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

Optimal. Leaf size=70 \[ \frac {a^2 x \log (a+b x)}{b^3 c \sqrt {c x^2}}-\frac {a x^2}{b^2 c \sqrt {c x^2}}+\frac {x^3}{2 b c \sqrt {c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 70, 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} \[ \frac {a^2 x \log (a+b x)}{b^3 c \sqrt {c x^2}}-\frac {a x^2}{b^2 c \sqrt {c x^2}}+\frac {x^3}{2 b c \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/((c*x^2)^(3/2)*(a + b*x)),x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 41, normalized size = 0.59 \[ \frac {x^3 \left (2 a^2 \log (a+b x)+b x (b x-2 a)\right )}{2 b^3 \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 42, normalized size = 0.60 \[ \frac {{\left (b^{2} x^{2} - 2 \, a b x + 2 \, a^{2} \log \left (b x + a\right )\right )} \sqrt {c x^{2}}}{2 \, b^{3} c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.96, size = 71, normalized size = 1.01 \[ \frac {\sqrt {c x^{2}} {\left (\frac {x}{b c} - \frac {2 \, a}{b^{2} c}\right )} - \frac {2 \, a^{2} \log \left ({\left | -{\left (\sqrt {c} x - \sqrt {c x^{2}}\right )} b \sqrt {c} - 2 \, a c \right |}\right )}{b^{3} \sqrt {c}}}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 40, normalized size = 0.57 \[ \frac {\left (b^{2} x^{2}+2 a^{2} \ln \left (b x +a \right )-2 a b x \right ) x^{3}}{2 \left (c \,x^{2}\right )^{\frac {3}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.66, size = 140, normalized size = 2.00 \[ \frac {x^{3}}{2 \, \sqrt {c x^{2}} b c} - \frac {a x^{2}}{\sqrt {c x^{2}} b^{2} c} + \frac {\left (-1\right )^{\frac {2 \, a c x}{b}} a^{2} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{3} c^{\frac {3}{2}}} - \frac {7 \, a^{2} x}{2 \, \sqrt {c x^{2}} b^{3} c} + \frac {a^{2} \log \left (b x\right )}{b^{3} c^{\frac {3}{2}}} + \frac {2 \, a^{3}}{\sqrt {c x^{2}} b^{4} c} - \frac {2 \, a^{3}}{b^{4} c^{\frac {3}{2}} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^3/(sqrt(c*x^2)*b*c) - a*x^2/(sqrt(c*x^2)*b^2*c) + (-1)^(2*a*c*x/b)*a^2*log(-2*a*c*x/(b*abs(b*x + a)))/(b
^3*c^(3/2)) - 7/2*a^2*x/(sqrt(c*x^2)*b^3*c) + a^2*log(b*x)/(b^3*c^(3/2)) + 2*a^3/(sqrt(c*x^2)*b^4*c) - 2*a^3/(
b^4*c^(3/2)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/((c*x^2)^(3/2)*(a + b*x)),x)

[Out]

int(x^5/((c*x^2)^(3/2)*(a + b*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{\left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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