\(\int \frac {x^3 (a+b x)^2}{\sqrt {c x^2}} \, dx\) [828]

Optimal result

Integrand size = 20, antiderivative size = 57 \[ \int \frac {x^3 (a+b x)^2}{\sqrt {c x^2}} \, dx=\frac {a^2 x^4}{3 \sqrt {c x^2}}+\frac {a b x^5}{2 \sqrt {c x^2}}+\frac {b^2 x^6}{5 \sqrt {c x^2}} \]

[Out]

1/3*a^2*x^4/(c*x^2)^(1/2)+1/2*a*b*x^5/(c*x^2)^(1/2)+1/5*b^2*x^6/(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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 45} \[ \int \frac {x^3 (a+b x)^2}{\sqrt {c x^2}} \, dx=\frac {a^2 x^4}{3 \sqrt {c x^2}}+\frac {a b x^5}{2 \sqrt {c x^2}}+\frac {b^2 x^6}{5 \sqrt {c x^2}} \]

[In]

Int[(x^3*(a + b*x)^2)/Sqrt[c*x^2],x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.61 \[ \int \frac {x^3 (a+b x)^2}{\sqrt {c x^2}} \, dx=\frac {x^4 \left (10 a^2+15 a b x+6 b^2 x^2\right )}{30 \sqrt {c x^2}} \]

[In]

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

[Out]

(x^4*(10*a^2 + 15*a*b*x + 6*b^2*x^2))/(30*Sqrt[c*x^2])

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.56

method	result	size
gosper	\(\frac {x^{4} \left (6 b^{2} x^{2}+15 a b x +10 a^{2}\right )}{30 \sqrt {c \,x^{2}}}\)	\(32\)
default	\(\frac {x^{4} \left (6 b^{2} x^{2}+15 a b x +10 a^{2}\right )}{30 \sqrt {c \,x^{2}}}\)	\(32\)
risch	\(\frac {a^{2} x^{4}}{3 \sqrt {c \,x^{2}}}+\frac {a b \,x^{5}}{2 \sqrt {c \,x^{2}}}+\frac {b^{2} x^{6}}{5 \sqrt {c \,x^{2}}}\)	\(46\)
trager	\(\frac {\left (6 b^{2} x^{4}+15 a b \,x^{3}+6 b^{2} x^{3}+10 a^{2} x^{2}+15 a b \,x^{2}+6 b^{2} x^{2}+10 a^{2} x +15 a b x +6 b^{2} x +10 a^{2}+15 a b +6 b^{2}\right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{30 c x}\)	\(97\)

[In]

int(x^3*(b*x+a)^2/(c*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/30*x^4*(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 = 36, normalized size of antiderivative = 0.63 \[ \int \frac {x^3 (a+b x)^2}{\sqrt {c x^2}} \, dx=\frac {{\left (6 \, b^{2} x^{4} + 15 \, a b x^{3} + 10 \, a^{2} x^{2}\right )} \sqrt {c x^{2}}}{30 \, c} \]

[In]

integrate(x^3*(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)/c

Sympy [A] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int \frac {x^3 (a+b x)^2}{\sqrt {c x^2}} \, dx=\frac {a^{2} x^{4}}{3 \sqrt {c x^{2}}} + \frac {a b x^{5}}{2 \sqrt {c x^{2}}} + \frac {b^{2} x^{6}}{5 \sqrt {c x^{2}}} \]

[In]

integrate(x**3*(b*x+a)**2/(c*x**2)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.95 \[ \int \frac {x^3 (a+b x)^2}{\sqrt {c x^2}} \, dx=\frac {\sqrt {c x^{2}} b^{2} x^{4}}{5 \, c} + \frac {\sqrt {c x^{2}} a b x^{3}}{2 \, c} + \frac {\sqrt {c x^{2}} a^{2} x^{2}}{3 \, c} \]

[In]

integrate(x^3*(b*x+a)^2/(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.74 \[ \int \frac {x^3 (a+b x)^2}{\sqrt {c x^2}} \, dx=\frac {6 \, b^{2} \sqrt {c} x^{5} + 15 \, a b \sqrt {c} x^{4} + 10 \, a^{2} \sqrt {c} x^{3}}{30 \, c \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^3*(b*x+a)^2/(c*x^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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