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

Optimal. Leaf size=78 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^2 (b d-a e)}{3 b^2}+\frac {e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3}{4 b^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {770, 21, 43} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^2 (b d-a e)}{3 b^2}+\frac {e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3}{4 b^2} \end {gather*}

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]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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])

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]

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

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Mathematica [A]  time = 0.02, size = 64, normalized size = 0.82 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (6 a^2 (2 d+e x)+4 a b x (3 d+2 e x)+b^2 x^2 (4 d+3 e x)\right )}{12 (a+b x)} \end {gather*}

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

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IntegrateAlgebraic [F]  time = 0.84, size = 0, normalized size = 0.00 \begin {gather*} \int (a+b x) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 48, normalized size = 0.62 \begin {gather*} \frac {1}{4} \, b^{2} e x^{4} + a^{2} d x + \frac {1}{3} \, {\left (b^{2} d + 2 \, a b e\right )} x^{3} + \frac {1}{2} \, {\left (2 \, a b d + a^{2} e\right )} x^{2} \end {gather*}

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

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giac [A]  time = 0.16, size = 88, normalized size = 1.13 \begin {gather*} \frac {1}{4} \, b^{2} x^{4} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, b^{2} d x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, a b x^{3} e \mathrm {sgn}\left (b x + a\right ) + a b d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a^{2} x^{2} e \mathrm {sgn}\left (b x + a\right ) + a^{2} d x \mathrm {sgn}\left (b x + a\right ) \end {gather*}

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

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maple [A]  time = 0.05, size = 66, normalized size = 0.85 \begin {gather*} \frac {\left (3 e \,b^{2} x^{3}+8 x^{2} a b e +4 x^{2} b^{2} d +6 a^{2} e x +12 a b d x +12 a^{2} d \right ) \sqrt {\left (b x +a \right )^{2}}\, x}{12 b x +12 a} \end {gather*}

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^2*e*x^3+8*a*b*e*x^2+4*b^2*d*x^2+6*a^2*e*x+12*a*b*d*x+12*a^2*d)*((b*x+a)^2)^(1/2)/(b*x+a)

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maxima [B]  time = 0.64, size = 251, normalized size = 3.22 \begin {gather*} \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d x + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} e x}{2 \, b} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d}{2 \, b} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} e}{2 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (b d + a e\right )} a x}{2 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} e x}{4 \, b} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (b d + a e\right )} a^{2}}{2 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a e}{12 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (b d + a e\right )}}{3 \, b^{2}} \end {gather*}

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]

1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*d*x + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*e*x/b + 1/2*sqrt(b^2*x^2 + 2*a
*b*x + a^2)*a^2*d/b + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^3*e/b^2 - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*(b*d + a
*e)*a*x/b + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*e*x/b - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*(b*d + a*e)*a^2/b^2
- 5/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*e/b^2 + 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*(b*d + a*e)/b^2

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mupad [B]  time = 2.50, size = 219, normalized size = 2.81 \begin {gather*} \frac {d\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{24\,b^3}+\frac {e\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b}-\frac {a^2\,e\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b}-\frac {5\,a\,e\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{96\,b^4}+\frac {a\,\left (a+b\,x\right )\,\left (3\,b\,d-a\,e+2\,b\,e\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(24*b^3) + (e*x*(a^2 + b^
2*x^2 + 2*a*b*x)^(3/2))/(4*b) - (a^2*e*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*b) - (5*a*e*(8*b^2*
(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(96*b^4) + (a*(a + b*x)*(3*b*d - a*
e + 2*b*e*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(6*b^2)

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sympy [A]  time = 0.11, size = 49, normalized size = 0.63 \begin {gather*} a^{2} d x + \frac {b^{2} e x^{4}}{4} + x^{3} \left (\frac {2 a b e}{3} + \frac {b^{2} d}{3}\right ) + x^{2} \left (\frac {a^{2} e}{2} + a b d\right ) \end {gather*}

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

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