\(\int \frac {1}{x^2 (a x+b x^2)^{5/2}} \, dx\) [68]

Optimal result

Integrand size = 17, antiderivative size = 106 \[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2}{7 a x^2 \left (a x+b x^2\right )^{3/2}}+\frac {4 b}{7 a^2 x \left (a x+b x^2\right )^{3/2}}-\frac {32 b^2 (a+2 b x)}{21 a^4 \left (a x+b x^2\right )^{3/2}}+\frac {256 b^3 (a+2 b x)}{21 a^6 \sqrt {a x+b x^2}} \]

[Out]

-2/7/a/x^2/(b*x^2+a*x)^(3/2)+4/7*b/a^2/x/(b*x^2+a*x)^(3/2)-32/21*b^2*(2*b*x+a)/a^4/(b*x^2+a*x)^(3/2)+256/21*b^
3*(2*b*x+a)/a^6/(b*x^2+a*x)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {672, 628, 627} \[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{5/2}} \, dx=\frac {256 b^3 (a+2 b x)}{21 a^6 \sqrt {a x+b x^2}}-\frac {32 b^2 (a+2 b x)}{21 a^4 \left (a x+b x^2\right )^{3/2}}+\frac {4 b}{7 a^2 x \left (a x+b x^2\right )^{3/2}}-\frac {2}{7 a x^2 \left (a x+b x^2\right )^{3/2}} \]

[In]

Int[1/(x^2*(a*x + b*x^2)^(5/2)),x]

[Out]

-2/(7*a*x^2*(a*x + b*x^2)^(3/2)) + (4*b)/(7*a^2*x*(a*x + b*x^2)^(3/2)) - (32*b^2*(a + 2*b*x))/(21*a^4*(a*x + b
*x^2)^(3/2)) + (256*b^3*(a + 2*b*x))/(21*a^6*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]

Rule 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-e)*(d + e*x)^m*((a
+ b*x + c*x^2)^(p + 1)/((m + p + 1)*(2*c*d - b*e))), x] + Dist[c*(Simplify[m + 2*p + 2]/((m + p + 1)*(2*c*d -
b*e))), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a
*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2}{7 a x^2 \left (a x+b x^2\right )^{3/2}}-\frac {(10 b) \int \frac {1}{x \left (a x+b x^2\right )^{5/2}} \, dx}{7 a} \\ & = -\frac {2}{7 a x^2 \left (a x+b x^2\right )^{3/2}}+\frac {4 b}{7 a^2 x \left (a x+b x^2\right )^{3/2}}+\frac {\left (16 b^2\right ) \int \frac {1}{\left (a x+b x^2\right )^{5/2}} \, dx}{7 a^2} \\ & = -\frac {2}{7 a x^2 \left (a x+b x^2\right )^{3/2}}+\frac {4 b}{7 a^2 x \left (a x+b x^2\right )^{3/2}}-\frac {32 b^2 (a+2 b x)}{21 a^4 \left (a x+b x^2\right )^{3/2}}-\frac {\left (128 b^3\right ) \int \frac {1}{\left (a x+b x^2\right )^{3/2}} \, dx}{21 a^4} \\ & = -\frac {2}{7 a x^2 \left (a x+b x^2\right )^{3/2}}+\frac {4 b}{7 a^2 x \left (a x+b x^2\right )^{3/2}}-\frac {32 b^2 (a+2 b x)}{21 a^4 \left (a x+b x^2\right )^{3/2}}+\frac {256 b^3 (a+2 b x)}{21 a^6 \sqrt {a x+b x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 \left (-3 a^5+6 a^4 b x-16 a^3 b^2 x^2+96 a^2 b^3 x^3+384 a b^4 x^4+256 b^5 x^5\right )}{21 a^6 x^2 (x (a+b x))^{3/2}} \]

[In]

Integrate[1/(x^2*(a*x + b*x^2)^(5/2)),x]

[Out]

(2*(-3*a^5 + 6*a^4*b*x - 16*a^3*b^2*x^2 + 96*a^2*b^3*x^3 + 384*a*b^4*x^4 + 256*b^5*x^5))/(21*a^6*x^2*(x*(a + b
*x))^(3/2))

Maple [A] (verified)

Time = 2.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(-\frac {2 \left (-\frac {256}{3} b^{5} x^{5}-128 a \,b^{4} x^{4}-32 a^{2} b^{3} x^{3}+\frac {16}{3} a^{3} b^{2} x^{2}-2 a^{4} b x +a^{5}\right )}{7 \sqrt {x \left (b x +a \right )}\, x^{3} \left (b x +a \right ) a^{6}}\)	\(75\)
gosper	\(-\frac {2 \left (b x +a \right ) \left (-256 b^{5} x^{5}-384 a \,b^{4} x^{4}-96 a^{2} b^{3} x^{3}+16 a^{3} b^{2} x^{2}-6 a^{4} b x +3 a^{5}\right )}{21 x \,a^{6} \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}\)	\(77\)
trager	\(-\frac {2 \left (-256 b^{5} x^{5}-384 a \,b^{4} x^{4}-96 a^{2} b^{3} x^{3}+16 a^{3} b^{2} x^{2}-6 a^{4} b x +3 a^{5}\right ) \sqrt {b \,x^{2}+a x}}{21 a^{6} x^{4} \left (b x +a \right )^{2}}\)	\(79\)
risch	\(-\frac {2 \left (b x +a \right ) \left (-158 b^{3} x^{3}+37 a \,b^{2} x^{2}-12 a^{2} b x +3 a^{3}\right )}{21 a^{6} x^{3} \sqrt {x \left (b x +a \right )}}+\frac {2 b^{4} \left (14 b x +15 a \right ) x}{3 \sqrt {x \left (b x +a \right )}\, \left (b x +a \right ) a^{6}}\)	\(87\)
default	\(-\frac {2}{7 a \,x^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {10 b \left (-\frac {2}{5 a x \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {8 b \left (-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\right )}{5 a}\right )}{7 a}\)	\(99\)

[In]

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

[Out]

-2/7*(-256/3*b^5*x^5-128*a*b^4*x^4-32*a^2*b^3*x^3+16/3*a^3*b^2*x^2-2*a^4*b*x+a^5)/(x*(b*x+a))^(1/2)/x^3/(b*x+a
)/a^6

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (256 \, b^{5} x^{5} + 384 \, a b^{4} x^{4} + 96 \, a^{2} b^{3} x^{3} - 16 \, a^{3} b^{2} x^{2} + 6 \, a^{4} b x - 3 \, a^{5}\right )} \sqrt {b x^{2} + a x}}{21 \, {\left (a^{6} b^{2} x^{6} + 2 \, a^{7} b x^{5} + a^{8} x^{4}\right )}} \]

[In]

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

[Out]

2/21*(256*b^5*x^5 + 384*a*b^4*x^4 + 96*a^2*b^3*x^3 - 16*a^3*b^2*x^2 + 6*a^4*b*x - 3*a^5)*sqrt(b*x^2 + a*x)/(a^
6*b^2*x^6 + 2*a^7*b*x^5 + a^8*x^4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{5/2}} \, dx=-\frac {64 \, b^{3} x}{21 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{4}} + \frac {512 \, b^{4} x}{21 \, \sqrt {b x^{2} + a x} a^{6}} - \frac {32 \, b^{2}}{21 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{3}} + \frac {256 \, b^{3}}{21 \, \sqrt {b x^{2} + a x} a^{5}} + \frac {4 \, b}{7 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2} x} - \frac {2}{7 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a x^{2}} \]

[In]

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

[Out]

-64/21*b^3*x/((b*x^2 + a*x)^(3/2)*a^4) + 512/21*b^4*x/(sqrt(b*x^2 + a*x)*a^6) - 32/21*b^2/((b*x^2 + a*x)^(3/2)
*a^3) + 256/21*b^3/(sqrt(b*x^2 + a*x)*a^5) + 4/7*b/((b*x^2 + a*x)^(3/2)*a^2*x) - 2/7/((b*x^2 + a*x)^(3/2)*a*x^
2)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.49 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x^2 \left (a x+b x^2\right )^{5/2}} \, dx=\frac {\sqrt {b\,x^2+a\,x}\,\left (\frac {256\,b^3}{21\,a^5}+\frac {512\,b^4\,x}{21\,a^6}\right )}{x\,\left (a+b\,x\right )}-\frac {\sqrt {b\,x^2+a\,x}\,\left (\frac {74\,b^2}{21\,a^3}+\frac {88\,b^3\,x}{21\,a^4}\right )}{x^2\,{\left (a+b\,x\right )}^2}-\frac {2\,\sqrt {b\,x^2+a\,x}}{7\,a^3\,x^4}+\frac {8\,b\,\sqrt {b\,x^2+a\,x}}{7\,a^4\,x^3} \]

[In]

int(1/(x^2*(a*x + b*x^2)^(5/2)),x)

[Out]

((a*x + b*x^2)^(1/2)*((256*b^3)/(21*a^5) + (512*b^4*x)/(21*a^6)))/(x*(a + b*x)) - ((a*x + b*x^2)^(1/2)*((74*b^
2)/(21*a^3) + (88*b^3*x)/(21*a^4)))/(x^2*(a + b*x)^2) - (2*(a*x + b*x^2)^(1/2))/(7*a^3*x^4) + (8*b*(a*x + b*x^
2)^(1/2))/(7*a^4*x^3)