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

Optimal. Leaf size=120 \[ \frac{2 x^{7/2} \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{7 (a+b x)}+\frac{2 a A x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac{2 b B x^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}{9 (a+b x)} \]

[Out]

(2*a*A*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a + b*x)) + (2*(A*b + a*B)*x^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*
x^2])/(7*(a + b*x)) + (2*b*B*x^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*(a + b*x))

________________________________________________________________________________________

Rubi [A]  time = 0.0428388, antiderivative size = 120, 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} \[ \frac{2 x^{7/2} \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{7 (a+b x)}+\frac{2 a A x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac{2 b B x^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}{9 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0261576, size = 51, normalized size = 0.42 \[ \frac{2 x^{5/2} \sqrt{(a+b x)^2} (9 a (7 A+5 B x)+5 b x (9 A+7 B x))}{315 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*Sqrt[(a + b*x)^2]*(9*a*(7*A + 5*B*x) + 5*b*x*(9*A + 7*B*x)))/(315*(a + b*x))

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 44, normalized size = 0.4 \begin{align*}{\frac{70\,Bb{x}^{2}+90\,Abx+90\,aBx+126\,aA}{315\,bx+315\,a}{x}^{{\frac{5}{2}}}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*x^(5/2)*(35*B*b*x^2+45*A*b*x+45*B*a*x+63*A*a)*((b*x+a)^2)^(1/2)/(b*x+a)

________________________________________________________________________________________

Maxima [A]  time = 1.02371, size = 47, normalized size = 0.39 \begin{align*} \frac{2}{63} \,{\left (7 \, b x^{2} + 9 \, a x\right )} B x^{\frac{5}{2}} + \frac{2}{35} \,{\left (5 \, b x^{2} + 7 \, a x\right )} A x^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(7*b*x^2 + 9*a*x)*B*x^(5/2) + 2/35*(5*b*x^2 + 7*a*x)*A*x^(3/2)

________________________________________________________________________________________

Fricas [A]  time = 1.53923, size = 84, normalized size = 0.7 \begin{align*} \frac{2}{315} \,{\left (35 \, B b x^{4} + 63 \, A a x^{2} + 45 \,{\left (B a + A b\right )} x^{3}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*B*b*x^4 + 63*A*a*x^2 + 45*(B*a + A*b)*x^3)*sqrt(x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.11989, size = 72, normalized size = 0.6 \begin{align*} \frac{2}{9} \, B b x^{\frac{9}{2}} \mathrm{sgn}\left (b x + a\right ) + \frac{2}{7} \, B a x^{\frac{7}{2}} \mathrm{sgn}\left (b x + a\right ) + \frac{2}{7} \, A b x^{\frac{7}{2}} \mathrm{sgn}\left (b x + a\right ) + \frac{2}{5} \, A a x^{\frac{5}{2}} \mathrm{sgn}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*B*b*x^(9/2)*sgn(b*x + a) + 2/7*B*a*x^(7/2)*sgn(b*x + a) + 2/7*A*b*x^(7/2)*sgn(b*x + a) + 2/5*A*a*x^(5/2)*s
gn(b*x + a)