3.1438 \(\int \frac{a+b x}{(c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=40 \[ \frac{2 (b c-a d)}{3 d^2 (c+d x)^{3/2}}-\frac{2 b}{d^2 \sqrt{c+d x}} \]

[Out]

(2*(b*c - a*d))/(3*d^2*(c + d*x)^(3/2)) - (2*b)/(d^2*Sqrt[c + d*x])

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Rubi [A]  time = 0.0139255, antiderivative size = 40, 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} \[ \frac{2 (b c-a d)}{3 d^2 (c+d x)^{3/2}}-\frac{2 b}{d^2 \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/(c + d*x)^(5/2),x]

[Out]

(2*(b*c - a*d))/(3*d^2*(c + d*x)^(3/2)) - (2*b)/(d^2*Sqrt[c + d*x])

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

Mathematica [A]  time = 0.0275788, size = 29, normalized size = 0.72 \[ -\frac{2 (a d+2 b c+3 b d x)}{3 d^2 (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/(c + d*x)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.003, size = 26, normalized size = 0.7 \begin{align*} -{\frac{6\,bdx+2\,ad+4\,bc}{3\,{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)^(5/2),x)

[Out]

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

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Maxima [A]  time = 0.949559, size = 38, normalized size = 0.95 \begin{align*} -\frac{2 \,{\left (3 \,{\left (d x + c\right )} b - b c + a d\right )}}{3 \,{\left (d x + c\right )}^{\frac{3}{2}} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.00135, size = 103, normalized size = 2.58 \begin{align*} -\frac{2 \,{\left (3 \, b d x + 2 \, b c + a d\right )} \sqrt{d x + c}}{3 \,{\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

-2/3*(3*b*d*x + 2*b*c + a*d)*sqrt(d*x + c)/(d^4*x^2 + 2*c*d^3*x + c^2*d^2)

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Sympy [A]  time = 1.07312, size = 124, normalized size = 3.1 \begin{align*} \begin{cases} - \frac{2 a d}{3 c d^{2} \sqrt{c + d x} + 3 d^{3} x \sqrt{c + d x}} - \frac{4 b c}{3 c d^{2} \sqrt{c + d x} + 3 d^{3} x \sqrt{c + d x}} - \frac{6 b d x}{3 c d^{2} \sqrt{c + d x} + 3 d^{3} x \sqrt{c + d x}} & \text{for}\: d \neq 0 \\\frac{a x + \frac{b x^{2}}{2}}{c^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)**(5/2),x)

[Out]

Piecewise((-2*a*d/(3*c*d**2*sqrt(c + d*x) + 3*d**3*x*sqrt(c + d*x)) - 4*b*c/(3*c*d**2*sqrt(c + d*x) + 3*d**3*x
*sqrt(c + d*x)) - 6*b*d*x/(3*c*d**2*sqrt(c + d*x) + 3*d**3*x*sqrt(c + d*x)), Ne(d, 0)), ((a*x + b*x**2/2)/c**(
5/2), True))

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Giac [A]  time = 1.05535, size = 38, normalized size = 0.95 \begin{align*} -\frac{2 \,{\left (3 \,{\left (d x + c\right )} b - b c + a d\right )}}{3 \,{\left (d x + c\right )}^{\frac{3}{2}} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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