\(\int \frac {(a+b x) (A+B x)}{x^{5/2}} \, dx\) [325]

Optimal result

Integrand size = 16, antiderivative size = 35 \[ \int \frac {(a+b x) (A+B x)}{x^{5/2}} \, dx=-\frac {2 a A}{3 x^{3/2}}-\frac {2 (A b+a B)}{\sqrt {x}}+2 b B \sqrt {x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, 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} \[ \int \frac {(a+b x) (A+B x)}{x^{5/2}} \, dx=-\frac {2 (a B+A b)}{\sqrt {x}}-\frac {2 a A}{3 x^{3/2}}+2 b B \sqrt {x} \]

[In]

Int[((a + b*x)*(A + B*x))/x^(5/2),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x) (A+B x)}{x^{5/2}} \, dx=-\frac {2 (3 b x (A-B x)+a (A+3 B x))}{3 x^{3/2}} \]

[In]

Integrate[((a + b*x)*(A + B*x))/x^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77

method	result	size
gosper	\(-\frac {2 \left (-3 b B \,x^{2}+3 A b x +3 B a x +A a \right )}{3 x^{\frac {3}{2}}}\)	\(27\)
trager	\(-\frac {2 \left (-3 b B \,x^{2}+3 A b x +3 B a x +A a \right )}{3 x^{\frac {3}{2}}}\)	\(27\)
risch	\(-\frac {2 \left (-3 b B \,x^{2}+3 A b x +3 B a x +A a \right )}{3 x^{\frac {3}{2}}}\)	\(27\)
derivativedivides	\(-\frac {2 a A}{3 x^{\frac {3}{2}}}-\frac {2 \left (A b +B a \right )}{\sqrt {x}}+2 b B \sqrt {x}\)	\(28\)
default	\(-\frac {2 a A}{3 x^{\frac {3}{2}}}-\frac {2 \left (A b +B a \right )}{\sqrt {x}}+2 b B \sqrt {x}\)	\(28\)

[In]

int((b*x+a)*(B*x+A)/x^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x) (A+B x)}{x^{5/2}} \, dx=\frac {2 \, {\left (3 \, B b x^{2} - A a - 3 \, {\left (B a + A b\right )} x\right )}}{3 \, x^{\frac {3}{2}}} \]

[In]

integrate((b*x+a)*(B*x+A)/x^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b x) (A+B x)}{x^{5/2}} \, dx=- \frac {2 A a}{3 x^{\frac {3}{2}}} - \frac {2 A b}{\sqrt {x}} - \frac {2 B a}{\sqrt {x}} + 2 B b \sqrt {x} \]

[In]

integrate((b*x+a)*(B*x+A)/x**(5/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x) (A+B x)}{x^{5/2}} \, dx=2 \, B b \sqrt {x} - \frac {2 \, {\left (A a + 3 \, {\left (B a + A b\right )} x\right )}}{3 \, x^{\frac {3}{2}}} \]

[In]

integrate((b*x+a)*(B*x+A)/x^(5/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x) (A+B x)}{x^{5/2}} \, dx=2 \, B b \sqrt {x} - \frac {2 \, {\left (3 \, B a x + 3 \, A b x + A a\right )}}{3 \, x^{\frac {3}{2}}} \]

[In]

integrate((b*x+a)*(B*x+A)/x^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x) (A+B x)}{x^{5/2}} \, dx=-\frac {2\,A\,a+6\,A\,b\,x+6\,B\,a\,x-6\,B\,b\,x^2}{3\,x^{3/2}} \]

[In]

int(((A + B*x)*(a + b*x))/x^(5/2),x)

[Out]

-(2*A*a + 6*A*b*x + 6*B*a*x - 6*B*b*x^2)/(3*x^(3/2))