\(\int \frac {x (a+b x)^2}{(c x^2)^{3/2}} \, dx\) [838]

Optimal result

Integrand size = 18, antiderivative size = 56 \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=-\frac {a^2}{c \sqrt {c x^2}}+\frac {b^2 x^2}{c \sqrt {c x^2}}+\frac {2 a b x \log (x)}{c \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 56, 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} \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=-\frac {a^2}{c \sqrt {c x^2}}+\frac {2 a b x \log (x)}{c \sqrt {c x^2}}+\frac {b^2 x^2}{c \sqrt {c x^2}} \]

[In]

Int[(x*(a + b*x)^2)/(c*x^2)^(3/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.62 \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=\frac {-a^2 x^2+b^2 x^4+2 a b x^3 \log (x)}{\left (c x^2\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.57

method	result	size
default	\(\frac {x^{2} \left (2 a b \ln \left (x \right ) x +b^{2} x^{2}-a^{2}\right )}{\left (c \,x^{2}\right )^{\frac {3}{2}}}\)	\(32\)
risch	\(-\frac {a^{2}}{c \sqrt {c \,x^{2}}}+\frac {b^{2} x^{2}}{c \sqrt {c \,x^{2}}}+\frac {2 a b x \ln \left (x \right )}{c \sqrt {c \,x^{2}}}\)	\(51\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.61 \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x \log \left (x\right ) - a^{2}\right )} \sqrt {c x^{2}}}{c^{2} x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=a^{2} \left (\begin {cases} \tilde {\infty } x^{2} & \text {for}\: c = 0 \\- \frac {1}{c \sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right ) + \frac {2 a b x^{3} \log {\left (x \right )}}{\left (c x^{2}\right )^{\frac {3}{2}}} + \frac {b^{2} x^{4}}{\left (c x^{2}\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.75 \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=\frac {b^{2} x^{2}}{\sqrt {c x^{2}} c} + \frac {2 \, a b \log \left (x\right )}{c^{\frac {3}{2}}} - \frac {a^{2}}{\sqrt {c x^{2}} c} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{3/2}} \, dx=\frac {\frac {b^{2} x}{\sqrt {c} \mathrm {sgn}\left (x\right )} + \frac {2 \, a b \log \left ({\left | x \right |}\right )}{\sqrt {c} \mathrm {sgn}\left (x\right )} - \frac {a^{2}}{\sqrt {c} x \mathrm {sgn}\left (x\right )}}{c} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((x*(a + b*x)^2)/(c*x^2)^(3/2),x)

[Out]

int((x*(a + b*x)^2)/(c*x^2)^(3/2), x)