3.1543 \(\int (d+e x)^4 \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=92 \[ \frac{b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^6}{6 e^2 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^5 (b d-a e)}{5 e^2 (a+b x)} \]

[Out]

-((b*d - a*e)*(d + e*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^2*(a + b*x)) + (b*(d + e*x)^6*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2])/(6*e^2*(a + b*x))

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Rubi [A]  time = 0.0400595, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {646, 43} \[ \frac{b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^6}{6 e^2 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^5 (b d-a e)}{5 e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

-((b*d - a*e)*(d + e*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^2*(a + b*x)) + (b*(d + e*x)^6*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2])/(6*e^2*(a + b*x))

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

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])

Rubi steps

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

Mathematica [A]  time = 0.0378056, size = 111, normalized size = 1.21 \[ \frac{x \sqrt{(a+b x)^2} \left (6 a \left (10 d^2 e^2 x^2+10 d^3 e x+5 d^4+5 d e^3 x^3+e^4 x^4\right )+b x \left (45 d^2 e^2 x^2+40 d^3 e x+15 d^4+24 d e^3 x^3+5 e^4 x^4\right )\right )}{30 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(x*Sqrt[(a + b*x)^2]*(6*a*(5*d^4 + 10*d^3*e*x + 10*d^2*e^2*x^2 + 5*d*e^3*x^3 + e^4*x^4) + b*x*(15*d^4 + 40*d^3
*e*x + 45*d^2*e^2*x^2 + 24*d*e^3*x^3 + 5*e^4*x^4)))/(30*(a + b*x))

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Maple [A]  time = 0.042, size = 114, normalized size = 1.2 \begin{align*}{\frac{x \left ( 5\,b{e}^{4}{x}^{5}+6\,{x}^{4}a{e}^{4}+24\,{x}^{4}bd{e}^{3}+30\,{x}^{3}ad{e}^{3}+45\,{x}^{3}b{d}^{2}{e}^{2}+60\,{x}^{2}a{d}^{2}{e}^{2}+40\,{x}^{2}b{d}^{3}e+60\,xa{d}^{3}e+15\,xb{d}^{4}+30\,a{d}^{4} \right ) }{30\,bx+30\,a}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*x*(5*b*e^4*x^5+6*a*e^4*x^4+24*b*d*e^3*x^4+30*a*d*e^3*x^3+45*b*d^2*e^2*x^3+60*a*d^2*e^2*x^2+40*b*d^3*e*x^2
+60*a*d^3*e*x+15*b*d^4*x+30*a*d^4)*((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((e*x+d)^4*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.76527, size = 212, normalized size = 2.3 \begin{align*} \frac{1}{6} \, b e^{4} x^{6} + a d^{4} x + \frac{1}{5} \,{\left (4 \, b d e^{3} + a e^{4}\right )} x^{5} + \frac{1}{2} \,{\left (3 \, b d^{2} e^{2} + 2 \, a d e^{3}\right )} x^{4} + \frac{2}{3} \,{\left (2 \, b d^{3} e + 3 \, a d^{2} e^{2}\right )} x^{3} + \frac{1}{2} \,{\left (b d^{4} + 4 \, a d^{3} e\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/6*b*e^4*x^6 + a*d^4*x + 1/5*(4*b*d*e^3 + a*e^4)*x^5 + 1/2*(3*b*d^2*e^2 + 2*a*d*e^3)*x^4 + 2/3*(2*b*d^3*e + 3
*a*d^2*e^2)*x^3 + 1/2*(b*d^4 + 4*a*d^3*e)*x^2

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Sympy [A]  time = 0.118562, size = 100, normalized size = 1.09 \begin{align*} a d^{4} x + \frac{b e^{4} x^{6}}{6} + x^{5} \left (\frac{a e^{4}}{5} + \frac{4 b d e^{3}}{5}\right ) + x^{4} \left (a d e^{3} + \frac{3 b d^{2} e^{2}}{2}\right ) + x^{3} \left (2 a d^{2} e^{2} + \frac{4 b d^{3} e}{3}\right ) + x^{2} \left (2 a d^{3} e + \frac{b d^{4}}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4*((b*x+a)**2)**(1/2),x)

[Out]

a*d**4*x + b*e**4*x**6/6 + x**5*(a*e**4/5 + 4*b*d*e**3/5) + x**4*(a*d*e**3 + 3*b*d**2*e**2/2) + x**3*(2*a*d**2
*e**2 + 4*b*d**3*e/3) + x**2*(2*a*d**3*e + b*d**4/2)

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Giac [B]  time = 1.24279, size = 207, normalized size = 2.25 \begin{align*} \frac{1}{6} \, b x^{6} e^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{4}{5} \, b d x^{5} e^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, b d^{2} x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{4}{3} \, b d^{3} x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, b d^{4} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{5} \, a x^{5} e^{4} \mathrm{sgn}\left (b x + a\right ) + a d x^{4} e^{3} \mathrm{sgn}\left (b x + a\right ) + 2 \, a d^{2} x^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) + 2 \, a d^{3} x^{2} e \mathrm{sgn}\left (b x + a\right ) + a d^{4} x \mathrm{sgn}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/6*b*x^6*e^4*sgn(b*x + a) + 4/5*b*d*x^5*e^3*sgn(b*x + a) + 3/2*b*d^2*x^4*e^2*sgn(b*x + a) + 4/3*b*d^3*x^3*e*s
gn(b*x + a) + 1/2*b*d^4*x^2*sgn(b*x + a) + 1/5*a*x^5*e^4*sgn(b*x + a) + a*d*x^4*e^3*sgn(b*x + a) + 2*a*d^2*x^3
*e^2*sgn(b*x + a) + 2*a*d^3*x^2*e*sgn(b*x + a) + a*d^4*x*sgn(b*x + a)