3.6.93 \(\int x^4 (A+B x) (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=210 \[ \frac {b^2 x^8 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{8 (a+b x)}+\frac {3 a b x^7 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{7 (a+b x)}+\frac {a^2 x^6 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{6 (a+b x)}+\frac {b^3 B x^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {a^3 A x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]

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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^8 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{8 (a+b x)}+\frac {3 a b x^7 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{7 (a+b x)}+\frac {a^2 x^6 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{6 (a+b x)}+\frac {a^3 A x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {b^3 B x^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

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Mathematica [A]  time = 0.03, size = 87, normalized size = 0.41 \begin {gather*} \frac {x^5 \sqrt {(a+b x)^2} \left (84 a^3 (6 A+5 B x)+180 a^2 b x (7 A+6 B x)+135 a b^2 x^2 (8 A+7 B x)+35 b^3 x^3 (9 A+8 B x)\right )}{2520 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(x^5*Sqrt[(a + b*x)^2]*(84*a^3*(6*A + 5*B*x) + 180*a^2*b*x*(7*A + 6*B*x) + 135*a*b^2*x^2*(8*A + 7*B*x) + 35*b^
3*x^3*(9*A + 8*B*x)))/(2520*(a + b*x))

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IntegrateAlgebraic [F]  time = 1.03, size = 0, normalized size = 0.00 \begin {gather*} \int x^4 (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^4*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

Defer[IntegrateAlgebraic][x^4*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2), x]

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fricas [A]  time = 0.39, size = 73, normalized size = 0.35 \begin {gather*} \frac {1}{9} \, B b^{3} x^{9} + \frac {1}{5} \, A a^{3} x^{5} + \frac {1}{8} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{8} + \frac {3}{7} \, {\left (B a^{2} b + A a b^{2}\right )} x^{7} + \frac {1}{6} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/9*B*b^3*x^9 + 1/5*A*a^3*x^5 + 1/8*(3*B*a*b^2 + A*b^3)*x^8 + 3/7*(B*a^2*b + A*a*b^2)*x^7 + 1/6*(B*a^3 + 3*A*a
^2*b)*x^6

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giac [A]  time = 0.21, size = 150, normalized size = 0.71 \begin {gather*} \frac {1}{9} \, B b^{3} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{8} \, B a b^{2} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{8} \, A b^{3} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{7} \, B a^{2} b x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{7} \, A a b^{2} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{6} \, B a^{3} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A a^{2} b x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, A a^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) - \frac {{\left (5 \, B a^{9} - 9 \, A a^{8} b\right )} \mathrm {sgn}\left (b x + a\right )}{2520 \, b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

1/9*B*b^3*x^9*sgn(b*x + a) + 3/8*B*a*b^2*x^8*sgn(b*x + a) + 1/8*A*b^3*x^8*sgn(b*x + a) + 3/7*B*a^2*b*x^7*sgn(b
*x + a) + 3/7*A*a*b^2*x^7*sgn(b*x + a) + 1/6*B*a^3*x^6*sgn(b*x + a) + 1/2*A*a^2*b*x^6*sgn(b*x + a) + 1/5*A*a^3
*x^5*sgn(b*x + a) - 1/2520*(5*B*a^9 - 9*A*a^8*b)*sgn(b*x + a)/b^6

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maple [A]  time = 0.05, size = 92, normalized size = 0.44 \begin {gather*} \frac {\left (280 b^{3} B \,x^{4}+315 A \,b^{3} x^{3}+945 x^{3} B a \,b^{2}+1080 x^{2} A a \,b^{2}+1080 B \,a^{2} b \,x^{2}+1260 x A \,a^{2} b +420 B \,a^{3} x +504 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x^{5}}{2520 \left (b x +a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/2520*x^5*(280*B*b^3*x^4+315*A*b^3*x^3+945*B*a*b^2*x^3+1080*A*a*b^2*x^2+1080*B*a^2*b*x^2+1260*A*a^2*b*x+420*B
*a^3*x+504*A*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [B]  time = 0.57, size = 361, normalized size = 1.72 \begin {gather*} \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B x^{4}}{9 \, b^{2}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a x^{3}}{72 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A x^{3}}{8 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{5} x}{4 \, b^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{4} x}{4 \, b^{4}} + \frac {37 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{2} x^{2}}{168 \, b^{4}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a x^{2}}{56 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{6}}{4 \, b^{6}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{5}}{4 \, b^{5}} - \frac {121 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{3} x}{504 \, b^{5}} + \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{2} x}{56 \, b^{4}} + \frac {125 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{4}}{504 \, b^{6}} - \frac {69 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{3}}{280 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/9*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*x^4/b^2 - 13/72*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a*x^3/b^3 + 1/8*(b^2*x
^2 + 2*a*b*x + a^2)^(5/2)*A*x^3/b^2 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^5*x/b^5 + 1/4*(b^2*x^2 + 2*a*b*x
+ a^2)^(3/2)*A*a^4*x/b^4 + 37/168*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^2*x^2/b^4 - 11/56*(b^2*x^2 + 2*a*b*x +
a^2)^(5/2)*A*a*x^2/b^3 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^6/b^6 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A
*a^5/b^5 - 121/504*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^3*x/b^5 + 13/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*a^2*x
/b^4 + 125/504*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^4/b^6 - 69/280*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*a^3/b^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^4\,\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^4*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int(x^4*(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^{4} \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**4*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(x**4*(A + B*x)*((a + b*x)**2)**(3/2), x)

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