3.1573 \(\int (d+e x) (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\)

Optimal. Leaf size=69 \[ \frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} (b d-a e)}{6 b^2}+\frac{e \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2} \]

[Out]

((b*d - a*e)*(a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(6*b^2) + (e*(a^2 + 2*a*b*x + b^2*x^2)^(7/2))/(7*b^2)

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Rubi [A]  time = 0.0235983, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {640, 609} \[ \frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} (b d-a e)}{6 b^2}+\frac{e \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

((b*d - a*e)*(a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(6*b^2) + (e*(a^2 + 2*a*b*x + b^2*x^2)^(7/2))/(7*b^2)

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 609

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p + 1
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rubi steps

\begin{align*} \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx &=\frac{e \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2}+\frac{\left (2 b^2 d-2 a b e\right ) \int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx}{2 b^2}\\ &=\frac{(b d-a e) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b^2}+\frac{e \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0453663, size = 121, normalized size = 1.75 \[ \frac{x \sqrt{(a+b x)^2} \left (35 a^3 b^2 x^2 (4 d+3 e x)+21 a^2 b^3 x^3 (5 d+4 e x)+35 a^4 b x (3 d+2 e x)+21 a^5 (2 d+e x)+7 a b^4 x^4 (6 d+5 e x)+b^5 x^5 (7 d+6 e x)\right )}{42 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(x*Sqrt[(a + b*x)^2]*(21*a^5*(2*d + e*x) + 35*a^4*b*x*(3*d + 2*e*x) + 35*a^3*b^2*x^2*(4*d + 3*e*x) + 21*a^2*b^
3*x^3*(5*d + 4*e*x) + 7*a*b^4*x^4*(6*d + 5*e*x) + b^5*x^5*(7*d + 6*e*x)))/(42*(a + b*x))

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Maple [B]  time = 0.16, size = 138, normalized size = 2. \begin{align*}{\frac{x \left ( 6\,e{b}^{5}{x}^{6}+35\,{x}^{5}ea{b}^{4}+7\,{x}^{5}d{b}^{5}+84\,{a}^{2}{b}^{3}e{x}^{4}+42\,a{b}^{4}d{x}^{4}+105\,{x}^{3}e{a}^{3}{b}^{2}+105\,{x}^{3}d{a}^{2}{b}^{3}+70\,{x}^{2}e{a}^{4}b+140\,{x}^{2}d{a}^{3}{b}^{2}+21\,xe{a}^{5}+105\,xd{a}^{4}b+42\,d{a}^{5} \right ) }{42\, \left ( bx+a \right ) ^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/42*x*(6*b^5*e*x^6+35*a*b^4*e*x^5+7*b^5*d*x^5+84*a^2*b^3*e*x^4+42*a*b^4*d*x^4+105*a^3*b^2*e*x^3+105*a^2*b^3*d
*x^3+70*a^4*b*e*x^2+140*a^3*b^2*d*x^2+21*a^5*e*x+105*a^4*b*d*x+42*a^5*d)*((b*x+a)^2)^(5/2)/(b*x+a)^5

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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)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51008, size = 247, normalized size = 3.58 \begin{align*} \frac{1}{7} \, b^{5} e x^{7} + a^{5} d x + \frac{1}{6} \,{\left (b^{5} d + 5 \, a b^{4} e\right )} x^{6} +{\left (a b^{4} d + 2 \, a^{2} b^{3} e\right )} x^{5} + \frac{5}{2} \,{\left (a^{2} b^{3} d + a^{3} b^{2} e\right )} x^{4} + \frac{5}{3} \,{\left (2 \, a^{3} b^{2} d + a^{4} b e\right )} x^{3} + \frac{1}{2} \,{\left (5 \, a^{4} b d + a^{5} e\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/7*b^5*e*x^7 + a^5*d*x + 1/6*(b^5*d + 5*a*b^4*e)*x^6 + (a*b^4*d + 2*a^2*b^3*e)*x^5 + 5/2*(a^2*b^3*d + a^3*b^2
*e)*x^4 + 5/3*(2*a^3*b^2*d + a^4*b*e)*x^3 + 1/2*(5*a^4*b*d + a^5*e)*x^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d + e x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((d + e*x)*((a + b*x)**2)**(5/2), x)

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Giac [B]  time = 1.24066, size = 269, normalized size = 3.9 \begin{align*} \frac{1}{7} \, b^{5} x^{7} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{6} \, b^{5} d x^{6} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{6} \, a b^{4} x^{6} e \mathrm{sgn}\left (b x + a\right ) + a b^{4} d x^{5} \mathrm{sgn}\left (b x + a\right ) + 2 \, a^{2} b^{3} x^{5} e \mathrm{sgn}\left (b x + a\right ) + \frac{5}{2} \, a^{2} b^{3} d x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{2} \, a^{3} b^{2} x^{4} e \mathrm{sgn}\left (b x + a\right ) + \frac{10}{3} \, a^{3} b^{2} d x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{3} \, a^{4} b x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{5}{2} \, a^{4} b d x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, a^{5} x^{2} e \mathrm{sgn}\left (b x + a\right ) + a^{5} d 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)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

1/7*b^5*x^7*e*sgn(b*x + a) + 1/6*b^5*d*x^6*sgn(b*x + a) + 5/6*a*b^4*x^6*e*sgn(b*x + a) + a*b^4*d*x^5*sgn(b*x +
a) + 2*a^2*b^3*x^5*e*sgn(b*x + a) + 5/2*a^2*b^3*d*x^4*sgn(b*x + a) + 5/2*a^3*b^2*x^4*e*sgn(b*x + a) + 10/3*a^
3*b^2*d*x^3*sgn(b*x + a) + 5/3*a^4*b*x^3*e*sgn(b*x + a) + 5/2*a^4*b*d*x^2*sgn(b*x + a) + 1/2*a^5*x^2*e*sgn(b*x
+ a) + a^5*d*x*sgn(b*x + a)