3.448 \(\int \frac{A+B x}{(a+b x)^{5/2}} \, dx\)

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

[Out]

(-2*(A*b - a*B))/(3*b^2*(a + b*x)^(3/2)) - (2*B)/(b^2*Sqrt[a + b*x])

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Rubi [A]  time = 0.0131098, 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} \[ -\frac{2 (A b-a B)}{3 b^2 (a+b x)^{3/2}}-\frac{2 B}{b^2 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(a + b*x)^(5/2),x]

[Out]

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

Mathematica [A]  time = 0.0181173, size = 29, normalized size = 0.72 \[ -\frac{2 (2 a B+A b+3 b B x)}{3 b^2 (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(a + b*x)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.002, size = 26, normalized size = 0.7 \begin{align*} -{\frac{6\,bBx+2\,Ab+4\,Ba}{3\,{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)^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.12953, size = 38, normalized size = 0.95 \begin{align*} -\frac{2 \,{\left (3 \,{\left (b x + a\right )} B - B a + A b\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.37413, size = 103, normalized size = 2.58 \begin{align*} -\frac{2 \,{\left (3 \, B b x + 2 \, B a + A b\right )} \sqrt{b x + a}}{3 \,{\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*B*b*x + 2*B*a + A*b)*sqrt(b*x + a)/(b^4*x^2 + 2*a*b^3*x + a^2*b^2)

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Sympy [A]  time = 1.22473, size = 124, normalized size = 3.1 \begin{align*} \begin{cases} - \frac{2 A b}{3 a b^{2} \sqrt{a + b x} + 3 b^{3} x \sqrt{a + b x}} - \frac{4 B a}{3 a b^{2} \sqrt{a + b x} + 3 b^{3} x \sqrt{a + b x}} - \frac{6 B b x}{3 a b^{2} \sqrt{a + b x} + 3 b^{3} x \sqrt{a + b x}} & \text{for}\: b \neq 0 \\\frac{A x + \frac{B x^{2}}{2}}{a^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)**(5/2),x)

[Out]

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

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Giac [A]  time = 1.16561, size = 38, normalized size = 0.95 \begin{align*} -\frac{2 \,{\left (3 \,{\left (b x + a\right )} B - B a + A b\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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