\(\int \frac {A+B x}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [726]

Optimal result

Integrand size = 26, antiderivative size = 71 \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {B}{3 b^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {A b-a B}{4 b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \]

[Out]

-1/3*B/b^2/(b^2*x^2+2*a*b*x+a^2)^(3/2)+1/4*(-A*b+B*a)/b^2/(b*x+a)/(b^2*x^2+2*a*b*x+a^2)^(3/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {654, 621} \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {A b-a B}{4 b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {B}{3 b^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

-1/3*B/(b^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) - (A*b - a*B)/(4*b^2*(a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))

Rule 621

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[2*((a + b*x + c*x^2)^(p + 1)/((2*p + 1)*(b + 2*
c*x))), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {B}{3 b^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {\left (2 A b^2-2 a b B\right ) \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx}{2 b^2} \\ & = -\frac {B}{3 b^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {A b-a B}{4 b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(345\) vs. \(2(71)=142\).

Time = 0.97 (sec) , antiderivative size = 345, normalized size of antiderivative = 4.86 \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {x \left (-3 \sqrt {a^2} A b^7 x^7-3 a^3 b^4 x^4 \sqrt {(a+b x)^2} (A+B x)+3 a^2 b^5 x^5 \sqrt {(a+b x)^2} (A+B x)+6 a^7 (2 A+B x) \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )+3 a b^6 x^6 \left (\sqrt {a^2} B x-A \sqrt {(a+b x)^2}\right )+a^5 b^2 x^2 \left (12 \sqrt {a^2} A-6 A \sqrt {(a+b x)^2}+B x \left (\sqrt {a^2}-3 \sqrt {(a+b x)^2}\right )\right )+3 a^4 b^3 x^3 \left (B x \sqrt {(a+b x)^2}+A \left (\sqrt {a^2}+\sqrt {(a+b x)^2}\right )\right )+2 a^6 b x \left (9 \sqrt {a^2} A-3 A \sqrt {(a+b x)^2}+B x \left (2 \sqrt {a^2}+\sqrt {(a+b x)^2}\right )\right )\right )}{12 a^8 (a+b x)^3 \left (a^2+a b x-\sqrt {a^2} \sqrt {(a+b x)^2}\right )} \]

[In]

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

[Out]

-1/12*(x*(-3*Sqrt[a^2]*A*b^7*x^7 - 3*a^3*b^4*x^4*Sqrt[(a + b*x)^2]*(A + B*x) + 3*a^2*b^5*x^5*Sqrt[(a + b*x)^2]
*(A + B*x) + 6*a^7*(2*A + B*x)*(Sqrt[a^2] - Sqrt[(a + b*x)^2]) + 3*a*b^6*x^6*(Sqrt[a^2]*B*x - A*Sqrt[(a + b*x)
^2]) + a^5*b^2*x^2*(12*Sqrt[a^2]*A - 6*A*Sqrt[(a + b*x)^2] + B*x*(Sqrt[a^2] - 3*Sqrt[(a + b*x)^2])) + 3*a^4*b^
3*x^3*(B*x*Sqrt[(a + b*x)^2] + A*(Sqrt[a^2] + Sqrt[(a + b*x)^2])) + 2*a^6*b*x*(9*Sqrt[a^2]*A - 3*A*Sqrt[(a + b
*x)^2] + B*x*(2*Sqrt[a^2] + Sqrt[(a + b*x)^2]))))/(a^8*(a + b*x)^3*(a^2 + a*b*x - Sqrt[a^2]*Sqrt[(a + b*x)^2])
)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.46

method	result	size
gosper	\(-\frac {\left (b x +a \right ) \left (4 B b x +3 A b +B a \right )}{12 b^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(33\)
default	\(-\frac {\left (b x +a \right ) \left (4 B b x +3 A b +B a \right )}{12 b^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(33\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B x}{3 b}-\frac {3 A b +B a}{12 b^{2}}\right )}{\left (b x +a \right )^{5}}\)	\(39\)

[In]

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

[Out]

-1/12*(b*x+a)/b^2*(4*B*b*x+3*A*b+B*a)/((b*x+a)^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {4 \, B b x + B a + 3 \, A b}{12 \, {\left (b^{6} x^{4} + 4 \, a b^{5} x^{3} + 6 \, a^{2} b^{4} x^{2} + 4 \, a^{3} b^{3} x + a^{4} b^{2}\right )}} \]

[In]

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

[Out]

-1/12*(4*B*b*x + B*a + 3*A*b)/(b^6*x^4 + 4*a*b^5*x^3 + 6*a^2*b^4*x^2 + 4*a^3*b^3*x + a^4*b^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {B}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {B a}{4 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {A}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} \]

[In]

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

[Out]

-1/3*B/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 1/4*B*a/(b^6*(x + a/b)^4) - 1/4*A/(b^5*(x + a/b)^4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.46 \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {4 \, B b x + B a + 3 \, A b}{12 \, {\left (b x + a\right )}^{4} b^{2} \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

-1/12*(4*B*b*x + B*a + 3*A*b)/((b*x + a)^4*b^2*sgn(b*x + a))

Mupad [B] (verification not implemented)

Time = 10.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.61 \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (3\,A\,b+B\,a+4\,B\,b\,x\right )}{12\,b^2\,{\left (a+b\,x\right )}^5} \]

[In]

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

[Out]

-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(3*A*b + B*a + 4*B*b*x))/(12*b^2*(a + b*x)^5)