3.9.29 \(\int \frac {x^2 (a+b x)^2}{\sqrt {c x^2}} \, dx\) [829]

Optimal. Leaf size=57 \[ \frac {a^2 x^3}{2 \sqrt {c x^2}}+\frac {2 a b x^4}{3 \sqrt {c x^2}}+\frac {b^2 x^5}{4 \sqrt {c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 45} \begin {gather*} \frac {a^2 x^3}{2 \sqrt {c x^2}}+\frac {2 a b x^4}{3 \sqrt {c x^2}}+\frac {b^2 x^5}{4 \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x)^2)/Sqrt[c*x^2],x]

[Out]

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

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Mathematica [A]
time = 0.00, size = 35, normalized size = 0.61 \begin {gather*} \frac {x^3 \left (6 a^2+8 a b x+3 b^2 x^2\right )}{12 \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(6*a^2 + 8*a*b*x + 3*b^2*x^2))/(12*Sqrt[c*x^2])

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Mathics [A]
time = 1.88, size = 31, normalized size = 0.54 \begin {gather*} \frac {x^3 \left (6 a^2+8 a b x+3 b^2 x^2\right )}{12 \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x ^ 3 (6 a ^ 2 + 8 a b x + 3 b ^ 2 x ^ 2) / (12 Sqrt[c x ^ 2])

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Maple [A]
time = 0.10, size = 32, normalized size = 0.56

method	result	size
gosper	\(\frac {x^{3} \left (3 x^{2} b^{2}+8 a b x +6 a^{2}\right )}{12 \sqrt {c \,x^{2}}}\)	\(32\)
default	\(\frac {x^{3} \left (3 x^{2} b^{2}+8 a b x +6 a^{2}\right )}{12 \sqrt {c \,x^{2}}}\)	\(32\)
risch	\(\frac {a^{2} x^{3}}{2 \sqrt {c \,x^{2}}}+\frac {2 a b \,x^{4}}{3 \sqrt {c \,x^{2}}}+\frac {b^{2} x^{5}}{4 \sqrt {c \,x^{2}}}\)	\(46\)
trager	\(\frac {\left (3 b^{2} x^{3}+8 a b \,x^{2}+3 x^{2} b^{2}+6 a^{2} x +8 a b x +3 b^{2} x +6 a^{2}+8 a b +3 b^{2}\right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{12 c x}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*x^3*(3*b^2*x^2+8*a*b*x+6*a^2)/(c*x^2)^(1/2)

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Maxima [A]
time = 0.26, size = 47, normalized size = 0.82 \begin {gather*} \frac {\sqrt {c x^{2}} b^{2} x^{3}}{4 \, c} + \frac {2 \, \sqrt {c x^{2}} a b x^{2}}{3 \, c} + \frac {a^{2} x^{2}}{2 \, \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.29, size = 34, normalized size = 0.60 \begin {gather*} \frac {{\left (3 \, b^{2} x^{3} + 8 \, a b x^{2} + 6 \, a^{2} x\right )} \sqrt {c x^{2}}}{12 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b^2*x^3 + 8*a*b*x^2 + 6*a^2*x)*sqrt(c*x^2)/c

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Sympy [A]
time = 0.25, size = 51, normalized size = 0.89 \begin {gather*} \frac {a^{2} x^{3}}{2 \sqrt {c x^{2}}} + \frac {2 a b x^{4}}{3 \sqrt {c x^{2}}} + \frac {b^{2} x^{5}}{4 \sqrt {c x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.00, size = 38, normalized size = 0.67 \begin {gather*} \frac {\frac {1}{4} b^{2} x^{4}+\frac {2}{3} a b x^{3}+\frac {1}{2} a^{2} x^{2}}{\sqrt {c} \mathrm {sign}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b^2*x^4 + 8*a*b*x^3 + 6*a^2*x^2)/(sqrt(c)*sgn(x))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2\,{\left (a+b\,x\right )}^2}{\sqrt {c\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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