\(\int \frac {(a+b x)^2}{x^4 (c x^2)^{5/2}} \, dx\) [851]

Optimal result

Integrand size = 20, antiderivative size = 66 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{5/2}} \, dx=-\frac {a^2}{8 c^2 x^7 \sqrt {c x^2}}-\frac {2 a b}{7 c^2 x^6 \sqrt {c x^2}}-\frac {b^2}{6 c^2 x^5 \sqrt {c x^2}} \]

[Out]

-1/8*a^2/c^2/x^7/(c*x^2)^(1/2)-2/7*a*b/c^2/x^6/(c*x^2)^(1/2)-1/6*b^2/c^2/x^5/(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} \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{5/2}} \, dx=-\frac {a^2}{8 c^2 x^7 \sqrt {c x^2}}-\frac {2 a b}{7 c^2 x^6 \sqrt {c x^2}}-\frac {b^2}{6 c^2 x^5 \sqrt {c x^2}} \]

[In]

Int[(a + b*x)^2/(x^4*(c*x^2)^(5/2)),x]

[Out]

-1/8*a^2/(c^2*x^7*Sqrt[c*x^2]) - (2*a*b)/(7*c^2*x^6*Sqrt[c*x^2]) - b^2/(6*c^2*x^5*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^9} \, dx}{c^2 \sqrt {c x^2}} \\ & = \frac {x \int \left (\frac {a^2}{x^9}+\frac {2 a b}{x^8}+\frac {b^2}{x^7}\right ) \, dx}{c^2 \sqrt {c x^2}} \\ & = -\frac {a^2}{8 c^2 x^7 \sqrt {c x^2}}-\frac {2 a b}{7 c^2 x^6 \sqrt {c x^2}}-\frac {b^2}{6 c^2 x^5 \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.53 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{5/2}} \, dx=\frac {-21 a^2-48 a b x-28 b^2 x^2}{168 x^3 \left (c x^2\right )^{5/2}} \]

[In]

Integrate[(a + b*x)^2/(x^4*(c*x^2)^(5/2)),x]

[Out]

(-21*a^2 - 48*a*b*x - 28*b^2*x^2)/(168*x^3*(c*x^2)^(5/2))

Maple [A] (verified)

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

method	result	size
gosper	\(-\frac {28 b^{2} x^{2}+48 a b x +21 a^{2}}{168 x^{3} \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(32\)
default	\(-\frac {28 b^{2} x^{2}+48 a b x +21 a^{2}}{168 x^{3} \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(32\)
risch	\(\frac {-\frac {1}{6} b^{2} x^{2}-\frac {2}{7} a b x -\frac {1}{8} a^{2}}{c^{2} x^{7} \sqrt {c \,x^{2}}}\)	\(34\)
trager	\(\frac {\left (-1+x \right ) \left (21 a^{2} x^{7}+48 a b \,x^{7}+28 b^{2} x^{7}+21 a^{2} x^{6}+48 a b \,x^{6}+28 b^{2} x^{6}+21 a^{2} x^{5}+48 a b \,x^{5}+28 b^{2} x^{5}+21 a^{2} x^{4}+48 a b \,x^{4}+28 b^{2} x^{4}+21 a^{2} x^{3}+48 a b \,x^{3}+28 b^{2} x^{3}+21 a^{2} x^{2}+48 a b \,x^{2}+28 b^{2} x^{2}+21 a^{2} x +48 a b x +21 a^{2}\right ) \sqrt {c \,x^{2}}}{168 c^{3} x^{9}}\)	\(174\)

[In]

int((b*x+a)^2/x^4/(c*x^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/168*(28*b^2*x^2+48*a*b*x+21*a^2)/x^3/(c*x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.52 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{5/2}} \, dx=-\frac {{\left (28 \, b^{2} x^{2} + 48 \, a b x + 21 \, a^{2}\right )} \sqrt {c x^{2}}}{168 \, c^{3} x^{9}} \]

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(5/2),x, algorithm="fricas")

[Out]

-1/168*(28*b^2*x^2 + 48*a*b*x + 21*a^2)*sqrt(c*x^2)/(c^3*x^9)

Sympy [A] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{5/2}} \, dx=- \frac {a^{2}}{8 x^{3} \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {2 a b}{7 x^{2} \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {b^{2}}{6 x \left (c x^{2}\right )^{\frac {5}{2}}} \]

[In]

integrate((b*x+a)**2/x**4/(c*x**2)**(5/2),x)

[Out]

-a**2/(8*x**3*(c*x**2)**(5/2)) - 2*a*b/(7*x**2*(c*x**2)**(5/2)) - b**2/(6*x*(c*x**2)**(5/2))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.50 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{5/2}} \, dx=-\frac {b^{2}}{6 \, c^{\frac {5}{2}} x^{6}} - \frac {2 \, a b}{7 \, c^{\frac {5}{2}} x^{7}} - \frac {a^{2}}{8 \, c^{\frac {5}{2}} x^{8}} \]

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(5/2),x, algorithm="maxima")

[Out]

-1/6*b^2/(c^(5/2)*x^6) - 2/7*a*b/(c^(5/2)*x^7) - 1/8*a^2/(c^(5/2)*x^8)

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.47 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{5/2}} \, dx=-\frac {28 \, b^{2} x^{2} + 48 \, a b x + 21 \, a^{2}}{168 \, c^{\frac {5}{2}} x^{8} \mathrm {sgn}\left (x\right )} \]

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(5/2),x, algorithm="giac")

[Out]

-1/168*(28*b^2*x^2 + 48*a*b*x + 21*a^2)/(c^(5/2)*x^8*sgn(x))

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.64 \[ \int \frac {(a+b x)^2}{x^4 \left (c x^2\right )^{5/2}} \, dx=-\frac {21\,a^2\,\sqrt {x^2}+28\,b^2\,x^2\,\sqrt {x^2}+48\,a\,b\,x\,\sqrt {x^2}}{168\,c^{5/2}\,x^9} \]

[In]

int((a + b*x)^2/(x^4*(c*x^2)^(5/2)),x)

[Out]

-(21*a^2*(x^2)^(1/2) + 28*b^2*x^2*(x^2)^(1/2) + 48*a*b*x*(x^2)^(1/2))/(168*c^(5/2)*x^9)