\(\int \frac {(c x^2)^{5/2} (a+b x)^2}{x} \, dx\) [822]

Optimal result

Integrand size = 20, antiderivative size = 66 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^2}{x} \, dx=\frac {1}{5} a^2 c^2 x^4 \sqrt {c x^2}+\frac {1}{3} a b c^2 x^5 \sqrt {c x^2}+\frac {1}{7} b^2 c^2 x^6 \sqrt {c x^2} \]

[Out]

1/5*a^2*c^2*x^4*(c*x^2)^(1/2)+1/3*a*b*c^2*x^5*(c*x^2)^(1/2)+1/7*b^2*c^2*x^6*(c*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 45} \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^2}{x} \, dx=\frac {1}{5} a^2 c^2 x^4 \sqrt {c x^2}+\frac {1}{3} a b c^2 x^5 \sqrt {c x^2}+\frac {1}{7} b^2 c^2 x^6 \sqrt {c x^2} \]

[In]

Int[((c*x^2)^(5/2)*(a + b*x)^2)/x,x]

[Out]

(a^2*c^2*x^4*Sqrt[c*x^2])/5 + (a*b*c^2*x^5*Sqrt[c*x^2])/3 + (b^2*c^2*x^6*Sqrt[c*x^2])/7

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)^2 \, dx}{x} \\ & = \frac {\left (c^2 \sqrt {c x^2}\right ) \int \left (a^2 x^4+2 a b x^5+b^2 x^6\right ) \, dx}{x} \\ & = \frac {1}{5} a^2 c^2 x^4 \sqrt {c x^2}+\frac {1}{3} a b c^2 x^5 \sqrt {c x^2}+\frac {1}{7} b^2 c^2 x^6 \sqrt {c x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.58 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^2}{x} \, dx=\frac {1}{105} c \left (c x^2\right )^{3/2} \left (21 a^2 x^2+35 a b x^3+15 b^2 x^4\right ) \]

[In]

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

[Out]

(c*(c*x^2)^(3/2)*(21*a^2*x^2 + 35*a*b*x^3 + 15*b^2*x^4))/105

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.44

method	result	size
gosper	\(\frac {\left (15 b^{2} x^{2}+35 a b x +21 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{105}\)	\(29\)
default	\(\frac {\left (15 b^{2} x^{2}+35 a b x +21 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{105}\)	\(29\)
risch	\(\frac {a^{2} c^{2} x^{4} \sqrt {c \,x^{2}}}{5}+\frac {a b \,c^{2} x^{5} \sqrt {c \,x^{2}}}{3}+\frac {b^{2} c^{2} x^{6} \sqrt {c \,x^{2}}}{7}\)	\(55\)
trager	\(\frac {c^{2} \left (15 b^{2} x^{6}+35 a b \,x^{5}+15 b^{2} x^{5}+21 a^{2} x^{4}+35 a b \,x^{4}+15 b^{2} x^{4}+21 a^{2} x^{3}+35 a b \,x^{3}+15 b^{2} x^{3}+21 a^{2} x^{2}+35 a b \,x^{2}+15 b^{2} x^{2}+21 a^{2} x +35 a b x +15 b^{2} x +21 a^{2}+35 a b +15 b^{2}\right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{105 x}\)	\(143\)

[In]

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

[Out]

1/105*(15*b^2*x^2+35*a*b*x+21*a^2)*(c*x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.64 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^2}{x} \, dx=\frac {1}{105} \, {\left (15 \, b^{2} c^{2} x^{6} + 35 \, a b c^{2} x^{5} + 21 \, a^{2} c^{2} x^{4}\right )} \sqrt {c x^{2}} \]

[In]

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

[Out]

1/105*(15*b^2*c^2*x^6 + 35*a*b*c^2*x^5 + 21*a^2*c^2*x^4)*sqrt(c*x^2)

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.67 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^2}{x} \, dx=\frac {a^{2} \left (c x^{2}\right )^{\frac {5}{2}}}{5} + \frac {a b x \left (c x^{2}\right )^{\frac {5}{2}}}{3} + \frac {b^{2} x^{2} \left (c x^{2}\right )^{\frac {5}{2}}}{7} \]

[In]

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

[Out]

a**2*(c*x**2)**(5/2)/5 + a*b*x*(c*x**2)**(5/2)/3 + b**2*x**2*(c*x**2)**(5/2)/7

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.61 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^2}{x} \, dx=\frac {1}{3} \, \left (c x^{2}\right )^{\frac {5}{2}} a b x + \frac {1}{5} \, \left (c x^{2}\right )^{\frac {5}{2}} a^{2} + \frac {\left (c x^{2}\right )^{\frac {7}{2}} b^{2}}{7 \, c} \]

[In]

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

[Out]

1/3*(c*x^2)^(5/2)*a*b*x + 1/5*(c*x^2)^(5/2)*a^2 + 1/7*(c*x^2)^(7/2)*b^2/c

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.67 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^2}{x} \, dx=\frac {1}{105} \, {\left (15 \, b^{2} c^{2} x^{7} \mathrm {sgn}\left (x\right ) + 35 \, a b c^{2} x^{6} \mathrm {sgn}\left (x\right ) + 21 \, a^{2} c^{2} x^{5} \mathrm {sgn}\left (x\right )\right )} \sqrt {c} \]

[In]

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

[Out]

1/105*(15*b^2*c^2*x^7*sgn(x) + 35*a*b*c^2*x^6*sgn(x) + 21*a^2*c^2*x^5*sgn(x))*sqrt(c)

Mupad [F(-1)]

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

[In]

int(((c*x^2)^(5/2)*(a + b*x)^2)/x,x)

[Out]

int(((c*x^2)^(5/2)*(a + b*x)^2)/x, x)