\(\int \frac {a+b x}{x^2 (c x^2)^{5/2}} \, dx\) [801]

Optimal result

Integrand size = 18, antiderivative size = 41 \[ \int \frac {a+b x}{x^2 \left (c x^2\right )^{5/2}} \, dx=-\frac {a}{6 c^2 x^5 \sqrt {c x^2}}-\frac {b}{5 c^2 x^4 \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 45} \[ \int \frac {a+b x}{x^2 \left (c x^2\right )^{5/2}} \, dx=-\frac {a}{6 c^2 x^5 \sqrt {c x^2}}-\frac {b}{5 c^2 x^4 \sqrt {c x^2}} \]

[In]

Int[(a + b*x)/(x^2*(c*x^2)^(5/2)),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.56 \[ \int \frac {a+b x}{x^2 \left (c x^2\right )^{5/2}} \, dx=-\frac {c x (5 a+6 b x)}{30 \left (c x^2\right )^{7/2}} \]

[In]

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

[Out]

-1/30*(c*x*(5*a + 6*b*x))/(c*x^2)^(7/2)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.51

method	result	size
gosper	\(-\frac {6 b x +5 a}{30 x \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(21\)
default	\(-\frac {6 b x +5 a}{30 x \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(21\)
risch	\(\frac {-\frac {b x}{5}-\frac {a}{6}}{c^{2} x^{5} \sqrt {c \,x^{2}}}\)	\(23\)
trager	\(\frac {\left (-1+x \right ) \left (5 a \,x^{5}+6 b \,x^{5}+5 a \,x^{4}+6 b \,x^{4}+5 a \,x^{3}+6 b \,x^{3}+5 a \,x^{2}+6 b \,x^{2}+5 a x +6 b x +5 a \right ) \sqrt {c \,x^{2}}}{30 c^{3} x^{7}}\)	\(79\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.56 \[ \int \frac {a+b x}{x^2 \left (c x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c x^{2}} {\left (6 \, b x + 5 \, a\right )}}{30 \, c^{3} x^{7}} \]

[In]

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

[Out]

-1/30*sqrt(c*x^2)*(6*b*x + 5*a)/(c^3*x^7)

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.63 \[ \int \frac {a+b x}{x^2 \left (c x^2\right )^{5/2}} \, dx=- \frac {a}{6 x \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {b}{5 \left (c x^{2}\right )^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.46 \[ \int \frac {a+b x}{x^2 \left (c x^2\right )^{5/2}} \, dx=-\frac {b}{5 \, c^{\frac {5}{2}} x^{5}} - \frac {a}{6 \, c^{\frac {5}{2}} x^{6}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.49 \[ \int \frac {a+b x}{x^2 \left (c x^2\right )^{5/2}} \, dx=-\frac {6 \, b x + 5 \, a}{30 \, c^{\frac {5}{2}} x^{6} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.63 \[ \int \frac {a+b x}{x^2 \left (c x^2\right )^{5/2}} \, dx=-\frac {5\,a\,\sqrt {x^2}+6\,b\,x\,\sqrt {x^2}}{30\,c^{5/2}\,x^7} \]

[In]

int((a + b*x)/(x^2*(c*x^2)^(5/2)),x)

[Out]

-(5*a*(x^2)^(1/2) + 6*b*x*(x^2)^(1/2))/(30*c^(5/2)*x^7)