3.9.39 \(\int \frac {(a+b x)^2}{(c x^2)^{3/2}} \, dx\) [839]

Optimal. Leaf size=58 \[ -\frac {2 a b}{c \sqrt {c x^2}}-\frac {a^2}{2 c x \sqrt {c x^2}}+\frac {b^2 x \log (x)}{c \sqrt {c x^2}} \]

[Out]

-2*a*b/c/(c*x^2)^(1/2)-1/2*a^2/c/x/(c*x^2)^(1/2)+b^2*x*ln(x)/c/(c*x^2)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {15, 45} \begin {gather*} -\frac {a^2}{2 c x \sqrt {c x^2}}-\frac {2 a b}{c \sqrt {c x^2}}+\frac {b^2 x \log (x)}{c \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(c*x^2)^(3/2),x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.00, size = 34, normalized size = 0.59 \begin {gather*} \frac {x \left (-a (a+4 b x)+2 b^2 x^2 \log (x)\right )}{2 \left (c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-(a*(a + 4*b*x)) + 2*b^2*x^2*Log[x]))/(2*(c*x^2)^(3/2))

________________________________________________________________________________________

Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded while calling a Python object} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

cought exception: maximum recursion depth exceeded while calling a Python object

________________________________________________________________________________________

Maple [A]
time = 0.11, size = 32, normalized size = 0.55

method result size
default \(\frac {x \left (2 b^{2} \ln \left (x \right ) x^{2}-4 a b x -a^{2}\right )}{2 \left (c \,x^{2}\right )^{\frac {3}{2}}}\) \(32\)
risch \(\frac {-\frac {1}{2} a^{2}-2 a b x}{c x \sqrt {c \,x^{2}}}+\frac {b^{2} x \ln \left (x \right )}{c \sqrt {c \,x^{2}}}\) \(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 35, normalized size = 0.60 \begin {gather*} \frac {b^{2} \log \left (x\right )}{c^{\frac {3}{2}}} - \frac {2 \, a b}{\sqrt {c x^{2}} c} - \frac {a^{2}}{2 \, c^{\frac {3}{2}} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*log(x)/c^(3/2) - 2*a*b/(sqrt(c*x^2)*c) - 1/2*a^2/(c^(3/2)*x^2)

________________________________________________________________________________________

Fricas [A]
time = 0.30, size = 36, normalized size = 0.62 \begin {gather*} \frac {{\left (2 \, b^{2} x^{2} \log \left (x\right ) - 4 \, a b x - a^{2}\right )} \sqrt {c x^{2}}}{2 \, c^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^2*x^2*log(x) - 4*a*b*x - a^2)*sqrt(c*x^2)/(c^2*x^3)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right )^{2}}{\left (c x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.00, size = 41, normalized size = 0.71 \begin {gather*} \frac {\frac {-4 a b x-a^{2}}{2 \left (x^{2} \mathrm {sign}\left (x\right )\right )}+\frac {b^{2} \ln \left |x\right |}{\mathrm {sign}\left (x\right )}}{\sqrt {c} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^2*log(abs(x))/sgn(x) - (4*a*b*x + a^2)/(x^2*sgn(x)))/c^(3/2)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^2}{{\left (c\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/(c*x^2)^(3/2),x)

[Out]

int((a + b*x)^2/(c*x^2)^(3/2), x)

________________________________________________________________________________________