3.1597 \(\int (b+2 c x) \sqrt{d+e x} (a+b x+c x^2) \, dx\)

Optimal. Leaf size=132 \[ \frac{2 (d+e x)^{5/2} \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{5 e^4}-\frac{2 (d+e x)^{3/2} (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{3 e^4}-\frac{6 c (d+e x)^{7/2} (2 c d-b e)}{7 e^4}+\frac{4 c^2 (d+e x)^{9/2}}{9 e^4} \]

[Out]

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

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Rubi [A]  time = 0.0700871, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.038, Rules used = {771} \[ \frac{2 (d+e x)^{5/2} \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{5 e^4}-\frac{2 (d+e x)^{3/2} (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{3 e^4}-\frac{6 c (d+e x)^{7/2} (2 c d-b e)}{7 e^4}+\frac{4 c^2 (d+e x)^{9/2}}{9 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

Mathematica [A]  time = 0.128145, size = 110, normalized size = 0.83 \[ \frac{2 (d+e x)^{3/2} \left (3 c e \left (14 a e (3 e x-2 d)+3 b \left (8 d^2-12 d e x+15 e^2 x^2\right )\right )+21 b e^2 (5 a e-2 b d+3 b e x)+c^2 \left (48 d^2 e x-32 d^3-60 d e^2 x^2+70 e^3 x^3\right )\right )}{315 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x)^(3/2)*(21*b*e^2*(-2*b*d + 5*a*e + 3*b*e*x) + c^2*(-32*d^3 + 48*d^2*e*x - 60*d*e^2*x^2 + 70*e^3*x^
3) + 3*c*e*(14*a*e*(-2*d + 3*e*x) + 3*b*(8*d^2 - 12*d*e*x + 15*e^2*x^2))))/(315*e^4)

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Maple [A]  time = 0.004, size = 123, normalized size = 0.9 \begin{align*}{\frac{140\,{c}^{2}{x}^{3}{e}^{3}+270\,bc{e}^{3}{x}^{2}-120\,{c}^{2}d{e}^{2}{x}^{2}+252\,ac{e}^{3}x+126\,{b}^{2}{e}^{3}x-216\,bcd{e}^{2}x+96\,{c}^{2}{d}^{2}ex+210\,ab{e}^{3}-168\,acd{e}^{2}-84\,{b}^{2}d{e}^{2}+144\,b{d}^{2}ce-64\,{c}^{2}{d}^{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((2*c*x+b)*(c*x^2+b*x+a)*(e*x+d)^(1/2),x)

[Out]

2/315*(e*x+d)^(3/2)*(70*c^2*e^3*x^3+135*b*c*e^3*x^2-60*c^2*d*e^2*x^2+126*a*c*e^3*x+63*b^2*e^3*x-108*b*c*d*e^2*
x+48*c^2*d^2*e*x+105*a*b*e^3-84*a*c*d*e^2-42*b^2*d*e^2+72*b*c*d^2*e-32*c^2*d^3)/e^4

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Maxima [A]  time = 0.977821, size = 163, normalized size = 1.23 \begin{align*} \frac{2 \,{\left (70 \,{\left (e x + d\right )}^{\frac{9}{2}} c^{2} - 135 \,{\left (2 \, c^{2} d - b c e\right )}{\left (e x + d\right )}^{\frac{7}{2}} + 63 \,{\left (6 \, c^{2} d^{2} - 6 \, b c d e +{\left (b^{2} + 2 \, a c\right )} e^{2}\right )}{\left (e x + d\right )}^{\frac{5}{2}} - 105 \,{\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e - a b e^{3} +{\left (b^{2} + 2 \, a c\right )} d e^{2}\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((2*c*x+b)*(c*x^2+b*x+a)*(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

2/315*(70*(e*x + d)^(9/2)*c^2 - 135*(2*c^2*d - b*c*e)*(e*x + d)^(7/2) + 63*(6*c^2*d^2 - 6*b*c*d*e + (b^2 + 2*a
*c)*e^2)*(e*x + d)^(5/2) - 105*(2*c^2*d^3 - 3*b*c*d^2*e - a*b*e^3 + (b^2 + 2*a*c)*d*e^2)*(e*x + d)^(3/2))/e^4

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Fricas [A]  time = 1.25928, size = 383, normalized size = 2.9 \begin{align*} \frac{2 \,{\left (70 \, c^{2} e^{4} x^{4} - 32 \, c^{2} d^{4} + 72 \, b c d^{3} e + 105 \, a b d e^{3} - 42 \,{\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + 5 \,{\left (2 \, c^{2} d e^{3} + 27 \, b c e^{4}\right )} x^{3} - 3 \,{\left (4 \, c^{2} d^{2} e^{2} - 9 \, b c d e^{3} - 21 \,{\left (b^{2} + 2 \, a c\right )} e^{4}\right )} x^{2} +{\left (16 \, c^{2} d^{3} e - 36 \, b c d^{2} e^{2} + 105 \, a b e^{4} + 21 \,{\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x\right )} \sqrt{e x + d}}{315 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(70*c^2*e^4*x^4 - 32*c^2*d^4 + 72*b*c*d^3*e + 105*a*b*d*e^3 - 42*(b^2 + 2*a*c)*d^2*e^2 + 5*(2*c^2*d*e^3
+ 27*b*c*e^4)*x^3 - 3*(4*c^2*d^2*e^2 - 9*b*c*d*e^3 - 21*(b^2 + 2*a*c)*e^4)*x^2 + (16*c^2*d^3*e - 36*b*c*d^2*e^
2 + 105*a*b*e^4 + 21*(b^2 + 2*a*c)*d*e^3)*x)*sqrt(e*x + d)/e^4

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Sympy [A]  time = 3.64201, size = 155, normalized size = 1.17 \begin{align*} \frac{2 \left (\frac{2 c^{2} \left (d + e x\right )^{\frac{9}{2}}}{9 e^{3}} + \frac{\left (d + e x\right )^{\frac{7}{2}} \left (3 b c e - 6 c^{2} d\right )}{7 e^{3}} + \frac{\left (d + e x\right )^{\frac{5}{2}} \left (2 a c e^{2} + b^{2} e^{2} - 6 b c d e + 6 c^{2} d^{2}\right )}{5 e^{3}} + \frac{\left (d + e x\right )^{\frac{3}{2}} \left (a b e^{3} - 2 a c d e^{2} - b^{2} d e^{2} + 3 b c d^{2} e - 2 c^{2} d^{3}\right )}{3 e^{3}}\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.6333, size = 230, normalized size = 1.74 \begin{align*} \frac{2}{315} \,{\left (21 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} b^{2} e^{\left (-1\right )} + 42 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} a c 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 )} b c e^{\left (-2\right )} + 2 \,{\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 )} c^{2} e^{\left (-3\right )} + 105 \,{\left (x e + d\right )}^{\frac{3}{2}} a b\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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