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

Optimal result

Integrand size = 17, antiderivative size = 58 \[ \int \frac {(a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=-\frac {2 a b}{c \sqrt {c x^2}}-\frac {a^2}{2 c x \sqrt {c x^2}}+\frac {b^2 x \log (x)}{c \sqrt {c x^2}} \]

[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] (verified)

Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {15, 45} \[ \int \frac {(a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=-\frac {a^2}{2 c x \sqrt {c x^2}}-\frac {2 a b}{c \sqrt {c x^2}}+\frac {b^2 x \log (x)}{c \sqrt {c x^2}} \]

[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*} \text {integral}& = \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] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.55 \[ \int \frac {(a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=\frac {-\frac {1}{2} a x (a+4 b x)+b^2 x^3 \log (x)}{\left (c x^2\right )^{3/2}} \]

[In]

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

[Out]

(-1/2*(a*x*(a + 4*b*x)) + b^2*x^3*Log[x])/(c*x^2)^(3/2)

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 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	\(\frac {-\frac {1}{2} a^{2}-2 a b x}{c x \sqrt {c \,x^{2}}}+\frac {b^{2} x \ln \left (x \right )}{c \sqrt {c \,x^{2}}}\)	\(44\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.62 \[ \int \frac {(a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=\frac {{\left (2 \, b^{2} x^{2} \log \left (x\right ) - 4 \, a b x - a^{2}\right )} \sqrt {c x^{2}}}{2 \, c^{2} x^{3}} \]

[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 [A] (verification not implemented)

Time = 1.62 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=- \frac {a^{2} x}{2 \left (c x^{2}\right )^{\frac {3}{2}}} + 2 a b \left (\begin {cases} \tilde {\infty } x^{2} & \text {for}\: c = 0 \\- \frac {1}{c \sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right ) + \frac {b^{2} x^{3} \log {\left (x \right )}}{\left (c x^{2}\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

-a**2*x/(2*(c*x**2)**(3/2)) + 2*a*b*Piecewise((zoo*x**2, Eq(c, 0)), (-1/(c*sqrt(c*x**2)), True)) + b**2*x**3*l
og(x)/(c*x**2)**(3/2)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.60 \[ \int \frac {(a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=\frac {b^{2} \log \left (x\right )}{c^{\frac {3}{2}}} - \frac {2 \, a b}{\sqrt {c x^{2}} c} - \frac {a^{2}}{2 \, c^{\frac {3}{2}} x^{2}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.72 \[ \int \frac {(a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=\frac {\frac {2 \, b^{2} \log \left ({\left | x \right |}\right )}{\sqrt {c} \mathrm {sgn}\left (x\right )} - \frac {4 \, a b x + a^{2}}{\sqrt {c} x^{2} \mathrm {sgn}\left (x\right )}}{2 \, c} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=\int \frac {{\left (a+b\,x\right )}^2}{{\left (c\,x^2\right )}^{3/2}} \,d x \]

[In]

int((a + b*x)^2/(c*x^2)^(3/2),x)

[Out]

int((a + b*x)^2/(c*x^2)^(3/2), x)