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

Optimal. Leaf size=90 \[ -\frac {2 b x \log (x)}{a^3 c \sqrt {c x^2}}+\frac {2 b x \log (a+b x)}{a^3 c \sqrt {c x^2}}-\frac {b x}{a^2 c \sqrt {c x^2} (a+b x)}-\frac {1}{a^2 c \sqrt {c x^2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 44} \begin {gather*} -\frac {b x}{a^2 c \sqrt {c x^2} (a+b x)}-\frac {2 b x \log (x)}{a^3 c \sqrt {c x^2}}+\frac {2 b x \log (a+b x)}{a^3 c \sqrt {c x^2}}-\frac {1}{a^2 c \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/((c*x^2)^(3/2)*(a + b*x)^2),x]

[Out]

-(1/(a^2*c*Sqrt[c*x^2])) - (b*x)/(a^2*c*Sqrt[c*x^2]*(a + b*x)) - (2*b*x*Log[x])/(a^3*c*Sqrt[c*x^2]) + (2*b*x*L
og[a + b*x])/(a^3*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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {x}{\left (c x^2\right )^{3/2} (a+b x)^2} \, dx &=\frac {x \int \frac {1}{x^2 (a+b x)^2} \, dx}{c \sqrt {c x^2}}\\ &=\frac {x \int \left (\frac {1}{a^2 x^2}-\frac {2 b}{a^3 x}+\frac {b^2}{a^2 (a+b x)^2}+\frac {2 b^2}{a^3 (a+b x)}\right ) \, dx}{c \sqrt {c x^2}}\\ &=-\frac {1}{a^2 c \sqrt {c x^2}}-\frac {b x}{a^2 c \sqrt {c x^2} (a+b x)}-\frac {2 b x \log (x)}{a^3 c \sqrt {c x^2}}+\frac {2 b x \log (a+b x)}{a^3 c \sqrt {c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 59, normalized size = 0.66 \begin {gather*} \frac {x^2 (-a (a+2 b x)-2 b x \log (x) (a+b x)+2 b x (a+b x) \log (a+b x))}{a^3 \left (c x^2\right )^{3/2} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.06, size = 61, normalized size = 0.68 \begin {gather*} \frac {-\frac {2 b x^3 \log (x)}{a^3}+\frac {2 b x^3 \log (a+b x)}{a^3}+\frac {-a x^2-2 b x^3}{a^2 (a+b x)}}{\left (c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.21, size = 66, normalized size = 0.73 \begin {gather*} -\frac {{\left (2 \, a b x + a^{2} - 2 \, {\left (b^{2} x^{2} + a b x\right )} \log \left (\frac {b x + a}{x}\right )\right )} \sqrt {c x^{2}}}{a^{3} b c^{2} x^{3} + a^{4} c^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a*b*x + a^2 - 2*(b^2*x^2 + a*b*x)*log((b*x + a)/x))*sqrt(c*x^2)/(a^3*b*c^2*x^3 + a^4*c^2*x^2)

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giac [A]  time = 1.11, size = 137, normalized size = 1.52 \begin {gather*} \frac {\frac {2 \, b^{2} \log \left ({\left | -\frac {a}{b x + a} + 1 \right |}\right )}{a^{3} \mathrm {sgn}\left (-\frac {b}{b x + a} + \frac {a b}{{\left (b x + a\right )}^{2}}\right )} + \frac {b^{2}}{{\left (b x + a\right )} a^{2} \mathrm {sgn}\left (-\frac {b}{b x + a} + \frac {a b}{{\left (b x + a\right )}^{2}}\right )} - \frac {b^{2}}{a^{3} {\left (\frac {a}{b x + a} - 1\right )} \mathrm {sgn}\left (-\frac {b}{b x + a} + \frac {a b}{{\left (b x + a\right )}^{2}}\right )}}{b c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 74, normalized size = 0.82 \begin {gather*} -\frac {\left (2 b^{2} x^{2} \ln \relax (x )-2 b^{2} x^{2} \ln \left (b x +a \right )+2 a b x \ln \relax (x )-2 a b x \ln \left (b x +a \right )+2 a b x +a^{2}\right ) x^{2}}{\left (c \,x^{2}\right )^{\frac {3}{2}} \left (b x +a \right ) a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.41, size = 79, normalized size = 0.88 \begin {gather*} \frac {1}{\sqrt {c x^{2}} a b c x + \sqrt {c x^{2}} a^{2} c} + \frac {2 \, \left (-1\right )^{\frac {2 \, a c x}{b}} b \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{a^{3} c^{\frac {3}{2}}} - \frac {2}{\sqrt {c x^{2}} a^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{{\left (c\,x^2\right )}^{3/2}\,{\left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((c*x^2)^(3/2)*(a + b*x)^2),x)

[Out]

int(x/((c*x^2)^(3/2)*(a + b*x)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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