\(\int \frac {A+B x}{(a+b x)^{5/2}} \, dx\) [448]

Optimal result

Integrand size = 15, antiderivative size = 40 \[ \int \frac {A+B x}{(a+b x)^{5/2}} \, dx=-\frac {2 (A b-a B)}{3 b^2 (a+b x)^{3/2}}-\frac {2 B}{b^2 \sqrt {a+b x}} \]

[Out]

-2/3*(A*b-B*a)/b^2/(b*x+a)^(3/2)-2*B/b^2/(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \[ \int \frac {A+B x}{(a+b x)^{5/2}} \, dx=-\frac {2 (A b-a B)}{3 b^2 (a+b x)^{3/2}}-\frac {2 B}{b^2 \sqrt {a+b x}} \]

[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 45

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*} \text {integral}& = \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] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.72 \[ \int \frac {A+B x}{(a+b x)^{5/2}} \, dx=-\frac {2 (A b+2 a B+3 b B x)}{3 b^2 (a+b x)^{3/2}} \]

[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))

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.65

method	result	size
gosper	\(-\frac {2 \left (3 b B x +A b +2 B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{2}}\)	\(26\)
trager	\(-\frac {2 \left (3 b B x +A b +2 B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{2}}\)	\(26\)
pseudoelliptic	\(-\frac {2 \left (\left (3 B x +A \right ) b +2 B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{2}}\)	\(26\)
derivativedivides	\(\frac {-\frac {2 B}{\sqrt {b x +a}}-\frac {2 \left (A b -B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{2}}\)	\(34\)
default	\(\frac {-\frac {2 B}{\sqrt {b x +a}}-\frac {2 \left (A b -B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{2}}\)	\(34\)

[In]

int((B*x+A)/(b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.15 \[ \int \frac {A+B x}{(a+b x)^{5/2}} \, dx=-\frac {2 \, {\left (3 \, B b x + 2 \, B a + A b\right )} \sqrt {b x + a}}{3 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (39) = 78\).

Time = 0.32 (sec) , antiderivative size = 124, normalized size of antiderivative = 3.10 \[ \int \frac {A+B x}{(a+b x)^{5/2}} \, dx=\begin {cases} - \frac {2 A b}{3 a b^{2} \sqrt {a + b x} + 3 b^{3} x \sqrt {a + b x}} - \frac {4 B a}{3 a b^{2} \sqrt {a + b x} + 3 b^{3} x \sqrt {a + b x}} - \frac {6 B b x}{3 a b^{2} \sqrt {a + b x} + 3 b^{3} x \sqrt {a + b x}} & \text {for}\: b \neq 0 \\\frac {A x + \frac {B x^{2}}{2}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.70 \[ \int \frac {A+B x}{(a+b x)^{5/2}} \, dx=-\frac {2 \, {\left (3 \, {\left (b x + a\right )} B - B a + A b\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.70 \[ \int \frac {A+B x}{(a+b x)^{5/2}} \, dx=-\frac {2 \, {\left (3 \, {\left (b x + a\right )} B - B a + A b\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2}} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.72 \[ \int \frac {A+B x}{(a+b x)^{5/2}} \, dx=-\frac {2\,A\,b-2\,B\,a+6\,B\,\left (a+b\,x\right )}{3\,b^2\,{\left (a+b\,x\right )}^{3/2}} \]

[In]

int((A + B*x)/(a + b*x)^(5/2),x)

[Out]

-(2*A*b - 2*B*a + 6*B*(a + b*x))/(3*b^2*(a + b*x)^(3/2))