\(\int x^{3/2} (a+b x) (A+B x) \, dx\) [321]

Optimal result

Integrand size = 16, antiderivative size = 39 \[ \int x^{3/2} (a+b x) (A+B x) \, dx=\frac {2}{5} a A x^{5/2}+\frac {2}{7} (A b+a B) x^{7/2}+\frac {2}{9} b B x^{9/2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {77} \[ \int x^{3/2} (a+b x) (A+B x) \, dx=\frac {2}{7} x^{7/2} (a B+A b)+\frac {2}{5} a A x^{5/2}+\frac {2}{9} b B x^{9/2} \]

[In]

Int[x^(3/2)*(a + b*x)*(A + B*x),x]

[Out]

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

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])

Rubi steps \begin{align*} \text {integral}& = \int \left (a A x^{3/2}+(A b+a B) x^{5/2}+b B x^{7/2}\right ) \, dx \\ & = \frac {2}{5} a A x^{5/2}+\frac {2}{7} (A b+a B) x^{7/2}+\frac {2}{9} b B x^{9/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int x^{3/2} (a+b x) (A+B x) \, dx=\frac {2}{315} x^{5/2} (9 a (7 A+5 B x)+5 b x (9 A+7 B x)) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.72

method	result	size
gosper	\(\frac {2 x^{\frac {5}{2}} \left (35 b B \,x^{2}+45 A b x +45 B a x +63 A a \right )}{315}\)	\(28\)
derivativedivides	\(\frac {2 a A \,x^{\frac {5}{2}}}{5}+\frac {2 \left (A b +B a \right ) x^{\frac {7}{2}}}{7}+\frac {2 b B \,x^{\frac {9}{2}}}{9}\)	\(28\)
default	\(\frac {2 a A \,x^{\frac {5}{2}}}{5}+\frac {2 \left (A b +B a \right ) x^{\frac {7}{2}}}{7}+\frac {2 b B \,x^{\frac {9}{2}}}{9}\)	\(28\)
trager	\(\frac {2 x^{\frac {5}{2}} \left (35 b B \,x^{2}+45 A b x +45 B a x +63 A a \right )}{315}\)	\(28\)
risch	\(\frac {2 x^{\frac {5}{2}} \left (35 b B \,x^{2}+45 A b x +45 B a x +63 A a \right )}{315}\)	\(28\)

[In]

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

[Out]

2/315*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 = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82 \[ \int x^{3/2} (a+b x) (A+B x) \, dx=\frac {2}{315} \, {\left (35 \, B b x^{4} + 63 \, A a x^{2} + 45 \, {\left (B a + A b\right )} x^{3}\right )} \sqrt {x} \]

[In]

integrate(x^(3/2)*(b*x+a)*(B*x+A),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 [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.18 \[ \int x^{3/2} (a+b x) (A+B x) \, dx=\frac {2 A a x^{\frac {5}{2}}}{5} + \frac {2 A b x^{\frac {7}{2}}}{7} + \frac {2 B a x^{\frac {7}{2}}}{7} + \frac {2 B b x^{\frac {9}{2}}}{9} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.69 \[ \int x^{3/2} (a+b x) (A+B x) \, dx=\frac {2}{9} \, B b x^{\frac {9}{2}} + \frac {2}{5} \, A a x^{\frac {5}{2}} + \frac {2}{7} \, {\left (B a + A b\right )} x^{\frac {7}{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int x^{3/2} (a+b x) (A+B x) \, dx=\frac {2}{9} \, B b x^{\frac {9}{2}} + \frac {2}{7} \, B a x^{\frac {7}{2}} + \frac {2}{7} \, A b x^{\frac {7}{2}} + \frac {2}{5} \, A a x^{\frac {5}{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.69 \[ \int x^{3/2} (a+b x) (A+B x) \, dx=\frac {2\,x^{5/2}\,\left (63\,A\,a+45\,A\,b\,x+45\,B\,a\,x+35\,B\,b\,x^2\right )}{315} \]

[In]

int(x^(3/2)*(A + B*x)*(a + b*x),x)

[Out]

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