3.9.96 \(\int \frac {\sqrt {c x^2}}{(a+b x)^2} \, dx\) [896]

Optimal. Leaf size=47 \[ \frac {a \sqrt {c x^2}}{b^2 x (a+b x)}+\frac {\sqrt {c x^2} \log (a+b x)}{b^2 x} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 47, 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 \sqrt {c x^2}}{b^2 x (a+b x)}+\frac {\sqrt {c x^2} \log (a+b x)}{b^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x^2]/(a + b*x)^2,x]

[Out]

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

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 {\sqrt {c x^2}}{(a+b x)^2} \, dx &=\frac {\sqrt {c x^2} \int \frac {x}{(a+b x)^2} \, dx}{x}\\ &=\frac {\sqrt {c x^2} \int \left (-\frac {a}{b (a+b x)^2}+\frac {1}{b (a+b x)}\right ) \, dx}{x}\\ &=\frac {a \sqrt {c x^2}}{b^2 x (a+b x)}+\frac {\sqrt {c x^2} \log (a+b x)}{b^2 x}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 36, normalized size = 0.77 \begin {gather*} \frac {c x (a+(a+b x) \log (a+b x))}{b^2 \sqrt {c x^2} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*x^2]/(a + b*x)^2,x]

[Out]

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

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Maple [A]
time = 0.14, size = 41, normalized size = 0.87

method	result	size
default	\(\frac {\sqrt {c \,x^{2}}\, \left (b \ln \left (b x +a \right ) x +a \ln \left (b x +a \right )+a \right )}{x \,b^{2} \left (b x +a \right )}\)	\(41\)
risch	\(\frac {a \sqrt {c \,x^{2}}}{b^{2} x \left (b x +a \right )}+\frac {\ln \left (b x +a \right ) \sqrt {c \,x^{2}}}{b^{2} x}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 79, normalized size = 1.68 \begin {gather*} \frac {\left (-1\right )^{\frac {2 \, c x}{b}} \sqrt {c} \log \left (\frac {2 \, c x}{b}\right )}{b^{2}} + \frac {\left (-1\right )^{\frac {2 \, a c x}{b}} \sqrt {c} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{2}} - \frac {\sqrt {c x^{2}}}{b^{2} x + a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.49, size = 38, normalized size = 0.81 \begin {gather*} \frac {\sqrt {c x^{2}} {\left ({\left (b x + a\right )} \log \left (b x + a\right ) + a\right )}}{b^{3} x^{2} + a b^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.79, size = 46, normalized size = 0.98 \begin {gather*} -\sqrt {c} {\left (\frac {{\left (\log \left ({\left | a \right |}\right ) + 1\right )} \mathrm {sgn}\left (x\right )}{b^{2}} - \frac {\log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (x\right )}{b^{2}} - \frac {a \mathrm {sgn}\left (x\right )}{{\left (b x + a\right )} b^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2)^(1/2)/(a + b*x)^2,x)

[Out]

int((c*x^2)^(1/2)/(a + b*x)^2, x)

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