\(\int x^3 (c x^2)^{3/2} (a+b x)^2 \, dx\) [812]

Optimal result

Integrand size = 20, antiderivative size = 60 \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{7} a^2 c x^6 \sqrt {c x^2}+\frac {1}{4} a b c x^7 \sqrt {c x^2}+\frac {1}{9} b^2 c x^8 \sqrt {c x^2} \]

[Out]

1/7*a^2*c*x^6*(c*x^2)^(1/2)+1/4*a*b*c*x^7*(c*x^2)^(1/2)+1/9*b^2*c*x^8*(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} \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{7} a^2 c x^6 \sqrt {c x^2}+\frac {1}{4} a b c x^7 \sqrt {c x^2}+\frac {1}{9} b^2 c x^8 \sqrt {c x^2} \]

[In]

Int[x^3*(c*x^2)^(3/2)*(a + b*x)^2,x]

[Out]

(a^2*c*x^6*Sqrt[c*x^2])/7 + (a*b*c*x^7*Sqrt[c*x^2])/4 + (b^2*c*x^8*Sqrt[c*x^2])/9

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^6 (a+b x)^2 \, dx}{x} \\ & = \frac {\left (c \sqrt {c x^2}\right ) \int \left (a^2 x^6+2 a b x^7+b^2 x^8\right ) \, dx}{x} \\ & = \frac {1}{7} a^2 c x^6 \sqrt {c x^2}+\frac {1}{4} a b c x^7 \sqrt {c x^2}+\frac {1}{9} b^2 c x^8 \sqrt {c x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.62 \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{252} \left (c x^2\right )^{3/2} \left (36 a^2 x^4+63 a b x^5+28 b^2 x^6\right ) \]

[In]

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

[Out]

((c*x^2)^(3/2)*(36*a^2*x^4 + 63*a*b*x^5 + 28*b^2*x^6))/252

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.53

method	result	size
gosper	\(\frac {x^{4} \left (28 b^{2} x^{2}+63 a b x +36 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{252}\)	\(32\)
default	\(\frac {x^{4} \left (28 b^{2} x^{2}+63 a b x +36 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{252}\)	\(32\)
risch	\(\frac {a^{2} c \,x^{6} \sqrt {c \,x^{2}}}{7}+\frac {a b c \,x^{7} \sqrt {c \,x^{2}}}{4}+\frac {b^{2} c \,x^{8} \sqrt {c \,x^{2}}}{9}\)	\(49\)
trager	\(\frac {c \left (28 b^{2} x^{8}+63 a b \,x^{7}+28 b^{2} x^{7}+36 a^{2} x^{6}+63 a b \,x^{6}+28 b^{2} x^{6}+36 a^{2} x^{5}+63 a b \,x^{5}+28 b^{2} x^{5}+36 a^{2} x^{4}+63 a b \,x^{4}+28 b^{2} x^{4}+36 a^{2} x^{3}+63 a b \,x^{3}+28 b^{2} x^{3}+36 a^{2} x^{2}+63 a b \,x^{2}+28 b^{2} x^{2}+36 a^{2} x +63 a b x +28 b^{2} x +36 a^{2}+63 a b +28 b^{2}\right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{252 x}\)	\(187\)

[In]

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

[Out]

1/252*x^4*(28*b^2*x^2+63*a*b*x+36*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 \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{252} \, {\left (28 \, b^{2} c x^{8} + 63 \, a b c x^{7} + 36 \, a^{2} c x^{6}\right )} \sqrt {c x^{2}} \]

[In]

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

[Out]

1/252*(28*b^2*c*x^8 + 63*a*b*c*x^7 + 36*a^2*c*x^6)*sqrt(c*x^2)

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.82 \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {a^{2} x^{4} \left (c x^{2}\right )^{\frac {3}{2}}}{7} + \frac {a b x^{5} \left (c x^{2}\right )^{\frac {3}{2}}}{4} + \frac {b^{2} x^{6} \left (c x^{2}\right )^{\frac {3}{2}}}{9} \]

[In]

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

[Out]

a**2*x**4*(c*x**2)**(3/2)/7 + a*b*x**5*(c*x**2)**(3/2)/4 + b**2*x**6*(c*x**2)**(3/2)/9

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {\left (c x^{2}\right )^{\frac {5}{2}} b^{2} x^{4}}{9 \, c} + \frac {\left (c x^{2}\right )^{\frac {5}{2}} a b x^{3}}{4 \, c} + \frac {\left (c x^{2}\right )^{\frac {5}{2}} a^{2} x^{2}}{7 \, c} \]

[In]

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

[Out]

1/9*(c*x^2)^(5/2)*b^2*x^4/c + 1/4*(c*x^2)^(5/2)*a*b*x^3/c + 1/7*(c*x^2)^(5/2)*a^2*x^2/c

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.58 \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{252} \, {\left (28 \, b^{2} x^{9} \mathrm {sgn}\left (x\right ) + 63 \, a b x^{8} \mathrm {sgn}\left (x\right ) + 36 \, a^{2} x^{7} \mathrm {sgn}\left (x\right )\right )} c^{\frac {3}{2}} \]

[In]

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

[Out]

1/252*(28*b^2*x^9*sgn(x) + 63*a*b*x^8*sgn(x) + 36*a^2*x^7*sgn(x))*c^(3/2)

Mupad [F(-1)]

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

[In]

int(x^3*(c*x^2)^(3/2)*(a + b*x)^2,x)

[Out]

int(x^3*(c*x^2)^(3/2)*(a + b*x)^2, x)