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

Optimal. Leaf size=41 \[ -\frac {a}{6 c^2 x^5 \sqrt {c x^2}}-\frac {b}{5 c^2 x^4 \sqrt {c x^2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 43} \begin {gather*} -\frac {a}{6 c^2 x^5 \sqrt {c x^2}}-\frac {b}{5 c^2 x^4 \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/(x^2*(c*x^2)^(5/2)),x]

[Out]

-a/(6*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 43

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*} \int \frac {a+b x}{x^2 \left (c x^2\right )^{5/2}} \, dx &=\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*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.66 \begin {gather*} -\frac {\sqrt {c x^2} (5 a+6 b x)}{30 c^3 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 24, normalized size = 0.59 \begin {gather*} \frac {-5 a-6 b x}{30 x \left (c x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a - 6*b*x)/(30*x*(c*x^2)^(5/2))

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fricas [A]  time = 1.03, size = 23, normalized size = 0.56 \begin {gather*} -\frac {\sqrt {c x^{2}} {\left (6 \, b x + 5 \, a\right )}}{30 \, c^{3} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.00, size = 21, normalized size = 0.51 \begin {gather*} -\frac {6 b x +5 a}{30 \left (c \,x^{2}\right )^{\frac {5}{2}} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.32, size = 19, normalized size = 0.46 \begin {gather*} -\frac {b}{5 \, c^{\frac {5}{2}} x^{5}} - \frac {a}{6 \, c^{\frac {5}{2}} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.16, size = 26, normalized size = 0.63 \begin {gather*} -\frac {5\,a\,\sqrt {x^2}+6\,b\,x\,\sqrt {x^2}}{30\,c^{5/2}\,x^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 1.34, size = 32, normalized size = 0.78 \begin {gather*} - \frac {a}{6 c^{\frac {5}{2}} x \left (x^{2}\right )^{\frac {5}{2}}} - \frac {b}{5 c^{\frac {5}{2}} \left (x^{2}\right )^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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