\(\int \sqrt {x} (a+b x)^{3/2} (A+B x) \, dx\) [492]

Optimal result

Integrand size = 20, antiderivative size = 159 \[ \int \sqrt {x} (a+b x)^{3/2} (A+B x) \, dx=\frac {a^2 (8 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{64 b^2}+\frac {a (8 A b-3 a B) x^{3/2} \sqrt {a+b x}}{32 b}+\frac {(8 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{24 b}+\frac {B x^{3/2} (a+b x)^{5/2}}{4 b}-\frac {a^3 (8 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{5/2}} \]

[Out]

1/24*(8*A*b-3*B*a)*x^(3/2)*(b*x+a)^(3/2)/b+1/4*B*x^(3/2)*(b*x+a)^(5/2)/b-1/64*a^3*(8*A*b-3*B*a)*arctanh(b^(1/2
)*x^(1/2)/(b*x+a)^(1/2))/b^(5/2)+1/32*a*(8*A*b-3*B*a)*x^(3/2)*(b*x+a)^(1/2)/b+1/64*a^2*(8*A*b-3*B*a)*x^(1/2)*(
b*x+a)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 159, 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 = {81, 52, 65, 223, 212} \[ \int \sqrt {x} (a+b x)^{3/2} (A+B x) \, dx=-\frac {a^3 (8 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{5/2}}+\frac {a^2 \sqrt {x} \sqrt {a+b x} (8 A b-3 a B)}{64 b^2}+\frac {a x^{3/2} \sqrt {a+b x} (8 A b-3 a B)}{32 b}+\frac {x^{3/2} (a+b x)^{3/2} (8 A b-3 a B)}{24 b}+\frac {B x^{3/2} (a+b x)^{5/2}}{4 b} \]

[In]

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

[Out]

(a^2*(8*A*b - 3*a*B)*Sqrt[x]*Sqrt[a + b*x])/(64*b^2) + (a*(8*A*b - 3*a*B)*x^(3/2)*Sqrt[a + b*x])/(32*b) + ((8*
A*b - 3*a*B)*x^(3/2)*(a + b*x)^(3/2))/(24*b) + (B*x^(3/2)*(a + b*x)^(5/2))/(4*b) - (a^3*(8*A*b - 3*a*B)*ArcTan
h[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(64*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)^{5/2}}{4 b}+\frac {\left (4 A b-\frac {3 a B}{2}\right ) \int \sqrt {x} (a+b x)^{3/2} \, dx}{4 b} \\ & = \frac {(8 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{24 b}+\frac {B x^{3/2} (a+b x)^{5/2}}{4 b}+\frac {(a (8 A b-3 a B)) \int \sqrt {x} \sqrt {a+b x} \, dx}{16 b} \\ & = \frac {a (8 A b-3 a B) x^{3/2} \sqrt {a+b x}}{32 b}+\frac {(8 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{24 b}+\frac {B x^{3/2} (a+b x)^{5/2}}{4 b}+\frac {\left (a^2 (8 A b-3 a B)\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{64 b} \\ & = \frac {a^2 (8 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{64 b^2}+\frac {a (8 A b-3 a B) x^{3/2} \sqrt {a+b x}}{32 b}+\frac {(8 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{24 b}+\frac {B x^{3/2} (a+b x)^{5/2}}{4 b}-\frac {\left (a^3 (8 A b-3 a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{128 b^2} \\ & = \frac {a^2 (8 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{64 b^2}+\frac {a (8 A b-3 a B) x^{3/2} \sqrt {a+b x}}{32 b}+\frac {(8 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{24 b}+\frac {B x^{3/2} (a+b x)^{5/2}}{4 b}-\frac {\left (a^3 (8 A b-3 a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{64 b^2} \\ & = \frac {a^2 (8 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{64 b^2}+\frac {a (8 A b-3 a B) x^{3/2} \sqrt {a+b x}}{32 b}+\frac {(8 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{24 b}+\frac {B x^{3/2} (a+b x)^{5/2}}{4 b}-\frac {\left (a^3 (8 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 )}{64 b^2} \\ & = \frac {a^2 (8 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{64 b^2}+\frac {a (8 A b-3 a B) x^{3/2} \sqrt {a+b x}}{32 b}+\frac {(8 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{24 b}+\frac {B x^{3/2} (a+b x)^{5/2}}{4 b}-\frac {a^3 (8 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.79 \[ \int \sqrt {x} (a+b x)^{3/2} (A+B x) \, dx=\frac {\sqrt {b} \sqrt {x} \sqrt {a+b x} \left (-9 a^3 B+6 a^2 b (4 A+B x)+16 b^3 x^2 (4 A+3 B x)+8 a b^2 x (14 A+9 B x)\right )+6 a^3 (-8 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{192 b^{5/2}} \]

[In]

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

[Out]

(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x]*(-9*a^3*B + 6*a^2*b*(4*A + B*x) + 16*b^3*x^2*(4*A + 3*B*x) + 8*a*b^2*x*(14*A +
9*B*x)) + 6*a^3*(-8*A*b + 3*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(192*b^(5/2))

Maple [A] (verified)

Time = 1.42 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {\left (48 b^{3} B \,x^{3}+64 A \,b^{3} x^{2}+72 B a \,b^{2} x^{2}+112 a \,b^{2} A x +6 a^{2} b B x +24 a^{2} b A -9 a^{3} B \right ) \sqrt {x}\, \sqrt {b x +a}}{192 b^{2}}-\frac {a^{3} \left (8 A b -3 B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{128 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(135\)
default	\(-\frac {\sqrt {b x +a}\, \sqrt {x}\, \left (-96 B \,b^{\frac {7}{2}} x^{3} \sqrt {x \left (b x +a \right )}-128 A \,b^{\frac {7}{2}} x^{2} \sqrt {x \left (b x +a \right )}-144 B a \,b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}-224 A \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} a x -12 B \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{2} x +24 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3} b -48 A \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{2}-9 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{4}+18 B \sqrt {x \left (b x +a \right )}\, \sqrt {b}\, a^{3}\right )}{384 b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}}\)	\(218\)

[In]

int((b*x+a)^(3/2)*(B*x+A)*x^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/192/b^2*(48*B*b^3*x^3+64*A*b^3*x^2+72*B*a*b^2*x^2+112*A*a*b^2*x+6*B*a^2*b*x+24*A*a^2*b-9*B*a^3)*x^(1/2)*(b*x
+a)^(1/2)-1/128*a^3/b^(5/2)*(8*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 = 0.24 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.57 \[ \int \sqrt {x} (a+b x)^{3/2} (A+B x) \, dx=\left [-\frac {3 \, {\left (3 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (48 \, B b^{4} x^{3} - 9 \, B a^{3} b + 24 \, A a^{2} b^{2} + 8 \, {\left (9 \, B a b^{3} + 8 \, A b^{4}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b^{2} + 56 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{384 \, b^{3}}, -\frac {3 \, {\left (3 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (48 \, B b^{4} x^{3} - 9 \, B a^{3} b + 24 \, A a^{2} b^{2} + 8 \, {\left (9 \, B a b^{3} + 8 \, A b^{4}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b^{2} + 56 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{192 \, b^{3}}\right ] \]

[In]

integrate((b*x+a)^(3/2)*(B*x+A)*x^(1/2),x, algorithm="fricas")

[Out]

[-1/384*(3*(3*B*a^4 - 8*A*a^3*b)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(48*B*b^4*x^3 -
9*B*a^3*b + 24*A*a^2*b^2 + 8*(9*B*a*b^3 + 8*A*b^4)*x^2 + 2*(3*B*a^2*b^2 + 56*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x)
)/b^3, -1/192*(3*(3*B*a^4 - 8*A*a^3*b)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (48*B*b^4*x^3 - 9
*B*a^3*b + 24*A*a^2*b^2 + 8*(9*B*a*b^3 + 8*A*b^4)*x^2 + 2*(3*B*a^2*b^2 + 56*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x))
/b^3]

Sympy [A] (verification not implemented)

Time = 1.41 (sec) , antiderivative size = 457, normalized size of antiderivative = 2.87 \[ \int \sqrt {x} (a+b x)^{3/2} (A+B x) \, dx=2 A a \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{8 b} + \sqrt {a + b x} \left (\frac {a \sqrt {x}}{8 b} + \frac {x^{\frac {3}{2}}}{4}\right ) & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {3}{2}}}{3} & \text {otherwise} \end {cases}\right ) + 2 A b \left (\begin {cases} \frac {a^{3} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{16 b^{2}} + \sqrt {a + b x} \left (- \frac {a^{2} \sqrt {x}}{16 b^{2}} + \frac {a x^{\frac {3}{2}}}{24 b} + \frac {x^{\frac {5}{2}}}{6}\right ) & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {5}{2}}}{5} & \text {otherwise} \end {cases}\right ) + 2 B a \left (\begin {cases} \frac {a^{3} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{16 b^{2}} + \sqrt {a + b x} \left (- \frac {a^{2} \sqrt {x}}{16 b^{2}} + \frac {a x^{\frac {3}{2}}}{24 b} + \frac {x^{\frac {5}{2}}}{6}\right ) & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {5}{2}}}{5} & \text {otherwise} \end {cases}\right ) + 2 B b \left (\begin {cases} - \frac {5 a^{4} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{128 b^{3}} + \sqrt {a + b x} \left (\frac {5 a^{3} \sqrt {x}}{128 b^{3}} - \frac {5 a^{2} x^{\frac {3}{2}}}{192 b^{2}} + \frac {a x^{\frac {5}{2}}}{48 b} + \frac {x^{\frac {7}{2}}}{8}\right ) & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {7}{2}}}{7} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((b*x+a)**(3/2)*(B*x+A)*x**(1/2),x)

[Out]

2*A*a*Piecewise((-a**2*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))/(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)) + 2*A*b*Piecewise((a**3*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (s
qrt(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*B*a*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*B*b*Piecew
ise((-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))/s
qrt(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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (125) = 250\).

Time = 0.21 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.81 \[ \int \sqrt {x} (a+b x)^{3/2} (A+B x) \, dx=\frac {1}{4} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B x + \frac {1}{2} \, \sqrt {b x^{2} + a x} A a x + \frac {5 \, \sqrt {b x^{2} + a x} B a^{2} x}{32 \, b} - \frac {5 \, B a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {5}{2}}} - \frac {A a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {3}{2}}} + \frac {5 \, \sqrt {b x^{2} + a x} B a^{3}}{64 \, b^{2}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{24 \, b} + \frac {\sqrt {b x^{2} + a x} A a^{2}}{4 \, b} - \frac {\sqrt {b x^{2} + a x} {\left (B a + A b\right )} a x}{4 \, b} + \frac {{\left (B a + A b\right )} a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {5}{2}}} - \frac {\sqrt {b x^{2} + a x} {\left (B a + A b\right )} a^{2}}{8 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} {\left (B a + A b\right )}}{3 \, b} \]

[In]

integrate((b*x+a)^(3/2)*(B*x+A)*x^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \sqrt {x} (a+b x)^{3/2} (A+B x) \, dx=\text {Timed out} \]

[In]

integrate((b*x+a)^(3/2)*(B*x+A)*x^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

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

[In]

int(x^(1/2)*(A + B*x)*(a + b*x)^(3/2),x)

[Out]

int(x^(1/2)*(A + B*x)*(a + b*x)^(3/2), x)