\(\int \frac {(c x^2)^{3/2} (a+b x)}{x^3} \, dx\) [770]

Optimal result

Integrand size = 18, antiderivative size = 29 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)}{x^3} \, dx=a c \sqrt {c x^2}+\frac {1}{2} b c x \sqrt {c x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {15} \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)}{x^3} \, dx=a c \sqrt {c x^2}+\frac {1}{2} b c x \sqrt {c x^2} \]

[In]

Int[((c*x^2)^(3/2)*(a + b*x))/x^3,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)}{x^3} \, dx=\frac {1}{2} c \sqrt {c x^2} (2 a+b x) \]

[In]

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

[Out]

(c*Sqrt[c*x^2]*(2*a + b*x))/2

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69

method	result	size
gosper	\(\frac {\left (b x +2 a \right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{2 x^{2}}\)	\(20\)
default	\(\frac {\left (b x +2 a \right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{2 x^{2}}\)	\(20\)
risch	\(a c \sqrt {c \,x^{2}}+\frac {b c x \sqrt {c \,x^{2}}}{2}\)	\(24\)
trager	\(\frac {c \left (b x +2 a +b \right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{2 x}\)	\(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.62 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)}{x^3} \, dx=\frac {1}{2} \, {\left (b c x + 2 \, a c\right )} \sqrt {c x^{2}} \]

[In]

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

[Out]

1/2*(b*c*x + 2*a*c)*sqrt(c*x^2)

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)}{x^3} \, dx=\frac {a \left (c x^{2}\right )^{\frac {3}{2}}}{x^{2}} + \frac {b \left (c x^{2}\right )^{\frac {3}{2}}}{2 x} \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)}{x^3} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((c*x^2)^(3/2)*(b*x+a)/x^3,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 = 17, normalized size of antiderivative = 0.59 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)}{x^3} \, dx=\frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) \]

[In]

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

[Out]

1/2*(b*x^2 + 2*a*x)*c^(3/2)*sgn(x)

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.48 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)}{x^3} \, dx=\frac {c^{3/2}\,\left |x\right |\,\left (2\,a+b\,x\right )}{2} \]

[In]

int(((c*x^2)^(3/2)*(a + b*x))/x^3,x)

[Out]

(c^(3/2)*abs(x)*(2*a + b*x))/2