\(\int \frac {(a+b x) (A+B x)}{x^{7/2}} \, dx\) [326]

Optimal result

Integrand size = 16, antiderivative size = 37 \[ \int \frac {(a+b x) (A+B x)}{x^{7/2}} \, dx=-\frac {2 a A}{5 x^{5/2}}-\frac {2 (A b+a B)}{3 x^{3/2}}-\frac {2 b B}{\sqrt {x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, 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} \[ \int \frac {(a+b x) (A+B x)}{x^{7/2}} \, dx=-\frac {2 (a B+A b)}{3 x^{3/2}}-\frac {2 a A}{5 x^{5/2}}-\frac {2 b B}{\sqrt {x}} \]

[In]

Int[((a + b*x)*(A + B*x))/x^(7/2),x]

[Out]

(-2*a*A)/(5*x^(5/2)) - (2*(A*b + a*B))/(3*x^(3/2)) - (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^{7/2}}+\frac {A b+a B}{x^{5/2}}+\frac {b B}{x^{3/2}}\right ) \, dx \\ & = -\frac {2 a A}{5 x^{5/2}}-\frac {2 (A b+a B)}{3 x^{3/2}}-\frac {2 b B}{\sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b x) (A+B x)}{x^{7/2}} \, dx=-\frac {2 \left (3 a A+5 A b x+5 a B x+15 b B x^2\right )}{15 x^{5/2}} \]

[In]

Integrate[((a + b*x)*(A + B*x))/x^(7/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76

method	result	size
gosper	\(-\frac {2 \left (15 b B \,x^{2}+5 A b x +5 B a x +3 A a \right )}{15 x^{\frac {5}{2}}}\)	\(28\)
derivativedivides	\(-\frac {2 a A}{5 x^{\frac {5}{2}}}-\frac {2 \left (A b +B a \right )}{3 x^{\frac {3}{2}}}-\frac {2 b B}{\sqrt {x}}\)	\(28\)
default	\(-\frac {2 a A}{5 x^{\frac {5}{2}}}-\frac {2 \left (A b +B a \right )}{3 x^{\frac {3}{2}}}-\frac {2 b B}{\sqrt {x}}\)	\(28\)
trager	\(-\frac {2 \left (15 b B \,x^{2}+5 A b x +5 B a x +3 A a \right )}{15 x^{\frac {5}{2}}}\)	\(28\)
risch	\(-\frac {2 \left (15 b B \,x^{2}+5 A b x +5 B a x +3 A a \right )}{15 x^{\frac {5}{2}}}\)	\(28\)

[In]

int((b*x+a)*(B*x+A)/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(a+b x) (A+B x)}{x^{7/2}} \, dx=-\frac {2 \, {\left (15 \, B b x^{2} + 3 \, A a + 5 \, {\left (B a + A b\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \]

[In]

integrate((b*x+a)*(B*x+A)/x^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b x) (A+B x)}{x^{7/2}} \, dx=- \frac {2 A a}{5 x^{\frac {5}{2}}} - \frac {2 A b}{3 x^{\frac {3}{2}}} - \frac {2 B a}{3 x^{\frac {3}{2}}} - \frac {2 B b}{\sqrt {x}} \]

[In]

integrate((b*x+a)*(B*x+A)/x**(7/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(a+b x) (A+B x)}{x^{7/2}} \, dx=-\frac {2 \, {\left (15 \, B b x^{2} + 3 \, A a + 5 \, {\left (B a + A b\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \]

[In]

integrate((b*x+a)*(B*x+A)/x^(7/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(a+b x) (A+B x)}{x^{7/2}} \, dx=-\frac {2 \, {\left (15 \, B b x^{2} + 5 \, B a x + 5 \, A b x + 3 \, A a\right )}}{15 \, x^{\frac {5}{2}}} \]

[In]

integrate((b*x+a)*(B*x+A)/x^(7/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {(a+b x) (A+B x)}{x^{7/2}} \, dx=-\frac {2\,B\,b\,x^2+\left (\frac {2\,A\,b}{3}+\frac {2\,B\,a}{3}\right )\,x+\frac {2\,A\,a}{5}}{x^{5/2}} \]

[In]

int(((A + B*x)*(a + b*x))/x^(7/2),x)

[Out]

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