\(\int \frac {(a+b x)^2 (A+B x)}{x^{3/2}} \, dx\) [332]

Optimal result

Integrand size = 18, antiderivative size = 59 \[ \int \frac {(a+b x)^2 (A+B x)}{x^{3/2}} \, dx=-\frac {2 a^2 A}{\sqrt {x}}+2 a (2 A b+a B) \sqrt {x}+\frac {2}{3} b (A b+2 a B) x^{3/2}+\frac {2}{5} b^2 B x^{5/2} \]

[Out]

2/3*b*(A*b+2*B*a)*x^(3/2)+2/5*b^2*B*x^(5/2)-2*a^2*A/x^(1/2)+2*a*(2*A*b+B*a)*x^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {77} \[ \int \frac {(a+b x)^2 (A+B x)}{x^{3/2}} \, dx=-\frac {2 a^2 A}{\sqrt {x}}+\frac {2}{3} b x^{3/2} (2 a B+A b)+2 a \sqrt {x} (a B+2 A b)+\frac {2}{5} b^2 B x^{5/2} \]

[In]

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

[Out]

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

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 (\frac {a^2 A}{x^{3/2}}+\frac {a (2 A b+a B)}{\sqrt {x}}+b (A b+2 a B) \sqrt {x}+b^2 B x^{3/2}\right ) \, dx \\ & = -\frac {2 a^2 A}{\sqrt {x}}+2 a (2 A b+a B) \sqrt {x}+\frac {2}{3} b (A b+2 a B) x^{3/2}+\frac {2}{5} b^2 B x^{5/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.83 \[ \int \frac {(a+b x)^2 (A+B x)}{x^{3/2}} \, dx=\frac {-30 a^2 (A-B x)+20 a b x (3 A+B x)+2 b^2 x^2 (5 A+3 B x)}{15 \sqrt {x}} \]

[In]

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

[Out]

(-30*a^2*(A - B*x) + 20*a*b*x*(3*A + B*x) + 2*b^2*x^2*(5*A + 3*B*x))/(15*Sqrt[x])

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.88

method	result	size
gosper	\(-\frac {2 \left (-3 b^{2} B \,x^{3}-5 A \,b^{2} x^{2}-10 B a b \,x^{2}-30 a A b x -15 a^{2} B x +15 a^{2} A \right )}{15 \sqrt {x}}\)	\(52\)
trager	\(-\frac {2 \left (-3 b^{2} B \,x^{3}-5 A \,b^{2} x^{2}-10 B a b \,x^{2}-30 a A b x -15 a^{2} B x +15 a^{2} A \right )}{15 \sqrt {x}}\)	\(52\)
risch	\(-\frac {2 \left (-3 b^{2} B \,x^{3}-5 A \,b^{2} x^{2}-10 B a b \,x^{2}-30 a A b x -15 a^{2} B x +15 a^{2} A \right )}{15 \sqrt {x}}\)	\(52\)
derivativedivides	\(\frac {2 b^{2} B \,x^{\frac {5}{2}}}{5}+\frac {2 A \,b^{2} x^{\frac {3}{2}}}{3}+\frac {4 B a b \,x^{\frac {3}{2}}}{3}+4 a b A \sqrt {x}+2 a^{2} B \sqrt {x}-\frac {2 a^{2} A}{\sqrt {x}}\)	\(54\)
default	\(\frac {2 b^{2} B \,x^{\frac {5}{2}}}{5}+\frac {2 A \,b^{2} x^{\frac {3}{2}}}{3}+\frac {4 B a b \,x^{\frac {3}{2}}}{3}+4 a b A \sqrt {x}+2 a^{2} B \sqrt {x}-\frac {2 a^{2} A}{\sqrt {x}}\)	\(54\)

[In]

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

[Out]

-2/15*(-3*B*b^2*x^3-5*A*b^2*x^2-10*B*a*b*x^2-30*A*a*b*x-15*B*a^2*x+15*A*a^2)/x^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x)^2 (A+B x)}{x^{3/2}} \, dx=\frac {2 \, {\left (3 \, B b^{2} x^{3} - 15 \, A a^{2} + 5 \, {\left (2 \, B a b + A b^{2}\right )} x^{2} + 15 \, {\left (B a^{2} + 2 \, A a b\right )} x\right )}}{15 \, \sqrt {x}} \]

[In]

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

[Out]

2/15*(3*B*b^2*x^3 - 15*A*a^2 + 5*(2*B*a*b + A*b^2)*x^2 + 15*(B*a^2 + 2*A*a*b)*x)/sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b x)^2 (A+B x)}{x^{3/2}} \, dx=- \frac {2 A a^{2}}{\sqrt {x}} + 4 A a b \sqrt {x} + \frac {2 A b^{2} x^{\frac {3}{2}}}{3} + 2 B a^{2} \sqrt {x} + \frac {4 B a b x^{\frac {3}{2}}}{3} + \frac {2 B b^{2} x^{\frac {5}{2}}}{5} \]

[In]

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

[Out]

-2*A*a**2/sqrt(x) + 4*A*a*b*sqrt(x) + 2*A*b**2*x**(3/2)/3 + 2*B*a**2*sqrt(x) + 4*B*a*b*x**(3/2)/3 + 2*B*b**2*x
**(5/2)/5

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x)^2 (A+B x)}{x^{3/2}} \, dx=\frac {2}{5} \, B b^{2} x^{\frac {5}{2}} - \frac {2 \, A a^{2}}{\sqrt {x}} + \frac {2}{3} \, {\left (2 \, B a b + A b^{2}\right )} x^{\frac {3}{2}} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} \sqrt {x} \]

[In]

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

[Out]

2/5*B*b^2*x^(5/2) - 2*A*a^2/sqrt(x) + 2/3*(2*B*a*b + A*b^2)*x^(3/2) + 2*(B*a^2 + 2*A*a*b)*sqrt(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x)^2 (A+B x)}{x^{3/2}} \, dx=\frac {2}{5} \, B b^{2} x^{\frac {5}{2}} + \frac {4}{3} \, B a b x^{\frac {3}{2}} + \frac {2}{3} \, A b^{2} x^{\frac {3}{2}} + 2 \, B a^{2} \sqrt {x} + 4 \, A a b \sqrt {x} - \frac {2 \, A a^{2}}{\sqrt {x}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x)^2 (A+B x)}{x^{3/2}} \, dx=\sqrt {x}\,\left (2\,B\,a^2+4\,A\,b\,a\right )+x^{3/2}\,\left (\frac {2\,A\,b^2}{3}+\frac {4\,B\,a\,b}{3}\right )-\frac {2\,A\,a^2}{\sqrt {x}}+\frac {2\,B\,b^2\,x^{5/2}}{5} \]

[In]

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

[Out]

x^(1/2)*(2*B*a^2 + 4*A*a*b) + x^(3/2)*((2*A*b^2)/3 + (4*B*a*b)/3) - (2*A*a^2)/x^(1/2) + (2*B*b^2*x^(5/2))/5