\(\int \frac {(a+b x^2) (A+B x^2)}{x^{5/2}} \, dx\) [349]

Optimal result

Integrand size = 20, antiderivative size = 37 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^{5/2}} \, dx=-\frac {2 a A}{3 x^{3/2}}+2 (A b+a B) \sqrt {x}+\frac {2}{5} b B x^{5/2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {459} \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^{5/2}} \, dx=2 \sqrt {x} (a B+A b)-\frac {2 a A}{3 x^{3/2}}+\frac {2}{5} b B x^{5/2} \]

[In]

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

[Out]

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

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^{5/2}} \, dx=-\frac {2 \left (5 a A-15 A b x^2-15 a B x^2-3 b B x^4\right )}{15 x^{3/2}} \]

[In]

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

[Out]

(-2*(5*a*A - 15*A*b*x^2 - 15*a*B*x^2 - 3*b*B*x^4))/(15*x^(3/2))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(\frac {2 b B \,x^{\frac {5}{2}}}{5}+2 A b \sqrt {x}+2 B a \sqrt {x}-\frac {2 a A}{3 x^{\frac {3}{2}}}\)	\(30\)
default	\(\frac {2 b B \,x^{\frac {5}{2}}}{5}+2 A b \sqrt {x}+2 B a \sqrt {x}-\frac {2 a A}{3 x^{\frac {3}{2}}}\)	\(30\)
gosper	\(-\frac {2 \left (-3 b B \,x^{4}-15 A b \,x^{2}-15 B a \,x^{2}+5 A a \right )}{15 x^{\frac {3}{2}}}\)	\(32\)
trager	\(-\frac {2 \left (-3 b B \,x^{4}-15 A b \,x^{2}-15 B a \,x^{2}+5 A a \right )}{15 x^{\frac {3}{2}}}\)	\(32\)
risch	\(-\frac {2 \left (-3 b B \,x^{4}-15 A b \,x^{2}-15 B a \,x^{2}+5 A a \right )}{15 x^{\frac {3}{2}}}\)	\(32\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^{5/2}} \, dx=\frac {2 \, {\left (3 \, B b x^{4} + 15 \, {\left (B a + A b\right )} x^{2} - 5 \, A a\right )}}{15 \, x^{\frac {3}{2}}} \]

[In]

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

[Out]

2/15*(3*B*b*x^4 + 15*(B*a + A*b)*x^2 - 5*A*a)/x^(3/2)

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^{5/2}} \, dx=- \frac {2 A a}{3 x^{\frac {3}{2}}} + 2 A b \sqrt {x} + 2 B a \sqrt {x} + \frac {2 B b x^{\frac {5}{2}}}{5} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^{5/2}} \, dx=\frac {2}{5} \, B b x^{\frac {5}{2}} + 2 \, {\left (B a + A b\right )} \sqrt {x} - \frac {2 \, A a}{3 \, x^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^{5/2}} \, dx=\frac {2}{5} \, B b x^{\frac {5}{2}} + 2 \, B a \sqrt {x} + 2 \, A b \sqrt {x} - \frac {2 \, A a}{3 \, x^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^{5/2}} \, dx=\frac {30\,A\,b\,x^2-10\,A\,a+30\,B\,a\,x^2+6\,B\,b\,x^4}{15\,x^{3/2}} \]

[In]

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

[Out]

(30*A*b*x^2 - 10*A*a + 30*B*a*x^2 + 6*B*b*x^4)/(15*x^(3/2))