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

Optimal. Leaf size=220 \[ \frac {2 b^2 x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{9 (a+b x)}+\frac {6 a b x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{7 (a+b x)}+\frac {2 a^2 x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{5 (a+b x)}+\frac {2 b^3 B x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {2 a^3 A x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)} \]

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Rubi [A]  time = 0.08, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {770, 76} \begin {gather*} \frac {2 b^2 x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{9 (a+b x)}+\frac {6 a b x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{7 (a+b x)}+\frac {2 a^2 x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{5 (a+b x)}+\frac {2 a^3 A x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {2 b^3 B x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*A*x^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (2*a^2*(3*A*b + a*B)*x^(5/2)*Sqrt[a^2 + 2*a*b*
x + b^2*x^2])/(5*(a + b*x)) + (6*a*b*(A*b + a*B)*x^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*(a + b*x)) + (2*b^2
*(A*b + 3*a*B)*x^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*(a + b*x)) + (2*b^3*B*x^(11/2)*Sqrt[a^2 + 2*a*b*x + b
^2*x^2])/(11*(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 \sqrt {x} (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 \sqrt {x} \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 \sqrt {x}+a^2 b^3 (3 A b+a B) x^{3/2}+3 a b^4 (A b+a B) x^{5/2}+b^5 (A b+3 a B) x^{7/2}+b^6 B x^{9/2}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {2 a^3 A x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {2 a^2 (3 A b+a B) x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {6 a b (A b+a B) x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {2 b^2 (A b+3 a B) x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {2 b^3 B x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 89, normalized size = 0.40 \begin {gather*} \frac {2 x^{3/2} \sqrt {(a+b x)^2} \left (231 a^3 (5 A+3 B x)+297 a^2 b x (7 A+5 B x)+165 a b^2 x^2 (9 A+7 B x)+35 b^3 x^3 (11 A+9 B x)\right )}{3465 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*Sqrt[(a + b*x)^2]*(231*a^3*(5*A + 3*B*x) + 297*a^2*b*x*(7*A + 5*B*x) + 165*a*b^2*x^2*(9*A + 7*B*x)
+ 35*b^3*x^3*(11*A + 9*B*x)))/(3465*(a + b*x))

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IntegrateAlgebraic [A]  time = 8.78, size = 115, normalized size = 0.52 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} \left (1155 a^3 A x^{3/2}+693 a^3 B x^{5/2}+2079 a^2 A b x^{5/2}+1485 a^2 b B x^{7/2}+1485 a A b^2 x^{7/2}+1155 a b^2 B x^{9/2}+385 A b^3 x^{9/2}+315 b^3 B x^{11/2}\right )}{3465 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(1155*a^3*A*x^(3/2) + 2079*a^2*A*b*x^(5/2) + 693*a^3*B*x^(5/2) + 1485*a*A*b^2*x^(7/2) + 1
485*a^2*b*B*x^(7/2) + 385*A*b^3*x^(9/2) + 1155*a*b^2*B*x^(9/2) + 315*b^3*B*x^(11/2)))/(3465*(a + b*x))

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fricas [A]  time = 0.41, size = 76, normalized size = 0.35 \begin {gather*} \frac {2}{3465} \, {\left (315 \, B b^{3} x^{5} + 1155 \, A a^{3} x + 385 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{4} + 1485 \, {\left (B a^{2} b + A a b^{2}\right )} x^{3} + 693 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{2}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)*x^(1/2),x, algorithm="fricas")

[Out]

2/3465*(315*B*b^3*x^5 + 1155*A*a^3*x + 385*(3*B*a*b^2 + A*b^3)*x^4 + 1485*(B*a^2*b + A*a*b^2)*x^3 + 693*(B*a^3
+ 3*A*a^2*b)*x^2)*sqrt(x)

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giac [A]  time = 0.33, size = 125, normalized size = 0.57 \begin {gather*} \frac {2}{11} \, B b^{3} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, B a b^{2} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{9} \, A b^{3} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{7} \, B a^{2} b x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{7} \, A a b^{2} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{5} \, B a^{3} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{5} \, A a^{2} b x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, A a^{3} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)*x^(1/2),x, algorithm="giac")

[Out]

2/11*B*b^3*x^(11/2)*sgn(b*x + a) + 2/3*B*a*b^2*x^(9/2)*sgn(b*x + a) + 2/9*A*b^3*x^(9/2)*sgn(b*x + a) + 6/7*B*a
^2*b*x^(7/2)*sgn(b*x + a) + 6/7*A*a*b^2*x^(7/2)*sgn(b*x + a) + 2/5*B*a^3*x^(5/2)*sgn(b*x + a) + 6/5*A*a^2*b*x^
(5/2)*sgn(b*x + a) + 2/3*A*a^3*x^(3/2)*sgn(b*x + a)

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maple [A]  time = 0.06, size = 92, normalized size = 0.42 \begin {gather*} \frac {2 \left (315 B \,b^{3} x^{4}+385 A \,b^{3} x^{3}+1155 B a \,b^{2} x^{3}+1485 A a \,b^{2} x^{2}+1485 B \,a^{2} b \,x^{2}+2079 A \,a^{2} b x +693 B \,a^{3} x +1155 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x^{\frac {3}{2}}}{3465 \left (b x +a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)*x^(1/2),x)

[Out]

2/3465*x^(3/2)*(315*B*b^3*x^4+385*A*b^3*x^3+1155*B*a*b^2*x^3+1485*A*a*b^2*x^2+1485*B*a^2*b*x^2+2079*A*a^2*b*x+
693*B*a^3*x+1155*A*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [A]  time = 0.62, size = 137, normalized size = 0.62 \begin {gather*} \frac {2}{315} \, {\left (5 \, {\left (7 \, b^{3} x^{2} + 9 \, a b^{2} x\right )} x^{\frac {5}{2}} + 18 \, {\left (5 \, a b^{2} x^{2} + 7 \, a^{2} b x\right )} x^{\frac {3}{2}} + 21 \, {\left (3 \, a^{2} b x^{2} + 5 \, a^{3} x\right )} \sqrt {x}\right )} A + \frac {2}{3465} \, {\left (35 \, {\left (9 \, b^{3} x^{2} + 11 \, a b^{2} x\right )} x^{\frac {7}{2}} + 110 \, {\left (7 \, a b^{2} x^{2} + 9 \, a^{2} b x\right )} x^{\frac {5}{2}} + 99 \, {\left (5 \, a^{2} b x^{2} + 7 \, a^{3} x\right )} x^{\frac {3}{2}}\right )} B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)*x^(1/2),x, algorithm="maxima")

[Out]

2/315*(5*(7*b^3*x^2 + 9*a*b^2*x)*x^(5/2) + 18*(5*a*b^2*x^2 + 7*a^2*b*x)*x^(3/2) + 21*(3*a^2*b*x^2 + 5*a^3*x)*s
qrt(x))*A + 2/3465*(35*(9*b^3*x^2 + 11*a*b^2*x)*x^(7/2) + 110*(7*a*b^2*x^2 + 9*a^2*b*x)*x^(5/2) + 99*(5*a^2*b*
x^2 + 7*a^3*x)*x^(3/2))*B

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

[Out]

int(x^(1/2)*(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 \sqrt {x} \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((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)*x**(1/2),x)

[Out]

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

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