\(\int x^{3/2} (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx\) [784]

Optimal result

Integrand size = 31, antiderivative size = 120 \[ \int x^{3/2} (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 a A x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {2 (A b+a B) x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {2 b B x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {784, 77} \[ \int x^{3/2} (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{7 (a+b x)}+\frac {2 a A x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {2 b B x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)} \]

[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 77

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 784

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*} \text {integral}& = \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] (verified)

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.42 \[ \int x^{3/2} (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 x^{5/2} \sqrt {(a+b x)^2} (9 a (7 A+5 B x)+5 b x (9 A+7 B x))}{315 (a+b x)} \]

[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 [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.28

method	result	size
default	\(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) x^{\frac {5}{2}} \left (35 B b \,x^{2}+45 A b x +45 a B x +63 a A \right )}{315}\)	\(34\)
gosper	\(\frac {2 x^{\frac {5}{2}} \left (35 B b \,x^{2}+45 A b x +45 a B x +63 a A \right ) \sqrt {\left (b x +a \right )^{2}}}{315 \left (b x +a \right )}\)	\(44\)
risch	\(\frac {2 x^{\frac {5}{2}} \left (35 B b \,x^{2}+45 A b x +45 a B x +63 a A \right ) \sqrt {\left (b x +a \right )^{2}}}{315 \left (b x +a \right )}\)	\(44\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.27 \[ \int x^{3/2} (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2}{315} \, {\left (35 \, B b x^{4} + 63 \, A a x^{2} + 45 \, {\left (B a + A b\right )} x^{3}\right )} \sqrt {x} \]

[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)]

Timed out. \[ \int x^{3/2} (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.29 \[ \int x^{3/2} (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2}{63} \, {\left (7 \, b x^{2} + 9 \, a x\right )} B x^{\frac {5}{2}} + \frac {2}{35} \, {\left (5 \, b x^{2} + 7 \, a x\right )} A x^{\frac {3}{2}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.44 \[ \int x^{3/2} (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2}{9} \, B b x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{7} \, B a x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{7} \, A b x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{5} \, A a x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) \]

[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)

Mupad [F(-1)]

Timed out. \[ \int x^{3/2} (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\int x^{3/2}\,\sqrt {{\left (a+b\,x\right )}^2}\,\left (A+B\,x\right ) \,d x \]

[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)