\(\int \frac {x (A+B x)}{(b x+c x^2)^{5/2}} \, dx\) [133]

Optimal result

Integrand size = 20, antiderivative size = 70 \[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 (b B-A c) x}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac {2 (b B-4 A c) (b+2 c x)}{3 b^3 c \sqrt {b x+c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {791, 627} \[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 (b+2 c x) (b B-4 A c)}{3 b^3 c \sqrt {b x+c x^2}}-\frac {2 x (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

[In]

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

[Out]

(-2*(b*B - A*c)*x)/(3*b*c*(b*x + c*x^2)^(3/2)) + (2*(b*B - 4*A*c)*(b + 2*c*x))/(3*b^3*c*Sqrt[b*x + c*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 791

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

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.79 \[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {x \left (2 b B x (3 b+2 c x)-2 A \left (3 b^2+12 b c x+8 c^2 x^2\right )\right )}{3 b^3 (x (b+c x))^{3/2}} \]

[In]

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

[Out]

(x*(2*b*B*x*(3*b + 2*c*x) - 2*A*(3*b^2 + 12*b*c*x + 8*c^2*x^2)))/(3*b^3*(x*(b + c*x))^(3/2))

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(-\frac {2 \left (\left (-B x +A \right ) b^{2}+4 c \left (-\frac {B x}{6}+A \right ) x b +\frac {8 A \,c^{2} x^{2}}{3}\right )}{\sqrt {x \left (c x +b \right )}\, \left (c x +b \right ) b^{3}}\)	\(53\)
gosper	\(-\frac {2 x^{2} \left (c x +b \right ) \left (8 A \,c^{2} x^{2}-2 B b c \,x^{2}+12 A b c x -3 b^{2} B x +3 A \,b^{2}\right )}{3 b^{3} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}\)	\(62\)
trager	\(-\frac {2 \left (8 A \,c^{2} x^{2}-2 B b c \,x^{2}+12 A b c x -3 b^{2} B x +3 A \,b^{2}\right ) \sqrt {c \,x^{2}+b x}}{3 b^{3} \left (c x +b \right )^{2} x}\)	\(64\)
risch	\(-\frac {2 A \left (c x +b \right )}{b^{3} \sqrt {x \left (c x +b \right )}}-\frac {2 \left (5 A \,c^{2} x -2 B b c x +6 A b c -3 B \,b^{2}\right ) x}{3 \sqrt {x \left (c x +b \right )}\, \left (c x +b \right ) b^{3}}\)	\(69\)
default	\(B \left (-\frac {x}{2 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )}{4 c}\right )+A \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )\)	\(168\)

[In]

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

[Out]

-2*((-B*x+A)*b^2+4*c*(-1/6*B*x+A)*x*b+8/3*A*c^2*x^2)/(x*(c*x+b))^(1/2)/(c*x+b)/b^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.10 \[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, A b^{2} - 2 \, {\left (B b c - 4 \, A c^{2}\right )} x^{2} - 3 \, {\left (B b^{2} - 4 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (b^{3} c^{2} x^{3} + 2 \, b^{4} c x^{2} + b^{5} x\right )}} \]

[In]

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

[Out]

-2/3*(3*A*b^2 - 2*(B*b*c - 4*A*c^2)*x^2 - 3*(B*b^2 - 4*A*b*c)*x)*sqrt(c*x^2 + b*x)/(b^3*c^2*x^3 + 2*b^4*c*x^2
+ b^5*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.59 \[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {4 \, B x}{3 \, \sqrt {c x^{2} + b x} b^{2}} + \frac {2 \, A x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} - \frac {2 \, B x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {16 \, A c x}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {8 \, A}{3 \, \sqrt {c x^{2} + b x} b^{2}} + \frac {2 \, B}{3 \, \sqrt {c x^{2} + b x} b c} \]

[In]

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

[Out]

4/3*B*x/(sqrt(c*x^2 + b*x)*b^2) + 2/3*A*x/((c*x^2 + b*x)^(3/2)*b) - 2/3*B*x/((c*x^2 + b*x)^(3/2)*c) - 16/3*A*c
*x/(sqrt(c*x^2 + b*x)*b^3) - 8/3*A/(sqrt(c*x^2 + b*x)*b^2) + 2/3*B/(sqrt(c*x^2 + b*x)*b*c)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.59 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90 \[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (-3\,B\,b^2\,x+3\,A\,b^2-2\,B\,b\,c\,x^2+12\,A\,b\,c\,x+8\,A\,c^2\,x^2\right )}{3\,b^3\,x\,{\left (b+c\,x\right )}^2} \]

[In]

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

[Out]

-(2*(b*x + c*x^2)^(1/2)*(3*A*b^2 + 8*A*c^2*x^2 - 3*B*b^2*x - 2*B*b*c*x^2 + 12*A*b*c*x))/(3*b^3*x*(b + c*x)^2)