\(\int \frac {x^2 (A+B x)}{\sqrt {a+b x}} \, dx\) [425]

Optimal result

Integrand size = 18, antiderivative size = 93 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x}} \, dx=\frac {2 a^2 (A b-a B) \sqrt {a+b x}}{b^4}-\frac {2 a (2 A b-3 a B) (a+b x)^{3/2}}{3 b^4}+\frac {2 (A b-3 a B) (a+b x)^{5/2}}{5 b^4}+\frac {2 B (a+b x)^{7/2}}{7 b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x}} \, dx=\frac {2 a^2 \sqrt {a+b x} (A b-a B)}{b^4}+\frac {2 (a+b x)^{5/2} (A b-3 a B)}{5 b^4}-\frac {2 a (a+b x)^{3/2} (2 A b-3 a B)}{3 b^4}+\frac {2 B (a+b x)^{7/2}}{7 b^4} \]

[In]

Int[(x^2*(A + B*x))/Sqrt[a + b*x],x]

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.73 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x}} \, dx=\frac {2 \sqrt {a+b x} \left (-48 a^3 B+8 a^2 b (7 A+3 B x)+3 b^3 x^2 (7 A+5 B x)-2 a b^2 x (14 A+9 B x)\right )}{105 b^4} \]

[In]

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

[Out]

(2*Sqrt[a + b*x]*(-48*a^3*B + 8*a^2*b*(7*A + 3*B*x) + 3*b^3*x^2*(7*A + 5*B*x) - 2*a*b^2*x*(14*A + 9*B*x)))/(10
5*b^4)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.62

method	result	size
pseudoelliptic	\(\frac {16 \left (\frac {3 x^{2} \left (\frac {5 B x}{7}+A \right ) b^{3}}{8}-\frac {x a \left (\frac {9 B x}{14}+A \right ) b^{2}}{2}+a^{2} \left (\frac {3 B x}{7}+A \right ) b -\frac {6 a^{3} B}{7}\right ) \sqrt {b x +a}}{15 b^{4}}\)	\(58\)
gosper	\(\frac {2 \sqrt {b x +a}\, \left (15 b^{3} B \,x^{3}+21 A \,b^{3} x^{2}-18 B a \,b^{2} x^{2}-28 a \,b^{2} A x +24 a^{2} b B x +56 a^{2} b A -48 a^{3} B \right )}{105 b^{4}}\)	\(71\)
trager	\(\frac {2 \sqrt {b x +a}\, \left (15 b^{3} B \,x^{3}+21 A \,b^{3} x^{2}-18 B a \,b^{2} x^{2}-28 a \,b^{2} A x +24 a^{2} b B x +56 a^{2} b A -48 a^{3} B \right )}{105 b^{4}}\)	\(71\)
risch	\(\frac {2 \sqrt {b x +a}\, \left (15 b^{3} B \,x^{3}+21 A \,b^{3} x^{2}-18 B a \,b^{2} x^{2}-28 a \,b^{2} A x +24 a^{2} b B x +56 a^{2} b A -48 a^{3} B \right )}{105 b^{4}}\)	\(71\)
derivativedivides	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 \left (A b -3 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a^{2} B -2 a \left (A b -B a \right )\right ) \left (b x +a \right )^{\frac {3}{2}}}{3}+2 a^{2} \left (A b -B a \right ) \sqrt {b x +a}}{b^{4}}\)	\(79\)
default	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 \left (A b -3 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a^{2} B -2 a \left (A b -B a \right )\right ) \left (b x +a \right )^{\frac {3}{2}}}{3}+2 a^{2} \left (A b -B a \right ) \sqrt {b x +a}}{b^{4}}\)	\(79\)

[In]

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

[Out]

16/15*(3/8*x^2*(5/7*B*x+A)*b^3-1/2*x*a*(9/14*B*x+A)*b^2+a^2*(3/7*B*x+A)*b-6/7*a^3*B)*(b*x+a)^(1/2)/b^4

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.77 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x}} \, dx=\frac {2 \, {\left (15 \, B b^{3} x^{3} - 48 \, B a^{3} + 56 \, A a^{2} b - 3 \, {\left (6 \, B a b^{2} - 7 \, A b^{3}\right )} x^{2} + 4 \, {\left (6 \, B a^{2} b - 7 \, A a b^{2}\right )} x\right )} \sqrt {b x + a}}{105 \, b^{4}} \]

[In]

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

[Out]

2/105*(15*B*b^3*x^3 - 48*B*a^3 + 56*A*a^2*b - 3*(6*B*a*b^2 - 7*A*b^3)*x^2 + 4*(6*B*a^2*b - 7*A*a*b^2)*x)*sqrt(
b*x + a)/b^4

Sympy [A] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.17 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x}} \, dx=\begin {cases} \frac {2 \left (\frac {B \left (a + b x\right )^{\frac {7}{2}}}{7 b} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (A b - 3 B a\right )}{5 b} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (- 2 A a b + 3 B a^{2}\right )}{3 b} + \frac {\sqrt {a + b x} \left (A a^{2} b - B a^{3}\right )}{b}\right )}{b^{3}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{3}}{3} + \frac {B x^{4}}{4}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(B*(a + b*x)**(7/2)/(7*b) + (a + b*x)**(5/2)*(A*b - 3*B*a)/(5*b) + (a + b*x)**(3/2)*(-2*A*a*b + 3
*B*a**2)/(3*b) + sqrt(a + b*x)*(A*a**2*b - B*a**3)/b)/b**3, Ne(b, 0)), ((A*x**3/3 + B*x**4/4)/sqrt(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x}} \, dx=\frac {2 \, {\left (15 \, {\left (b x + a\right )}^{\frac {7}{2}} B - 21 \, {\left (3 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {5}{2}} + 35 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} {\left (b x + a\right )}^{\frac {3}{2}} - 105 \, {\left (B a^{3} - A a^{2} b\right )} \sqrt {b x + a}\right )}}{105 \, b^{4}} \]

[In]

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

[Out]

2/105*(15*(b*x + a)^(7/2)*B - 21*(3*B*a - A*b)*(b*x + a)^(5/2) + 35*(3*B*a^2 - 2*A*a*b)*(b*x + a)^(3/2) - 105*
(B*a^3 - A*a^2*b)*sqrt(b*x + a))/b^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x}} \, dx=\frac {2 \, {\left (\frac {7 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} A}{b^{2}} + \frac {3 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} B}{b^{3}}\right )}}{105 \, b} \]

[In]

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

[Out]

2/105*(7*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*A/b^2 + 3*(5*(b*x + a)^(7/2) - 21*(
b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*B/b^3)/b

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.91 \[ \int \frac {x^2 (A+B x)}{\sqrt {a+b x}} \, dx=\frac {\left (6\,B\,a^2-4\,A\,a\,b\right )\,{\left (a+b\,x\right )}^{3/2}}{3\,b^4}+\frac {2\,B\,{\left (a+b\,x\right )}^{7/2}}{7\,b^4}+\frac {\left (2\,A\,b-6\,B\,a\right )\,{\left (a+b\,x\right )}^{5/2}}{5\,b^4}-\frac {\left (2\,B\,a^3-2\,A\,a^2\,b\right )\,\sqrt {a+b\,x}}{b^4} \]

[In]

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

[Out]

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