\(\int \frac {x^5}{(c x^2)^{3/2} (a+b x)^2} \, dx\) [917]

Optimal result

Integrand size = 20, antiderivative size = 73 \[ \int \frac {x^5}{\left (c x^2\right )^{3/2} (a+b x)^2} \, dx=\frac {x^2}{b^2 c \sqrt {c x^2}}-\frac {a^2 x}{b^3 c \sqrt {c x^2} (a+b x)}-\frac {2 a x \log (a+b x)}{b^3 c \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 45} \[ \int \frac {x^5}{\left (c x^2\right )^{3/2} (a+b x)^2} \, dx=-\frac {a^2 x}{b^3 c \sqrt {c x^2} (a+b x)}-\frac {2 a x \log (a+b x)}{b^3 c \sqrt {c x^2}}+\frac {x^2}{b^2 c \sqrt {c x^2}} \]

[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 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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.74 \[ \int \frac {x^5}{\left (c x^2\right )^{3/2} (a+b x)^2} \, dx=\frac {x^3 \left (-a^2+a b x+b^2 x^2-2 a (a+b x) \log (a+b x)\right )}{b^3 \left (c x^2\right )^{3/2} (a+b x)} \]

[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))

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.85

method	result	size
default	\(-\frac {x^{3} \left (2 \ln \left (b x +a \right ) x a b -b^{2} x^{2}+2 a^{2} \ln \left (b x +a \right )-a b x +a^{2}\right )}{\left (c \,x^{2}\right )^{\frac {3}{2}} b^{3} \left (b x +a \right )}\)	\(62\)
risch	\(\frac {x^{2}}{b^{2} c \sqrt {c \,x^{2}}}-\frac {a^{2} x}{b^{3} c \left (b x +a \right ) \sqrt {c \,x^{2}}}-\frac {2 a x \ln \left (b x +a \right )}{b^{3} c \sqrt {c \,x^{2}}}\)	\(68\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86 \[ \int \frac {x^5}{\left (c x^2\right )^{3/2} (a+b x)^2} \, dx=\frac {{\left (b^{2} x^{2} + a b x - a^{2} - 2 \, {\left (a b x + a^{2}\right )} \log \left (b x + a\right )\right )} \sqrt {c x^{2}}}{b^{4} c^{2} x^{2} + a b^{3} c^{2} x} \]

[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)

Sympy [F]

\[ \int \frac {x^5}{\left (c x^2\right )^{3/2} (a+b x)^2} \, dx=\int \frac {x^{5}}{\left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{2}}\, dx \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (67) = 134\).

Time = 0.21 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.04 \[ \int \frac {x^5}{\left (c x^2\right )^{3/2} (a+b x)^2} \, dx=\frac {a^{3}}{\sqrt {c x^{2}} b^{5} c x + \sqrt {c x^{2}} a b^{4} c} + \frac {x^{2}}{\sqrt {c x^{2}} b^{2} c} - \frac {2 \, \left (-1\right )^{\frac {2 \, a c x}{b}} a \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{3} c^{\frac {3}{2}}} + \frac {2 \, a x}{\sqrt {c x^{2}} b^{3} c} - \frac {2 \, a \log \left (b x\right )}{b^{3} c^{\frac {3}{2}}} - \frac {5 \, a^{2}}{\sqrt {c x^{2}} b^{4} c} + \frac {4 \, a^{2}}{b^{4} c^{\frac {3}{2}} x} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int \frac {x^5}{\left (c x^2\right )^{3/2} (a+b x)^2} \, dx=\frac {\frac {{\left (2 \, a \log \left ({\left | a \right |}\right ) + a\right )} \mathrm {sgn}\left (x\right )}{b^{3} \sqrt {c}} + \frac {x}{b^{2} \sqrt {c} \mathrm {sgn}\left (x\right )} - \frac {2 \, a \log \left ({\left | b x + a \right |}\right )}{b^{3} \sqrt {c} \mathrm {sgn}\left (x\right )} - \frac {a^{2}}{{\left (b x + a\right )} b^{3} \sqrt {c} \mathrm {sgn}\left (x\right )}}{c} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{\left (c x^2\right )^{3/2} (a+b x)^2} \, dx=\int \frac {x^5}{{\left (c\,x^2\right )}^{3/2}\,{\left (a+b\,x\right )}^2} \,d x \]

[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)