\(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{5/2}} \, dx\) [788]

Optimal result

Integrand size = 31, antiderivative size = 116 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{5/2}} \, dx=-\frac {2 a A \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}-\frac {2 (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)}+\frac {2 b B \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {784, 77} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{5/2}} \, dx=-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{\sqrt {x} (a+b x)}-\frac {2 a A \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}+\frac {2 b B \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \]

[In]

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

[Out]

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

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 784

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.40 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{5/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} (3 b x (A-B x)+a (A+3 B x))}{3 x^{3/2} (a+b x)} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.28

method	result	size
default	\(-\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (-3 B b \,x^{2}+3 A b x +3 a B x +a A \right )}{3 x^{\frac {3}{2}}}\)	\(33\)
gosper	\(-\frac {2 \left (-3 B b \,x^{2}+3 A b x +3 a B x +a A \right ) \sqrt {\left (b x +a \right )^{2}}}{3 x^{\frac {3}{2}} \left (b x +a \right )}\)	\(43\)
risch	\(-\frac {2 \left (-3 B b \,x^{2}+3 A b x +3 a B x +a A \right ) \sqrt {\left (b x +a \right )^{2}}}{3 x^{\frac {3}{2}} \left (b x +a \right )}\)	\(43\)

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/x^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.23 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{5/2}} \, dx=\frac {2 \, {\left (3 \, B b x^{2} - A a - 3 \, {\left (B a + A b\right )} x\right )}}{3 \, x^{\frac {3}{2}}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/x**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.28 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{5/2}} \, dx=\frac {2 \, {\left (b x^{2} - a x\right )} B}{x^{\frac {3}{2}}} - \frac {2 \, {\left (3 \, b x^{2} + a x\right )} A}{3 \, x^{\frac {5}{2}}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(5/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.44 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{5/2}} \, dx=2 \, B b \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - \frac {2 \, {\left (3 \, B a x \mathrm {sgn}\left (b x + a\right ) + 3 \, A b x \mathrm {sgn}\left (b x + a\right ) + A a \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, x^{\frac {3}{2}}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.06 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.47 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{5/2}} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,A\,a}{3\,b}-2\,B\,x^2+\frac {x\,\left (6\,A\,b+6\,B\,a\right )}{3\,b}\right )}{x^{5/2}+\frac {a\,x^{3/2}}{b}} \]

[In]

int((((a + b*x)^2)^(1/2)*(A + B*x))/x^(5/2),x)

[Out]

-(((a + b*x)^2)^(1/2)*((2*A*a)/(3*b) - 2*B*x^2 + (x*(6*A*b + 6*B*a))/(3*b)))/(x^(5/2) + (a*x^(3/2))/b)