\(\int \frac {(a+b x)^2}{x^4 (c x^2)^{3/2}} \, dx\) [843]

Optimal result

Integrand size = 20, antiderivative size = 66 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{3/2}} \, dx=-\frac {a^2}{6 c x^5 \sqrt {c x^2}}-\frac {2 a b}{5 c x^4 \sqrt {c x^2}}-\frac {b^2}{4 c x^3 \sqrt {c x^2}} \]

[Out]

-1/6*a^2/c/x^5/(c*x^2)^(1/2)-2/5*a*b/c/x^4/(c*x^2)^(1/2)-1/4*b^2/c/x^3/(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} \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{3/2}} \, dx=-\frac {a^2}{6 c x^5 \sqrt {c x^2}}-\frac {2 a b}{5 c x^4 \sqrt {c x^2}}-\frac {b^2}{4 c x^3 \sqrt {c x^2}} \]

[In]

Int[(a + b*x)^2/(x^4*(c*x^2)^(3/2)),x]

[Out]

-1/6*a^2/(c*x^5*Sqrt[c*x^2]) - (2*a*b)/(5*c*x^4*Sqrt[c*x^2]) - b^2/(4*c*x^3*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^7} \, dx}{c \sqrt {c x^2}} \\ & = \frac {x \int \left (\frac {a^2}{x^7}+\frac {2 a b}{x^6}+\frac {b^2}{x^5}\right ) \, dx}{c \sqrt {c x^2}} \\ & = -\frac {a^2}{6 c x^5 \sqrt {c x^2}}-\frac {2 a b}{5 c x^4 \sqrt {c x^2}}-\frac {b^2}{4 c x^3 \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.53 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{3/2}} \, dx=\frac {-10 a^2-24 a b x-15 b^2 x^2}{60 x^3 \left (c x^2\right )^{3/2}} \]

[In]

Integrate[(a + b*x)^2/(x^4*(c*x^2)^(3/2)),x]

[Out]

(-10*a^2 - 24*a*b*x - 15*b^2*x^2)/(60*x^3*(c*x^2)^(3/2))

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.48

method	result	size
gosper	\(-\frac {15 b^{2} x^{2}+24 a b x +10 a^{2}}{60 x^{3} \left (c \,x^{2}\right )^{\frac {3}{2}}}\)	\(32\)
default	\(-\frac {15 b^{2} x^{2}+24 a b x +10 a^{2}}{60 x^{3} \left (c \,x^{2}\right )^{\frac {3}{2}}}\)	\(32\)
risch	\(\frac {-\frac {1}{4} b^{2} x^{2}-\frac {2}{5} a b x -\frac {1}{6} a^{2}}{c \,x^{5} \sqrt {c \,x^{2}}}\)	\(34\)
trager	\(\frac {\left (-1+x \right ) \left (10 a^{2} x^{5}+24 a b \,x^{5}+15 b^{2} x^{5}+10 a^{2} x^{4}+24 a b \,x^{4}+15 b^{2} x^{4}+10 a^{2} x^{3}+24 a b \,x^{3}+15 b^{2} x^{3}+10 a^{2} x^{2}+24 a b \,x^{2}+15 b^{2} x^{2}+10 a^{2} x +24 a b x +10 a^{2}\right ) \sqrt {c \,x^{2}}}{60 c^{2} x^{7}}\)	\(128\)

[In]

int((b*x+a)^2/x^4/(c*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/60*(15*b^2*x^2+24*a*b*x+10*a^2)/x^3/(c*x^2)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.52 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{3/2}} \, dx=-\frac {{\left (15 \, b^{2} x^{2} + 24 \, a b x + 10 \, a^{2}\right )} \sqrt {c x^{2}}}{60 \, c^{2} x^{7}} \]

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(3/2),x, algorithm="fricas")

[Out]

-1/60*(15*b^2*x^2 + 24*a*b*x + 10*a^2)*sqrt(c*x^2)/(c^2*x^7)

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{3/2}} \, dx=- \frac {a^{2}}{6 x^{3} \left (c x^{2}\right )^{\frac {3}{2}}} - \frac {2 a b}{5 x^{2} \left (c x^{2}\right )^{\frac {3}{2}}} - \frac {b^{2}}{4 x \left (c x^{2}\right )^{\frac {3}{2}}} \]

[In]

integrate((b*x+a)**2/x**4/(c*x**2)**(3/2),x)

[Out]

-a**2/(6*x**3*(c*x**2)**(3/2)) - 2*a*b/(5*x**2*(c*x**2)**(3/2)) - b**2/(4*x*(c*x**2)**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.50 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{3/2}} \, dx=-\frac {b^{2}}{4 \, c^{\frac {3}{2}} x^{4}} - \frac {2 \, a b}{5 \, c^{\frac {3}{2}} x^{5}} - \frac {a^{2}}{6 \, c^{\frac {3}{2}} x^{6}} \]

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(3/2),x, algorithm="maxima")

[Out]

-1/4*b^2/(c^(3/2)*x^4) - 2/5*a*b/(c^(3/2)*x^5) - 1/6*a^2/(c^(3/2)*x^6)

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.47 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{3/2}} \, dx=-\frac {15 \, b^{2} x^{2} + 24 \, a b x + 10 \, a^{2}}{60 \, c^{\frac {3}{2}} x^{6} \mathrm {sgn}\left (x\right )} \]

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(3/2),x, algorithm="giac")

[Out]

-1/60*(15*b^2*x^2 + 24*a*b*x + 10*a^2)/(c^(3/2)*x^6*sgn(x))

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.64 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{3/2}} \, dx=-\frac {10\,a^2\,\sqrt {x^2}+15\,b^2\,x^2\,\sqrt {x^2}+24\,a\,b\,x\,\sqrt {x^2}}{60\,c^{3/2}\,x^7} \]

[In]

int((a + b*x)^2/(x^4*(c*x^2)^(3/2)),x)

[Out]

-(10*a^2*(x^2)^(1/2) + 15*b^2*x^2*(x^2)^(1/2) + 24*a*b*x*(x^2)^(1/2))/(60*c^(3/2)*x^7)