### 3.165 $$\int x^3 (a^2+2 a b x+b^2 x^2)^{5/2} \, dx$$

Optimal. Leaf size=144 $\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^8}{9 b^4}-\frac{3 a \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^4}+\frac{3 a^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^4}-\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^4}$

[Out]

-(a^3*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*b^4) + (3*a^2*(a + b*x)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
7*b^4) - (3*a*(a + b*x)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*b^4) + ((a + b*x)^8*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
/(9*b^4)

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Rubi [A]  time = 0.0519569, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {646, 43} $\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^8}{9 b^4}-\frac{3 a \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^4}+\frac{3 a^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^4}-\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^4}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

-(a^3*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*b^4) + (3*a^2*(a + b*x)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
7*b^4) - (3*a*(a + b*x)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*b^4) + ((a + b*x)^8*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
/(9*b^4)

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 x^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int x^3 \left (a b+b^2 x\right )^5 \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (-\frac{a^3 \left (a b+b^2 x\right )^5}{b^3}+\frac{3 a^2 \left (a b+b^2 x\right )^6}{b^4}-\frac{3 a \left (a b+b^2 x\right )^7}{b^5}+\frac{\left (a b+b^2 x\right )^8}{b^6}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=-\frac{a^3 (a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{6 b^4}+\frac{3 a^2 (a+b x)^6 \sqrt{a^2+2 a b x+b^2 x^2}}{7 b^4}-\frac{3 a (a+b x)^7 \sqrt{a^2+2 a b x+b^2 x^2}}{8 b^4}+\frac{(a+b x)^8 \sqrt{a^2+2 a b x+b^2 x^2}}{9 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0190502, size = 77, normalized size = 0.53 $\frac{x^4 \sqrt{(a+b x)^2} \left (840 a^3 b^2 x^2+720 a^2 b^3 x^3+504 a^4 b x+126 a^5+315 a b^4 x^4+56 b^5 x^5\right )}{504 (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^3*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(x^4*Sqrt[(a + b*x)^2]*(126*a^5 + 504*a^4*b*x + 840*a^3*b^2*x^2 + 720*a^2*b^3*x^3 + 315*a*b^4*x^4 + 56*b^5*x^5
))/(504*(a + b*x))

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Maple [A]  time = 0.188, size = 74, normalized size = 0.5 \begin{align*}{\frac{{x}^{4} \left ( 56\,{b}^{5}{x}^{5}+315\,a{b}^{4}{x}^{4}+720\,{a}^{2}{b}^{3}{x}^{3}+840\,{a}^{3}{b}^{2}{x}^{2}+504\,{a}^{4}bx+126\,{a}^{5} \right ) }{504\, \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(x^3*(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/504*x^4*(56*b^5*x^5+315*a*b^4*x^4+720*a^2*b^3*x^3+840*a^3*b^2*x^2+504*a^4*b*x+126*a^5)*((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(x^3*(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 = 1.7779, size = 124, normalized size = 0.86 \begin{align*} \frac{1}{9} \, b^{5} x^{9} + \frac{5}{8} \, a b^{4} x^{8} + \frac{10}{7} \, a^{2} b^{3} x^{7} + \frac{5}{3} \, a^{3} b^{2} x^{6} + a^{4} b x^{5} + \frac{1}{4} \, a^{5} x^{4} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/9*b^5*x^9 + 5/8*a*b^4*x^8 + 10/7*a^2*b^3*x^7 + 5/3*a^3*b^2*x^6 + a^4*b*x^5 + 1/4*a^5*x^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \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(x**3*(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(x**3*((a + b*x)**2)**(5/2), x)

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Giac [A]  time = 1.34331, size = 143, normalized size = 0.99 \begin{align*} \frac{1}{9} \, b^{5} x^{9} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{8} \, a b^{4} x^{8} \mathrm{sgn}\left (b x + a\right ) + \frac{10}{7} \, a^{2} b^{3} x^{7} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{3} \, a^{3} b^{2} x^{6} \mathrm{sgn}\left (b x + a\right ) + a^{4} b x^{5} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, a^{5} x^{4} \mathrm{sgn}\left (b x + a\right ) - \frac{a^{9} \mathrm{sgn}\left (b x + a\right )}{504 \, b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

1/9*b^5*x^9*sgn(b*x + a) + 5/8*a*b^4*x^8*sgn(b*x + a) + 10/7*a^2*b^3*x^7*sgn(b*x + a) + 5/3*a^3*b^2*x^6*sgn(b*
x + a) + a^4*b*x^5*sgn(b*x + a) + 1/4*a^5*x^4*sgn(b*x + a) - 1/504*a^9*sgn(b*x + a)/b^4