\(\int \frac {(c x^2)^{5/2} (a+b x)}{x^2} \, dx\) [777]

Optimal result

Integrand size = 18, antiderivative size = 41 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^2} \, dx=\frac {1}{4} a c^2 x^3 \sqrt {c x^2}+\frac {1}{5} b c^2 x^4 \sqrt {c x^2} \]

[Out]

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

[In]

Int[((c*x^2)^(5/2)*(a + b*x))/x^2,x]

[Out]

(a*c^2*x^3*Sqrt[c*x^2])/4 + (b*c^2*x^4*Sqrt[c*x^2])/5

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^3 (a+b x) \, dx}{x} \\ & = \frac {\left (c^2 \sqrt {c x^2}\right ) \int \left (a x^3+b x^4\right ) \, dx}{x} \\ & = \frac {1}{4} a c^2 x^3 \sqrt {c x^2}+\frac {1}{5} b c^2 x^4 \sqrt {c x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.56 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^2} \, dx=\frac {1}{20} c x \left (c x^2\right )^{3/2} (5 a+4 b x) \]

[In]

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

[Out]

(c*x*(c*x^2)^(3/2)*(5*a + 4*b*x))/20

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.51

method	result	size
gosper	\(\frac {\left (4 b x +5 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{20 x}\)	\(21\)
default	\(\frac {\left (4 b x +5 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{20 x}\)	\(21\)
risch	\(\frac {a \,c^{2} x^{3} \sqrt {c \,x^{2}}}{4}+\frac {b \,c^{2} x^{4} \sqrt {c \,x^{2}}}{5}\)	\(34\)
trager	\(\frac {c^{2} \left (4 b \,x^{4}+5 a \,x^{3}+4 b \,x^{3}+5 a \,x^{2}+4 b \,x^{2}+5 a x +4 b x +5 a +4 b \right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{20 x}\)	\(64\)

[In]

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

[Out]

1/20/x*(4*b*x+5*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 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^2} \, dx=\frac {1}{20} \, {\left (4 \, b c^{2} x^{4} + 5 \, a c^{2} x^{3}\right )} \sqrt {c x^{2}} \]

[In]

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

[Out]

1/20*(4*b*c^2*x^4 + 5*a*c^2*x^3)*sqrt(c*x^2)

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.59 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^2} \, dx=\frac {a \left (c x^{2}\right )^{\frac {5}{2}}}{4 x} + \frac {b \left (c x^{2}\right )^{\frac {5}{2}}}{5} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.59 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^2} \, dx=\frac {1}{5} \, \left (c x^{2}\right )^{\frac {5}{2}} b + \frac {\left (c x^{2}\right )^{\frac {5}{2}} a}{4 \, x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.68 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^2} \, dx=\frac {1}{20} \, {\left (4 \, b c^{2} x^{5} \mathrm {sgn}\left (x\right ) + 5 \, a c^{2} x^{4} \mathrm {sgn}\left (x\right )\right )} \sqrt {c} \]

[In]

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

[Out]

1/20*(4*b*c^2*x^5*sgn(x) + 5*a*c^2*x^4*sgn(x))*sqrt(c)

Mupad [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.61 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)}{x^2} \, dx=\frac {c^{5/2}\,\left (4\,b\,\sqrt {x^{10}}+5\,a\,x^3\,\sqrt {x^2}\right )}{20} \]

[In]

int(((c*x^2)^(5/2)*(a + b*x))/x^2,x)

[Out]

(c^(5/2)*(4*b*(x^10)^(1/2) + 5*a*x^3*(x^2)^(1/2)))/20