\(\int \frac {x^{7/2} (A+B x)}{\sqrt {a+b x}} \, dx\) [514]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 20, antiderivative size = 192 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a+b x}} \, dx=-\frac {7 a^3 (10 A b-9 a B) \sqrt {x} \sqrt {a+b x}}{128 b^5}+\frac {7 a^2 (10 A b-9 a B) x^{3/2} \sqrt {a+b x}}{192 b^4}-\frac {7 a (10 A b-9 a B) x^{5/2} \sqrt {a+b x}}{240 b^3}+\frac {(10 A b-9 a B) x^{7/2} \sqrt {a+b x}}{40 b^2}+\frac {B x^{9/2} \sqrt {a+b x}}{5 b}+\frac {7 a^4 (10 A b-9 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{11/2}} \]

[Out]

7/128*a^4*(10*A*b-9*B*a)*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(11/2)+7/192*a^2*(10*A*b-9*B*a)*x^(3/2)*(b*x
+a)^(1/2)/b^4-7/240*a*(10*A*b-9*B*a)*x^(5/2)*(b*x+a)^(1/2)/b^3+1/40*(10*A*b-9*B*a)*x^(7/2)*(b*x+a)^(1/2)/b^2+1
/5*B*x^(9/2)*(b*x+a)^(1/2)/b-7/128*a^3*(10*A*b-9*B*a)*x^(1/2)*(b*x+a)^(1/2)/b^5

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {81, 52, 65, 223, 212} \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {7 a^4 (10 A b-9 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{11/2}}-\frac {7 a^3 \sqrt {x} \sqrt {a+b x} (10 A b-9 a B)}{128 b^5}+\frac {7 a^2 x^{3/2} \sqrt {a+b x} (10 A b-9 a B)}{192 b^4}-\frac {7 a x^{5/2} \sqrt {a+b x} (10 A b-9 a B)}{240 b^3}+\frac {x^{7/2} \sqrt {a+b x} (10 A b-9 a B)}{40 b^2}+\frac {B x^{9/2} \sqrt {a+b x}}{5 b} \]

[In]

Int[(x^(7/2)*(A + B*x))/Sqrt[a + b*x],x]

[Out]

(-7*a^3*(10*A*b - 9*a*B)*Sqrt[x]*Sqrt[a + b*x])/(128*b^5) + (7*a^2*(10*A*b - 9*a*B)*x^(3/2)*Sqrt[a + b*x])/(19
2*b^4) - (7*a*(10*A*b - 9*a*B)*x^(5/2)*Sqrt[a + b*x])/(240*b^3) + ((10*A*b - 9*a*B)*x^(7/2)*Sqrt[a + b*x])/(40
*b^2) + (B*x^(9/2)*Sqrt[a + b*x])/(5*b) + (7*a^4*(10*A*b - 9*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(1
28*b^(11/2))

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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 {B x^{9/2} \sqrt {a+b x}}{5 b}+\frac {\left (5 A b-\frac {9 a B}{2}\right ) \int \frac {x^{7/2}}{\sqrt {a+b x}} \, dx}{5 b} \\ & = \frac {(10 A b-9 a B) x^{7/2} \sqrt {a+b x}}{40 b^2}+\frac {B x^{9/2} \sqrt {a+b x}}{5 b}-\frac {(7 a (10 A b-9 a B)) \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx}{80 b^2} \\ & = -\frac {7 a (10 A b-9 a B) x^{5/2} \sqrt {a+b x}}{240 b^3}+\frac {(10 A b-9 a B) x^{7/2} \sqrt {a+b x}}{40 b^2}+\frac {B x^{9/2} \sqrt {a+b x}}{5 b}+\frac {\left (7 a^2 (10 A b-9 a B)\right ) \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{96 b^3} \\ & = \frac {7 a^2 (10 A b-9 a B) x^{3/2} \sqrt {a+b x}}{192 b^4}-\frac {7 a (10 A b-9 a B) x^{5/2} \sqrt {a+b x}}{240 b^3}+\frac {(10 A b-9 a B) x^{7/2} \sqrt {a+b x}}{40 b^2}+\frac {B x^{9/2} \sqrt {a+b x}}{5 b}-\frac {\left (7 a^3 (10 A b-9 a B)\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{128 b^4} \\ & = -\frac {7 a^3 (10 A b-9 a B) \sqrt {x} \sqrt {a+b x}}{128 b^5}+\frac {7 a^2 (10 A b-9 a B) x^{3/2} \sqrt {a+b x}}{192 b^4}-\frac {7 a (10 A b-9 a B) x^{5/2} \sqrt {a+b x}}{240 b^3}+\frac {(10 A b-9 a B) x^{7/2} \sqrt {a+b x}}{40 b^2}+\frac {B x^{9/2} \sqrt {a+b x}}{5 b}+\frac {\left (7 a^4 (10 A b-9 a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{256 b^5} \\ & = -\frac {7 a^3 (10 A b-9 a B) \sqrt {x} \sqrt {a+b x}}{128 b^5}+\frac {7 a^2 (10 A b-9 a B) x^{3/2} \sqrt {a+b x}}{192 b^4}-\frac {7 a (10 A b-9 a B) x^{5/2} \sqrt {a+b x}}{240 b^3}+\frac {(10 A b-9 a B) x^{7/2} \sqrt {a+b x}}{40 b^2}+\frac {B x^{9/2} \sqrt {a+b x}}{5 b}+\frac {\left (7 a^4 (10 A b-9 a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{128 b^5} \\ & = -\frac {7 a^3 (10 A b-9 a B) \sqrt {x} \sqrt {a+b x}}{128 b^5}+\frac {7 a^2 (10 A b-9 a B) x^{3/2} \sqrt {a+b x}}{192 b^4}-\frac {7 a (10 A b-9 a B) x^{5/2} \sqrt {a+b x}}{240 b^3}+\frac {(10 A b-9 a B) x^{7/2} \sqrt {a+b x}}{40 b^2}+\frac {B x^{9/2} \sqrt {a+b x}}{5 b}+\frac {\left (7 a^4 (10 A b-9 a B)\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^5} \\ & = -\frac {7 a^3 (10 A b-9 a B) \sqrt {x} \sqrt {a+b x}}{128 b^5}+\frac {7 a^2 (10 A b-9 a B) x^{3/2} \sqrt {a+b x}}{192 b^4}-\frac {7 a (10 A b-9 a B) x^{5/2} \sqrt {a+b x}}{240 b^3}+\frac {(10 A b-9 a B) x^{7/2} \sqrt {a+b x}}{40 b^2}+\frac {B x^{9/2} \sqrt {a+b x}}{5 b}+\frac {7 a^4 (10 A b-9 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{11/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.76 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {\sqrt {x} \sqrt {a+b x} \left (945 a^4 B-210 a^3 b (5 A+3 B x)+96 b^4 x^3 (5 A+4 B x)+28 a^2 b^2 x (25 A+18 B x)-16 a b^3 x^2 (35 A+27 B x)\right )}{1920 b^5}+\frac {7 a^4 (-10 A b+9 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{64 b^{11/2}} \]

[In]

Integrate[(x^(7/2)*(A + B*x))/Sqrt[a + b*x],x]

[Out]

(Sqrt[x]*Sqrt[a + b*x]*(945*a^4*B - 210*a^3*b*(5*A + 3*B*x) + 96*b^4*x^3*(5*A + 4*B*x) + 28*a^2*b^2*x*(25*A +
18*B*x) - 16*a*b^3*x^2*(35*A + 27*B*x)))/(1920*b^5) + (7*a^4*(-10*A*b + 9*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt
[a] - Sqrt[a + b*x])])/(64*b^(11/2))

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.83

method result size
risch \(-\frac {\left (-384 B \,x^{4} b^{4}-480 A \,x^{3} b^{4}+432 B \,x^{3} a \,b^{3}+560 A \,x^{2} a \,b^{3}-504 B \,x^{2} a^{2} b^{2}-700 A x \,a^{2} b^{2}+630 B x \,a^{3} b +1050 A \,a^{3} b -945 B \,a^{4}\right ) \sqrt {x}\, \sqrt {b x +a}}{1920 b^{5}}+\frac {7 a^{4} \left (10 A b -9 B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{256 b^{\frac {11}{2}} \sqrt {x}\, \sqrt {b x +a}}\) \(159\)
default \(\frac {\sqrt {x}\, \sqrt {b x +a}\, \left (768 B \,b^{\frac {9}{2}} x^{4} \sqrt {x \left (b x +a \right )}+960 A \,b^{\frac {9}{2}} x^{3} \sqrt {x \left (b x +a \right )}-864 B a \,b^{\frac {7}{2}} x^{3} \sqrt {x \left (b x +a \right )}-1120 A a \,b^{\frac {7}{2}} x^{2} \sqrt {x \left (b x +a \right )}+1008 B \,a^{2} b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+1400 A \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, a^{2} x -1260 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a^{3} x +1050 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{4} b -2100 A \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a^{3}-945 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{5}+1890 B \sqrt {b}\, \sqrt {x \left (b x +a \right )}\, a^{4}\right )}{3840 b^{\frac {11}{2}} \sqrt {x \left (b x +a \right )}}\) \(260\)

[In]

int(x^(7/2)*(B*x+A)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/1920*(-384*B*b^4*x^4-480*A*b^4*x^3+432*B*a*b^3*x^3+560*A*a*b^3*x^2-504*B*a^2*b^2*x^2-700*A*a^2*b^2*x+630*B*
a^3*b*x+1050*A*a^3*b-945*B*a^4)*x^(1/2)*(b*x+a)^(1/2)/b^5+7/256*a^4*(10*A*b-9*B*a)/b^(11/2)*ln((1/2*a+b*x)/b^(
1/2)+(b*x^2+a*x)^(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 = 296, normalized size of antiderivative = 1.54 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a+b x}} \, dx=\left [-\frac {105 \, {\left (9 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (384 \, B b^{5} x^{4} + 945 \, B a^{4} b - 1050 \, A a^{3} b^{2} - 48 \, {\left (9 \, B a b^{4} - 10 \, A b^{5}\right )} x^{3} + 56 \, {\left (9 \, B a^{2} b^{3} - 10 \, A a b^{4}\right )} x^{2} - 70 \, {\left (9 \, B a^{3} b^{2} - 10 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{3840 \, b^{6}}, \frac {105 \, {\left (9 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (384 \, B b^{5} x^{4} + 945 \, B a^{4} b - 1050 \, A a^{3} b^{2} - 48 \, {\left (9 \, B a b^{4} - 10 \, A b^{5}\right )} x^{3} + 56 \, {\left (9 \, B a^{2} b^{3} - 10 \, A a b^{4}\right )} x^{2} - 70 \, {\left (9 \, B a^{3} b^{2} - 10 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{1920 \, b^{6}}\right ] \]

[In]

integrate(x^(7/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/3840*(105*(9*B*a^5 - 10*A*a^4*b)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(384*B*b^5*x
^4 + 945*B*a^4*b - 1050*A*a^3*b^2 - 48*(9*B*a*b^4 - 10*A*b^5)*x^3 + 56*(9*B*a^2*b^3 - 10*A*a*b^4)*x^2 - 70*(9*
B*a^3*b^2 - 10*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^6, 1/1920*(105*(9*B*a^5 - 10*A*a^4*b)*sqrt(-b)*arctan(sq
rt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (384*B*b^5*x^4 + 945*B*a^4*b - 1050*A*a^3*b^2 - 48*(9*B*a*b^4 - 10*A*b^5)*
x^3 + 56*(9*B*a^2*b^3 - 10*A*a*b^4)*x^2 - 70*(9*B*a^3*b^2 - 10*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^6]

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a+b x}} \, dx=\text {Timed out} \]

[In]

integrate(x**(7/2)*(B*x+A)/(b*x+a)**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.31 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {\sqrt {b x^{2} + a x} B x^{4}}{5 \, b} - \frac {9 \, \sqrt {b x^{2} + a x} B a x^{3}}{40 \, b^{2}} + \frac {\sqrt {b x^{2} + a x} A x^{3}}{4 \, b} + \frac {21 \, \sqrt {b x^{2} + a x} B a^{2} x^{2}}{80 \, b^{3}} - \frac {7 \, \sqrt {b x^{2} + a x} A a x^{2}}{24 \, b^{2}} - \frac {21 \, \sqrt {b x^{2} + a x} B a^{3} x}{64 \, b^{4}} + \frac {35 \, \sqrt {b x^{2} + a x} A a^{2} x}{96 \, b^{3}} - \frac {63 \, B a^{5} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{256 \, b^{\frac {11}{2}}} + \frac {35 \, A a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {9}{2}}} + \frac {63 \, \sqrt {b x^{2} + a x} B a^{4}}{128 \, b^{5}} - \frac {35 \, \sqrt {b x^{2} + a x} A a^{3}}{64 \, b^{4}} \]

[In]

integrate(x^(7/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

1/5*sqrt(b*x^2 + a*x)*B*x^4/b - 9/40*sqrt(b*x^2 + a*x)*B*a*x^3/b^2 + 1/4*sqrt(b*x^2 + a*x)*A*x^3/b + 21/80*sqr
t(b*x^2 + a*x)*B*a^2*x^2/b^3 - 7/24*sqrt(b*x^2 + a*x)*A*a*x^2/b^2 - 21/64*sqrt(b*x^2 + a*x)*B*a^3*x/b^4 + 35/9
6*sqrt(b*x^2 + a*x)*A*a^2*x/b^3 - 63/256*B*a^5*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(11/2) + 35/128*
A*a^4*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(9/2) + 63/128*sqrt(b*x^2 + a*x)*B*a^4/b^5 - 35/64*sqrt(b
*x^2 + a*x)*A*a^3/b^4

Giac [A] (verification not implemented)

none

Time = 152.11 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.34 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a+b x}} \, dx=-\frac {\frac {10 \, {\left (\frac {105 \, a^{4} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {5}{2}}} - {\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {6 \, {\left (b x + a\right )}}{b^{3}} - \frac {25 \, a}{b^{3}}\right )} + \frac {163 \, a^{2}}{b^{3}}\right )} - \frac {279 \, a^{3}}{b^{3}}\right )} \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a}\right )} A {\left | b \right |}}{b^{2}} - \frac {3 \, {\left (\frac {315 \, a^{5} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {7}{2}}} + {\left (2 \, {\left (4 \, {\left (b x + a\right )} {\left (2 \, {\left (b x + a\right )} {\left (\frac {8 \, {\left (b x + a\right )}}{b^{4}} - \frac {41 \, a}{b^{4}}\right )} + \frac {171 \, a^{2}}{b^{4}}\right )} - \frac {745 \, a^{3}}{b^{4}}\right )} {\left (b x + a\right )} + \frac {965 \, a^{4}}{b^{4}}\right )} \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a}\right )} B {\left | b \right |}}{b^{2}}}{1920 \, b} \]

[In]

integrate(x^(7/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

-1/1920*(10*(105*a^4*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/b^(5/2) - (2*(b*x + a)*(4*(b*x
 + a)*(6*(b*x + a)/b^3 - 25*a/b^3) + 163*a^2/b^3) - 279*a^3/b^3)*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a))*A*abs(
b)/b^2 - 3*(315*a^5*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/b^(7/2) + (2*(4*(b*x + a)*(2*(b
*x + a)*(8*(b*x + a)/b^4 - 41*a/b^4) + 171*a^2/b^4) - 745*a^3/b^4)*(b*x + a) + 965*a^4/b^4)*sqrt((b*x + a)*b -
 a*b)*sqrt(b*x + a))*B*abs(b)/b^2)/b

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a+b x}} \, dx=\int \frac {x^{7/2}\,\left (A+B\,x\right )}{\sqrt {a+b\,x}} \,d x \]

[In]

int((x^(7/2)*(A + B*x))/(a + b*x)^(1/2),x)

[Out]

int((x^(7/2)*(A + B*x))/(a + b*x)^(1/2), x)