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

Optimal. Leaf size=42 \[ \frac{2 (a+b x)^{3/2} (A b-a B)}{3 b^2}+\frac{2 B (a+b x)^{5/2}}{5 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0167864, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {43} \[ \frac{2 (a+b x)^{3/2} (A b-a B)}{3 b^2}+\frac{2 B (a+b x)^{5/2}}{5 b^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]*(A + B*x),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0177103, size = 30, normalized size = 0.71 \[ \frac{2 (a+b x)^{3/2} (-2 a B+5 A b+3 b B x)}{15 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x]*(A + B*x),x]

[Out]

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

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Maple [A]  time = 0.002, size = 27, normalized size = 0.6 \begin{align*}{\frac{6\,bBx+10\,Ab-4\,Ba}{15\,{b}^{2}} \left ( bx+a \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.19657, size = 45, normalized size = 1.07 \begin{align*} \frac{2 \,{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} B - 5 \,{\left (B a - A b\right )}{\left (b x + a\right )}^{\frac{3}{2}}\right )}}{15 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.34161, size = 108, normalized size = 2.57 \begin{align*} \frac{2 \,{\left (3 \, B b^{2} x^{2} - 2 \, B a^{2} + 5 \, A a b +{\left (B a b + 5 \, A b^{2}\right )} x\right )} \sqrt{b x + a}}{15 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*B*b^2*x^2 - 2*B*a^2 + 5*A*a*b + (B*a*b + 5*A*b^2)*x)*sqrt(b*x + a)/b^2

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Sympy [A]  time = 1.96798, size = 36, normalized size = 0.86 \begin{align*} \frac{2 \left (\frac{B \left (a + b x\right )^{\frac{5}{2}}}{5 b} + \frac{\left (a + b x\right )^{\frac{3}{2}} \left (A b - B a\right )}{3 b}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14161, size = 55, normalized size = 1.31 \begin{align*} \frac{2 \,{\left (5 \,{\left (b x + a\right )}^{\frac{3}{2}} A + \frac{{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a\right )} B}{b}\right )}}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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