### 3.1623 $$\int (d+e x)^{3/2} (a^2+2 a b x+b^2 x^2) \, dx$$

Optimal. Leaf size=71 $-\frac{4 b (d+e x)^{7/2} (b d-a e)}{7 e^3}+\frac{2 (d+e x)^{5/2} (b d-a e)^2}{5 e^3}+\frac{2 b^2 (d+e x)^{9/2}}{9 e^3}$

[Out]

(2*(b*d - a*e)^2*(d + e*x)^(5/2))/(5*e^3) - (4*b*(b*d - a*e)*(d + e*x)^(7/2))/(7*e^3) + (2*b^2*(d + e*x)^(9/2)
)/(9*e^3)

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Rubi [A]  time = 0.0228057, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {27, 43} $-\frac{4 b (d+e x)^{7/2} (b d-a e)}{7 e^3}+\frac{2 (d+e x)^{5/2} (b d-a e)^2}{5 e^3}+\frac{2 b^2 (d+e x)^{9/2}}{9 e^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(b*d - a*e)^2*(d + e*x)^(5/2))/(5*e^3) - (4*b*(b*d - a*e)*(d + e*x)^(7/2))/(7*e^3) + (2*b^2*(d + e*x)^(9/2)
)/(9*e^3)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

Mathematica [A]  time = 0.0368846, size = 61, normalized size = 0.86 $\frac{2 (d+e x)^{5/2} \left (63 a^2 e^2+18 a b e (5 e x-2 d)+b^2 \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )}{315 e^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(d + e*x)^(5/2)*(63*a^2*e^2 + 18*a*b*e*(-2*d + 5*e*x) + b^2*(8*d^2 - 20*d*e*x + 35*e^2*x^2)))/(315*e^3)

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Maple [A]  time = 0.046, size = 63, normalized size = 0.9 \begin{align*}{\frac{70\,{b}^{2}{x}^{2}{e}^{2}+180\,xab{e}^{2}-40\,x{b}^{2}de+126\,{a}^{2}{e}^{2}-72\,abde+16\,{b}^{2}{d}^{2}}{315\,{e}^{3}} \left ( ex+d \right ) ^{{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(e*x+d)^(5/2)*(35*b^2*e^2*x^2+90*a*b*e^2*x-20*b^2*d*e*x+63*a^2*e^2-36*a*b*d*e+8*b^2*d^2)/e^3

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Maxima [A]  time = 1.03479, size = 92, normalized size = 1.3 \begin{align*} \frac{2 \,{\left (35 \,{\left (e x + d\right )}^{\frac{9}{2}} b^{2} - 90 \,{\left (b^{2} d - a b e\right )}{\left (e x + d\right )}^{\frac{7}{2}} + 63 \,{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )}{\left (e x + d\right )}^{\frac{5}{2}}\right )}}{315 \, e^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*(e*x + d)^(9/2)*b^2 - 90*(b^2*d - a*b*e)*(e*x + d)^(7/2) + 63*(b^2*d^2 - 2*a*b*d*e + a^2*e^2)*(e*x +
d)^(5/2))/e^3

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Fricas [B]  time = 1.53832, size = 300, normalized size = 4.23 \begin{align*} \frac{2 \,{\left (35 \, b^{2} e^{4} x^{4} + 8 \, b^{2} d^{4} - 36 \, a b d^{3} e + 63 \, a^{2} d^{2} e^{2} + 10 \,{\left (5 \, b^{2} d e^{3} + 9 \, a b e^{4}\right )} x^{3} + 3 \,{\left (b^{2} d^{2} e^{2} + 48 \, a b d e^{3} + 21 \, a^{2} e^{4}\right )} x^{2} - 2 \,{\left (2 \, b^{2} d^{3} e - 9 \, a b d^{2} e^{2} - 63 \, a^{2} d e^{3}\right )} x\right )} \sqrt{e x + d}}{315 \, e^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^2*e^4*x^4 + 8*b^2*d^4 - 36*a*b*d^3*e + 63*a^2*d^2*e^2 + 10*(5*b^2*d*e^3 + 9*a*b*e^4)*x^3 + 3*(b^2*
d^2*e^2 + 48*a*b*d*e^3 + 21*a^2*e^4)*x^2 - 2*(2*b^2*d^3*e - 9*a*b*d^2*e^2 - 63*a^2*d*e^3)*x)*sqrt(e*x + d)/e^3

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Sympy [A]  time = 9.39779, size = 240, normalized size = 3.38 \begin{align*} a^{2} d \left (\begin{cases} \sqrt{d} x & \text{for}\: e = 0 \\\frac{2 \left (d + e x\right )^{\frac{3}{2}}}{3 e} & \text{otherwise} \end{cases}\right ) + \frac{2 a^{2} \left (- \frac{d \left (d + e x\right )^{\frac{3}{2}}}{3} + \frac{\left (d + e x\right )^{\frac{5}{2}}}{5}\right )}{e} + \frac{4 a b d \left (- \frac{d \left (d + e x\right )^{\frac{3}{2}}}{3} + \frac{\left (d + e x\right )^{\frac{5}{2}}}{5}\right )}{e^{2}} + \frac{4 a b \left (\frac{d^{2} \left (d + e x\right )^{\frac{3}{2}}}{3} - \frac{2 d \left (d + e x\right )^{\frac{5}{2}}}{5} + \frac{\left (d + e x\right )^{\frac{7}{2}}}{7}\right )}{e^{2}} + \frac{2 b^{2} d \left (\frac{d^{2} \left (d + e x\right )^{\frac{3}{2}}}{3} - \frac{2 d \left (d + e x\right )^{\frac{5}{2}}}{5} + \frac{\left (d + e x\right )^{\frac{7}{2}}}{7}\right )}{e^{3}} + \frac{2 b^{2} \left (- \frac{d^{3} \left (d + e x\right )^{\frac{3}{2}}}{3} + \frac{3 d^{2} \left (d + e x\right )^{\frac{5}{2}}}{5} - \frac{3 d \left (d + e x\right )^{\frac{7}{2}}}{7} + \frac{\left (d + e x\right )^{\frac{9}{2}}}{9}\right )}{e^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*d*Piecewise((sqrt(d)*x, Eq(e, 0)), (2*(d + e*x)**(3/2)/(3*e), True)) + 2*a**2*(-d*(d + e*x)**(3/2)/3 + (d
+ e*x)**(5/2)/5)/e + 4*a*b*d*(-d*(d + e*x)**(3/2)/3 + (d + e*x)**(5/2)/5)/e**2 + 4*a*b*(d**2*(d + e*x)**(3/2)
/3 - 2*d*(d + e*x)**(5/2)/5 + (d + e*x)**(7/2)/7)/e**2 + 2*b**2*d*(d**2*(d + e*x)**(3/2)/3 - 2*d*(d + e*x)**(5
/2)/5 + (d + e*x)**(7/2)/7)/e**3 + 2*b**2*(-d**3*(d + e*x)**(3/2)/3 + 3*d**2*(d + e*x)**(5/2)/5 - 3*d*(d + e*x
)**(7/2)/7 + (d + e*x)**(9/2)/9)/e**3

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Giac [B]  time = 1.21849, size = 288, normalized size = 4.06 \begin{align*} \frac{2}{315} \,{\left (42 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} a b d e^{\left (-1\right )} + 3 \,{\left (15 \,{\left (x e + d\right )}^{\frac{7}{2}} - 42 \,{\left (x e + d\right )}^{\frac{5}{2}} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{2}\right )} b^{2} d e^{\left (-2\right )} + 105 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} d + 6 \,{\left (15 \,{\left (x e + d\right )}^{\frac{7}{2}} - 42 \,{\left (x e + d\right )}^{\frac{5}{2}} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{2}\right )} a b e^{\left (-1\right )} +{\left (35 \,{\left (x e + d\right )}^{\frac{9}{2}} - 135 \,{\left (x e + d\right )}^{\frac{7}{2}} d + 189 \,{\left (x e + d\right )}^{\frac{5}{2}} d^{2} - 105 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{3}\right )} b^{2} e^{\left (-2\right )} + 21 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} a^{2}\right )} e^{\left (-1\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(42*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*a*b*d*e^(-1) + 3*(15*(x*e + d)^(7/2) - 42*(x*e + d)^(5/2)*
d + 35*(x*e + d)^(3/2)*d^2)*b^2*d*e^(-2) + 105*(x*e + d)^(3/2)*a^2*d + 6*(15*(x*e + d)^(7/2) - 42*(x*e + d)^(5
/2)*d + 35*(x*e + d)^(3/2)*d^2)*a*b*e^(-1) + (35*(x*e + d)^(9/2) - 135*(x*e + d)^(7/2)*d + 189*(x*e + d)^(5/2)
*d^2 - 105*(x*e + d)^(3/2)*d^3)*b^2*e^(-2) + 21*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*a^2)*e^(-1)