\(\int \frac {x^3 (a+b x)}{(c x^2)^{5/2}} \, dx\) [796]

Optimal result

Integrand size = 18, antiderivative size = 33 \[ \int \frac {x^3 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {a}{c^2 \sqrt {c x^2}}+\frac {b x \log (x)}{c^2 \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 45} \[ \int \frac {x^3 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=\frac {b x \log (x)}{c^2 \sqrt {c x^2}}-\frac {a}{c^2 \sqrt {c x^2}} \]

[In]

Int[(x^3*(a + b*x))/(c*x^2)^(5/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=\frac {-a x^4+b x^5 \log (x)}{\left (c x^2\right )^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64

method	result	size
default	\(\frac {x^{4} \left (b \ln \left (x \right ) x -a \right )}{\left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(21\)
risch	\(-\frac {a}{c^{2} \sqrt {c \,x^{2}}}+\frac {b x \ln \left (x \right )}{c^{2} \sqrt {c \,x^{2}}}\)	\(30\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70 \[ \int \frac {x^3 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c x^{2}} {\left (b x \log \left (x\right ) - a\right )}}{c^{3} x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {x^3 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=- \frac {a x^{4}}{\left (c x^{2}\right )^{\frac {5}{2}}} + \frac {b x^{5} \log {\left (x \right )}}{\left (c x^{2}\right )^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {a x^{2}}{\left (c x^{2}\right )^{\frac {3}{2}} c} + \frac {b \log \left (x\right )}{c^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=\frac {b \log \left ({\left | x \right |}\right )}{c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} - \frac {a}{c^{\frac {5}{2}} x \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((x^3*(a + b*x))/(c*x^2)^(5/2),x)

[Out]

int((x^3*(a + b*x))/(c*x^2)^(5/2), x)