3.799 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x^{9/2}} \, dx\)

Optimal. Leaf size=214 \[ -\frac{2 a^2 \sqrt{a^2+2 a b x+b^2 x^2} (a B+3 A b)}{5 x^{5/2} (a+b x)}-\frac{2 a b \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{x^{3/2} (a+b x)}-\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (3 a B+A b)}{\sqrt{x} (a+b x)}-\frac{2 a^3 A \sqrt{a^2+2 a b x+b^2 x^2}}{7 x^{7/2} (a+b x)}+\frac{2 b^3 B \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

[Out]

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

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Rubi [A]  time = 0.0800435, antiderivative size = 214, normalized size of antiderivative = 1., 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} \[ -\frac{2 a^2 \sqrt{a^2+2 a b x+b^2 x^2} (a B+3 A b)}{5 x^{5/2} (a+b x)}-\frac{2 a b \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{x^{3/2} (a+b x)}-\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (3 a B+A b)}{\sqrt{x} (a+b x)}-\frac{2 a^3 A \sqrt{a^2+2 a b x+b^2 x^2}}{7 x^{7/2} (a+b x)}+\frac{2 b^3 B \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^(9/2),x]

[Out]

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

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]

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])

Rubi steps

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

Mathematica [A]  time = 0.0346976, size = 84, normalized size = 0.39 \[ -\frac{2 \sqrt{(a+b x)^2} \left (7 a^2 b x (3 A+5 B x)+a^3 (5 A+7 B x)+35 a b^2 x^2 (A+3 B x)+35 b^3 x^3 (A-B x)\right )}{35 x^{7/2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^(9/2),x]

[Out]

(-2*Sqrt[(a + b*x)^2]*(35*b^3*x^3*(A - B*x) + 35*a*b^2*x^2*(A + 3*B*x) + 7*a^2*b*x*(3*A + 5*B*x) + a^3*(5*A +
7*B*x)))/(35*x^(7/2)*(a + b*x))

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Maple [A]  time = 0.006, size = 92, normalized size = 0.4 \begin{align*} -{\frac{-70\,B{x}^{4}{b}^{3}+70\,A{b}^{3}{x}^{3}+210\,B{x}^{3}a{b}^{2}+70\,A{x}^{2}a{b}^{2}+70\,B{x}^{2}{a}^{2}b+42\,A{a}^{2}bx+14\,{a}^{3}Bx+10\,A{a}^{3}}{35\, \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}{x}^{-{\frac{7}{2}}}} \end{align*}

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^(9/2),x)

[Out]

-2/35*(-35*B*b^3*x^4+35*A*b^3*x^3+105*B*a*b^2*x^3+35*A*a*b^2*x^2+35*B*a^2*b*x^2+21*A*a^2*b*x+7*B*a^3*x+5*A*a^3
)*((b*x+a)^2)^(3/2)/x^(7/2)/(b*x+a)^3

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Maxima [A]  time = 1.02415, size = 181, normalized size = 0.85 \begin{align*} \frac{2}{15} \, B{\left (\frac{15 \,{\left (b^{3} x^{2} - a b^{2} x\right )}}{x^{\frac{3}{2}}} - \frac{10 \,{\left (3 \, a b^{2} x^{2} + a^{2} b x\right )}}{x^{\frac{5}{2}}} - \frac{5 \, a^{2} b x^{2} + 3 \, a^{3} x}{x^{\frac{7}{2}}}\right )} - \frac{2}{105} \, A{\left (\frac{35 \,{\left (3 \, b^{3} x^{2} + a b^{2} x\right )}}{x^{\frac{5}{2}}} + \frac{14 \,{\left (5 \, a b^{2} x^{2} + 3 \, a^{2} b x\right )}}{x^{\frac{7}{2}}} + \frac{3 \,{\left (7 \, a^{2} b x^{2} + 5 \, a^{3} x\right )}}{x^{\frac{9}{2}}}\right )} \end{align*}

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^(9/2),x, algorithm="maxima")

[Out]

2/15*B*(15*(b^3*x^2 - a*b^2*x)/x^(3/2) - 10*(3*a*b^2*x^2 + a^2*b*x)/x^(5/2) - (5*a^2*b*x^2 + 3*a^3*x)/x^(7/2))
- 2/105*A*(35*(3*b^3*x^2 + a*b^2*x)/x^(5/2) + 14*(5*a*b^2*x^2 + 3*a^2*b*x)/x^(7/2) + 3*(7*a^2*b*x^2 + 5*a^3*x
)/x^(9/2))

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Fricas [A]  time = 1.36096, size = 166, normalized size = 0.78 \begin{align*} \frac{2 \,{\left (35 \, B b^{3} x^{4} - 5 \, A a^{3} - 35 \,{\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} - 35 \,{\left (B a^{2} b + A a b^{2}\right )} x^{2} - 7 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{35 \, x^{\frac{7}{2}}} \end{align*}

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^(9/2),x, algorithm="fricas")

[Out]

2/35*(35*B*b^3*x^4 - 5*A*a^3 - 35*(3*B*a*b^2 + A*b^3)*x^3 - 35*(B*a^2*b + A*a*b^2)*x^2 - 7*(B*a^3 + 3*A*a^2*b)
*x)/x^(7/2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.14654, size = 167, normalized size = 0.78 \begin{align*} 2 \, B b^{3} \sqrt{x} \mathrm{sgn}\left (b x + a\right ) - \frac{2 \,{\left (105 \, B a b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + 35 \, A b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + 35 \, B a^{2} b x^{2} \mathrm{sgn}\left (b x + a\right ) + 35 \, A a b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 7 \, B a^{3} x \mathrm{sgn}\left (b x + a\right ) + 21 \, A a^{2} b x \mathrm{sgn}\left (b x + a\right ) + 5 \, A a^{3} \mathrm{sgn}\left (b x + a\right )\right )}}{35 \, x^{\frac{7}{2}}} \end{align*}

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^(9/2),x, algorithm="giac")

[Out]

2*B*b^3*sqrt(x)*sgn(b*x + a) - 2/35*(105*B*a*b^2*x^3*sgn(b*x + a) + 35*A*b^3*x^3*sgn(b*x + a) + 35*B*a^2*b*x^2
*sgn(b*x + a) + 35*A*a*b^2*x^2*sgn(b*x + a) + 7*B*a^3*x*sgn(b*x + a) + 21*A*a^2*b*x*sgn(b*x + a) + 5*A*a^3*sgn
(b*x + a))/x^(7/2)