3.2045 \(\int (a+b x) \sqrt{d+e x} (a^2+2 a b x+b^2 x^2) \, dx\)

Optimal. Leaf size=100 \[ -\frac{6 b^2 (d+e x)^{7/2} (b d-a e)}{7 e^4}+\frac{6 b (d+e x)^{5/2} (b d-a e)^2}{5 e^4}-\frac{2 (d+e x)^{3/2} (b d-a e)^3}{3 e^4}+\frac{2 b^3 (d+e x)^{9/2}}{9 e^4} \]

[Out]

(-2*(b*d - a*e)^3*(d + e*x)^(3/2))/(3*e^4) + (6*b*(b*d - a*e)^2*(d + e*x)^(5/2))/(5*e^4) - (6*b^2*(b*d - a*e)*
(d + e*x)^(7/2))/(7*e^4) + (2*b^3*(d + e*x)^(9/2))/(9*e^4)

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Rubi [A]  time = 0.032099, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {27, 43} \[ -\frac{6 b^2 (d+e x)^{7/2} (b d-a e)}{7 e^4}+\frac{6 b (d+e x)^{5/2} (b d-a e)^2}{5 e^4}-\frac{2 (d+e x)^{3/2} (b d-a e)^3}{3 e^4}+\frac{2 b^3 (d+e x)^{9/2}}{9 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d - a*e)^3*(d + e*x)^(3/2))/(3*e^4) + (6*b*(b*d - a*e)^2*(d + e*x)^(5/2))/(5*e^4) - (6*b^2*(b*d - a*e)*
(d + e*x)^(7/2))/(7*e^4) + (2*b^3*(d + e*x)^(9/2))/(9*e^4)

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 (a+b x) \sqrt{d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx &=\int (a+b x)^3 \sqrt{d+e x} \, dx\\ &=\int \left (\frac{(-b d+a e)^3 \sqrt{d+e x}}{e^3}+\frac{3 b (b d-a e)^2 (d+e x)^{3/2}}{e^3}-\frac{3 b^2 (b d-a e) (d+e x)^{5/2}}{e^3}+\frac{b^3 (d+e x)^{7/2}}{e^3}\right ) \, dx\\ &=-\frac{2 (b d-a e)^3 (d+e x)^{3/2}}{3 e^4}+\frac{6 b (b d-a e)^2 (d+e x)^{5/2}}{5 e^4}-\frac{6 b^2 (b d-a e) (d+e x)^{7/2}}{7 e^4}+\frac{2 b^3 (d+e x)^{9/2}}{9 e^4}\\ \end{align*}

Mathematica [A]  time = 0.0568004, size = 79, normalized size = 0.79 \[ \frac{2 (d+e x)^{3/2} \left (-135 b^2 (d+e x)^2 (b d-a e)+189 b (d+e x) (b d-a e)^2-105 (b d-a e)^3+35 b^3 (d+e x)^3\right )}{315 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x)^(3/2)*(-105*(b*d - a*e)^3 + 189*b*(b*d - a*e)^2*(d + e*x) - 135*b^2*(b*d - a*e)*(d + e*x)^2 + 35*
b^3*(d + e*x)^3))/(315*e^4)

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Maple [A]  time = 0.007, size = 116, normalized size = 1.2 \begin{align*}{\frac{70\,{x}^{3}{b}^{3}{e}^{3}+270\,{x}^{2}a{b}^{2}{e}^{3}-60\,{x}^{2}{b}^{3}d{e}^{2}+378\,x{a}^{2}b{e}^{3}-216\,xa{b}^{2}d{e}^{2}+48\,x{b}^{3}{d}^{2}e+210\,{e}^{3}{a}^{3}-252\,d{e}^{2}{a}^{2}b+144\,a{d}^{2}e{b}^{2}-32\,{d}^{3}{b}^{3}}{315\,{e}^{4}} \left ( ex+d \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)*(e*x+d)^(1/2),x)

[Out]

2/315*(e*x+d)^(3/2)*(35*b^3*e^3*x^3+135*a*b^2*e^3*x^2-30*b^3*d*e^2*x^2+189*a^2*b*e^3*x-108*a*b^2*d*e^2*x+24*b^
3*d^2*e*x+105*a^3*e^3-126*a^2*b*d*e^2+72*a*b^2*d^2*e-16*b^3*d^3)/e^4

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Maxima [A]  time = 0.962159, size = 159, normalized size = 1.59 \begin{align*} \frac{2 \,{\left (35 \,{\left (e x + d\right )}^{\frac{9}{2}} b^{3} - 135 \,{\left (b^{3} d - a b^{2} e\right )}{\left (e x + d\right )}^{\frac{7}{2}} + 189 \,{\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )}{\left (e x + d\right )}^{\frac{5}{2}} - 105 \,{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )}{\left (e x + d\right )}^{\frac{3}{2}}\right )}}{315 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*(e*x + d)^(9/2)*b^3 - 135*(b^3*d - a*b^2*e)*(e*x + d)^(7/2) + 189*(b^3*d^2 - 2*a*b^2*d*e + a^2*b*e^2
)*(e*x + d)^(5/2) - 105*(b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*(e*x + d)^(3/2))/e^4

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Fricas [A]  time = 1.42544, size = 359, normalized size = 3.59 \begin{align*} \frac{2 \,{\left (35 \, b^{3} e^{4} x^{4} - 16 \, b^{3} d^{4} + 72 \, a b^{2} d^{3} e - 126 \, a^{2} b d^{2} e^{2} + 105 \, a^{3} d e^{3} + 5 \,{\left (b^{3} d e^{3} + 27 \, a b^{2} e^{4}\right )} x^{3} - 3 \,{\left (2 \, b^{3} d^{2} e^{2} - 9 \, a b^{2} d e^{3} - 63 \, a^{2} b e^{4}\right )} x^{2} +{\left (8 \, b^{3} d^{3} e - 36 \, a b^{2} d^{2} e^{2} + 63 \, a^{2} b d e^{3} + 105 \, a^{3} e^{4}\right )} x\right )} \sqrt{e x + d}}{315 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)*(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*b^3*e^4*x^4 - 16*b^3*d^4 + 72*a*b^2*d^3*e - 126*a^2*b*d^2*e^2 + 105*a^3*d*e^3 + 5*(b^3*d*e^3 + 27*a*
b^2*e^4)*x^3 - 3*(2*b^3*d^2*e^2 - 9*a*b^2*d*e^3 - 63*a^2*b*e^4)*x^2 + (8*b^3*d^3*e - 36*a*b^2*d^2*e^2 + 63*a^2
*b*d*e^3 + 105*a^3*e^4)*x)*sqrt(e*x + d)/e^4

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Sympy [A]  time = 3.33208, size = 146, normalized size = 1.46 \begin{align*} \frac{2 \left (\frac{b^{3} \left (d + e x\right )^{\frac{9}{2}}}{9 e^{3}} + \frac{\left (d + e x\right )^{\frac{7}{2}} \left (3 a b^{2} e - 3 b^{3} d\right )}{7 e^{3}} + \frac{\left (d + e x\right )^{\frac{5}{2}} \left (3 a^{2} b e^{2} - 6 a b^{2} d e + 3 b^{3} d^{2}\right )}{5 e^{3}} + \frac{\left (d + e x\right )^{\frac{3}{2}} \left (a^{3} e^{3} - 3 a^{2} b d e^{2} + 3 a b^{2} d^{2} e - b^{3} d^{3}\right )}{3 e^{3}}\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b**3*(d + e*x)**(9/2)/(9*e**3) + (d + e*x)**(7/2)*(3*a*b**2*e - 3*b**3*d)/(7*e**3) + (d + e*x)**(5/2)*(3*a*
*2*b*e**2 - 6*a*b**2*d*e + 3*b**3*d**2)/(5*e**3) + (d + e*x)**(3/2)*(a**3*e**3 - 3*a**2*b*d*e**2 + 3*a*b**2*d*
*2*e - b**3*d**3)/(3*e**3))/e

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Giac [A]  time = 1.11792, size = 196, normalized size = 1.96 \begin{align*} \frac{2}{315} \,{\left (63 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} a^{2} b e^{\left (-1\right )} + 9 \,{\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^{2} e^{\left (-2\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^{3} e^{\left (-3\right )} + 105 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{3}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)*(e*x+d)^(1/2),x, algorithm="giac")

[Out]

2/315*(63*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*a^2*b*e^(-1) + 9*(15*(x*e + d)^(7/2) - 42*(x*e + d)^(5/2)*
d + 35*(x*e + d)^(3/2)*d^2)*a*b^2*e^(-2) + (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^3*e^(-3) + 105*(x*e + d)^(3/2)*a^3)*e^(-1)