\(\int \frac {1}{(a x+b x^2)^{5/2}} \, dx\) [66]

Optimal result

Integrand size = 13, antiderivative size = 54 \[ \int \frac {1}{\left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2 (a+2 b x)}{3 a^2 \left (a x+b x^2\right )^{3/2}}+\frac {16 b (a+2 b x)}{3 a^4 \sqrt {a x+b x^2}} \]

[Out]

-2/3*(2*b*x+a)/a^2/(b*x^2+a*x)^(3/2)+16/3*b*(2*b*x+a)/a^4/(b*x^2+a*x)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {628, 627} \[ \int \frac {1}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {16 b (a+2 b x)}{3 a^4 \sqrt {a x+b x^2}}-\frac {2 (a+2 b x)}{3 a^2 \left (a x+b x^2\right )^{3/2}} \]

[In]

Int[(a*x + b*x^2)^(-5/2),x]

[Out]

(-2*(a + 2*b*x))/(3*a^2*(a*x + b*x^2)^(3/2)) + (16*b*(a + 2*b*x))/(3*a^4*Sqrt[a*x + b*x^2])

Rule 627

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1
)*(b^2 - 4*a*c))), x] - Dist[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (a+2 b x)}{3 a^2 \left (a x+b x^2\right )^{3/2}}-\frac {(8 b) \int \frac {1}{\left (a x+b x^2\right )^{3/2}} \, dx}{3 a^2} \\ & = -\frac {2 (a+2 b x)}{3 a^2 \left (a x+b x^2\right )^{3/2}}+\frac {16 b (a+2 b x)}{3 a^4 \sqrt {a x+b x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {-2 a^3+12 a^2 b x+48 a b^2 x^2+32 b^3 x^3}{3 a^4 (x (a+b x))^{3/2}} \]

[In]

Integrate[(a*x + b*x^2)^(-5/2),x]

[Out]

(-2*a^3 + 12*a^2*b*x + 48*a*b^2*x^2 + 32*b^3*x^3)/(3*a^4*(x*(a + b*x))^(3/2))

Maple [A] (verified)

Time = 2.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\)	\(47\)
pseudoelliptic	\(-\frac {2 \left (2 b x +a \right ) \left (-8 b^{2} x^{2}-8 a b x +a^{2}\right )}{3 \sqrt {x \left (b x +a \right )}\, x \left (b x +a \right ) a^{4}}\)	\(48\)
gosper	\(-\frac {2 x \left (b x +a \right ) \left (-16 b^{3} x^{3}-24 a \,b^{2} x^{2}-6 a^{2} b x +a^{3}\right )}{3 a^{4} \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}\)	\(51\)
trager	\(-\frac {2 \left (-16 b^{3} x^{3}-24 a \,b^{2} x^{2}-6 a^{2} b x +a^{3}\right ) \sqrt {b \,x^{2}+a x}}{3 a^{4} x^{2} \left (b x +a \right )^{2}}\)	\(55\)
risch	\(-\frac {2 \left (b x +a \right ) \left (-8 b x +a \right )}{3 a^{4} x \sqrt {x \left (b x +a \right )}}+\frac {2 b^{2} \left (8 b x +9 a \right ) x}{3 \sqrt {x \left (b x +a \right )}\, \left (b x +a \right ) a^{4}}\)	\(63\)

[In]

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

[Out]

-2/3*(2*b*x+a)/a^2/(b*x^2+a*x)^(3/2)+16/3*b*(2*b*x+a)/a^4/(b*x^2+a*x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.33 \[ \int \frac {1}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (16 \, b^{3} x^{3} + 24 \, a b^{2} x^{2} + 6 \, a^{2} b x - a^{3}\right )} \sqrt {b x^{2} + a x}}{3 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}} \]

[In]

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

[Out]

2/3*(16*b^3*x^3 + 24*a*b^2*x^2 + 6*a^2*b*x - a^3)*sqrt(b*x^2 + a*x)/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2)

Sympy [F]

\[ \int \frac {1}{\left (a x+b x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (a x + b x^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral((a*x + b*x**2)**(-5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.33 \[ \int \frac {1}{\left (a x+b x^2\right )^{5/2}} \, dx=-\frac {4 \, b x}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}} + \frac {32 \, b^{2} x}{3 \, \sqrt {b x^{2} + a x} a^{4}} - \frac {2}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a} + \frac {16 \, b}{3 \, \sqrt {b x^{2} + a x} a^{3}} \]

[In]

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

[Out]

-4/3*b*x/((b*x^2 + a*x)^(3/2)*a^2) + 32/3*b^2*x/(sqrt(b*x^2 + a*x)*a^4) - 2/3/((b*x^2 + a*x)^(3/2)*a) + 16/3*b
/(sqrt(b*x^2 + a*x)*a^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (2 \, {\left (4 \, x {\left (\frac {2 \, b^{3} x}{a^{4}} + \frac {3 \, b^{2}}{a^{3}}\right )} + \frac {3 \, b}{a^{2}}\right )} x - \frac {1}{a}\right )}}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {\left (2\,a+4\,b\,x\right )\,\left (-a^2+8\,a\,b\,x+8\,b^2\,x^2\right )}{3\,a^4\,{\left (b\,x^2+a\,x\right )}^{3/2}} \]

[In]

int(1/(a*x + b*x^2)^(5/2),x)

[Out]

((2*a + 4*b*x)*(8*b^2*x^2 - a^2 + 8*a*b*x))/(3*a^4*(a*x + b*x^2)^(3/2))