\(\int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx\) [503]

Optimal result

Integrand size = 20, antiderivative size = 192 \[ \int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx=\frac {a^3 (10 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{128 b^2}+\frac {a^2 (10 A b-3 a B) x^{3/2} \sqrt {a+b x}}{64 b}+\frac {a (10 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{48 b}+\frac {(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac {B x^{3/2} (a+b x)^{7/2}}{5 b}-\frac {a^4 (10 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{5/2}} \]

[Out]

1/48*a*(10*A*b-3*B*a)*x^(3/2)*(b*x+a)^(3/2)/b+1/40*(10*A*b-3*B*a)*x^(3/2)*(b*x+a)^(5/2)/b+1/5*B*x^(3/2)*(b*x+a
)^(7/2)/b-1/128*a^4*(10*A*b-3*B*a)*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(5/2)+1/64*a^2*(10*A*b-3*B*a)*x^(3
/2)*(b*x+a)^(1/2)/b+1/128*a^3*(10*A*b-3*B*a)*x^(1/2)*(b*x+a)^(1/2)/b^2

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} \[ \int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx=-\frac {a^4 (10 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{5/2}}+\frac {a^3 \sqrt {x} \sqrt {a+b x} (10 A b-3 a B)}{128 b^2}+\frac {a^2 x^{3/2} \sqrt {a+b x} (10 A b-3 a B)}{64 b}+\frac {a x^{3/2} (a+b x)^{3/2} (10 A b-3 a B)}{48 b}+\frac {x^{3/2} (a+b x)^{5/2} (10 A b-3 a B)}{40 b}+\frac {B x^{3/2} (a+b x)^{7/2}}{5 b} \]

[In]

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

[Out]

(a^3*(10*A*b - 3*a*B)*Sqrt[x]*Sqrt[a + b*x])/(128*b^2) + (a^2*(10*A*b - 3*a*B)*x^(3/2)*Sqrt[a + b*x])/(64*b) +
(a*(10*A*b - 3*a*B)*x^(3/2)*(a + b*x)^(3/2))/(48*b) + ((10*A*b - 3*a*B)*x^(3/2)*(a + b*x)^(5/2))/(40*b) + (B*
x^(3/2)*(a + b*x)^(7/2))/(5*b) - (a^4*(10*A*b - 3*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(128*b^(5/2))

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^{3/2} (a+b x)^{7/2}}{5 b}+\frac {\left (5 A b-\frac {3 a B}{2}\right ) \int \sqrt {x} (a+b x)^{5/2} \, dx}{5 b} \\ & = \frac {(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac {B x^{3/2} (a+b x)^{7/2}}{5 b}+\frac {(a (10 A b-3 a B)) \int \sqrt {x} (a+b x)^{3/2} \, dx}{16 b} \\ & = \frac {a (10 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{48 b}+\frac {(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac {B x^{3/2} (a+b x)^{7/2}}{5 b}+\frac {\left (a^2 (10 A b-3 a B)\right ) \int \sqrt {x} \sqrt {a+b x} \, dx}{32 b} \\ & = \frac {a^2 (10 A b-3 a B) x^{3/2} \sqrt {a+b x}}{64 b}+\frac {a (10 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{48 b}+\frac {(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac {B x^{3/2} (a+b x)^{7/2}}{5 b}+\frac {\left (a^3 (10 A b-3 a B)\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{128 b} \\ & = \frac {a^3 (10 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{128 b^2}+\frac {a^2 (10 A b-3 a B) x^{3/2} \sqrt {a+b x}}{64 b}+\frac {a (10 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{48 b}+\frac {(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac {B x^{3/2} (a+b x)^{7/2}}{5 b}-\frac {\left (a^4 (10 A b-3 a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{256 b^2} \\ & = \frac {a^3 (10 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{128 b^2}+\frac {a^2 (10 A b-3 a B) x^{3/2} \sqrt {a+b x}}{64 b}+\frac {a (10 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{48 b}+\frac {(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac {B x^{3/2} (a+b x)^{7/2}}{5 b}-\frac {\left (a^4 (10 A b-3 a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{128 b^2} \\ & = \frac {a^3 (10 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{128 b^2}+\frac {a^2 (10 A b-3 a B) x^{3/2} \sqrt {a+b x}}{64 b}+\frac {a (10 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{48 b}+\frac {(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac {B x^{3/2} (a+b x)^{7/2}}{5 b}-\frac {\left (a^4 (10 A b-3 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^2} \\ & = \frac {a^3 (10 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{128 b^2}+\frac {a^2 (10 A b-3 a B) x^{3/2} \sqrt {a+b x}}{64 b}+\frac {a (10 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{48 b}+\frac {(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac {B x^{3/2} (a+b x)^{7/2}}{5 b}-\frac {a^4 (10 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.91 \[ \int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx=\frac {\sqrt {b} \sqrt {x} \sqrt {a+b x} \left (-45 a^4 B+30 a^3 b (5 A+B x)+96 b^4 x^3 (5 A+4 B x)+16 a b^3 x^2 (85 A+63 B x)+4 a^2 b^2 x (295 A+186 B x)\right )+300 a^4 A b \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )+90 a^5 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{1920 b^{5/2}} \]

[In]

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

[Out]

(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x]*(-45*a^4*B + 30*a^3*b*(5*A + B*x) + 96*b^4*x^3*(5*A + 4*B*x) + 16*a*b^3*x^2*(85
*A + 63*B*x) + 4*a^2*b^2*x*(295*A + 186*B*x)) + 300*a^4*A*b*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqrt[a + b*x]
)] + 90*a^5*B*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(1920*b^(5/2))

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {\left (384 B \,x^{4} b^{4}+480 A \,x^{3} b^{4}+1008 B \,x^{3} a \,b^{3}+1360 A \,x^{2} a \,b^{3}+744 B \,x^{2} a^{2} b^{2}+1180 A x \,a^{2} b^{2}+30 B x \,a^{3} b +150 A \,a^{3} b -45 B \,a^{4}\right ) \sqrt {x}\, \sqrt {b x +a}}{1920 b^{2}}-\frac {a^{4} \left (10 A b -3 B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{256 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(159\)
default	\(-\frac {\sqrt {b x +a}\, \sqrt {x}\, \left (-768 B \,b^{\frac {9}{2}} x^{4} \sqrt {x \left (b x +a \right )}-960 A \,b^{\frac {9}{2}} x^{3} \sqrt {x \left (b x +a \right )}-2016 B a \,b^{\frac {7}{2}} x^{3} \sqrt {x \left (b x +a \right )}-2720 A a \,b^{\frac {7}{2}} x^{2} \sqrt {x \left (b x +a \right )}-1488 B \,a^{2} b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}-2360 A \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, a^{2} x -60 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a^{3} x +150 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{4} b -300 A \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a^{3}-45 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{5}+90 B \sqrt {b}\, \sqrt {x \left (b x +a \right )}\, a^{4}\right )}{3840 b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}}\)	\(260\)

[In]

int((b*x+a)^(5/2)*(B*x+A)*x^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/1920/b^2*(384*B*b^4*x^4+480*A*b^4*x^3+1008*B*a*b^3*x^3+1360*A*a*b^3*x^2+744*B*a^2*b^2*x^2+1180*A*a^2*b^2*x+3
0*B*a^3*b*x+150*A*a^3*b-45*B*a^4)*x^(1/2)*(b*x+a)^(1/2)-1/256*a^4/b^(5/2)*(10*A*b-3*B*a)*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 = 0.23 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.55 \[ \int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx=\left [-\frac {15 \, {\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (384 \, B b^{5} x^{4} - 45 \, B a^{4} b + 150 \, A a^{3} b^{2} + 48 \, {\left (21 \, B a b^{4} + 10 \, A b^{5}\right )} x^{3} + 8 \, {\left (93 \, B a^{2} b^{3} + 170 \, A a b^{4}\right )} x^{2} + 10 \, {\left (3 \, B a^{3} b^{2} + 118 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{3840 \, b^{3}}, -\frac {15 \, {\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (384 \, B b^{5} x^{4} - 45 \, B a^{4} b + 150 \, A a^{3} b^{2} + 48 \, {\left (21 \, B a b^{4} + 10 \, A b^{5}\right )} x^{3} + 8 \, {\left (93 \, B a^{2} b^{3} + 170 \, A a b^{4}\right )} x^{2} + 10 \, {\left (3 \, B a^{3} b^{2} + 118 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{1920 \, b^{3}}\right ] \]

[In]

integrate((b*x+a)^(5/2)*(B*x+A)*x^(1/2),x, algorithm="fricas")

[Out]

[-1/3840*(15*(3*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 - 45*B*a^4*b + 150*A*a^3*b^2 + 48*(21*B*a*b^4 + 10*A*b^5)*x^3 + 8*(93*B*a^2*b^3 + 170*A*a*b^4)*x^2 + 10*(3*B
*a^3*b^2 + 118*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^3, -1/1920*(15*(3*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 - 45*B*a^4*b + 150*A*a^3*b^2 + 48*(21*B*a*b^4 + 10*A*b^5)*x
^3 + 8*(93*B*a^2*b^3 + 170*A*a*b^4)*x^2 + 10*(3*B*a^3*b^2 + 118*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^3]

Sympy [A] (verification not implemented)

Time = 2.72 (sec) , antiderivative size = 748, normalized size of antiderivative = 3.90 \[ \int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx=\text {Too large to display} \]

[In]

integrate((b*x+a)**(5/2)*(B*x+A)*x**(1/2),x)

[Out]

2*A*a**2*Piecewise((-a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*l
og(sqrt(x))/sqrt(b*x), True))/(8*b) + sqrt(a + b*x)*(a*sqrt(x)/(8*b) + x**(3/2)/4), Ne(b, 0)), (sqrt(a)*x**(3/
2)/3, True)) + 4*A*a*b*Piecewise((a**3*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)
), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(16*b**2) + sqrt(a + b*x)*(-a**2*sqrt(x)/(16*b**2) + a*x**(3/2)/(24
*b) + x**(5/2)/6), Ne(b, 0)), (sqrt(a)*x**(5/2)/5, True)) + 2*A*b**2*Piecewise((-5*a**4*Piecewise((log(2*sqrt(
b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(128*b**3) + sqrt(
a + b*x)*(5*a**3*sqrt(x)/(128*b**3) - 5*a**2*x**(3/2)/(192*b**2) + a*x**(5/2)/(48*b) + x**(7/2)/8), Ne(b, 0)),
(sqrt(a)*x**(7/2)/7, True)) + 2*B*a**2*Piecewise((a**3*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/
sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(16*b**2) + sqrt(a + b*x)*(-a**2*sqrt(x)/(16*b**2)
+ a*x**(3/2)/(24*b) + x**(5/2)/6), Ne(b, 0)), (sqrt(a)*x**(5/2)/5, True)) + 4*B*a*b*Piecewise((-5*a**4*Piecew
ise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(1
28*b**3) + sqrt(a + b*x)*(5*a**3*sqrt(x)/(128*b**3) - 5*a**2*x**(3/2)/(192*b**2) + a*x**(5/2)/(48*b) + x**(7/2
)/8), Ne(b, 0)), (sqrt(a)*x**(7/2)/7, True)) + 2*B*b**2*Piecewise((7*a**5*Piecewise((log(2*sqrt(b)*sqrt(a + b*
x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(256*b**4) + sqrt(a + b*x)*(-7*a
**4*sqrt(x)/(256*b**4) + 7*a**3*x**(3/2)/(384*b**3) - 7*a**2*x**(5/2)/(480*b**2) + a*x**(7/2)/(80*b) + x**(9/2
)/10), Ne(b, 0)), (sqrt(a)*x**(9/2)/9, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 485 vs. \(2 (152) = 304\).

Time = 0.21 (sec) , antiderivative size = 485, normalized size of antiderivative = 2.53 \[ \int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx=\frac {1}{5} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B b x^{2} - \frac {7}{40} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a x + \frac {1}{2} \, \sqrt {b x^{2} + a x} A a^{2} x - \frac {7 \, \sqrt {b x^{2} + a x} B a^{3} x}{64 \, b} + \frac {7 \, B a^{5} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{256 \, b^{\frac {5}{2}}} - \frac {A a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {3}{2}}} - \frac {7 \, \sqrt {b x^{2} + a x} B a^{4}}{128 \, b^{2}} + \frac {7 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a^{2}}{48 \, b} + \frac {\sqrt {b x^{2} + a x} A a^{3}}{4 \, b} + \frac {5 \, {\left (2 \, B a b + A b^{2}\right )} \sqrt {b x^{2} + a x} a^{2} x}{32 \, b^{2}} + \frac {{\left (2 \, B a b + A b^{2}\right )} {\left (b x^{2} + a x\right )}^{\frac {3}{2}} x}{4 \, b} - \frac {{\left (B a^{2} + 2 \, A a b\right )} \sqrt {b x^{2} + a x} a x}{4 \, b} - \frac {5 \, {\left (2 \, B a b + A b^{2}\right )} a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {7}{2}}} + \frac {{\left (B a^{2} + 2 \, A a b\right )} a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {5}{2}}} + \frac {5 \, {\left (2 \, B a b + A b^{2}\right )} \sqrt {b x^{2} + a x} a^{3}}{64 \, b^{3}} - \frac {5 \, {\left (2 \, B a b + A b^{2}\right )} {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a}{24 \, b^{2}} - \frac {{\left (B a^{2} + 2 \, A a b\right )} \sqrt {b x^{2} + a x} a^{2}}{8 \, b^{2}} + \frac {{\left (B a^{2} + 2 \, A a b\right )} {\left (b x^{2} + a x\right )}^{\frac {3}{2}}}{3 \, b} \]

[In]

integrate((b*x+a)^(5/2)*(B*x+A)*x^(1/2),x, algorithm="maxima")

[Out]

1/5*(b*x^2 + a*x)^(3/2)*B*b*x^2 - 7/40*(b*x^2 + a*x)^(3/2)*B*a*x + 1/2*sqrt(b*x^2 + a*x)*A*a^2*x - 7/64*sqrt(b
*x^2 + a*x)*B*a^3*x/b + 7/256*B*a^5*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2) - 1/8*A*a^4*log(2*b*x
+ a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2) - 7/128*sqrt(b*x^2 + a*x)*B*a^4/b^2 + 7/48*(b*x^2 + a*x)^(3/2)*B*a
^2/b + 1/4*sqrt(b*x^2 + a*x)*A*a^3/b + 5/32*(2*B*a*b + A*b^2)*sqrt(b*x^2 + a*x)*a^2*x/b^2 + 1/4*(2*B*a*b + A*b
^2)*(b*x^2 + a*x)^(3/2)*x/b - 1/4*(B*a^2 + 2*A*a*b)*sqrt(b*x^2 + a*x)*a*x/b - 5/128*(2*B*a*b + A*b^2)*a^4*log(
2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) + 1/16*(B*a^2 + 2*A*a*b)*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a
*x)*sqrt(b))/b^(5/2) + 5/64*(2*B*a*b + A*b^2)*sqrt(b*x^2 + a*x)*a^3/b^3 - 5/24*(2*B*a*b + A*b^2)*(b*x^2 + a*x)
^(3/2)*a/b^2 - 1/8*(B*a^2 + 2*A*a*b)*sqrt(b*x^2 + a*x)*a^2/b^2 + 1/3*(B*a^2 + 2*A*a*b)*(b*x^2 + a*x)^(3/2)/b

Giac [F(-1)]

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

[In]

integrate((b*x+a)^(5/2)*(B*x+A)*x^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

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

[In]

int(x^(1/2)*(A + B*x)*(a + b*x)^(5/2),x)

[Out]

int(x^(1/2)*(A + B*x)*(a + b*x)^(5/2), x)