\(\int \frac {(a+b x)^2}{x^3 (c x^2)^{3/2}} \, dx\) [842]

Optimal result

Integrand size = 20, antiderivative size = 66 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{3/2}} \, dx=-\frac {a^2}{5 c x^4 \sqrt {c x^2}}-\frac {a b}{2 c x^3 \sqrt {c x^2}}-\frac {b^2}{3 c x^2 \sqrt {c x^2}} \]

[Out]

-1/5*a^2/c/x^4/(c*x^2)^(1/2)-1/2*a*b/c/x^3/(c*x^2)^(1/2)-1/3*b^2/c/x^2/(c*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 45} \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{3/2}} \, dx=-\frac {a^2}{5 c x^4 \sqrt {c x^2}}-\frac {a b}{2 c x^3 \sqrt {c x^2}}-\frac {b^2}{3 c x^2 \sqrt {c x^2}} \]

[In]

Int[(a + b*x)^2/(x^3*(c*x^2)^(3/2)),x]

[Out]

-1/5*a^2/(c*x^4*Sqrt[c*x^2]) - (a*b)/(2*c*x^3*Sqrt[c*x^2]) - b^2/(3*c*x^2*Sqrt[c*x^2])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {x \int \frac {(a+b x)^2}{x^6} \, dx}{c \sqrt {c x^2}} \\ & = \frac {x \int \left (\frac {a^2}{x^6}+\frac {2 a b}{x^5}+\frac {b^2}{x^4}\right ) \, dx}{c \sqrt {c x^2}} \\ & = -\frac {a^2}{5 c x^4 \sqrt {c x^2}}-\frac {a b}{2 c x^3 \sqrt {c x^2}}-\frac {b^2}{3 c x^2 \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.53 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{3/2}} \, dx=\frac {-6 a^2-15 a b x-10 b^2 x^2}{30 x^2 \left (c x^2\right )^{3/2}} \]

[In]

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

[Out]

(-6*a^2 - 15*a*b*x - 10*b^2*x^2)/(30*x^2*(c*x^2)^(3/2))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.48

method	result	size
gosper	\(-\frac {10 b^{2} x^{2}+15 a b x +6 a^{2}}{30 x^{2} \left (c \,x^{2}\right )^{\frac {3}{2}}}\)	\(32\)
default	\(-\frac {10 b^{2} x^{2}+15 a b x +6 a^{2}}{30 x^{2} \left (c \,x^{2}\right )^{\frac {3}{2}}}\)	\(32\)
risch	\(\frac {-\frac {1}{3} b^{2} x^{2}-\frac {1}{2} a b x -\frac {1}{5} a^{2}}{c \,x^{4} \sqrt {c \,x^{2}}}\)	\(34\)
trager	\(\frac {\left (-1+x \right ) \left (6 a^{2} x^{4}+15 a b \,x^{4}+10 b^{2} x^{4}+6 a^{2} x^{3}+15 a b \,x^{3}+10 b^{2} x^{3}+6 a^{2} x^{2}+15 a b \,x^{2}+10 b^{2} x^{2}+6 a^{2} x +15 a b x +6 a^{2}\right ) \sqrt {c \,x^{2}}}{30 c^{2} x^{6}}\)	\(105\)

[In]

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

[Out]

-1/30*(10*b^2*x^2+15*a*b*x+6*a^2)/x^2/(c*x^2)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.52 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{3/2}} \, dx=-\frac {{\left (10 \, b^{2} x^{2} + 15 \, a b x + 6 \, a^{2}\right )} \sqrt {c x^{2}}}{30 \, c^{2} x^{6}} \]

[In]

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

[Out]

-1/30*(10*b^2*x^2 + 15*a*b*x + 6*a^2)*sqrt(c*x^2)/(c^2*x^6)

Sympy [A] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.70 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{3/2}} \, dx=- \frac {a^{2}}{5 x^{2} \left (c x^{2}\right )^{\frac {3}{2}}} - \frac {a b}{2 x \left (c x^{2}\right )^{\frac {3}{2}}} - \frac {b^{2}}{3 \left (c x^{2}\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

-a**2/(5*x**2*(c*x**2)**(3/2)) - a*b/(2*x*(c*x**2)**(3/2)) - b**2/(3*(c*x**2)**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.50 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{3/2}} \, dx=-\frac {b^{2}}{3 \, c^{\frac {3}{2}} x^{3}} - \frac {a b}{2 \, c^{\frac {3}{2}} x^{4}} - \frac {a^{2}}{5 \, c^{\frac {3}{2}} x^{5}} \]

[In]

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

[Out]

-1/3*b^2/(c^(3/2)*x^3) - 1/2*a*b/(c^(3/2)*x^4) - 1/5*a^2/(c^(3/2)*x^5)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.47 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{3/2}} \, dx=-\frac {10 \, b^{2} x^{2} + 15 \, a b x + 6 \, a^{2}}{30 \, c^{\frac {3}{2}} x^{5} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-1/30*(10*b^2*x^2 + 15*a*b*x + 6*a^2)/(c^(3/2)*x^5*sgn(x))

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.64 \[ \int \frac {(a+b x)^2}{x^3 \left (c x^2\right )^{3/2}} \, dx=-\frac {6\,a^2\,\sqrt {x^2}+10\,b^2\,x^2\,\sqrt {x^2}+15\,a\,b\,x\,\sqrt {x^2}}{30\,c^{3/2}\,x^6} \]

[In]

int((a + b*x)^2/(x^3*(c*x^2)^(3/2)),x)

[Out]

-(6*a^2*(x^2)^(1/2) + 10*b^2*x^2*(x^2)^(1/2) + 15*a*b*x*(x^2)^(1/2))/(30*c^(3/2)*x^6)