3.6.95 \(\int x^2 (A+B x) (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=210 \[ \frac {b^2 x^6 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{6 (a+b x)}+\frac {3 a b x^5 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{5 (a+b x)}+\frac {a^2 x^4 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{4 (a+b x)}+\frac {b^3 B x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {a^3 A x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)} \]

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Rubi [A]  time = 0.09, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {770, 76} \begin {gather*} \frac {b^2 x^6 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{6 (a+b x)}+\frac {3 a b x^5 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{5 (a+b x)}+\frac {a^2 x^4 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{4 (a+b x)}+\frac {a^3 A x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {b^3 B x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*A*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (a^2*(3*A*b + a*B)*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/(4*(a + b*x)) + (3*a*b*(A*b + a*B)*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a + b*x)) + (b^2*(A*b + 3*a*B)*x^6
*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*(a + b*x)) + (b^3*B*x^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*(a + b*x))

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

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

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Mathematica [A]  time = 0.03, size = 87, normalized size = 0.41 \begin {gather*} \frac {x^3 \sqrt {(a+b x)^2} \left (35 a^3 (4 A+3 B x)+63 a^2 b x (5 A+4 B x)+42 a b^2 x^2 (6 A+5 B x)+10 b^3 x^3 (7 A+6 B x)\right )}{420 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*Sqrt[(a + b*x)^2]*(35*a^3*(4*A + 3*B*x) + 63*a^2*b*x*(5*A + 4*B*x) + 42*a*b^2*x^2*(6*A + 5*B*x) + 10*b^3*
x^3*(7*A + 6*B*x)))/(420*(a + b*x))

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IntegrateAlgebraic [F]  time = 0.94, size = 0, normalized size = 0.00 \begin {gather*} \int x^2 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^2*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

Defer[IntegrateAlgebraic][x^2*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2), x]

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fricas [A]  time = 0.41, size = 73, normalized size = 0.35 \begin {gather*} \frac {1}{7} \, B b^{3} x^{7} + \frac {1}{3} \, A a^{3} x^{3} + \frac {1}{6} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{6} + \frac {3}{5} \, {\left (B a^{2} b + A a b^{2}\right )} x^{5} + \frac {1}{4} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 150, normalized size = 0.71 \begin {gather*} \frac {1}{7} \, B b^{3} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B a b^{2} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{6} \, A b^{3} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, B a^{2} b x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, A a b^{2} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, B a^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, A a^{2} b x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, A a^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) - \frac {{\left (3 \, B a^{7} - 7 \, A a^{6} b\right )} \mathrm {sgn}\left (b x + a\right )}{420 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*B*b^3*x^7*sgn(b*x + a) + 1/2*B*a*b^2*x^6*sgn(b*x + a) + 1/6*A*b^3*x^6*sgn(b*x + a) + 3/5*B*a^2*b*x^5*sgn(b
*x + a) + 3/5*A*a*b^2*x^5*sgn(b*x + a) + 1/4*B*a^3*x^4*sgn(b*x + a) + 3/4*A*a^2*b*x^4*sgn(b*x + a) + 1/3*A*a^3
*x^3*sgn(b*x + a) - 1/420*(3*B*a^7 - 7*A*a^6*b)*sgn(b*x + a)/b^4

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maple [A]  time = 0.05, size = 92, normalized size = 0.44 \begin {gather*} \frac {\left (60 b^{3} B \,x^{4}+70 A \,b^{3} x^{3}+210 x^{3} B a \,b^{2}+252 x^{2} A a \,b^{2}+252 B \,a^{2} b \,x^{2}+315 x A \,a^{2} b +105 B \,a^{3} x +140 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x^{3}}{420 \left (b x +a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/420*x^3*(60*B*b^3*x^4+70*A*b^3*x^3+210*B*a*b^2*x^3+252*A*a*b^2*x^2+252*B*a^2*b*x^2+315*A*a^2*b*x+105*B*a^3*x
+140*A*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [A]  time = 0.54, size = 241, normalized size = 1.15 \begin {gather*} -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{3} x}{4 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{2} x}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B x^{2}}{7 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{4}}{4 \, b^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{3}}{4 \, b^{3}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a x}{14 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A x}{6 \, b^{2}} + \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{2}}{70 \, b^{4}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a}{30 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

-1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^3*x/b^3 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a^2*x/b^2 + 1/7*(b^2*
x^2 + 2*a*b*x + a^2)^(5/2)*B*x^2/b^2 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^4/b^4 + 1/4*(b^2*x^2 + 2*a*b*x
+ a^2)^(3/2)*A*a^3/b^3 - 3/14*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a*x/b^3 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*
A*x/b^2 + 17/70*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^2/b^4 - 7/30*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*a/b^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int(x^2*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

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

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