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

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

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Rubi [A]  time = 0.02, antiderivative size = 42, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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*}

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Mathematica [A]  time = 0.03, size = 30, normalized size = 0.71 \begin {gather*} \frac {2 (a+b x)^{3/2} (-2 a B+5 A b+3 b B x)}{15 b^2} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.02, size = 33, normalized size = 0.79 \begin {gather*} \frac {2 (a+b x)^{3/2} (3 B (a+b x)-5 a B+5 A b)}{15 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.03, size = 46, normalized size = 1.10 \begin {gather*} \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 {gather*}

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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giac [B]  time = 1.23, size = 100, normalized size = 2.38 \begin {gather*} \frac {2 \, {\left (15 \, \sqrt {b x + a} A a + 5 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} A + \frac {5 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} B a}{b} + \frac {{\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 )} B}{b}\right )}}{15 \, b} \end {gather*}

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

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maple [A]  time = 0.00, size = 27, normalized size = 0.64 \begin {gather*} \frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (3 B b x +5 A b -2 B a \right )}{15 b^{2}} \end {gather*}

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 = 0.91, size = 33, normalized size = 0.79 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.05, size = 29, normalized size = 0.69 \begin {gather*} \frac {2\,{\left (a+b\,x\right )}^{3/2}\,\left (5\,A\,b-5\,B\,a+3\,B\,\left (a+b\,x\right )\right )}{15\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)*(a + b*x)^(1/2),x)

[Out]

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

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sympy [A]  time = 2.16, size = 36, normalized size = 0.86 \begin {gather*} \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 {gather*}

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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