3.1216 \(\int (A+B x) \sqrt{d+e x} (b x+c x^2) \, dx\)

Optimal. Leaf size=126 \[ -\frac{2 (d+e x)^{7/2} (-A c e-b B e+3 B c d)}{7 e^4}+\frac{2 (d+e x)^{5/2} (B d (3 c d-2 b e)-A e (2 c d-b e))}{5 e^4}-\frac{2 d (d+e x)^{3/2} (B d-A e) (c d-b e)}{3 e^4}+\frac{2 B c (d+e x)^{9/2}}{9 e^4} \]

[Out]

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

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Rubi [A]  time = 0.07656, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.042, Rules used = {771} \[ -\frac{2 (d+e x)^{7/2} (-A c e-b B e+3 B c d)}{7 e^4}+\frac{2 (d+e x)^{5/2} (B d (3 c d-2 b e)-A e (2 c d-b e))}{5 e^4}-\frac{2 d (d+e x)^{3/2} (B d-A e) (c d-b e)}{3 e^4}+\frac{2 B c (d+e x)^{9/2}}{9 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.110265, size = 113, normalized size = 0.9 \[ \frac{2 (d+e x)^{3/2} \left (3 A e \left (7 b e (3 e x-2 d)+c \left (8 d^2-12 d e x+15 e^2 x^2\right )\right )+B \left (3 b e \left (8 d^2-12 d e x+15 e^2 x^2\right )+c \left (24 d^2 e x-16 d^3-30 d e^2 x^2+35 e^3 x^3\right )\right )\right )}{315 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x)^(3/2)*(3*A*e*(7*b*e*(-2*d + 3*e*x) + c*(8*d^2 - 12*d*e*x + 15*e^2*x^2)) + B*(3*b*e*(8*d^2 - 12*d*
e*x + 15*e^2*x^2) + c*(-16*d^3 + 24*d^2*e*x - 30*d*e^2*x^2 + 35*e^3*x^3))))/(315*e^4)

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Maple [A]  time = 0.006, size = 121, normalized size = 1. \begin{align*} -{\frac{-70\,Bc{x}^{3}{e}^{3}-90\,Ac{e}^{3}{x}^{2}-90\,Bb{e}^{3}{x}^{2}+60\,Bcd{e}^{2}{x}^{2}-126\,Ab{e}^{3}x+72\,Acd{e}^{2}x+72\,Bbd{e}^{2}x-48\,Bc{d}^{2}ex+84\,Abd{e}^{2}-48\,Ac{d}^{2}e-48\,Bb{d}^{2}e+32\,Bc{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((B*x+A)*(e*x+d)^(1/2)*(c*x^2+b*x),x)

[Out]

-2/315*(e*x+d)^(3/2)*(-35*B*c*e^3*x^3-45*A*c*e^3*x^2-45*B*b*e^3*x^2+30*B*c*d*e^2*x^2-63*A*b*e^3*x+36*A*c*d*e^2
*x+36*B*b*d*e^2*x-24*B*c*d^2*e*x+42*A*b*d*e^2-24*A*c*d^2*e-24*B*b*d^2*e+16*B*c*d^3)/e^4

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Maxima [A]  time = 1.05792, size = 151, normalized size = 1.2 \begin{align*} \frac{2 \,{\left (35 \,{\left (e x + d\right )}^{\frac{9}{2}} B c - 45 \,{\left (3 \, B c d -{\left (B b + A c\right )} e\right )}{\left (e x + d\right )}^{\frac{7}{2}} + 63 \,{\left (3 \, B c d^{2} + A b e^{2} - 2 \,{\left (B b + A c\right )} d e\right )}{\left (e x + d\right )}^{\frac{5}{2}} - 105 \,{\left (B c d^{3} + A b d e^{2} -{\left (B b + A c\right )} d^{2} e\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)*(e*x+d)^(1/2)*(c*x^2+b*x),x, algorithm="maxima")

[Out]

2/315*(35*(e*x + d)^(9/2)*B*c - 45*(3*B*c*d - (B*b + A*c)*e)*(e*x + d)^(7/2) + 63*(3*B*c*d^2 + A*b*e^2 - 2*(B*
b + A*c)*d*e)*(e*x + d)^(5/2) - 105*(B*c*d^3 + A*b*d*e^2 - (B*b + A*c)*d^2*e)*(e*x + d)^(3/2))/e^4

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Fricas [A]  time = 1.62525, size = 340, normalized size = 2.7 \begin{align*} \frac{2 \,{\left (35 \, B c e^{4} x^{4} - 16 \, B c d^{4} - 42 \, A b d^{2} e^{2} + 24 \,{\left (B b + A c\right )} d^{3} e + 5 \,{\left (B c d e^{3} + 9 \,{\left (B b + A c\right )} e^{4}\right )} x^{3} - 3 \,{\left (2 \, B c d^{2} e^{2} - 21 \, A b e^{4} - 3 \,{\left (B b + A c\right )} d e^{3}\right )} x^{2} +{\left (8 \, B c d^{3} e + 21 \, A b d e^{3} - 12 \,{\left (B b + A c\right )} d^{2} e^{2}\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)*(e*x+d)^(1/2)*(c*x^2+b*x),x, algorithm="fricas")

[Out]

2/315*(35*B*c*e^4*x^4 - 16*B*c*d^4 - 42*A*b*d^2*e^2 + 24*(B*b + A*c)*d^3*e + 5*(B*c*d*e^3 + 9*(B*b + A*c)*e^4)
*x^3 - 3*(2*B*c*d^2*e^2 - 21*A*b*e^4 - 3*(B*b + A*c)*d*e^3)*x^2 + (8*B*c*d^3*e + 21*A*b*d*e^3 - 12*(B*b + A*c)
*d^2*e^2)*x)*sqrt(e*x + d)/e^4

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Sympy [A]  time = 4.77643, size = 146, normalized size = 1.16 \begin{align*} \frac{2 \left (\frac{B c \left (d + e x\right )^{\frac{9}{2}}}{9 e^{3}} + \frac{\left (d + e x\right )^{\frac{7}{2}} \left (A c e + B b e - 3 B c d\right )}{7 e^{3}} + \frac{\left (d + e x\right )^{\frac{5}{2}} \left (A b e^{2} - 2 A c d e - 2 B b d e + 3 B c d^{2}\right )}{5 e^{3}} + \frac{\left (d + e x\right )^{\frac{3}{2}} \left (- A b d e^{2} + A c d^{2} e + B b d^{2} e - B c 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)*(e*x+d)**(1/2)*(c*x**2+b*x),x)

[Out]

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

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Giac [A]  time = 1.2587, size = 227, normalized size = 1.8 \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 )} A b 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 b e^{\left (-2\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 )} A c 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 c e^{\left (-3\right )}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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