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

Optimal. Leaf size=60 \[ \frac {1}{6} a^2 c x^5 \sqrt {c x^2}+\frac {2}{7} a b c x^6 \sqrt {c x^2}+\frac {1}{8} b^2 c x^7 \sqrt {c x^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 60, 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, 43} \begin {gather*} \frac {1}{6} a^2 c x^5 \sqrt {c x^2}+\frac {2}{7} a b c x^6 \sqrt {c x^2}+\frac {1}{8} b^2 c x^7 \sqrt {c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(c*x^2)^(3/2)*(a + b*x)^2,x]

[Out]

(a^2*c*x^5*Sqrt[c*x^2])/6 + (2*a*b*c*x^6*Sqrt[c*x^2])/7 + (b^2*c*x^7*Sqrt[c*x^2])/8

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 43

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 x^2 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx &=\frac {\left (c \sqrt {c x^2}\right ) \int x^5 (a+b x)^2 \, dx}{x}\\ &=\frac {\left (c \sqrt {c x^2}\right ) \int \left (a^2 x^5+2 a b x^6+b^2 x^7\right ) \, dx}{x}\\ &=\frac {1}{6} a^2 c x^5 \sqrt {c x^2}+\frac {2}{7} a b c x^6 \sqrt {c x^2}+\frac {1}{8} b^2 c x^7 \sqrt {c x^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 0.58 \begin {gather*} \frac {1}{168} x^3 \left (c x^2\right )^{3/2} \left (28 a^2+48 a b x+21 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(c*x^2)^(3/2)*(28*a^2 + 48*a*b*x + 21*b^2*x^2))/168

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IntegrateAlgebraic [A]  time = 0.03, size = 35, normalized size = 0.58 \begin {gather*} \frac {1}{168} x^3 \left (c x^2\right )^{3/2} \left (28 a^2+48 a b x+21 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(c*x^2)^(3/2)*(28*a^2 + 48*a*b*x + 21*b^2*x^2))/168

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fricas [A]  time = 1.14, size = 36, normalized size = 0.60 \begin {gather*} \frac {1}{168} \, {\left (21 \, b^{2} c x^{7} + 48 \, a b c x^{6} + 28 \, a^{2} c x^{5}\right )} \sqrt {c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/168*(21*b^2*c*x^7 + 48*a*b*c*x^6 + 28*a^2*c*x^5)*sqrt(c*x^2)

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giac [A]  time = 1.08, size = 35, normalized size = 0.58 \begin {gather*} \frac {1}{168} \, {\left (21 \, b^{2} x^{8} \mathrm {sgn}\relax (x) + 48 \, a b x^{7} \mathrm {sgn}\relax (x) + 28 \, a^{2} x^{6} \mathrm {sgn}\relax (x)\right )} c^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/168*(21*b^2*x^8*sgn(x) + 48*a*b*x^7*sgn(x) + 28*a^2*x^6*sgn(x))*c^(3/2)

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maple [A]  time = 0.00, size = 32, normalized size = 0.53 \begin {gather*} \frac {\left (21 b^{2} x^{2}+48 a b x +28 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {3}{2}} x^{3}}{168} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/168*x^3*(21*b^2*x^2+48*a*b*x+28*a^2)*(c*x^2)^(3/2)

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maxima [A]  time = 1.33, size = 52, normalized size = 0.87 \begin {gather*} \frac {\left (c x^{2}\right )^{\frac {5}{2}} b^{2} x^{3}}{8 \, c} + \frac {2 \, \left (c x^{2}\right )^{\frac {5}{2}} a b x^{2}}{7 \, c} + \frac {\left (c x^{2}\right )^{\frac {5}{2}} a^{2} x}{6 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(c*x^2)^(5/2)*b^2*x^3/c + 2/7*(c*x^2)^(5/2)*a*b*x^2/c + 1/6*(c*x^2)^(5/2)*a^2*x/c

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^2\,{\left (c\,x^2\right )}^{3/2}\,{\left (a+b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*x^2)^(3/2)*(a + b*x)^2,x)

[Out]

int(x^2*(c*x^2)^(3/2)*(a + b*x)^2, x)

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sympy [A]  time = 1.22, size = 61, normalized size = 1.02 \begin {gather*} \frac {a^{2} c^{\frac {3}{2}} x^{3} \left (x^{2}\right )^{\frac {3}{2}}}{6} + \frac {2 a b c^{\frac {3}{2}} x^{4} \left (x^{2}\right )^{\frac {3}{2}}}{7} + \frac {b^{2} c^{\frac {3}{2}} x^{5} \left (x^{2}\right )^{\frac {3}{2}}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*c**(3/2)*x**3*(x**2)**(3/2)/6 + 2*a*b*c**(3/2)*x**4*(x**2)**(3/2)/7 + b**2*c**(3/2)*x**5*(x**2)**(3/2)/8

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