\(\int \frac {1}{x^3 (b x+c x^2)^{3/2}} \, dx\) [59]

Optimal result

Integrand size = 17, antiderivative size = 103 \[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2}{7 b x^3 \sqrt {b x+c x^2}}+\frac {16 c}{35 b^2 x^2 \sqrt {b x+c x^2}}-\frac {32 c^2}{35 b^3 x \sqrt {b x+c x^2}}+\frac {128 c^3 (b+2 c x)}{35 b^5 \sqrt {b x+c x^2}} \]

[Out]

-2/7/b/x^3/(c*x^2+b*x)^(1/2)+16/35*c/b^2/x^2/(c*x^2+b*x)^(1/2)-32/35*c^2/b^3/x/(c*x^2+b*x)^(1/2)+128/35*c^3*(2
*c*x+b)/b^5/(c*x^2+b*x)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {672, 627} \[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {128 c^3 (b+2 c x)}{35 b^5 \sqrt {b x+c x^2}}-\frac {32 c^2}{35 b^3 x \sqrt {b x+c x^2}}+\frac {16 c}{35 b^2 x^2 \sqrt {b x+c x^2}}-\frac {2}{7 b x^3 \sqrt {b x+c x^2}} \]

[In]

Int[1/(x^3*(b*x + c*x^2)^(3/2)),x]

[Out]

-2/(7*b*x^3*Sqrt[b*x + c*x^2]) + (16*c)/(35*b^2*x^2*Sqrt[b*x + c*x^2]) - (32*c^2)/(35*b^3*x*Sqrt[b*x + c*x^2])
+ (128*c^3*(b + 2*c*x))/(35*b^5*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 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 b x^3 \sqrt {b x+c x^2}}-\frac {(8 c) \int \frac {1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx}{7 b} \\ & = -\frac {2}{7 b x^3 \sqrt {b x+c x^2}}+\frac {16 c}{35 b^2 x^2 \sqrt {b x+c x^2}}+\frac {\left (48 c^2\right ) \int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx}{35 b^2} \\ & = -\frac {2}{7 b x^3 \sqrt {b x+c x^2}}+\frac {16 c}{35 b^2 x^2 \sqrt {b x+c x^2}}-\frac {32 c^2}{35 b^3 x \sqrt {b x+c x^2}}-\frac {\left (64 c^3\right ) \int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx}{35 b^3} \\ & = -\frac {2}{7 b x^3 \sqrt {b x+c x^2}}+\frac {16 c}{35 b^2 x^2 \sqrt {b x+c x^2}}-\frac {32 c^2}{35 b^3 x \sqrt {b x+c x^2}}+\frac {128 c^3 (b+2 c x)}{35 b^5 \sqrt {b x+c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \left (-5 b^4+8 b^3 c x-16 b^2 c^2 x^2+64 b c^3 x^3+128 c^4 x^4\right )}{35 b^5 x^3 \sqrt {x (b+c x)}} \]

[In]

Integrate[1/(x^3*(b*x + c*x^2)^(3/2)),x]

[Out]

(2*(-5*b^4 + 8*b^3*c*x - 16*b^2*c^2*x^2 + 64*b*c^3*x^3 + 128*c^4*x^4))/(35*b^5*x^3*Sqrt[x*(b + c*x)])

Maple [A] (verified)

Time = 1.86 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.57

method	result	size
pseudoelliptic	\(\frac {\frac {256}{35} c^{4} x^{4}+\frac {128}{35} b \,c^{3} x^{3}-\frac {32}{35} b^{2} c^{2} x^{2}+\frac {16}{35} b^{3} c x -\frac {2}{7} b^{4}}{x^{3} \sqrt {x \left (c x +b \right )}\, b^{5}}\)	\(59\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (-128 c^{4} x^{4}-64 b \,c^{3} x^{3}+16 b^{2} c^{2} x^{2}-8 b^{3} c x +5 b^{4}\right )}{35 x^{2} b^{5} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\)	\(66\)
trager	\(-\frac {2 \left (-128 c^{4} x^{4}-64 b \,c^{3} x^{3}+16 b^{2} c^{2} x^{2}-8 b^{3} c x +5 b^{4}\right ) \sqrt {c \,x^{2}+b x}}{35 \left (c x +b \right ) b^{5} x^{4}}\)	\(68\)
risch	\(-\frac {2 \left (c x +b \right ) \left (-93 c^{3} x^{3}+29 b \,c^{2} x^{2}-13 b^{2} c x +5 b^{3}\right )}{35 b^{5} x^{3} \sqrt {x \left (c x +b \right )}}+\frac {2 c^{4} x}{\sqrt {x \left (c x +b \right )}\, b^{5}}\)	\(72\)
default	\(-\frac {2}{7 b \,x^{3} \sqrt {c \,x^{2}+b x}}-\frac {8 c \left (-\frac {2}{5 b \,x^{2} \sqrt {c \,x^{2}+b x}}-\frac {6 c \left (-\frac {2}{3 b x \sqrt {c \,x^{2}+b x}}+\frac {8 c \left (2 c x +b \right )}{3 b^{3} \sqrt {c \,x^{2}+b x}}\right )}{5 b}\right )}{7 b}\)	\(96\)

[In]

int(1/x^3/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/35*(128*c^4*x^4+64*b*c^3*x^3-16*b^2*c^2*x^2+8*b^3*c*x-5*b^4)/x^3/(x*(c*x+b))^(1/2)/b^5

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.70 \[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (128 \, c^{4} x^{4} + 64 \, b c^{3} x^{3} - 16 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 5 \, b^{4}\right )} \sqrt {c x^{2} + b x}}{35 \, {\left (b^{5} c x^{5} + b^{6} x^{4}\right )}} \]

[In]

integrate(1/x^3/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

2/35*(128*c^4*x^4 + 64*b*c^3*x^3 - 16*b^2*c^2*x^2 + 8*b^3*c*x - 5*b^4)*sqrt(c*x^2 + b*x)/(b^5*c*x^5 + b^6*x^4)

Sympy [F]

\[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{x^{3} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/x**3/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(1/(x**3*(x*(b + c*x))**(3/2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {256 \, c^{4} x}{35 \, \sqrt {c x^{2} + b x} b^{5}} + \frac {128 \, c^{3}}{35 \, \sqrt {c x^{2} + b x} b^{4}} - \frac {32 \, c^{2}}{35 \, \sqrt {c x^{2} + b x} b^{3} x} + \frac {16 \, c}{35 \, \sqrt {c x^{2} + b x} b^{2} x^{2}} - \frac {2}{7 \, \sqrt {c x^{2} + b x} b x^{3}} \]

[In]

integrate(1/x^3/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

256/35*c^4*x/(sqrt(c*x^2 + b*x)*b^5) + 128/35*c^3/(sqrt(c*x^2 + b*x)*b^4) - 32/35*c^2/(sqrt(c*x^2 + b*x)*b^3*x
) + 16/35*c/(sqrt(c*x^2 + b*x)*b^2*x^2) - 2/7/(sqrt(c*x^2 + b*x)*b*x^3)

Giac [F]

\[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{3}} \,d x } \]

[In]

integrate(1/x^3/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((c*x^2 + b*x)^(3/2)*x^3), x)

Mupad [B] (verification not implemented)

Time = 9.14 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c\,x^2+b\,x}\,\left (\frac {186\,c^3}{35\,b^4}+\frac {256\,c^4\,x}{35\,b^5}\right )}{x\,\left (b+c\,x\right )}-\frac {58\,c^2\,\sqrt {c\,x^2+b\,x}}{35\,b^4\,x^2}-\frac {2\,\sqrt {c\,x^2+b\,x}}{7\,b^2\,x^4}+\frac {26\,c\,\sqrt {c\,x^2+b\,x}}{35\,b^3\,x^3} \]

[In]

int(1/(x^3*(b*x + c*x^2)^(3/2)),x)

[Out]

((b*x + c*x^2)^(1/2)*((186*c^3)/(35*b^4) + (256*c^4*x)/(35*b^5)))/(x*(b + c*x)) - (58*c^2*(b*x + c*x^2)^(1/2))
/(35*b^4*x^2) - (2*(b*x + c*x^2)^(1/2))/(7*b^2*x^4) + (26*c*(b*x + c*x^2)^(1/2))/(35*b^3*x^3)