\(\int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx\) [527]

Optimal result

Integrand size = 20, antiderivative size = 167 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}-\frac {5 a (6 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{8 b^4}+\frac {5 (6 A b-7 a B) x^{3/2} \sqrt {a+b x}}{12 b^3}-\frac {(6 A b-7 a B) x^{5/2} \sqrt {a+b x}}{3 a b^2}+\frac {5 a^2 (6 A b-7 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{9/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {79, 52, 65, 223, 212} \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {5 a^2 (6 A b-7 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{9/2}}-\frac {5 a \sqrt {x} \sqrt {a+b x} (6 A b-7 a B)}{8 b^4}+\frac {5 x^{3/2} \sqrt {a+b x} (6 A b-7 a B)}{12 b^3}-\frac {x^{5/2} \sqrt {a+b x} (6 A b-7 a B)}{3 a b^2}+\frac {2 x^{7/2} (A b-a B)}{a b \sqrt {a+b x}} \]

[In]

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

[Out]

(2*(A*b - a*B)*x^(7/2))/(a*b*Sqrt[a + b*x]) - (5*a*(6*A*b - 7*a*B)*Sqrt[x]*Sqrt[a + b*x])/(8*b^4) + (5*(6*A*b
- 7*a*B)*x^(3/2)*Sqrt[a + b*x])/(12*b^3) - ((6*A*b - 7*a*B)*x^(5/2)*Sqrt[a + b*x])/(3*a*b^2) + (5*a^2*(6*A*b -
7*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(8*b^(9/2))

Rule 52

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] + Dist[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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(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] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}-\frac {\left (2 \left (3 A b-\frac {7 a B}{2}\right )\right ) \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx}{a b} \\ & = \frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}-\frac {(6 A b-7 a B) x^{5/2} \sqrt {a+b x}}{3 a b^2}+\frac {(5 (6 A b-7 a B)) \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{6 b^2} \\ & = \frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}+\frac {5 (6 A b-7 a B) x^{3/2} \sqrt {a+b x}}{12 b^3}-\frac {(6 A b-7 a B) x^{5/2} \sqrt {a+b x}}{3 a b^2}-\frac {(5 a (6 A b-7 a B)) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{8 b^3} \\ & = \frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}-\frac {5 a (6 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{8 b^4}+\frac {5 (6 A b-7 a B) x^{3/2} \sqrt {a+b x}}{12 b^3}-\frac {(6 A b-7 a B) x^{5/2} \sqrt {a+b x}}{3 a b^2}+\frac {\left (5 a^2 (6 A b-7 a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{16 b^4} \\ & = \frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}-\frac {5 a (6 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{8 b^4}+\frac {5 (6 A b-7 a B) x^{3/2} \sqrt {a+b x}}{12 b^3}-\frac {(6 A b-7 a B) x^{5/2} \sqrt {a+b x}}{3 a b^2}+\frac {\left (5 a^2 (6 A b-7 a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{8 b^4} \\ & = \frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}-\frac {5 a (6 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{8 b^4}+\frac {5 (6 A b-7 a B) x^{3/2} \sqrt {a+b x}}{12 b^3}-\frac {(6 A b-7 a B) x^{5/2} \sqrt {a+b x}}{3 a b^2}+\frac {\left (5 a^2 (6 A b-7 a B)\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^4} \\ & = \frac {2 (A b-a B) x^{7/2}}{a b \sqrt {a+b x}}-\frac {5 a (6 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{8 b^4}+\frac {5 (6 A b-7 a B) x^{3/2} \sqrt {a+b x}}{12 b^3}-\frac {(6 A b-7 a B) x^{5/2} \sqrt {a+b x}}{3 a b^2}+\frac {5 a^2 (6 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.75 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {x} \left (105 a^3 B+4 b^3 x^2 (3 A+2 B x)-2 a b^2 x (15 A+7 B x)+a^2 (-90 A b+35 b B x)\right )}{24 b^4 \sqrt {a+b x}}+\frac {5 a^2 (-6 A b+7 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{4 b^{9/2}} \]

[In]

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

[Out]

(Sqrt[x]*(105*a^3*B + 4*b^3*x^2*(3*A + 2*B*x) - 2*a*b^2*x*(15*A + 7*B*x) + a^2*(-90*A*b + 35*b*B*x)))/(24*b^4*
Sqrt[a + b*x]) + (5*a^2*(-6*A*b + 7*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqrt[a + b*x])])/(4*b^(9/2))

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.11

method	result	size
risch	\(-\frac {\left (-8 b^{2} B \,x^{2}-12 A \,b^{2} x +22 B a b x +42 a b A -57 a^{2} B \right ) \sqrt {x}\, \sqrt {b x +a}}{24 b^{4}}+\frac {a^{2} \left (30 A \sqrt {b}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )-\frac {35 B a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{\sqrt {b}}-\frac {32 \left (A b -B a \right ) \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{b \left (x +\frac {a}{b}\right )}\right ) \sqrt {x \left (b x +a \right )}}{16 b^{4} \sqrt {x}\, \sqrt {b x +a}}\)	\(186\)
default	\(\frac {\left (16 B \,b^{\frac {7}{2}} x^{3} \sqrt {x \left (b x +a \right )}+24 A \,b^{\frac {7}{2}} x^{2} \sqrt {x \left (b x +a \right )}-28 B a \,b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+90 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b^{2} x -60 A \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} a x -105 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3} b x +70 B \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{2} x +90 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3} b -180 A \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{2}-105 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{4}+210 B \sqrt {x \left (b x +a \right )}\, \sqrt {b}\, a^{3}\right ) \sqrt {x}}{48 b^{\frac {9}{2}} \sqrt {x \left (b x +a \right )}\, \sqrt {b x +a}}\)	\(288\)

[In]

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

[Out]

-1/24*(-8*B*b^2*x^2-12*A*b^2*x+22*B*a*b*x+42*A*a*b-57*B*a^2)*x^(1/2)*(b*x+a)^(1/2)/b^4+1/16*a^2/b^4*(30*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)-32*(A*b
-B*a)/b/(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)

none

Time = 0.24 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.83 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\left [-\frac {15 \, {\left (7 \, B a^{4} - 6 \, A a^{3} b + {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (8 \, B b^{4} x^{3} + 105 \, B a^{3} b - 90 \, A a^{2} b^{2} - 2 \, {\left (7 \, B a b^{3} - 6 \, A b^{4}\right )} x^{2} + 5 \, {\left (7 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{48 \, {\left (b^{6} x + a b^{5}\right )}}, \frac {15 \, {\left (7 \, B a^{4} - 6 \, A a^{3} b + {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (8 \, B b^{4} x^{3} + 105 \, B a^{3} b - 90 \, A a^{2} b^{2} - 2 \, {\left (7 \, B a b^{3} - 6 \, A b^{4}\right )} x^{2} + 5 \, {\left (7 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, {\left (b^{6} x + a b^{5}\right )}}\right ] \]

[In]

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

[Out]

[-1/48*(15*(7*B*a^4 - 6*A*a^3*b + (7*B*a^3*b - 6*A*a^2*b^2)*x)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqr
t(x) + a) - 2*(8*B*b^4*x^3 + 105*B*a^3*b - 90*A*a^2*b^2 - 2*(7*B*a*b^3 - 6*A*b^4)*x^2 + 5*(7*B*a^2*b^2 - 6*A*a
*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^6*x + a*b^5), 1/24*(15*(7*B*a^4 - 6*A*a^3*b + (7*B*a^3*b - 6*A*a^2*b^2)*x)*
sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (8*B*b^4*x^3 + 105*B*a^3*b - 90*A*a^2*b^2 - 2*(7*B*a*b^3
- 6*A*b^4)*x^2 + 5*(7*B*a^2*b^2 - 6*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^6*x + a*b^5)]

Sympy [A] (verification not implemented)

Time = 76.71 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.46 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=A \left (- \frac {15 a^{\frac {3}{2}} \sqrt {x}}{4 b^{3} \sqrt {1 + \frac {b x}{a}}} - \frac {5 \sqrt {a} x^{\frac {3}{2}}}{4 b^{2} \sqrt {1 + \frac {b x}{a}}} + \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 b^{\frac {7}{2}}} + \frac {x^{\frac {5}{2}}}{2 \sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) + B \left (\frac {35 a^{\frac {5}{2}} \sqrt {x}}{8 b^{4} \sqrt {1 + \frac {b x}{a}}} + \frac {35 a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 b^{3} \sqrt {1 + \frac {b x}{a}}} - \frac {7 \sqrt {a} x^{\frac {5}{2}}}{12 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {35 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 b^{\frac {9}{2}}} + \frac {x^{\frac {7}{2}}}{3 \sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) \]

[In]

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

[Out]

A*(-15*a**(3/2)*sqrt(x)/(4*b**3*sqrt(1 + b*x/a)) - 5*sqrt(a)*x**(3/2)/(4*b**2*sqrt(1 + b*x/a)) + 15*a**2*asinh
(sqrt(b)*sqrt(x)/sqrt(a))/(4*b**(7/2)) + x**(5/2)/(2*sqrt(a)*b*sqrt(1 + b*x/a))) + B*(35*a**(5/2)*sqrt(x)/(8*b
**4*sqrt(1 + b*x/a)) + 35*a**(3/2)*x**(3/2)/(24*b**3*sqrt(1 + b*x/a)) - 7*sqrt(a)*x**(5/2)/(12*b**2*sqrt(1 + b
*x/a)) - 35*a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(9/2)) + x**(7/2)/(3*sqrt(a)*b*sqrt(1 + b*x/a)))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.27 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {B x^{4}}{3 \, \sqrt {b x^{2} + a x} b} - \frac {7 \, B a x^{3}}{12 \, \sqrt {b x^{2} + a x} b^{2}} + \frac {A x^{3}}{2 \, \sqrt {b x^{2} + a x} b} + \frac {35 \, B a^{2} x^{2}}{24 \, \sqrt {b x^{2} + a x} b^{3}} - \frac {5 \, A a x^{2}}{4 \, \sqrt {b x^{2} + a x} b^{2}} + \frac {35 \, B a^{3} x}{8 \, \sqrt {b x^{2} + a x} b^{4}} - \frac {15 \, A a^{2} x}{4 \, \sqrt {b x^{2} + a x} b^{3}} - \frac {35 \, B a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {9}{2}}} + \frac {15 \, A a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {7}{2}}} \]

[In]

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

[Out]

1/3*B*x^4/(sqrt(b*x^2 + a*x)*b) - 7/12*B*a*x^3/(sqrt(b*x^2 + a*x)*b^2) + 1/2*A*x^3/(sqrt(b*x^2 + a*x)*b) + 35/
24*B*a^2*x^2/(sqrt(b*x^2 + a*x)*b^3) - 5/4*A*a*x^2/(sqrt(b*x^2 + a*x)*b^2) + 35/8*B*a^3*x/(sqrt(b*x^2 + a*x)*b
^4) - 15/4*A*a^2*x/(sqrt(b*x^2 + a*x)*b^3) - 35/16*B*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(9/2)
+ 15/8*A*a^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2)

Giac [A] (verification not implemented)

none

Time = 15.61 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.24 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {1}{24} \, \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} B {\left | b \right |}}{b^{6}} - \frac {19 \, B a b^{17} {\left | b \right |} - 6 \, A b^{18} {\left | b \right |}}{b^{23}}\right )} + \frac {3 \, {\left (29 \, B a^{2} b^{17} {\left | b \right |} - 18 \, A a b^{18} {\left | b \right |}\right )}}{b^{23}}\right )} + \frac {5 \, {\left (7 \, B a^{3} {\left | b \right |} - 6 \, A a^{2} b {\left | b \right |}\right )} \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{16 \, b^{\frac {11}{2}}} + \frac {4 \, {\left (B a^{4} {\left | b \right |} - A a^{3} b {\left | b \right |}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} b^{\frac {9}{2}}} \]

[In]

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

[Out]

1/24*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)*B*abs(b)/b^6 - (19*B*a*b^17*abs(b) - 6*A*
b^18*abs(b))/b^23) + 3*(29*B*a^2*b^17*abs(b) - 18*A*a*b^18*abs(b))/b^23) + 5/16*(7*B*a^3*abs(b) - 6*A*a^2*b*ab
s(b))*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/b^(11/2) + 4*(B*a^4*abs(b) - A*a^3*b*abs(b))/((
(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)*b^(9/2))

Mupad [F(-1)]

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

[In]

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

[Out]

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