3.14.16 \(\int (d+e x) (9+12 x+4 x^2)^{5/2} \, dx\)

Optimal. Leaf size=50 \[ \frac {1}{24} (2 x+3) \left (4 x^2+12 x+9\right )^{5/2} (2 d-3 e)+\frac {1}{28} e \left (4 x^2+12 x+9\right )^{7/2} \]

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Rubi [A]  time = 0.01, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {640, 609} \begin {gather*} \frac {1}{24} (2 x+3) \left (4 x^2+12 x+9\right )^{5/2} (2 d-3 e)+\frac {1}{28} e \left (4 x^2+12 x+9\right )^{7/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(9 + 12*x + 4*x^2)^(5/2),x]

[Out]

((2*d - 3*e)*(3 + 2*x)*(9 + 12*x + 4*x^2)^(5/2))/24 + (e*(9 + 12*x + 4*x^2)^(7/2))/28

Rule 609

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p + 1
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rule 640

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

Rubi steps

\begin {align*} \int (d+e x) \left (9+12 x+4 x^2\right )^{5/2} \, dx &=\frac {1}{28} e \left (9+12 x+4 x^2\right )^{7/2}+\frac {1}{2} (2 d-3 e) \int \left (9+12 x+4 x^2\right )^{5/2} \, dx\\ &=\frac {1}{24} (2 d-3 e) (3+2 x) \left (9+12 x+4 x^2\right )^{5/2}+\frac {1}{28} e \left (9+12 x+4 x^2\right )^{7/2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 81, normalized size = 1.62 \begin {gather*} \frac {x \sqrt {(2 x+3)^2} \left (14 d \left (16 x^5+144 x^4+540 x^3+1080 x^2+1215 x+729\right )+3 e x \left (64 x^5+560 x^4+2016 x^3+3780 x^2+3780 x+1701\right )\right )}{42 (2 x+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)*(9 + 12*x + 4*x^2)^(5/2),x]

[Out]

(x*Sqrt[(3 + 2*x)^2]*(14*d*(729 + 1215*x + 1080*x^2 + 540*x^3 + 144*x^4 + 16*x^5) + 3*e*x*(1701 + 3780*x + 378
0*x^2 + 2016*x^3 + 560*x^4 + 64*x^5)))/(42*(3 + 2*x))

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IntegrateAlgebraic [F]  time = 0.51, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x) \left (9+12 x+4 x^2\right )^{5/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(d + e*x)*(9 + 12*x + 4*x^2)^(5/2),x]

[Out]

Defer[IntegrateAlgebraic][(d + e*x)*(9 + 12*x + 4*x^2)^(5/2), x]

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fricas [A]  time = 0.39, size = 69, normalized size = 1.38 \begin {gather*} \frac {32}{7} \, e x^{7} + \frac {8}{3} \, {\left (2 \, d + 15 \, e\right )} x^{6} + 48 \, {\left (d + 3 \, e\right )} x^{5} + 90 \, {\left (2 \, d + 3 \, e\right )} x^{4} + 90 \, {\left (4 \, d + 3 \, e\right )} x^{3} + \frac {81}{2} \, {\left (10 \, d + 3 \, e\right )} x^{2} + 243 \, d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(4*x^2+12*x+9)^(5/2),x, algorithm="fricas")

[Out]

32/7*e*x^7 + 8/3*(2*d + 15*e)*x^6 + 48*(d + 3*e)*x^5 + 90*(2*d + 3*e)*x^4 + 90*(4*d + 3*e)*x^3 + 81/2*(10*d +
3*e)*x^2 + 243*d*x

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giac [B]  time = 0.16, size = 165, normalized size = 3.30 \begin {gather*} \frac {32}{7} \, x^{7} e \mathrm {sgn}\left (2 \, x + 3\right ) + \frac {16}{3} \, d x^{6} \mathrm {sgn}\left (2 \, x + 3\right ) + 40 \, x^{6} e \mathrm {sgn}\left (2 \, x + 3\right ) + 48 \, d x^{5} \mathrm {sgn}\left (2 \, x + 3\right ) + 144 \, x^{5} e \mathrm {sgn}\left (2 \, x + 3\right ) + 180 \, d x^{4} \mathrm {sgn}\left (2 \, x + 3\right ) + 270 \, x^{4} e \mathrm {sgn}\left (2 \, x + 3\right ) + 360 \, d x^{3} \mathrm {sgn}\left (2 \, x + 3\right ) + 270 \, x^{3} e \mathrm {sgn}\left (2 \, x + 3\right ) + 405 \, d x^{2} \mathrm {sgn}\left (2 \, x + 3\right ) + \frac {243}{2} \, x^{2} e \mathrm {sgn}\left (2 \, x + 3\right ) + 243 \, d x \mathrm {sgn}\left (2 \, x + 3\right ) + \frac {243}{56} \, {\left (14 \, d - 3 \, e\right )} \mathrm {sgn}\left (2 \, x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(4*x^2+12*x+9)^(5/2),x, algorithm="giac")

[Out]

32/7*x^7*e*sgn(2*x + 3) + 16/3*d*x^6*sgn(2*x + 3) + 40*x^6*e*sgn(2*x + 3) + 48*d*x^5*sgn(2*x + 3) + 144*x^5*e*
sgn(2*x + 3) + 180*d*x^4*sgn(2*x + 3) + 270*x^4*e*sgn(2*x + 3) + 360*d*x^3*sgn(2*x + 3) + 270*x^3*e*sgn(2*x +
3) + 405*d*x^2*sgn(2*x + 3) + 243/2*x^2*e*sgn(2*x + 3) + 243*d*x*sgn(2*x + 3) + 243/56*(14*d - 3*e)*sgn(2*x +
3)

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maple [B]  time = 0.05, size = 86, normalized size = 1.72 \begin {gather*} \frac {\left (192 e \,x^{6}+224 x^{5} d +1680 x^{5} e +2016 d \,x^{4}+6048 x^{4} e +7560 d \,x^{3}+11340 x^{3} e +15120 d \,x^{2}+11340 e \,x^{2}+17010 d x +5103 e x +10206 d \right ) \left (\left (2 x +3\right )^{2}\right )^{\frac {5}{2}} x}{42 \left (2 x +3\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(4*x^2+12*x+9)^(5/2),x)

[Out]

1/42*x*(192*e*x^6+224*d*x^5+1680*e*x^5+2016*d*x^4+6048*e*x^4+7560*d*x^3+11340*e*x^3+15120*d*x^2+11340*e*x^2+17
010*d*x+5103*e*x+10206*d)*((2*x+3)^2)^(5/2)/(2*x+3)^5

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maxima [A]  time = 2.39, size = 78, normalized size = 1.56 \begin {gather*} \frac {1}{28} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {7}{2}} e + \frac {1}{6} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {5}{2}} d x - \frac {1}{4} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {5}{2}} e x + \frac {1}{4} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {5}{2}} d - \frac {3}{8} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {5}{2}} e \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(4*x^2+12*x+9)^(5/2),x, algorithm="maxima")

[Out]

1/28*(4*x^2 + 12*x + 9)^(7/2)*e + 1/6*(4*x^2 + 12*x + 9)^(5/2)*d*x - 1/4*(4*x^2 + 12*x + 9)^(5/2)*e*x + 1/4*(4
*x^2 + 12*x + 9)^(5/2)*d - 3/8*(4*x^2 + 12*x + 9)^(5/2)*e

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \left (d+e\,x\right )\,{\left (4\,x^2+12\,x+9\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)*(12*x + 4*x^2 + 9)^(5/2),x)

[Out]

int((d + e*x)*(12*x + 4*x^2 + 9)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d + e x\right ) \left (\left (2 x + 3\right )^{2}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(4*x**2+12*x+9)**(5/2),x)

[Out]

Integral((d + e*x)*((2*x + 3)**2)**(5/2), x)

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