\(\int \frac {(c x^2)^{5/2} (a+b x)}{x^4} \, dx\) [779]

Optimal result

Integrand size = 18, antiderivative size = 39 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^4} \, dx=\frac {1}{2} a c^2 x \sqrt {c x^2}+\frac {1}{3} b c^2 x^2 \sqrt {c x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, 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} \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^4} \, dx=\frac {1}{2} a c^2 x \sqrt {c x^2}+\frac {1}{3} b c^2 x^2 \sqrt {c x^2} \]

[In]

Int[((c*x^2)^(5/2)*(a + b*x))/x^4,x]

[Out]

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

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 (a+b x) \, dx}{x} \\ & = \frac {\left (c^2 \sqrt {c x^2}\right ) \int \left (a x+b x^2\right ) \, dx}{x} \\ & = \frac {1}{2} a c^2 x \sqrt {c x^2}+\frac {1}{3} b c^2 x^2 \sqrt {c x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.64 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^4} \, dx=\frac {1}{6} c^2 x \sqrt {c x^2} (3 a+2 b x) \]

[In]

Integrate[((c*x^2)^(5/2)*(a + b*x))/x^4,x]

[Out]

(c^2*x*Sqrt[c*x^2]*(3*a + 2*b*x))/6

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.54

method	result	size
gosper	\(\frac {\left (2 b x +3 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{6 x^{3}}\)	\(21\)
default	\(\frac {\left (2 b x +3 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{6 x^{3}}\)	\(21\)
risch	\(\frac {a \,c^{2} x \sqrt {c \,x^{2}}}{2}+\frac {b \,c^{2} x^{2} \sqrt {c \,x^{2}}}{3}\)	\(32\)
trager	\(\frac {c^{2} \left (2 b \,x^{2}+3 a x +2 b x +3 a +2 b \right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{6 x}\)	\(40\)

[In]

int((c*x^2)^(5/2)*(b*x+a)/x^4,x,method=_RETURNVERBOSE)

[Out]

1/6/x^3*(2*b*x+3*a)*(c*x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.67 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^4} \, dx=\frac {1}{6} \, {\left (2 \, b c^{2} x^{2} + 3 \, a c^{2} x\right )} \sqrt {c x^{2}} \]

[In]

integrate((c*x^2)^(5/2)*(b*x+a)/x^4,x, algorithm="fricas")

[Out]

1/6*(2*b*c^2*x^2 + 3*a*c^2*x)*sqrt(c*x^2)

Sympy [A] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^4} \, dx=\frac {a \left (c x^{2}\right )^{\frac {5}{2}}}{2 x^{3}} + \frac {b \left (c x^{2}\right )^{\frac {5}{2}}}{3 x^{2}} \]

[In]

integrate((c*x**2)**(5/2)*(b*x+a)/x**4,x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^4} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((c*x^2)^(5/2)*(b*x+a)/x^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.72 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^4} \, dx=\frac {1}{6} \, {\left (2 \, b c^{2} x^{3} \mathrm {sgn}\left (x\right ) + 3 \, a c^{2} x^{2} \mathrm {sgn}\left (x\right )\right )} \sqrt {c} \]

[In]

integrate((c*x^2)^(5/2)*(b*x+a)/x^4,x, algorithm="giac")

[Out]

1/6*(2*b*c^2*x^3*sgn(x) + 3*a*c^2*x^2*sgn(x))*sqrt(c)

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.51 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^4} \, dx=\frac {c^{5/2}\,\left (2\,b\,\sqrt {x^6}+3\,a\,x\,\left |x\right |\right )}{6} \]

[In]

int(((c*x^2)^(5/2)*(a + b*x))/x^4,x)

[Out]

(c^(5/2)*(2*b*(x^6)^(1/2) + 3*a*x*abs(x)))/6