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]

2/5*a*A*x^(5/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/7*(A*b+B*a)*x^(7/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/9*b*B*x^(9/2)*((b*
x+a)^2)^(1/2)/(b*x+a)

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Rubi [A]  time = 0.04, antiderivative size = 120, 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} \[ \frac {2 x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{7 (a+b x)}+\frac {2 a A x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {2 b B x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)} \]

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 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) \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*}

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Mathematica [A]  time = 0.03, 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))

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fricas [A]  time = 0.85, size = 32, normalized size = 0.27 \[ \frac {2}{315} \, {\left (35 \, B b x^{4} + 63 \, A a x^{2} + 45 \, {\left (B a + A b\right )} x^{3}\right )} \sqrt {x} \]

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)

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giac [A]  time = 0.21, size = 53, normalized size = 0.44 \[ \frac {2}{9} \, B b x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{7} \, B a x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{7} \, A b x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{5} \, A a x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) \]

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)

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maple [A]  time = 0.05, size = 44, normalized size = 0.37 \[ \frac {2 \left (35 B b \,x^{2}+45 A b x +45 B a x +63 A a \right ) \sqrt {\left (b x +a \right )^{2}}\, x^{\frac {5}{2}}}{315 \left (b x +a \right )} \]

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)

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maxima [A]  time = 0.52, size = 35, normalized size = 0.29 \[ \frac {2}{63} \, {\left (7 \, b x^{2} + 9 \, a x\right )} B x^{\frac {5}{2}} + \frac {2}{35} \, {\left (5 \, b x^{2} + 7 \, a x\right )} A x^{\frac {3}{2}} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{3/2}\,\sqrt {{\left (a+b\,x\right )}^2}\,\left (A+B\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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