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

Optimal. Leaf size=220 \[ \frac {2 b^2 x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{11 (a+b x)}+\frac {2 a b x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{3 (a+b x)}+\frac {2 a^2 x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{7 (a+b x)}+\frac {2 b^3 B x^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}}{13 (a+b x)}+\frac {2 a^3 A x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]

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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^{11/2} \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{11 (a+b x)}+\frac {2 a b x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{3 (a+b x)}+\frac {2 a^2 x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{7 (a+b x)}+\frac {2 a^3 A x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {2 b^3 B x^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}}{13 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*A*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a + b*x)) + (2*a^2*(3*A*b + a*B)*x^(7/2)*Sqrt[a^2 + 2*a*b*
x + b^2*x^2])/(7*(a + b*x)) + (2*a*b*(A*b + a*B)*x^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (2*b^2
*(A*b + 3*a*B)*x^(11/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(11*(a + b*x)) + (2*b^3*B*x^(13/2)*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/(13*(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^{3/2} (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^{3/2} \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^{3/2}+a^2 b^3 (3 A b+a B) x^{5/2}+3 a b^4 (A b+a B) x^{7/2}+b^5 (A b+3 a B) x^{9/2}+b^6 B x^{11/2}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {2 a^3 A x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {2 a^2 (3 A b+a B) x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {2 a b (A b+a B) x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {2 b^2 (A b+3 a B) x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {2 b^3 B x^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}}{13 (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^{5/2} \sqrt {(a+b x)^2} \left (429 a^3 (7 A+5 B x)+715 a^2 b x (9 A+7 B x)+455 a b^2 x^2 (11 A+9 B x)+105 b^3 x^3 (13 A+11 B x)\right )}{15015 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*Sqrt[(a + b*x)^2]*(429*a^3*(7*A + 5*B*x) + 715*a^2*b*x*(9*A + 7*B*x) + 455*a*b^2*x^2*(11*A + 9*B*x)
+ 105*b^3*x^3*(13*A + 11*B*x)))/(15015*(a + b*x))

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IntegrateAlgebraic [A]  time = 10.28, size = 115, normalized size = 0.52 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} \left (3003 a^3 A x^{5/2}+2145 a^3 B x^{7/2}+6435 a^2 A b x^{7/2}+5005 a^2 b B x^{9/2}+5005 a A b^2 x^{9/2}+4095 a b^2 B x^{11/2}+1365 A b^3 x^{11/2}+1155 b^3 B x^{13/2}\right )}{15015 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(3003*a^3*A*x^(5/2) + 6435*a^2*A*b*x^(7/2) + 2145*a^3*B*x^(7/2) + 5005*a*A*b^2*x^(9/2) +
5005*a^2*b*B*x^(9/2) + 1365*A*b^3*x^(11/2) + 4095*a*b^2*B*x^(11/2) + 1155*b^3*B*x^(13/2)))/(15015*(a + b*x))

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fricas [A]  time = 0.43, size = 78, normalized size = 0.35 \begin {gather*} \frac {2}{15015} \, {\left (1155 \, B b^{3} x^{6} + 3003 \, A a^{3} x^{2} + 1365 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{5} + 5005 \, {\left (B a^{2} b + A a b^{2}\right )} x^{4} + 2145 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{3}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15015*(1155*B*b^3*x^6 + 3003*A*a^3*x^2 + 1365*(3*B*a*b^2 + A*b^3)*x^5 + 5005*(B*a^2*b + A*a*b^2)*x^4 + 2145*
(B*a^3 + 3*A*a^2*b)*x^3)*sqrt(x)

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giac [A]  time = 0.16, size = 125, normalized size = 0.57 \begin {gather*} \frac {2}{13} \, B b^{3} x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{11} \, B a b^{2} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{11} \, A b^{3} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, B a^{2} b x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, A a b^{2} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{7} \, B a^{3} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{7} \, A a^{2} b x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{5} \, A a^{3} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/13*B*b^3*x^(13/2)*sgn(b*x + a) + 6/11*B*a*b^2*x^(11/2)*sgn(b*x + a) + 2/11*A*b^3*x^(11/2)*sgn(b*x + a) + 2/3
*B*a^2*b*x^(9/2)*sgn(b*x + a) + 2/3*A*a*b^2*x^(9/2)*sgn(b*x + a) + 2/7*B*a^3*x^(7/2)*sgn(b*x + a) + 6/7*A*a^2*
b*x^(7/2)*sgn(b*x + a) + 2/5*A*a^3*x^(5/2)*sgn(b*x + a)

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maple [A]  time = 0.05, size = 92, normalized size = 0.42 \begin {gather*} \frac {2 \left (1155 B \,b^{3} x^{4}+1365 A \,b^{3} x^{3}+4095 B a \,b^{2} x^{3}+5005 A a \,b^{2} x^{2}+5005 B \,a^{2} b \,x^{2}+6435 A \,a^{2} b x +2145 B \,a^{3} x +3003 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x^{\frac {5}{2}}}{15015 \left (b x +a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15015*x^(5/2)*(1155*B*b^3*x^4+1365*A*b^3*x^3+4095*B*a*b^2*x^3+5005*A*a*b^2*x^2+5005*B*a^2*b*x^2+6435*A*a^2*b
*x+2145*B*a^3*x+3003*A*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [A]  time = 0.54, size = 137, normalized size = 0.62 \begin {gather*} \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 )} A + \frac {2}{9009} \, {\left (63 \, {\left (11 \, b^{3} x^{2} + 13 \, a b^{2} x\right )} x^{\frac {9}{2}} + 182 \, {\left (9 \, a b^{2} x^{2} + 11 \, a^{2} b x\right )} x^{\frac {7}{2}} + 143 \, {\left (7 \, a^{2} b x^{2} + 9 \, a^{3} x\right )} x^{\frac {5}{2}}\right )} B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))*A + 2/9009*(63*(11*b^3*x^2 + 13*a*b^2*x)*x^(9/2) + 182*(9*a*b^2*x^2 + 11*a^2*b*x)*x^(7/2) + 143*(7
*a^2*b*x^2 + 9*a^3*x)*x^(5/2))*B

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

[Out]

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

[Out]

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

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