3.1787 \(\int (A+B x) \sqrt{d+e x} (a^2+2 a b x+b^2 x^2) \, dx\)

Optimal. Leaf size=128 \[ -\frac{2 b (d+e x)^{7/2} (-2 a B e-A b e+3 b B d)}{7 e^4}+\frac{2 (d+e x)^{5/2} (b d-a e) (-a B e-2 A b e+3 b B d)}{5 e^4}-\frac{2 (d+e x)^{3/2} (b d-a e)^2 (B d-A e)}{3 e^4}+\frac{2 b^2 B (d+e x)^{9/2}}{9 e^4} \]

[Out]

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

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Rubi [A]  time = 0.0560255, antiderivative size = 128, 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, 77} \[ -\frac{2 b (d+e x)^{7/2} (-2 a B e-A b e+3 b B d)}{7 e^4}+\frac{2 (d+e x)^{5/2} (b d-a e) (-a B e-2 A b e+3 b B d)}{5 e^4}-\frac{2 (d+e x)^{3/2} (b d-a e)^2 (B d-A e)}{3 e^4}+\frac{2 b^2 B (d+e x)^{9/2}}{9 e^4} \]

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)^2*(B*d - A*e)*(d + e*x)^(3/2))/(3*e^4) + (2*(b*d - a*e)*(3*b*B*d - 2*A*b*e - a*B*e)*(d + e*x)^
(5/2))/(5*e^4) - (2*b*(3*b*B*d - A*b*e - 2*a*B*e)*(d + e*x)^(7/2))/(7*e^4) + (2*b^2*B*(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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A]  time = 0.107137, size = 107, normalized size = 0.84 \[ \frac{2 (d+e x)^{3/2} \left (-45 b (d+e x)^2 (-2 a B e-A b e+3 b B d)+63 (d+e x) (b d-a e) (-a B e-2 A b e+3 b B d)-105 (b d-a e)^2 (B d-A e)+35 b^2 B (d+e x)^3\right )}{315 e^4} \]

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)^2*(B*d - A*e) + 63*(b*d - a*e)*(3*b*B*d - 2*A*b*e - a*B*e)*(d + e*x) - 45
*b*(3*b*B*d - A*b*e - 2*a*B*e)*(d + e*x)^2 + 35*b^2*B*(d + e*x)^3))/(315*e^4)

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Maple [A]  time = 0.007, size = 169, normalized size = 1.3 \begin{align*}{\frac{70\,{b}^{2}B{x}^{3}{e}^{3}+90\,A{b}^{2}{e}^{3}{x}^{2}+180\,Bab{e}^{3}{x}^{2}-60\,B{b}^{2}d{e}^{2}{x}^{2}+252\,Axab{e}^{3}-72\,Ax{b}^{2}d{e}^{2}+126\,Bx{a}^{2}{e}^{3}-144\,Bxabd{e}^{2}+48\,B{b}^{2}{d}^{2}ex+210\,A{a}^{2}{e}^{3}-168\,Aabd{e}^{2}+48\,A{b}^{2}{d}^{2}e-84\,B{a}^{2}d{e}^{2}+96\,Bab{d}^{2}e-32\,{b}^{2}B{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)*(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*b^2*e^3*x^3+45*A*b^2*e^3*x^2+90*B*a*b*e^3*x^2-30*B*b^2*d*e^2*x^2+126*A*a*b*e^3*x-36*
A*b^2*d*e^2*x+63*B*a^2*e^3*x-72*B*a*b*d*e^2*x+24*B*b^2*d^2*e*x+105*A*a^2*e^3-84*A*a*b*d*e^2+24*A*b^2*d^2*e-42*
B*a^2*d*e^2+48*B*a*b*d^2*e-16*B*b^2*d^3)/e^4

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Maxima [A]  time = 0.988486, size = 215, normalized size = 1.68 \begin{align*} \frac{2 \,{\left (35 \,{\left (e x + d\right )}^{\frac{9}{2}} B b^{2} - 45 \,{\left (3 \, B b^{2} d -{\left (2 \, B a b + A b^{2}\right )} e\right )}{\left (e x + d\right )}^{\frac{7}{2}} + 63 \,{\left (3 \, B b^{2} d^{2} - 2 \,{\left (2 \, B a b + A b^{2}\right )} d e +{\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )}{\left (e x + d\right )}^{\frac{5}{2}} - 105 \,{\left (B b^{2} d^{3} - A a^{2} e^{3} -{\left (2 \, B a b + A b^{2}\right )} d^{2} e +{\left (B a^{2} + 2 \, A a b\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((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*b^2 - 45*(3*B*b^2*d - (2*B*a*b + A*b^2)*e)*(e*x + d)^(7/2) + 63*(3*B*b^2*d^2 - 2*(
2*B*a*b + A*b^2)*d*e + (B*a^2 + 2*A*a*b)*e^2)*(e*x + d)^(5/2) - 105*(B*b^2*d^3 - A*a^2*e^3 - (2*B*a*b + A*b^2)
*d^2*e + (B*a^2 + 2*A*a*b)*d*e^2)*(e*x + d)^(3/2))/e^4

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Fricas [A]  time = 1.22776, size = 489, normalized size = 3.82 \begin{align*} \frac{2 \,{\left (35 \, B b^{2} e^{4} x^{4} - 16 \, B b^{2} d^{4} + 105 \, A a^{2} d e^{3} + 24 \,{\left (2 \, B a b + A b^{2}\right )} d^{3} e - 42 \,{\left (B a^{2} + 2 \, A a b\right )} d^{2} e^{2} + 5 \,{\left (B b^{2} d e^{3} + 9 \,{\left (2 \, B a b + A b^{2}\right )} e^{4}\right )} x^{3} - 3 \,{\left (2 \, B b^{2} d^{2} e^{2} - 3 \,{\left (2 \, B a b + A b^{2}\right )} d e^{3} - 21 \,{\left (B a^{2} + 2 \, A a b\right )} e^{4}\right )} x^{2} +{\left (8 \, B b^{2} d^{3} e + 105 \, A a^{2} e^{4} - 12 \,{\left (2 \, B a b + A b^{2}\right )} d^{2} e^{2} + 21 \,{\left (B a^{2} + 2 \, A a b\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((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*b^2*e^4*x^4 - 16*B*b^2*d^4 + 105*A*a^2*d*e^3 + 24*(2*B*a*b + A*b^2)*d^3*e - 42*(B*a^2 + 2*A*a*b)*d
^2*e^2 + 5*(B*b^2*d*e^3 + 9*(2*B*a*b + A*b^2)*e^4)*x^3 - 3*(2*B*b^2*d^2*e^2 - 3*(2*B*a*b + A*b^2)*d*e^3 - 21*(
B*a^2 + 2*A*a*b)*e^4)*x^2 + (8*B*b^2*d^3*e + 105*A*a^2*e^4 - 12*(2*B*a*b + A*b^2)*d^2*e^2 + 21*(B*a^2 + 2*A*a*
b)*d*e^3)*x)*sqrt(e*x + d)/e^4

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Sympy [A]  time = 4.04624, size = 201, normalized size = 1.57 \begin{align*} \frac{2 \left (\frac{B b^{2} \left (d + e x\right )^{\frac{9}{2}}}{9 e^{3}} + \frac{\left (d + e x\right )^{\frac{7}{2}} \left (A b^{2} e + 2 B a b e - 3 B b^{2} d\right )}{7 e^{3}} + \frac{\left (d + e x\right )^{\frac{5}{2}} \left (2 A a b e^{2} - 2 A b^{2} d e + B a^{2} e^{2} - 4 B a b d e + 3 B b^{2} d^{2}\right )}{5 e^{3}} + \frac{\left (d + e x\right )^{\frac{3}{2}} \left (A a^{2} e^{3} - 2 A a b d e^{2} + A b^{2} d^{2} e - B a^{2} d e^{2} + 2 B a b d^{2} e - B b^{2} 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*b**2*(d + e*x)**(9/2)/(9*e**3) + (d + e*x)**(7/2)*(A*b**2*e + 2*B*a*b*e - 3*B*b**2*d)/(7*e**3) + (d + e*x
)**(5/2)*(2*A*a*b*e**2 - 2*A*b**2*d*e + B*a**2*e**2 - 4*B*a*b*d*e + 3*B*b**2*d**2)/(5*e**3) + (d + e*x)**(3/2)
*(A*a**2*e**3 - 2*A*a*b*d*e**2 + A*b**2*d**2*e - B*a**2*d*e**2 + 2*B*a*b*d**2*e - B*b**2*d**3)/(3*e**3))/e

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Giac [A]  time = 1.15161, size = 294, normalized size = 2.3 \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 a^{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 a b e^{\left (-1\right )} + 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 )} B a 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 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 b^{2} e^{\left (-3\right )} + 105 \,{\left (x e + d\right )}^{\frac{3}{2}} A a^{2}\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*(21*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*B*a^2*e^(-1) + 42*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d
)*A*a*b*e^(-1) + 6*(15*(x*e + d)^(7/2) - 42*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*B*a*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*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*b^2*e^(-3) + 105*(x*e + d)^(3/2)*A*a^2)*e^
(-1)