\(\int x^2 (c x^2)^{3/2} (a+b x)^2 \, dx\) [813]

Optimal result

Integrand size = 20, antiderivative size = 60 \[ \int x^2 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{6} a^2 c x^5 \sqrt {c x^2}+\frac {2}{7} a b c x^6 \sqrt {c x^2}+\frac {1}{8} b^2 c x^7 \sqrt {c x^2} \]

[Out]

1/6*a^2*c*x^5*(c*x^2)^(1/2)+2/7*a*b*c*x^6*(c*x^2)^(1/2)+1/8*b^2*c*x^7*(c*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 60, 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} \[ \int x^2 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{6} a^2 c x^5 \sqrt {c x^2}+\frac {2}{7} a b c x^6 \sqrt {c x^2}+\frac {1}{8} b^2 c x^7 \sqrt {c x^2} \]

[In]

Int[x^2*(c*x^2)^(3/2)*(a + b*x)^2,x]

[Out]

(a^2*c*x^5*Sqrt[c*x^2])/6 + (2*a*b*c*x^6*Sqrt[c*x^2])/7 + (b^2*c*x^7*Sqrt[c*x^2])/8

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 \sqrt {c x^2}\right ) \int x^5 (a+b x)^2 \, dx}{x} \\ & = \frac {\left (c \sqrt {c x^2}\right ) \int \left (a^2 x^5+2 a b x^6+b^2 x^7\right ) \, dx}{x} \\ & = \frac {1}{6} a^2 c x^5 \sqrt {c x^2}+\frac {2}{7} a b c x^6 \sqrt {c x^2}+\frac {1}{8} b^2 c x^7 \sqrt {c x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.62 \[ \int x^2 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{168} \left (c x^2\right )^{3/2} \left (28 a^2 x^3+48 a b x^4+21 b^2 x^5\right ) \]

[In]

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

[Out]

((c*x^2)^(3/2)*(28*a^2*x^3 + 48*a*b*x^4 + 21*b^2*x^5))/168

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.53

method	result	size
gosper	\(\frac {x^{3} \left (21 b^{2} x^{2}+48 a b x +28 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{168}\)	\(32\)
default	\(\frac {x^{3} \left (21 b^{2} x^{2}+48 a b x +28 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{168}\)	\(32\)
risch	\(\frac {a^{2} c \,x^{5} \sqrt {c \,x^{2}}}{6}+\frac {2 a b c \,x^{6} \sqrt {c \,x^{2}}}{7}+\frac {b^{2} c \,x^{7} \sqrt {c \,x^{2}}}{8}\)	\(49\)
trager	\(\frac {c \left (21 b^{2} x^{7}+48 a b \,x^{6}+21 b^{2} x^{6}+28 a^{2} x^{5}+48 a b \,x^{5}+21 b^{2} x^{5}+28 a^{2} x^{4}+48 a b \,x^{4}+21 b^{2} x^{4}+28 a^{2} x^{3}+48 a b \,x^{3}+21 b^{2} x^{3}+28 a^{2} x^{2}+48 a b \,x^{2}+21 b^{2} x^{2}+28 a^{2} x +48 a b x +21 b^{2} x +28 a^{2}+48 a b +21 b^{2}\right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{168 x}\)	\(164\)

[In]

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

[Out]

1/168*x^3*(21*b^2*x^2+48*a*b*x+28*a^2)*(c*x^2)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.60 \[ \int x^2 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{168} \, {\left (21 \, b^{2} c x^{7} + 48 \, a b c x^{6} + 28 \, a^{2} c x^{5}\right )} \sqrt {c x^{2}} \]

[In]

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

[Out]

1/168*(21*b^2*c*x^7 + 48*a*b*c*x^6 + 28*a^2*c*x^5)*sqrt(c*x^2)

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85 \[ \int x^2 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {a^{2} x^{3} \left (c x^{2}\right )^{\frac {3}{2}}}{6} + \frac {2 a b x^{4} \left (c x^{2}\right )^{\frac {3}{2}}}{7} + \frac {b^{2} x^{5} \left (c x^{2}\right )^{\frac {3}{2}}}{8} \]

[In]

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

[Out]

a**2*x**3*(c*x**2)**(3/2)/6 + 2*a*b*x**4*(c*x**2)**(3/2)/7 + b**2*x**5*(c*x**2)**(3/2)/8

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.87 \[ \int x^2 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {\left (c x^{2}\right )^{\frac {5}{2}} b^{2} x^{3}}{8 \, c} + \frac {2 \, \left (c x^{2}\right )^{\frac {5}{2}} a b x^{2}}{7 \, c} + \frac {\left (c x^{2}\right )^{\frac {5}{2}} a^{2} x}{6 \, c} \]

[In]

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

[Out]

1/8*(c*x^2)^(5/2)*b^2*x^3/c + 2/7*(c*x^2)^(5/2)*a*b*x^2/c + 1/6*(c*x^2)^(5/2)*a^2*x/c

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.58 \[ \int x^2 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{168} \, {\left (21 \, b^{2} x^{8} \mathrm {sgn}\left (x\right ) + 48 \, a b x^{7} \mathrm {sgn}\left (x\right ) + 28 \, a^{2} x^{6} \mathrm {sgn}\left (x\right )\right )} c^{\frac {3}{2}} \]

[In]

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

[Out]

1/168*(21*b^2*x^8*sgn(x) + 48*a*b*x^7*sgn(x) + 28*a^2*x^6*sgn(x))*c^(3/2)

Mupad [F(-1)]

Timed out. \[ \int x^2 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\int x^2\,{\left (c\,x^2\right )}^{3/2}\,{\left (a+b\,x\right )}^2 \,d x \]

[In]

int(x^2*(c*x^2)^(3/2)*(a + b*x)^2,x)

[Out]

int(x^2*(c*x^2)^(3/2)*(a + b*x)^2, x)