\(\int x \sqrt {c x^2} (a+b x)^2 \, dx\) [806]

Optimal result

Integrand size = 18, antiderivative size = 57 \[ \int x \sqrt {c x^2} (a+b x)^2 \, dx=\frac {1}{3} a^2 x^2 \sqrt {c x^2}+\frac {1}{2} a b x^3 \sqrt {c x^2}+\frac {1}{5} b^2 x^4 \sqrt {c x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 57, 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} \[ \int x \sqrt {c x^2} (a+b x)^2 \, dx=\frac {1}{3} a^2 x^2 \sqrt {c x^2}+\frac {1}{2} a b x^3 \sqrt {c x^2}+\frac {1}{5} b^2 x^4 \sqrt {c x^2} \]

[In]

Int[x*Sqrt[c*x^2]*(a + b*x)^2,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.61 \[ \int x \sqrt {c x^2} (a+b x)^2 \, dx=\frac {1}{30} x^2 \sqrt {c x^2} \left (10 a^2+15 a b x+6 b^2 x^2\right ) \]

[In]

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

[Out]

(x^2*Sqrt[c*x^2]*(10*a^2 + 15*a*b*x + 6*b^2*x^2))/30

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.56

method	result	size
gosper	\(\frac {x^{2} \left (6 b^{2} x^{2}+15 a b x +10 a^{2}\right ) \sqrt {c \,x^{2}}}{30}\)	\(32\)
default	\(\frac {x^{2} \left (6 b^{2} x^{2}+15 a b x +10 a^{2}\right ) \sqrt {c \,x^{2}}}{30}\)	\(32\)
risch	\(\frac {a^{2} x^{2} \sqrt {c \,x^{2}}}{3}+\frac {a b \,x^{3} \sqrt {c \,x^{2}}}{2}+\frac {b^{2} x^{4} \sqrt {c \,x^{2}}}{5}\)	\(46\)
trager	\(\frac {\left (6 b^{2} x^{4}+15 a b \,x^{3}+6 b^{2} x^{3}+10 a^{2} x^{2}+15 a b \,x^{2}+6 b^{2} x^{2}+10 a^{2} x +15 a b x +6 b^{2} x +10 a^{2}+15 a b +6 b^{2}\right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{30 x}\)	\(94\)

[In]

int(x*(b*x+a)^2*(c*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/30*x^2*(6*b^2*x^2+15*a*b*x+10*a^2)*(c*x^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.58 \[ \int x \sqrt {c x^2} (a+b x)^2 \, dx=\frac {1}{30} \, {\left (6 \, b^{2} x^{4} + 15 \, a b x^{3} + 10 \, a^{2} x^{2}\right )} \sqrt {c x^{2}} \]

[In]

integrate(x*(b*x+a)^2*(c*x^2)^(1/2),x, algorithm="fricas")

[Out]

1/30*(6*b^2*x^4 + 15*a*b*x^3 + 10*a^2*x^2)*sqrt(c*x^2)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int x \sqrt {c x^2} (a+b x)^2 \, dx=\frac {a^{2} x^{2} \sqrt {c x^{2}}}{3} + \frac {a b x^{3} \sqrt {c x^{2}}}{2} + \frac {b^{2} x^{4} \sqrt {c x^{2}}}{5} \]

[In]

integrate(x*(b*x+a)**2*(c*x**2)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int x \sqrt {c x^2} (a+b x)^2 \, dx=\frac {\left (c x^{2}\right )^{\frac {3}{2}} b^{2} x^{2}}{5 \, c} + \frac {\left (c x^{2}\right )^{\frac {3}{2}} a b x}{2 \, c} + \frac {\left (c x^{2}\right )^{\frac {3}{2}} a^{2}}{3 \, c} \]

[In]

integrate(x*(b*x+a)^2*(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.61 \[ \int x \sqrt {c x^2} (a+b x)^2 \, dx=\frac {1}{30} \, {\left (6 \, b^{2} x^{5} \mathrm {sgn}\left (x\right ) + 15 \, a b x^{4} \mathrm {sgn}\left (x\right ) + 10 \, a^{2} x^{3} \mathrm {sgn}\left (x\right )\right )} \sqrt {c} \]

[In]

integrate(x*(b*x+a)^2*(c*x^2)^(1/2),x, algorithm="giac")

[Out]

1/30*(6*b^2*x^5*sgn(x) + 15*a*b*x^4*sgn(x) + 10*a^2*x^3*sgn(x))*sqrt(c)

Mupad [F(-1)]

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

[In]

int(x*(c*x^2)^(1/2)*(a + b*x)^2,x)

[Out]

int(x*(c*x^2)^(1/2)*(a + b*x)^2, x)