3.390 \(\int x \sqrt{a+b x} (A+B x) \, dx\)

Optimal. Leaf size=67 \[ \frac{2 (a+b x)^{5/2} (A b-2 a B)}{5 b^3}-\frac{2 a (a+b x)^{3/2} (A b-a B)}{3 b^3}+\frac{2 B (a+b x)^{7/2}}{7 b^3} \]

[Out]

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

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Rubi [A]  time = 0.0273428, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {77} \[ \frac{2 (a+b x)^{5/2} (A b-2 a B)}{5 b^3}-\frac{2 a (a+b x)^{3/2} (A b-a B)}{3 b^3}+\frac{2 B (a+b x)^{7/2}}{7 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a + b*x]*(A + B*x),x]

[Out]

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

Rule 77

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

Mathematica [A]  time = 0.0320046, size = 49, normalized size = 0.73 \[ \frac{2 (a+b x)^{3/2} \left (8 a^2 B-2 a b (7 A+6 B x)+3 b^2 x (7 A+5 B x)\right )}{105 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(3/2)*(8*a^2*B + 3*b^2*x*(7*A + 5*B*x) - 2*a*b*(7*A + 6*B*x)))/(105*b^3)

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Maple [A]  time = 0.003, size = 47, normalized size = 0.7 \begin{align*} -{\frac{-30\,{b}^{2}B{x}^{2}-42\,{b}^{2}Ax+24\,abBx+28\,Aab-16\,B{a}^{2}}{105\,{b}^{3}} \left ( bx+a \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(b*x+a)^(3/2)*(-15*B*b^2*x^2-21*A*b^2*x+12*B*a*b*x+14*A*a*b-8*B*a^2)/b^3

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Maxima [A]  time = 1.16689, size = 73, normalized size = 1.09 \begin{align*} \frac{2 \,{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} B - 21 \,{\left (2 \, B a - A b\right )}{\left (b x + a\right )}^{\frac{5}{2}} + 35 \,{\left (B a^{2} - A a b\right )}{\left (b x + a\right )}^{\frac{3}{2}}\right )}}{105 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.33934, size = 161, normalized size = 2.4 \begin{align*} \frac{2 \,{\left (15 \, B b^{3} x^{3} + 8 \, B a^{3} - 14 \, A a^{2} b + 3 \,{\left (B a b^{2} + 7 \, A b^{3}\right )} x^{2} -{\left (4 \, B a^{2} b - 7 \, A a b^{2}\right )} x\right )} \sqrt{b x + a}}{105 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*B*b^3*x^3 + 8*B*a^3 - 14*A*a^2*b + 3*(B*a*b^2 + 7*A*b^3)*x^2 - (4*B*a^2*b - 7*A*a*b^2)*x)*sqrt(b*x +
a)/b^3

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Sympy [A]  time = 2.45543, size = 63, normalized size = 0.94 \begin{align*} \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 - 2 B a\right )}{5 b} + \frac{\left (a + b x\right )^{\frac{3}{2}} \left (- A a b + B a^{2}\right )}{3 b}\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23154, size = 93, normalized size = 1.39 \begin{align*} \frac{2 \,{\left (\frac{7 \,{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a\right )} A}{b} + \frac{{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2}\right )} B}{b^{2}}\right )}}{105 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(7*(3*(b*x + a)^(5/2) - 5*(b*x + a)^(3/2)*a)*A/b + (15*(b*x + a)^(7/2) - 42*(b*x + a)^(5/2)*a + 35*(b*x
+ a)^(3/2)*a^2)*B/b^2)/b