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

Optimal. Leaf size=18 \[ \frac {2}{5} \left (a+b x+c x^2\right )^{5/2} \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {629} \begin {gather*} \frac {2}{5} \left (a+b x+c x^2\right )^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x + c*x^2)^(5/2))/5

Rule 629

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d*(a + b*x + c*x^2)^(p +
1))/(b*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.94 \begin {gather*} \frac {2}{5} (a+x (b+c x))^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + x*(b + c*x))^(5/2))/5

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IntegrateAlgebraic [A]  time = 0.02, size = 18, normalized size = 1.00 \begin {gather*} \frac {2}{5} \left (a+b x+c x^2\right )^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x + c*x^2)^(5/2))/5

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fricas [B]  time = 0.42, size = 49, normalized size = 2.72 \begin {gather*} \frac {2}{5} \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \sqrt {c x^{2} + b x + a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 14, normalized size = 0.78 \begin {gather*} \frac {2}{5} \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(c*x^2 + b*x + a)^(5/2)

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maple [A]  time = 0.04, size = 15, normalized size = 0.83 \begin {gather*} \frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 14, normalized size = 0.78 \begin {gather*} \frac {2}{5} \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(c*x^2 + b*x + a)^(5/2)

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mupad [B]  time = 1.92, size = 14, normalized size = 0.78 \begin {gather*} \frac {2\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(a + b*x + c*x^2)^(5/2))/5

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sympy [B]  time = 0.61, size = 136, normalized size = 7.56 \begin {gather*} \frac {2 a^{2} \sqrt {a + b x + c x^{2}}}{5} + \frac {4 a b x \sqrt {a + b x + c x^{2}}}{5} + \frac {4 a c x^{2} \sqrt {a + b x + c x^{2}}}{5} + \frac {2 b^{2} x^{2} \sqrt {a + b x + c x^{2}}}{5} + \frac {4 b c x^{3} \sqrt {a + b x + c x^{2}}}{5} + \frac {2 c^{2} x^{4} \sqrt {a + b x + c x^{2}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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