\(\int \frac {a+b x}{x^4 (c x^2)^{5/2}} \, dx\) [803]

Optimal result

Integrand size = 18, antiderivative size = 41 \[ \int \frac {a+b x}{x^4 \left (c x^2\right )^{5/2}} \, dx=-\frac {a}{8 c^2 x^7 \sqrt {c x^2}}-\frac {b}{7 c^2 x^6 \sqrt {c x^2}} \]

[Out]

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

[In]

Int[(a + b*x)/(x^4*(c*x^2)^(5/2)),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.59 \[ \int \frac {a+b x}{x^4 \left (c x^2\right )^{5/2}} \, dx=\frac {-7 a-8 b x}{56 x^3 \left (c x^2\right )^{5/2}} \]

[In]

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

[Out]

(-7*a - 8*b*x)/(56*x^3*(c*x^2)^(5/2))

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.51

method	result	size
gosper	\(-\frac {8 b x +7 a}{56 x^{3} \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(21\)
default	\(-\frac {8 b x +7 a}{56 x^{3} \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(21\)
risch	\(\frac {-\frac {b x}{7}-\frac {a}{8}}{c^{2} x^{7} \sqrt {c \,x^{2}}}\)	\(23\)
trager	\(\frac {\left (-1+x \right ) \left (7 a \,x^{7}+8 b \,x^{7}+7 a \,x^{6}+8 b \,x^{6}+7 a \,x^{5}+8 b \,x^{5}+7 a \,x^{4}+8 b \,x^{4}+7 a \,x^{3}+8 b \,x^{3}+7 a \,x^{2}+8 b \,x^{2}+7 a x +8 b x +7 a \right ) \sqrt {c \,x^{2}}}{56 c^{3} x^{9}}\)	\(103\)

[In]

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

[Out]

-1/56*(8*b*x+7*a)/x^3/(c*x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.56 \[ \int \frac {a+b x}{x^4 \left (c x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c x^{2}} {\left (8 \, b x + 7 \, a\right )}}{56 \, c^{3} x^{9}} \]

[In]

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

[Out]

-1/56*sqrt(c*x^2)*(8*b*x + 7*a)/(c^3*x^9)

Sympy [A] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \frac {a+b x}{x^4 \left (c x^2\right )^{5/2}} \, dx=- \frac {a}{8 x^{3} \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {b}{7 x^{2} \left (c x^{2}\right )^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.46 \[ \int \frac {a+b x}{x^4 \left (c x^2\right )^{5/2}} \, dx=-\frac {b}{7 \, c^{\frac {5}{2}} x^{7}} - \frac {a}{8 \, c^{\frac {5}{2}} x^{8}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.49 \[ \int \frac {a+b x}{x^4 \left (c x^2\right )^{5/2}} \, dx=-\frac {8 \, b x + 7 \, a}{56 \, c^{\frac {5}{2}} x^{8} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-1/56*(8*b*x + 7*a)/(c^(5/2)*x^8*sgn(x))

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.63 \[ \int \frac {a+b x}{x^4 \left (c x^2\right )^{5/2}} \, dx=-\frac {7\,a\,\sqrt {x^2}+8\,b\,x\,\sqrt {x^2}}{56\,c^{5/2}\,x^9} \]

[In]

int((a + b*x)/(x^4*(c*x^2)^(5/2)),x)

[Out]

-(7*a*(x^2)^(1/2) + 8*b*x*(x^2)^(1/2))/(56*c^(5/2)*x^9)