\(\int \frac {(a+b x)^2}{x^4 \sqrt {c x^2}} \, dx\) [835]

Optimal result

Integrand size = 20, antiderivative size = 57 \[ \int \frac {(a+b x)^2}{x^4 \sqrt {c x^2}} \, dx=-\frac {a^2}{4 x^3 \sqrt {c x^2}}-\frac {2 a b}{3 x^2 \sqrt {c x^2}}-\frac {b^2}{2 x \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 57, 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} \[ \int \frac {(a+b x)^2}{x^4 \sqrt {c x^2}} \, dx=-\frac {a^2}{4 x^3 \sqrt {c x^2}}-\frac {2 a b}{3 x^2 \sqrt {c x^2}}-\frac {b^2}{2 x \sqrt {c x^2}} \]

[In]

Int[(a + b*x)^2/(x^4*Sqrt[c*x^2]),x]

[Out]

-1/4*a^2/(x^3*Sqrt[c*x^2]) - (2*a*b)/(3*x^2*Sqrt[c*x^2]) - b^2/(2*x*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^5} \, dx}{\sqrt {c x^2}} \\ & = \frac {x \int \left (\frac {a^2}{x^5}+\frac {2 a b}{x^4}+\frac {b^2}{x^3}\right ) \, dx}{\sqrt {c x^2}} \\ & = -\frac {a^2}{4 x^3 \sqrt {c x^2}}-\frac {2 a b}{3 x^2 \sqrt {c x^2}}-\frac {b^2}{2 x \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.61 \[ \int \frac {(a+b x)^2}{x^4 \sqrt {c x^2}} \, dx=-\frac {3 a^2+8 a b x+6 b^2 x^2}{12 x^3 \sqrt {c x^2}} \]

[In]

Integrate[(a + b*x)^2/(x^4*Sqrt[c*x^2]),x]

[Out]

-1/12*(3*a^2 + 8*a*b*x + 6*b^2*x^2)/(x^3*Sqrt[c*x^2])

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.54

method	result	size
risch	\(\frac {-\frac {1}{2} b^{2} x^{2}-\frac {2}{3} a b x -\frac {1}{4} a^{2}}{x^{3} \sqrt {c \,x^{2}}}\)	\(31\)
gosper	\(-\frac {6 b^{2} x^{2}+8 a b x +3 a^{2}}{12 x^{3} \sqrt {c \,x^{2}}}\)	\(32\)
default	\(-\frac {6 b^{2} x^{2}+8 a b x +3 a^{2}}{12 x^{3} \sqrt {c \,x^{2}}}\)	\(32\)
trager	\(\frac {\left (-1+x \right ) \left (3 a^{2} x^{3}+8 a b \,x^{3}+6 b^{2} x^{3}+3 a^{2} x^{2}+8 a b \,x^{2}+6 b^{2} x^{2}+3 a^{2} x +8 a b x +3 a^{2}\right ) \sqrt {c \,x^{2}}}{12 c \,x^{5}}\)	\(82\)

[In]

int((b*x+a)^2/x^4/(c*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.60 \[ \int \frac {(a+b x)^2}{x^4 \sqrt {c x^2}} \, dx=-\frac {{\left (6 \, b^{2} x^{2} + 8 \, a b x + 3 \, a^{2}\right )} \sqrt {c x^{2}}}{12 \, c x^{5}} \]

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(1/2),x, algorithm="fricas")

[Out]

-1/12*(6*b^2*x^2 + 8*a*b*x + 3*a^2)*sqrt(c*x^2)/(c*x^5)

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b x)^2}{x^4 \sqrt {c x^2}} \, dx=- \frac {a^{2}}{4 x^{3} \sqrt {c x^{2}}} - \frac {2 a b}{3 x^{2} \sqrt {c x^{2}}} - \frac {b^{2}}{2 x \sqrt {c x^{2}}} \]

[In]

integrate((b*x+a)**2/x**4/(c*x**2)**(1/2),x)

[Out]

-a**2/(4*x**3*sqrt(c*x**2)) - 2*a*b/(3*x**2*sqrt(c*x**2)) - b**2/(2*x*sqrt(c*x**2))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.58 \[ \int \frac {(a+b x)^2}{x^4 \sqrt {c x^2}} \, dx=-\frac {b^{2}}{2 \, \sqrt {c} x^{2}} - \frac {2 \, a b}{3 \, \sqrt {c} x^{3}} - \frac {a^{2}}{4 \, \sqrt {c} x^{4}} \]

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.54 \[ \int \frac {(a+b x)^2}{x^4 \sqrt {c x^2}} \, dx=-\frac {6 \, b^{2} x^{2} + 8 \, a b x + 3 \, a^{2}}{12 \, \sqrt {c} x^{4} \mathrm {sgn}\left (x\right )} \]

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(1/2),x, algorithm="giac")

[Out]

-1/12*(6*b^2*x^2 + 8*a*b*x + 3*a^2)/(sqrt(c)*x^4*sgn(x))

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b x)^2}{x^4 \sqrt {c x^2}} \, dx=-\frac {3\,a^2\,\sqrt {x^2}+6\,b^2\,x^2\,\sqrt {x^2}+8\,a\,b\,x\,\sqrt {x^2}}{12\,\sqrt {c}\,x^5} \]

[In]

int((a + b*x)^2/(x^4*(c*x^2)^(1/2)),x)

[Out]

-(3*a^2*(x^2)^(1/2) + 6*b^2*x^2*(x^2)^(1/2) + 8*a*b*x*(x^2)^(1/2))/(12*c^(1/2)*x^5)