\(\int \frac {(c x^2)^{5/2} (a+b x)}{x} \, dx\) [776]

   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 = 18, antiderivative size = 41 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x} \, dx=\frac {1}{5} a c^2 x^4 \sqrt {c x^2}+\frac {1}{6} b c^2 x^5 \sqrt {c x^2} \]

[Out]

1/5*a*c^2*x^4*(c*x^2)^(1/2)+1/6*b*c^2*x^5*(c*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, 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} \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x} \, dx=\frac {1}{5} a c^2 x^4 \sqrt {c x^2}+\frac {1}{6} b c^2 x^5 \sqrt {c x^2} \]

[In]

Int[((c*x^2)^(5/2)*(a + b*x))/x,x]

[Out]

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

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 {\left (c^2 \sqrt {c x^2}\right ) \int x^4 (a+b x) \, dx}{x} \\ & = \frac {\left (c^2 \sqrt {c x^2}\right ) \int \left (a x^4+b x^5\right ) \, dx}{x} \\ & = \frac {1}{5} a c^2 x^4 \sqrt {c x^2}+\frac {1}{6} b c^2 x^5 \sqrt {c x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.61 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x} \, dx=\frac {1}{30} c x^2 \left (c x^2\right )^{3/2} (6 a+5 b x) \]

[In]

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

[Out]

(c*x^2*(c*x^2)^(3/2)*(6*a + 5*b*x))/30

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.44

method result size
gosper \(\frac {\left (5 b x +6 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{30}\) \(18\)
default \(\frac {\left (5 b x +6 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{30}\) \(18\)
risch \(\frac {a \,c^{2} x^{4} \sqrt {c \,x^{2}}}{5}+\frac {b \,c^{2} x^{5} \sqrt {c \,x^{2}}}{6}\) \(34\)
trager \(\frac {c^{2} \left (5 b \,x^{5}+6 a \,x^{4}+5 b \,x^{4}+6 a \,x^{3}+5 b \,x^{3}+6 a \,x^{2}+5 b \,x^{2}+6 a x +5 b x +6 a +5 b \right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{30 x}\) \(76\)

[In]

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

[Out]

1/30*(5*b*x+6*a)*(c*x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.68 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x} \, dx=\frac {1}{30} \, {\left (5 \, b c^{2} x^{5} + 6 \, a c^{2} x^{4}\right )} \sqrt {c x^{2}} \]

[In]

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

[Out]

1/30*(5*b*c^2*x^5 + 6*a*c^2*x^4)*sqrt(c*x^2)

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.59 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x} \, dx=\frac {a \left (c x^{2}\right )^{\frac {5}{2}}}{5} + \frac {b x \left (c x^{2}\right )^{\frac {5}{2}}}{6} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.54 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x} \, dx=\frac {1}{6} \, \left (c x^{2}\right )^{\frac {5}{2}} b x + \frac {1}{5} \, \left (c x^{2}\right )^{\frac {5}{2}} a \]

[In]

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

[Out]

1/6*(c*x^2)^(5/2)*b*x + 1/5*(c*x^2)^(5/2)*a

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.68 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x} \, dx=\frac {1}{30} \, {\left (5 \, b c^{2} x^{6} \mathrm {sgn}\left (x\right ) + 6 \, a c^{2} x^{5} \mathrm {sgn}\left (x\right )\right )} \sqrt {c} \]

[In]

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

[Out]

1/30*(5*b*c^2*x^6*sgn(x) + 6*a*c^2*x^5*sgn(x))*sqrt(c)

Mupad [F(-1)]

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

[In]

int(((c*x^2)^(5/2)*(a + b*x))/x,x)

[Out]

int(((c*x^2)^(5/2)*(a + b*x))/x, x)