3.9.7 \(\int \sqrt {c x^2} (a+b x)^2 \, dx\) [807]

Optimal. Leaf size=55 \[ \frac {1}{2} a^2 x \sqrt {c x^2}+\frac {2}{3} a b x^2 \sqrt {c x^2}+\frac {1}{4} b^2 x^3 \sqrt {c x^2} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {15, 45} \begin {gather*} \frac {1}{2} a^2 x \sqrt {c x^2}+\frac {2}{3} a b x^2 \sqrt {c x^2}+\frac {1}{4} b^2 x^3 \sqrt {c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x^2]*(a + b*x)^2,x]

[Out]

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

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

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Mathematica [A]
time = 0.00, size = 33, normalized size = 0.60 \begin {gather*} \frac {1}{12} x \sqrt {c x^2} \left (6 a^2+8 a b x+3 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*x^2]*(a + b*x)^2,x]

[Out]

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

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Mathics [A]
time = 1.80, size = 29, normalized size = 0.53 \begin {gather*} \frac {x \left (6 a^2+8 a b x+3 b^2 x^2\right ) \sqrt {c x^2}}{12} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 30, normalized size = 0.55

method	result	size
gosper	\(\frac {x \left (3 x^{2} b^{2}+8 a b x +6 a^{2}\right ) \sqrt {c \,x^{2}}}{12}\)	\(30\)
default	\(\frac {x \left (3 x^{2} b^{2}+8 a b x +6 a^{2}\right ) \sqrt {c \,x^{2}}}{12}\)	\(30\)
risch	\(\frac {a^{2} x \sqrt {c \,x^{2}}}{2}+\frac {2 a b \,x^{2} \sqrt {c \,x^{2}}}{3}+\frac {b^{2} x^{3} \sqrt {c \,x^{2}}}{4}\)	\(44\)
trager	\(\frac {\left (3 b^{2} x^{3}+8 a b \,x^{2}+3 x^{2} b^{2}+6 a^{2} x +8 a b x +3 b^{2} x +6 a^{2}+8 a b +3 b^{2}\right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{12 x}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 44, normalized size = 0.80 \begin {gather*} \frac {1}{2} \, \sqrt {c x^{2}} a^{2} x + \frac {\left (c x^{2}\right )^{\frac {3}{2}} b^{2} x}{4 \, c} + \frac {2 \, \left (c x^{2}\right )^{\frac {3}{2}} a b}{3 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.29, size = 31, normalized size = 0.56 \begin {gather*} \frac {1}{12} \, {\left (3 \, b^{2} x^{3} + 8 \, a b x^{2} + 6 \, a^{2} x\right )} \sqrt {c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)

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Sympy [A]
time = 0.11, size = 49, normalized size = 0.89 \begin {gather*} \frac {a^{2} x \sqrt {c x^{2}}}{2} + \frac {2 a b x^{2} \sqrt {c x^{2}}}{3} + \frac {b^{2} x^{3} \sqrt {c x^{2}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.00, size = 40, normalized size = 0.73 \begin {gather*} \sqrt {c} \left (\frac {1}{2} a^{2} x^{2} \mathrm {sign}\left (x\right )+\frac {1}{4} b^{2} x^{4} \mathrm {sign}\left (x\right )+\frac {2}{3} a b x^{3} \mathrm {sign}\left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2)^(1/2)*(a + b*x)^2,x)

[Out]

int((c*x^2)^(1/2)*(a + b*x)^2, x)

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