, \sqrt {b}\, a^{3}\right ) \sqrt {x}}{24 b^{\frac {9}{2}} \sqrt {x \left (b x +a \right )}\, \left (b x +a \right )^{\frac {3}{2}}}\)	\(362\)

Input:

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

Output:

1/4*(2*B*b*x+4*A*b-11*B*a)*x^(1/2)*(b*x+a)^(1/2)/b^4-1/8*a/b^4*(20*A*b^(1/ 
2)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))-35*B*a*ln((1/2*a+b*x)/b^(1/2) 
+(b*x^2+a*x)^(1/2))/b^(1/2)-16*(3*A*b-4*B*a)/b/(x+a/b)*(b*(x+a/b)^2-(x+a/b 
)*a)^(1/2)+8*a^2*(A*b-B*a)/b^2*(2/3/a/(x+a/b)^2*(b*(x+a/b)^2-(x+a/b)*a)^(1 
/2)+4/3*b/a^2/(x+a/b)*(b*(x+a/b)^2-(x+a/b)*a)^(1/2)))*(x*(b*x+a))^(1/2)/x^ 
(1/2)/(b*x+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 370, normalized size of antiderivative = 2.43 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{5/2}} \, dx=\left [-\frac {15 \, {\left (7 \, B a^{4} - 4 \, A a^{3} b + {\left (7 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{2} + 2 \, {\left (7 \, B a^{3} b - 4 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (6 \, B b^{4} x^{3} - 105 \, B a^{3} b + 60 \, A a^{2} b^{2} - 3 \, {\left (7 \, B a b^{3} - 4 \, A b^{4}\right )} x^{2} - 20 \, {\left (7 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}, -\frac {15 \, {\left (7 \, B a^{4} - 4 \, A a^{3} b + {\left (7 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{2} + 2 \, {\left (7 \, B a^{3} b - 4 \, A a^{2} b^{2}\right )} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) - {\left (6 \, B b^{4} x^{3} - 105 \, B a^{3} b + 60 \, A a^{2} b^{2} - 3 \, {\left (7 \, B a b^{3} - 4 \, A b^{4}\right )} x^{2} - 20 \, {\left (7 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{12 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}\right ] \] Input:

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

Output:

[-1/24*(15*(7*B*a^4 - 4*A*a^3*b + (7*B*a^2*b^2 - 4*A*a*b^3)*x^2 + 2*(7*B*a 
^3*b - 4*A*a^2*b^2)*x)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) 
 + a) - 2*(6*B*b^4*x^3 - 105*B*a^3*b + 60*A*a^2*b^2 - 3*(7*B*a*b^3 - 4*A*b 
^4)*x^2 - 20*(7*B*a^2*b^2 - 4*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^7*x^2 
+ 2*a*b^6*x + a^2*b^5), -1/12*(15*(7*B*a^4 - 4*A*a^3*b + (7*B*a^2*b^2 - 4* 
A*a*b^3)*x^2 + 2*(7*B*a^3*b - 4*A*a^2*b^2)*x)*sqrt(-b)*arctan(sqrt(-b)*sqr 
t(x)/sqrt(b*x + a)) - (6*B*b^4*x^3 - 105*B*a^3*b + 60*A*a^2*b^2 - 3*(7*B*a 
*b^3 - 4*A*b^4)*x^2 - 20*(7*B*a^2*b^2 - 4*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x 
))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 517 vs. \(2 (118) = 236\).

Time = 0.07 (sec) , antiderivative size = 517, normalized size of antiderivative = 3.40 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{5/2}} \, dx=-\frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B a}{b^{6} x^{4} + 4 \, a b^{5} x^{3} + 6 \, a^{2} b^{4} x^{2} + 4 \, a^{3} b^{3} x + a^{4} b^{2}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a^{2}}{6 \, {\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}} + \frac {5 \, \sqrt {b x^{2} + a x} B a^{3}}{6 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} A}{b^{5} x^{4} + 4 \, a b^{4} x^{3} + 6 \, a^{2} b^{3} x^{2} + 4 \, a^{3} b^{2} x + a^{4} b} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B}{2 \, {\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{6 \, {\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} - \frac {5 \, \sqrt {b x^{2} + a x} A a^{2}}{6 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} - \frac {115 \, \sqrt {b x^{2} + a x} B a^{2}}{12 \, {\left (b^{5} x + a b^{4}\right )}} + \frac {35 \, \sqrt {b x^{2} + a x} A a}{6 \, {\left (b^{4} x + a b^{3}\right )}} + \frac {35 \, B a^{2} \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{8 \, b^{\frac {9}{2}}} - \frac {5 \, A a \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{2 \, b^{\frac {7}{2}}} \] Input:

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

Output:

-(b*x^2 + a*x)^(5/2)*B*a/(b^6*x^4 + 4*a*b^5*x^3 + 6*a^2*b^4*x^2 + 4*a^3*b^ 
3*x + a^4*b^2) - 5/6*(b*x^2 + a*x)^(3/2)*B*a^2/(b^6*x^3 + 3*a*b^5*x^2 + 3* 
a^2*b^4*x + a^3*b^3) + 5/6*sqrt(b*x^2 + a*x)*B*a^3/(b^6*x^2 + 2*a*b^5*x + 
a^2*b^4) + (b*x^2 + a*x)^(5/2)*A/(b^5*x^4 + 4*a*b^4*x^3 + 6*a^2*b^3*x^2 + 
4*a^3*b^2*x + a^4*b) + 1/2*(b*x^2 + a*x)^(5/2)*B/(b^5*x^3 + 3*a*b^4*x^2 + 
3*a^2*b^3*x + a^3*b^2) + 5/6*(b*x^2 + a*x)^(3/2)*A*a/(b^5*x^3 + 3*a*b^4*x^ 
2 + 3*a^2*b^3*x + a^3*b^2) - 5/4*(b*x^2 + a*x)^(3/2)*B*a/(b^5*x^2 + 2*a*b^ 
4*x + a^2*b^3) - 5/6*sqrt(b*x^2 + a*x)*A*a^2/(b^5*x^2 + 2*a*b^4*x + a^2*b^ 
3) - 115/12*sqrt(b*x^2 + a*x)*B*a^2/(b^5*x + a*b^4) + 35/6*sqrt(b*x^2 + a* 
x)*A*a/(b^4*x + a*b^3) + 35/8*B*a^2*log(2*x + a/b + 2*sqrt(b*x^2 + a*x)/sq 
rt(b))/b^(9/2) - 5/2*A*a*log(2*x + a/b + 2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(7 
/2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (118) = 236\).

Time = 15.64 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.20 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {1}{4} \, \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} B {\left | b \right |}}{b^{6}} - \frac {13 \, B a b^{11} {\left | b \right |} - 4 \, A b^{12} {\left | b \right |}}{b^{17}}\right )} - \frac {5 \, {\left (7 \, B a^{2} {\left | b \right |} - 4 \, A a b {\left | b \right |}\right )} \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{8 \, b^{\frac {11}{2}}} - \frac {4 \, {\left (12 \, B a^{3} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} {\left | b \right |} + 18 \, B a^{4} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b {\left | b \right |} - 9 \, A a^{2} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} b {\left | b \right |} + 10 \, B a^{5} b^{2} {\left | b \right |} - 12 \, A a^{3} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{2} {\left | b \right |} - 7 \, A a^{4} b^{3} {\left | b \right |}\right )}}{3 \, {\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3} b^{\frac {9}{2}}} \] Input:

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

Output:

1/4*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)*B*abs(b)/b^6 - (13* 
B*a*b^11*abs(b) - 4*A*b^12*abs(b))/b^17) - 5/8*(7*B*a^2*abs(b) - 4*A*a*b*a 
bs(b))*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/b^(11/2) - 
 4/3*(12*B*a^3*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*abs(b) 
+ 18*B*a^4*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b*abs(b) - 
9*A*a^2*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*b*abs(b) + 10* 
B*a^5*b^2*abs(b) - 12*A*a^3*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a* 
b))^2*b^2*abs(b) - 7*A*a^4*b^3*abs(b))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b* 
x + a)*b - a*b))^2 + a*b)^3*b^(9/2))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.56 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {15 \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{2}-10 \sqrt {b}\, \sqrt {b x +a}\, a^{2}-15 \sqrt {x}\, a^{2} b -5 \sqrt {x}\, a \,b^{2} x +2 \sqrt {x}\, b^{3} x^{2}}{4 \sqrt {b x +a}\, b^{4}} \] Input:

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

Output:

(15*sqrt(b)*sqrt(a + b*x)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a 
**2 - 10*sqrt(b)*sqrt(a + b*x)*a**2 - 15*sqrt(x)*a**2*b - 5*sqrt(x)*a*b**2 
*x + 2*sqrt(x)*b**3*x**2)/(4*sqrt(a + b*x)*b**4)