3.1379 \(\int (a+b x) \sqrt{c+d x} \, dx\)

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

[Out]

(-2*(b*c - a*d)*(c + d*x)^(3/2))/(3*d^2) + (2*b*(c + d*x)^(5/2))/(5*d^2)

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Rubi [A]  time = 0.0139514, 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} \[ \frac{2 b (c+d x)^{5/2}}{5 d^2}-\frac{2 (c+d x)^{3/2} (b c-a d)}{3 d^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)*Sqrt[c + d*x],x]

[Out]

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

Mathematica [A]  time = 0.0173893, size = 30, normalized size = 0.71 \[ \frac{2 (c+d x)^{3/2} (5 a d-2 b c+3 b d x)}{15 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)*Sqrt[c + d*x],x]

[Out]

(2*(c + d*x)^(3/2)*(-2*b*c + 5*a*d + 3*b*d*x))/(15*d^2)

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Maple [A]  time = 0.002, size = 27, normalized size = 0.6 \begin{align*}{\frac{6\,bdx+10\,ad-4\,bc}{15\,{d}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(d*x+c)^(1/2),x)

[Out]

2/15*(d*x+c)^(3/2)*(3*b*d*x+5*a*d-2*b*c)/d^2

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Maxima [A]  time = 0.942432, size = 45, normalized size = 1.07 \begin{align*} \frac{2 \,{\left (3 \,{\left (d x + c\right )}^{\frac{5}{2}} b - 5 \,{\left (b c - a d\right )}{\left (d x + c\right )}^{\frac{3}{2}}\right )}}{15 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*(d*x + c)^(5/2)*b - 5*(b*c - a*d)*(d*x + c)^(3/2))/d^2

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Fricas [A]  time = 2.10574, size = 108, normalized size = 2.57 \begin{align*} \frac{2 \,{\left (3 \, b d^{2} x^{2} - 2 \, b c^{2} + 5 \, a c d +{\left (b c d + 5 \, a d^{2}\right )} x\right )} \sqrt{d x + c}}{15 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*b*d^2*x^2 - 2*b*c^2 + 5*a*c*d + (b*c*d + 5*a*d^2)*x)*sqrt(d*x + c)/d^2

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Sympy [A]  time = 1.48833, size = 36, normalized size = 0.86 \begin{align*} \frac{2 \left (\frac{b \left (c + d x\right )^{\frac{5}{2}}}{5 d} + \frac{\left (c + d x\right )^{\frac{3}{2}} \left (a d - b c\right )}{3 d}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(d*x+c)**(1/2),x)

[Out]

2*(b*(c + d*x)**(5/2)/(5*d) + (c + d*x)**(3/2)*(a*d - b*c)/(3*d))/d

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Giac [A]  time = 1.05429, size = 55, normalized size = 1.31 \begin{align*} \frac{2 \,{\left (5 \,{\left (d x + c\right )}^{\frac{3}{2}} a + \frac{{\left (3 \,{\left (d x + c\right )}^{\frac{5}{2}} - 5 \,{\left (d x + c\right )}^{\frac{3}{2}} c\right )} b}{d}\right )}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(5*(d*x + c)^(3/2)*a + (3*(d*x + c)^(5/2) - 5*(d*x + c)^(3/2)*c)*b/d)/d