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

Optimal. Leaf size=210 \[ \frac {b^2 x^7 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{7 (a+b x)}+\frac {a b x^6 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{2 (a+b x)}+\frac {a^2 x^5 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{5 (a+b x)}+\frac {b^3 B x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{8 (a+b x)}+\frac {a^3 A x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \]

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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^7 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{7 (a+b x)}+\frac {a b x^6 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{2 (a+b x)}+\frac {a^2 x^5 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{5 (a+b x)}+\frac {a^3 A x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {b^3 B x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{8 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 87, normalized size = 0.41 \begin {gather*} \frac {x^4 \sqrt {(a+b x)^2} \left (14 a^3 (5 A+4 B x)+28 a^2 b x (6 A+5 B x)+20 a b^2 x^2 (7 A+6 B x)+5 b^3 x^3 (8 A+7 B x)\right )}{280 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*Sqrt[(a + b*x)^2]*(14*a^3*(5*A + 4*B*x) + 28*a^2*b*x*(6*A + 5*B*x) + 20*a*b^2*x^2*(7*A + 6*B*x) + 5*b^3*x
^3*(8*A + 7*B*x)))/(280*(a + b*x))

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IntegrateAlgebraic [F]  time = 0.97, size = 0, normalized size = 0.00 \begin {gather*} \int x^3 (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^3*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

Defer[IntegrateAlgebraic][x^3*(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}{8} \, B b^{3} x^{8} + \frac {1}{4} \, A a^{3} x^{4} + \frac {1}{7} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{7} + \frac {1}{2} \, {\left (B a^{2} b + A a b^{2}\right )} x^{6} + \frac {1}{5} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*b^3*x^8 + 1/4*A*a^3*x^4 + 1/7*(3*B*a*b^2 + A*b^3)*x^7 + 1/2*(B*a^2*b + A*a*b^2)*x^6 + 1/5*(B*a^3 + 3*A*a
^2*b)*x^5

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giac [A]  time = 0.20, size = 149, normalized size = 0.71 \begin {gather*} \frac {1}{8} \, B b^{3} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{7} \, B a b^{2} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{7} \, A b^{3} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B a^{2} b x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A a b^{2} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, B a^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, A a^{2} b x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A a^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (B a^{8} - 2 \, A a^{7} b\right )} \mathrm {sgn}\left (b x + a\right )}{280 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*b^3*x^8*sgn(b*x + a) + 3/7*B*a*b^2*x^7*sgn(b*x + a) + 1/7*A*b^3*x^7*sgn(b*x + a) + 1/2*B*a^2*b*x^6*sgn(b
*x + a) + 1/2*A*a*b^2*x^6*sgn(b*x + a) + 1/5*B*a^3*x^5*sgn(b*x + a) + 3/5*A*a^2*b*x^5*sgn(b*x + a) + 1/4*A*a^3
*x^4*sgn(b*x + a) + 1/280*(B*a^8 - 2*A*a^7*b)*sgn(b*x + a)/b^5

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maple [A]  time = 0.05, size = 92, normalized size = 0.44 \begin {gather*} \frac {\left (35 b^{3} B \,x^{4}+40 A \,b^{3} x^{3}+120 x^{3} B a \,b^{2}+140 x^{2} A a \,b^{2}+140 B \,a^{2} b \,x^{2}+168 x A \,a^{2} b +56 B \,a^{3} x +70 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x^{4}}{280 \left (b x +a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/280*x^4*(35*B*b^3*x^4+40*A*b^3*x^3+120*B*a*b^2*x^3+140*A*a*b^2*x^2+140*B*a^2*b*x^2+168*A*a^2*b*x+56*B*a^3*x+
70*A*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [B]  time = 0.57, size = 301, normalized size = 1.43 \begin {gather*} \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B x^{3}}{8 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{4} x}{4 \, b^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{3} x}{4 \, b^{3}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a x^{2}}{56 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A x^{2}}{7 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{5}}{4 \, b^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{4}}{4 \, b^{4}} + \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{2} x}{56 \, b^{4}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a x}{14 \, b^{3}} - \frac {69 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{3}}{280 \, b^{5}} + \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{2}}{70 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*x^3/b^2 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^4*x/b^4 - 1/4*(b^2*x^2
+ 2*a*b*x + a^2)^(3/2)*A*a^3*x/b^3 - 11/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a*x^2/b^3 + 1/7*(b^2*x^2 + 2*a*b
*x + a^2)^(5/2)*A*x^2/b^2 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^5/b^5 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2
)*A*a^4/b^4 + 13/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^2*x/b^4 - 3/14*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*a*x/b
^3 - 69/280*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^3/b^5 + 17/70*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*a^2/b^4

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

[Out]

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

[Out]

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

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