3.9.75 \(\int \frac {x^5}{(c x^2)^{3/2} (a+b x)^2} \, dx\)

Optimal. Leaf size=73 \[ -\frac {a^2 x}{b^3 c \sqrt {c x^2} (a+b x)}-\frac {2 a x \log (a+b x)}{b^3 c \sqrt {c x^2}}+\frac {x^2}{b^2 c \sqrt {c x^2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 43} \begin {gather*} -\frac {a^2 x}{b^3 c \sqrt {c x^2} (a+b x)}-\frac {2 a x \log (a+b x)}{b^3 c \sqrt {c x^2}}+\frac {x^2}{b^2 c \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/((c*x^2)^(3/2)*(a + b*x)^2),x]

[Out]

x^2/(b^2*c*Sqrt[c*x^2]) - (a^2*x)/(b^3*c*Sqrt[c*x^2]*(a + b*x)) - (2*a*x*Log[a + b*x])/(b^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 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 {x^5}{\left (c x^2\right )^{3/2} (a+b x)^2} \, dx &=\frac {x \int \frac {x^2}{(a+b x)^2} \, dx}{c \sqrt {c x^2}}\\ &=\frac {x \int \left (\frac {1}{b^2}+\frac {a^2}{b^2 (a+b x)^2}-\frac {2 a}{b^2 (a+b x)}\right ) \, dx}{c \sqrt {c x^2}}\\ &=\frac {x^2}{b^2 c \sqrt {c x^2}}-\frac {a^2 x}{b^3 c \sqrt {c x^2} (a+b x)}-\frac {2 a x \log (a+b x)}{b^3 c \sqrt {c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 54, normalized size = 0.74 \begin {gather*} \frac {x^3 \left (-a^2+a b x-2 a (a+b x) \log (a+b x)+b^2 x^2\right )}{b^3 \left (c x^2\right )^{3/2} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/((c*x^2)^(3/2)*(a + b*x)^2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.06, size = 59, normalized size = 0.81 \begin {gather*} \frac {\frac {-a^2 x^3+a b x^4+b^2 x^5}{b^3 (a+b x)}-\frac {2 a x^3 \log (a+b x)}{b^3}}{\left (c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^5/((c*x^2)^(3/2)*(a + b*x)^2),x]

[Out]

((-(a^2*x^3) + a*b*x^4 + b^2*x^5)/(b^3*(a + b*x)) - (2*a*x^3*Log[a + b*x])/b^3)/(c*x^2)^(3/2)

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fricas [A]  time = 1.00, size = 63, normalized size = 0.86 \begin {gather*} \frac {{\left (b^{2} x^{2} + a b x - a^{2} - 2 \, {\left (a b x + a^{2}\right )} \log \left (b x + a\right )\right )} \sqrt {c x^{2}}}{b^{4} c^{2} x^{2} + a b^{3} c^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^2)^(3/2)/(b*x+a)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.26, size = 127, normalized size = 1.74 \begin {gather*} -\frac {\frac {2 \, a \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{3} \mathrm {sgn}\left (-\frac {b}{b x + a} + \frac {a b}{{\left (b x + a\right )}^{2}}\right )} + \frac {b x + a}{b^{3} \mathrm {sgn}\left (-\frac {b}{b x + a} + \frac {a b}{{\left (b x + a\right )}^{2}}\right )} - \frac {a^{2}}{{\left (b x + a\right )} b^{3} \mathrm {sgn}\left (-\frac {b}{b x + a} + \frac {a b}{{\left (b x + a\right )}^{2}}\right )}}{c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^2)^(3/2)/(b*x+a)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 62, normalized size = 0.85 \begin {gather*} -\frac {\left (2 a b x \ln \left (b x +a \right )-b^{2} x^{2}+2 a^{2} \ln \left (b x +a \right )-a b x +a^{2}\right ) x^{3}}{\left (c \,x^{2}\right )^{\frac {3}{2}} \left (b x +a \right ) b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(c*x^2)^(3/2)/(b*x+a)^2,x)

[Out]

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

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maxima [B]  time = 1.61, size = 149, normalized size = 2.04 \begin {gather*} \frac {a^{3}}{\sqrt {c x^{2}} b^{5} c x + \sqrt {c x^{2}} a b^{4} c} + \frac {x^{2}}{\sqrt {c x^{2}} b^{2} c} - \frac {2 \, \left (-1\right )^{\frac {2 \, a c x}{b}} a \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{3} c^{\frac {3}{2}}} + \frac {2 \, a x}{\sqrt {c x^{2}} b^{3} c} - \frac {2 \, a \log \left (b x\right )}{b^{3} c^{\frac {3}{2}}} - \frac {5 \, a^{2}}{\sqrt {c x^{2}} b^{4} c} + \frac {4 \, a^{2}}{b^{4} c^{\frac {3}{2}} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^2)^(3/2)/(b*x+a)^2,x, algorithm="maxima")

[Out]

a^3/(sqrt(c*x^2)*b^5*c*x + sqrt(c*x^2)*a*b^4*c) + x^2/(sqrt(c*x^2)*b^2*c) - 2*(-1)^(2*a*c*x/b)*a*log(-2*a*c*x/
(b*abs(b*x + a)))/(b^3*c^(3/2)) + 2*a*x/(sqrt(c*x^2)*b^3*c) - 2*a*log(b*x)/(b^3*c^(3/2)) - 5*a^2/(sqrt(c*x^2)*
b^4*c) + 4*a^2/(b^4*c^(3/2)*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5}{{\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^5/((c*x^2)^(3/2)*(a + b*x)^2),x)

[Out]

int(x^5/((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^{5}}{\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**5/(c*x**2)**(3/2)/(b*x+a)**2,x)

[Out]

Integral(x**5/((c*x**2)**(3/2)*(a + b*x)**2), x)

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