\(\int \frac {(a+b x)^2}{\sqrt {c x^2}} \, dx\) [831]

Optimal result

Integrand size = 17, antiderivative size = 52 \[ \int \frac {(a+b x)^2}{\sqrt {c x^2}} \, dx=\frac {2 a b x^2}{\sqrt {c x^2}}+\frac {b^2 x^3}{2 \sqrt {c x^2}}+\frac {a^2 x \log (x)}{\sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 52, 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} \[ \int \frac {(a+b x)^2}{\sqrt {c x^2}} \, dx=\frac {a^2 x \log (x)}{\sqrt {c x^2}}+\frac {2 a b x^2}{\sqrt {c x^2}}+\frac {b^2 x^3}{2 \sqrt {c x^2}} \]

[In]

Int[(a + b*x)^2/Sqrt[c*x^2],x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.62 \[ \int \frac {(a+b x)^2}{\sqrt {c x^2}} \, dx=\frac {x \left (b x (4 a+b x)+2 a^2 \log (x)\right )}{2 \sqrt {c x^2}} \]

[In]

Integrate[(a + b*x)^2/Sqrt[c*x^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {x \left (b^{2} x^{2}+2 a^{2} \ln \left (x \right )+4 a b x \right )}{2 \sqrt {c \,x^{2}}}\)	\(31\)
risch	\(\frac {x b \left (\frac {1}{2} b \,x^{2}+2 a x \right )}{\sqrt {c \,x^{2}}}+\frac {a^{2} x \ln \left (x \right )}{\sqrt {c \,x^{2}}}\)	\(37\)

[In]

int((b*x+a)^2/(c*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.67 \[ \int \frac {(a+b x)^2}{\sqrt {c x^2}} \, dx=\frac {{\left (b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} \log \left (x\right )\right )} \sqrt {c x^{2}}}{2 \, c x} \]

[In]

integrate((b*x+a)^2/(c*x^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b x)^2}{\sqrt {c x^2}} \, dx=\begin {cases} \frac {a^{2} x \log {\left (x \right )}}{\sqrt {c x^{2}}} + \sqrt {c x^{2}} \cdot \left (\frac {2 a b}{c} + \frac {b^{2} x}{2 c}\right ) & \text {for}\: c \neq 0 \\\tilde {\infty } \left (\begin {cases} a^{2} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{3}}{3 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((b*x+a)**2/(c*x**2)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.67 \[ \int \frac {(a+b x)^2}{\sqrt {c x^2}} \, dx=\frac {b^{2} x^{2}}{2 \, \sqrt {c}} + \frac {a^{2} \log \left (x\right )}{\sqrt {c}} + \frac {2 \, \sqrt {c x^{2}} a b}{c} \]

[In]

integrate((b*x+a)^2/(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.83 \[ \int \frac {(a+b x)^2}{\sqrt {c x^2}} \, dx=\frac {a^{2} \log \left ({\left | x \right |}\right )}{\sqrt {c} \mathrm {sgn}\left (x\right )} + \frac {b^{2} c^{\frac {3}{2}} x^{2} \mathrm {sgn}\left (x\right ) + 4 \, a b c^{\frac {3}{2}} x \mathrm {sgn}\left (x\right )}{2 \, c^{2}} \]

[In]

integrate((b*x+a)^2/(c*x^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + b*x)^2/(c*x^2)^(1/2),x)

[Out]

int((a + b*x)^2/(c*x^2)^(1/2), x)