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

Optimal. Leaf size=52 \[ \frac {a^2 x \log (x)}{\sqrt {c x^2}}+\frac {2 a b x^2}{\sqrt {c x^2}}+\frac {b^2 x^3}{2 \sqrt {c x^2}} \]

[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)

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Rubi [A]  time = 0.01, antiderivative size = 52, 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, 43} \[ \frac {a^2 x \log (x)}{\sqrt {c x^2}}+\frac {2 a b x^2}{\sqrt {c x^2}}+\frac {b^2 x^3}{2 \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[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 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)^2}{\sqrt {c x^2}} \, dx &=\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*}

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Mathematica [A]  time = 0.00, size = 32, normalized size = 0.62 \[ \frac {x \left (2 a^2 \log (x)+b x (4 a+b x)\right )}{2 \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.41, size = 35, normalized size = 0.67 \[ \frac {{\left (b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} \log \relax (x)\right )} \sqrt {c x^{2}}}{2 \, c x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 1.12, size = 50, normalized size = 0.96 \[ -\frac {a^{2} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2}} \right |}\right )}{\sqrt {c}} + \frac {1}{2} \, \sqrt {c x^{2}} {\left (\frac {b^{2} x}{c} + \frac {4 \, a b}{c}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 31, normalized size = 0.60 \[ \frac {\left (b^{2} x^{2}+2 a^{2} \ln \relax (x )+4 a b x \right ) x}{2 \sqrt {c \,x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.35, size = 35, normalized size = 0.67 \[ \frac {b^{2} x^{2}}{2 \, \sqrt {c}} + \frac {a^{2} \log \relax (x)}{\sqrt {c}} + \frac {2 \, \sqrt {c x^{2}} a b}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (a+b\,x\right )}^2}{\sqrt {c\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/(c*x^2)^(1/2),x)

[Out]

int((a + b*x)^2/(c*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right )^{2}}{\sqrt {c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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