\(\int \frac {x^3 (a+b x)^2}{(c x^2)^{5/2}} \, dx\) [844]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 20, antiderivative size = 56 \[ \int \frac {x^3 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=-\frac {a^2}{c^2 \sqrt {c x^2}}+\frac {b^2 x^2}{c^2 \sqrt {c x^2}}+\frac {2 a b x \log (x)}{c^2 \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 56, 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} \[ \int \frac {x^3 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=-\frac {a^2}{c^2 \sqrt {c x^2}}+\frac {2 a b x \log (x)}{c^2 \sqrt {c x^2}}+\frac {b^2 x^2}{c^2 \sqrt {c x^2}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.62 \[ \int \frac {x^3 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=\frac {-a^2 x^4+b^2 x^6+2 a b x^5 \log (x)}{\left (c x^2\right )^{5/2}} \]

[In]

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

[Out]

(-(a^2*x^4) + b^2*x^6 + 2*a*b*x^5*Log[x])/(c*x^2)^(5/2)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.57

method result size
default \(\frac {x^{4} \left (2 a b \ln \left (x \right ) x +b^{2} x^{2}-a^{2}\right )}{\left (c \,x^{2}\right )^{\frac {5}{2}}}\) \(32\)
risch \(-\frac {a^{2}}{c^{2} \sqrt {c \,x^{2}}}+\frac {b^{2} x^{2}}{c^{2} \sqrt {c \,x^{2}}}+\frac {2 a b x \ln \left (x \right )}{c^{2} \sqrt {c \,x^{2}}}\) \(51\)

[In]

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

[Out]

x^4*(2*a*b*ln(x)*x+b^2*x^2-a^2)/(c*x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.61 \[ \int \frac {x^3 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x \log \left (x\right ) - a^{2}\right )} \sqrt {c x^{2}}}{c^{3} x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.88 \[ \int \frac {x^3 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=- \frac {a^{2} x^{4}}{\left (c x^{2}\right )^{\frac {5}{2}}} + \frac {2 a b x^{5} \log {\left (x \right )}}{\left (c x^{2}\right )^{\frac {5}{2}}} + \frac {b^{2} x^{6}}{\left (c x^{2}\right )^{\frac {5}{2}}} \]

[In]

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

[Out]

-a**2*x**4/(c*x**2)**(5/2) + 2*a*b*x**5*log(x)/(c*x**2)**(5/2) + b**2*x**6/(c*x**2)**(5/2)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.80 \[ \int \frac {x^3 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=\frac {b^{2} x^{4}}{\left (c x^{2}\right )^{\frac {3}{2}} c} - \frac {a^{2} x^{2}}{\left (c x^{2}\right )^{\frac {3}{2}} c} + \frac {2 \, a b \log \left (x\right )}{c^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.75 \[ \int \frac {x^3 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=\frac {b^{2} x}{c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} + \frac {2 \, a b \log \left ({\left | x \right |}\right )}{c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} - \frac {a^{2}}{c^{\frac {5}{2}} x \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

b^2*x/(c^(5/2)*sgn(x)) + 2*a*b*log(abs(x))/(c^(5/2)*sgn(x)) - a^2/(c^(5/2)*x*sgn(x))

Mupad [F(-1)]

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

[In]

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

[Out]

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