\(\int x^{3/2} (a+b x)^{3/2} (A+B x) \, dx\) [491]

Optimal result

Integrand size = 20, antiderivative size = 192 \[ \int x^{3/2} (a+b x)^{3/2} (A+B x) \, dx=-\frac {3 a^3 (2 A b-a B) \sqrt {x} \sqrt {a+b x}}{128 b^3}+\frac {a^2 (2 A b-a B) x^{3/2} \sqrt {a+b x}}{64 b^2}+\frac {a (2 A b-a B) x^{5/2} \sqrt {a+b x}}{16 b}+\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {3 a^4 (2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{7/2}} \]

[Out]

1/8*(2*A*b-B*a)*x^(5/2)*(b*x+a)^(3/2)/b+1/5*B*x^(5/2)*(b*x+a)^(5/2)/b+3/128*a^4*(2*A*b-B*a)*arctanh(b^(1/2)*x^
(1/2)/(b*x+a)^(1/2))/b^(7/2)+1/64*a^2*(2*A*b-B*a)*x^(3/2)*(b*x+a)^(1/2)/b^2+1/16*a*(2*A*b-B*a)*x^(5/2)*(b*x+a)
^(1/2)/b-3/128*a^3*(2*A*b-B*a)*x^(1/2)*(b*x+a)^(1/2)/b^3

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} \[ \int x^{3/2} (a+b x)^{3/2} (A+B x) \, dx=\frac {3 a^4 (2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{7/2}}-\frac {3 a^3 \sqrt {x} \sqrt {a+b x} (2 A b-a B)}{128 b^3}+\frac {a^2 x^{3/2} \sqrt {a+b x} (2 A b-a B)}{64 b^2}+\frac {a x^{5/2} \sqrt {a+b x} (2 A b-a B)}{16 b}+\frac {x^{5/2} (a+b x)^{3/2} (2 A b-a B)}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b} \]

[In]

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

[Out]

(-3*a^3*(2*A*b - a*B)*Sqrt[x]*Sqrt[a + b*x])/(128*b^3) + (a^2*(2*A*b - a*B)*x^(3/2)*Sqrt[a + b*x])/(64*b^2) +
(a*(2*A*b - a*B)*x^(5/2)*Sqrt[a + b*x])/(16*b) + ((2*A*b - a*B)*x^(5/2)*(a + b*x)^(3/2))/(8*b) + (B*x^(5/2)*(a
+ b*x)^(5/2))/(5*b) + (3*a^4*(2*A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(128*b^(7/2))

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^{5/2} (a+b x)^{5/2}}{5 b}+\frac {\left (5 A b-\frac {5 a B}{2}\right ) \int x^{3/2} (a+b x)^{3/2} \, dx}{5 b} \\ & = \frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {(3 a (2 A b-a B)) \int x^{3/2} \sqrt {a+b x} \, dx}{16 b} \\ & = \frac {a (2 A b-a B) x^{5/2} \sqrt {a+b x}}{16 b}+\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {\left (a^2 (2 A b-a B)\right ) \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{32 b} \\ & = \frac {a^2 (2 A b-a B) x^{3/2} \sqrt {a+b x}}{64 b^2}+\frac {a (2 A b-a B) x^{5/2} \sqrt {a+b x}}{16 b}+\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}-\frac {\left (3 a^3 (2 A b-a B)\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{128 b^2} \\ & = -\frac {3 a^3 (2 A b-a B) \sqrt {x} \sqrt {a+b x}}{128 b^3}+\frac {a^2 (2 A b-a B) x^{3/2} \sqrt {a+b x}}{64 b^2}+\frac {a (2 A b-a B) x^{5/2} \sqrt {a+b x}}{16 b}+\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {\left (3 a^4 (2 A b-a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{256 b^3} \\ & = -\frac {3 a^3 (2 A b-a B) \sqrt {x} \sqrt {a+b x}}{128 b^3}+\frac {a^2 (2 A b-a B) x^{3/2} \sqrt {a+b x}}{64 b^2}+\frac {a (2 A b-a B) x^{5/2} \sqrt {a+b x}}{16 b}+\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {\left (3 a^4 (2 A b-a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{128 b^3} \\ & = -\frac {3 a^3 (2 A b-a B) \sqrt {x} \sqrt {a+b x}}{128 b^3}+\frac {a^2 (2 A b-a B) x^{3/2} \sqrt {a+b x}}{64 b^2}+\frac {a (2 A b-a B) x^{5/2} \sqrt {a+b x}}{16 b}+\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {\left (3 a^4 (2 A b-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^3} \\ & = -\frac {3 a^3 (2 A b-a B) \sqrt {x} \sqrt {a+b x}}{128 b^3}+\frac {a^2 (2 A b-a B) x^{3/2} \sqrt {a+b x}}{64 b^2}+\frac {a (2 A b-a B) x^{5/2} \sqrt {a+b x}}{16 b}+\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {3 a^4 (2 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.74 \[ \int x^{3/2} (a+b x)^{3/2} (A+B x) \, dx=\frac {\sqrt {x} \sqrt {a+b x} \left (15 a^4 B-10 a^3 b (3 A+B x)+4 a^2 b^2 x (5 A+2 B x)+32 b^4 x^3 (5 A+4 B x)+16 a b^3 x^2 (15 A+11 B x)\right )}{640 b^3}+\frac {3 a^4 (-2 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{64 b^{7/2}} \]

[In]

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

[Out]

(Sqrt[x]*Sqrt[a + b*x]*(15*a^4*B - 10*a^3*b*(3*A + B*x) + 4*a^2*b^2*x*(5*A + 2*B*x) + 32*b^4*x^3*(5*A + 4*B*x)
+ 16*a*b^3*x^2*(15*A + 11*B*x)))/(640*b^3) + (3*a^4*(-2*A*b + a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqrt[
a + b*x])])/(64*b^(7/2))

Maple [A] (verified)

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

method	result	size
risch	\(-\frac {\left (-128 B \,x^{4} b^{4}-160 A \,x^{3} b^{4}-176 B \,x^{3} a \,b^{3}-240 A \,x^{2} a \,b^{3}-8 B \,x^{2} a^{2} b^{2}-20 A x \,a^{2} b^{2}+10 B x \,a^{3} b +30 A \,a^{3} b -15 B \,a^{4}\right ) \sqrt {x}\, \sqrt {b x +a}}{640 b^{3}}+\frac {3 a^{4} \left (2 A b -B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{256 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(159\)
default	\(\frac {\sqrt {x}\, \sqrt {b x +a}\, \left (256 B \,b^{\frac {9}{2}} x^{4} \sqrt {x \left (b x +a \right )}+320 A \,b^{\frac {9}{2}} x^{3} \sqrt {x \left (b x +a \right )}+352 B a \,b^{\frac {7}{2}} x^{3} \sqrt {x \left (b x +a \right )}+480 A a \,b^{\frac {7}{2}} x^{2} \sqrt {x \left (b x +a \right )}+16 B \,a^{2} b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+40 A \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, a^{2} x -20 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a^{3} x +30 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{4} b -60 A \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a^{3}-15 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{5}+30 B \sqrt {b}\, \sqrt {x \left (b x +a \right )}\, a^{4}\right )}{1280 b^{\frac {7}{2}} \sqrt {x \left (b x +a \right )}}\)	\(260\)

[In]

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

[Out]

-1/640/b^3*(-128*B*b^4*x^4-160*A*b^4*x^3-176*B*a*b^3*x^3-240*A*a*b^3*x^2-8*B*a^2*b^2*x^2-20*A*a^2*b^2*x+10*B*a
^3*b*x+30*A*a^3*b-15*B*a^4)*x^(1/2)*(b*x+a)^(1/2)+3/256*a^4/b^(7/2)*(2*A*b-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 = 290, normalized size of antiderivative = 1.51 \[ \int x^{3/2} (a+b x)^{3/2} (A+B x) \, dx=\left [-\frac {15 \, {\left (B a^{5} - 2 \, A a^{4} b\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (128 \, B b^{5} x^{4} + 15 \, B a^{4} b - 30 \, A a^{3} b^{2} + 16 \, {\left (11 \, B a b^{4} + 10 \, A b^{5}\right )} x^{3} + 8 \, {\left (B a^{2} b^{3} + 30 \, A a b^{4}\right )} x^{2} - 10 \, {\left (B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{1280 \, b^{4}}, \frac {15 \, {\left (B a^{5} - 2 \, A a^{4} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (128 \, B b^{5} x^{4} + 15 \, B a^{4} b - 30 \, A a^{3} b^{2} + 16 \, {\left (11 \, B a b^{4} + 10 \, A b^{5}\right )} x^{3} + 8 \, {\left (B a^{2} b^{3} + 30 \, A a b^{4}\right )} x^{2} - 10 \, {\left (B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{640 \, b^{4}}\right ] \]

[In]

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

[Out]

[-1/1280*(15*(B*a^5 - 2*A*a^4*b)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(128*B*b^5*x^4 +
15*B*a^4*b - 30*A*a^3*b^2 + 16*(11*B*a*b^4 + 10*A*b^5)*x^3 + 8*(B*a^2*b^3 + 30*A*a*b^4)*x^2 - 10*(B*a^3*b^2 -
2*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^4, 1/640*(15*(B*a^5 - 2*A*a^4*b)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(
-b)/(b*sqrt(x))) + (128*B*b^5*x^4 + 15*B*a^4*b - 30*A*a^3*b^2 + 16*(11*B*a*b^4 + 10*A*b^5)*x^3 + 8*(B*a^2*b^3
+ 30*A*a*b^4)*x^2 - 10*(B*a^3*b^2 - 2*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^4]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (173) = 346\).

Time = 150.37 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.85 \[ \int x^{3/2} (a+b x)^{3/2} (A+B x) \, dx=- \frac {3 A a^{\frac {7}{2}} \sqrt {x}}{64 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {A a^{\frac {5}{2}} x^{\frac {3}{2}}}{64 b \sqrt {1 + \frac {b x}{a}}} + \frac {13 A a^{\frac {3}{2}} x^{\frac {5}{2}}}{32 \sqrt {1 + \frac {b x}{a}}} + \frac {5 A \sqrt {a} b x^{\frac {7}{2}}}{8 \sqrt {1 + \frac {b x}{a}}} + \frac {3 A a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {5}{2}}} + \frac {A b^{2} x^{\frac {9}{2}}}{4 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} + \frac {3 B a^{\frac {9}{2}} \sqrt {x}}{128 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {B a^{\frac {7}{2}} x^{\frac {3}{2}}}{128 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {B a^{\frac {5}{2}} x^{\frac {5}{2}}}{320 b \sqrt {1 + \frac {b x}{a}}} + \frac {23 B a^{\frac {3}{2}} x^{\frac {7}{2}}}{80 \sqrt {1 + \frac {b x}{a}}} + \frac {19 B \sqrt {a} b x^{\frac {9}{2}}}{40 \sqrt {1 + \frac {b x}{a}}} - \frac {3 B a^{5} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{128 b^{\frac {7}{2}}} + \frac {B b^{2} x^{\frac {11}{2}}}{5 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]

[In]

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

[Out]

-3*A*a**(7/2)*sqrt(x)/(64*b**2*sqrt(1 + b*x/a)) - A*a**(5/2)*x**(3/2)/(64*b*sqrt(1 + b*x/a)) + 13*A*a**(3/2)*x
**(5/2)/(32*sqrt(1 + b*x/a)) + 5*A*sqrt(a)*b*x**(7/2)/(8*sqrt(1 + b*x/a)) + 3*A*a**4*asinh(sqrt(b)*sqrt(x)/sqr
t(a))/(64*b**(5/2)) + A*b**2*x**(9/2)/(4*sqrt(a)*sqrt(1 + b*x/a)) + 3*B*a**(9/2)*sqrt(x)/(128*b**3*sqrt(1 + b*
x/a)) + B*a**(7/2)*x**(3/2)/(128*b**2*sqrt(1 + b*x/a)) - B*a**(5/2)*x**(5/2)/(320*b*sqrt(1 + b*x/a)) + 23*B*a*
*(3/2)*x**(7/2)/(80*sqrt(1 + b*x/a)) + 19*B*sqrt(a)*b*x**(9/2)/(40*sqrt(1 + b*x/a)) - 3*B*a**5*asinh(sqrt(b)*s
qrt(x)/sqrt(a))/(128*b**(7/2)) + B*b**2*x**(11/2)/(5*sqrt(a)*sqrt(1 + b*x/a))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.23 \[ \int x^{3/2} (a+b x)^{3/2} (A+B x) \, dx=\frac {1}{4} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} A x + \frac {3 \, \sqrt {b x^{2} + a x} B a^{3} x}{64 \, b^{2}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a x}{8 \, b} - \frac {3 \, \sqrt {b x^{2} + a x} A a^{2} x}{32 \, b} - \frac {3 \, B a^{5} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{256 \, b^{\frac {7}{2}}} + \frac {3 \, A a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {5}{2}}} + \frac {3 \, \sqrt {b x^{2} + a x} B a^{4}}{128 \, b^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a^{2}}{16 \, b^{2}} - \frac {3 \, \sqrt {b x^{2} + a x} A a^{3}}{64 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B}{5 \, b} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{8 \, b} \]

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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