3.1711 \(\int (A+B x) (d+e x) \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=164 \[ \frac{x^3 \sqrt{a^2+2 a b x+b^2 x^2} (a B e+A b e+b B d)}{3 (a+b x)}+\frac{x^2 \sqrt{a^2+2 a b x+b^2 x^2} (a A e+a B d+A b d)}{2 (a+b x)}+\frac{a A d x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{b B e x^4 \sqrt{a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \]

[Out]

(a*A*d*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + ((A*b*d + a*B*d + a*A*e)*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/(2*(a + b*x)) + ((b*B*d + A*b*e + a*B*e)*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (b*B*e*x^4*Sqrt[
a^2 + 2*a*b*x + b^2*x^2])/(4*(a + b*x))

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Rubi [A]  time = 0.0845222, antiderivative size = 164, 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 = {770, 77} \[ \frac{x^3 \sqrt{a^2+2 a b x+b^2 x^2} (a B e+A b e+b B d)}{3 (a+b x)}+\frac{x^2 \sqrt{a^2+2 a b x+b^2 x^2} (a A e+a B d+A b d)}{2 (a+b x)}+\frac{a A d x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{b B e x^4 \sqrt{a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*A*d*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + ((A*b*d + a*B*d + a*A*e)*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/(2*(a + b*x)) + ((b*B*d + A*b*e + a*B*e)*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (b*B*e*x^4*Sqrt[
a^2 + 2*a*b*x + b^2*x^2])/(4*(a + b*x))

Rule 770

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

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

Mathematica [A]  time = 0.0371268, size = 74, normalized size = 0.45 \[ \frac{x \sqrt{(a+b x)^2} (2 a (3 A (2 d+e x)+B x (3 d+2 e x))+b x (A (6 d+4 e x)+B x (4 d+3 e x)))}{12 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(2*a*(3*A*(2*d + e*x) + B*x*(3*d + 2*e*x)) + b*x*(B*x*(4*d + 3*e*x) + A*(6*d + 4*e*x))))/
(12*(a + b*x))

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Maple [A]  time = 0.004, size = 76, normalized size = 0.5 \begin{align*}{\frac{x \left ( 3\,bBe{x}^{3}+4\,{x}^{2}Abe+4\,{x}^{2}aBe+4\,{x}^{2}Bbd+6\,xaAe+6\,xAbd+6\,xBad+12\,aAd \right ) }{12\,bx+12\,a}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*x*(3*B*b*e*x^3+4*A*b*e*x^2+4*B*a*e*x^2+4*B*b*d*x^2+6*A*a*e*x+6*A*b*d*x+6*B*a*d*x+12*A*a*d)*((b*x+a)^2)^(1
/2)/(b*x+a)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60249, size = 126, normalized size = 0.77 \begin{align*} \frac{1}{4} \, B b e x^{4} + A a d x + \frac{1}{3} \,{\left (B b d +{\left (B a + A b\right )} e\right )} x^{3} + \frac{1}{2} \,{\left (A a e +{\left (B a + A b\right )} d\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*B*b*e*x^4 + A*a*d*x + 1/3*(B*b*d + (B*a + A*b)*e)*x^3 + 1/2*(A*a*e + (B*a + A*b)*d)*x^2

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Sympy [A]  time = 0.10072, size = 63, normalized size = 0.38 \begin{align*} A a d x + \frac{B b e x^{4}}{4} + x^{3} \left (\frac{A b e}{3} + \frac{B a e}{3} + \frac{B b d}{3}\right ) + x^{2} \left (\frac{A a e}{2} + \frac{A b d}{2} + \frac{B a d}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a*d*x + B*b*e*x**4/4 + x**3*(A*b*e/3 + B*a*e/3 + B*b*d/3) + x**2*(A*a*e/2 + A*b*d/2 + B*a*d/2)

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Giac [A]  time = 1.13302, size = 154, normalized size = 0.94 \begin{align*} \frac{1}{4} \, B b x^{4} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, B b d x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, B a x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, A b x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, B a d x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, A b d x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, A a x^{2} e \mathrm{sgn}\left (b x + a\right ) + A a d x \mathrm{sgn}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*B*b*x^4*e*sgn(b*x + a) + 1/3*B*b*d*x^3*sgn(b*x + a) + 1/3*B*a*x^3*e*sgn(b*x + a) + 1/3*A*b*x^3*e*sgn(b*x +
a) + 1/2*B*a*d*x^2*sgn(b*x + a) + 1/2*A*b*d*x^2*sgn(b*x + a) + 1/2*A*a*x^2*e*sgn(b*x + a) + A*a*d*x*sgn(b*x +
a)