\(\int \frac {(A+B x) (b x+c x^2)}{x^{9/2}} \, dx\) [148]

Optimal result

Integrand size = 20, antiderivative size = 37 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{x^{9/2}} \, dx=-\frac {2 A b}{5 x^{5/2}}-\frac {2 (b B+A c)}{3 x^{3/2}}-\frac {2 B c}{\sqrt {x}} \]

[Out]

-2/5*A*b/x^(5/2)-2/3*(A*c+B*b)/x^(3/2)-2*B*c/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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {779} \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{x^{9/2}} \, dx=-\frac {2 (A c+b B)}{3 x^{3/2}}-\frac {2 A b}{5 x^{5/2}}-\frac {2 B c}{\sqrt {x}} \]

[In]

Int[((A + B*x)*(b*x + c*x^2))/x^(9/2),x]

[Out]

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

Rule 779

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{x^{9/2}} \, dx=-\frac {2 \left (3 A b+5 b B x+5 A c x+15 B c x^2\right )}{15 x^{5/2}} \]

[In]

Integrate[((A + B*x)*(b*x + c*x^2))/x^(9/2),x]

[Out]

(-2*(3*A*b + 5*b*B*x + 5*A*c*x + 15*B*c*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	\(-\frac {2 \left (15 B c \,x^{2}+5 A c x +5 B b x +3 A b \right )}{15 x^{\frac {5}{2}}}\)	\(28\)
derivativedivides	\(-\frac {2 A b}{5 x^{\frac {5}{2}}}-\frac {2 \left (A c +B b \right )}{3 x^{\frac {3}{2}}}-\frac {2 B c}{\sqrt {x}}\)	\(28\)
default	\(-\frac {2 A b}{5 x^{\frac {5}{2}}}-\frac {2 \left (A c +B b \right )}{3 x^{\frac {3}{2}}}-\frac {2 B c}{\sqrt {x}}\)	\(28\)
trager	\(-\frac {2 \left (15 B c \,x^{2}+5 A c x +5 B b x +3 A b \right )}{15 x^{\frac {5}{2}}}\)	\(28\)
risch	\(-\frac {2 \left (15 B c \,x^{2}+5 A c x +5 B b x +3 A b \right )}{15 x^{\frac {5}{2}}}\)	\(28\)

[In]

int((B*x+A)*(c*x^2+b*x)/x^(9/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{x^{9/2}} \, dx=-\frac {2 \, {\left (15 \, B c x^{2} + 3 \, A b + 5 \, {\left (B b + A c\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x)/x^(9/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.24 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{x^{9/2}} \, dx=- \frac {2 A b}{5 x^{\frac {5}{2}}} - \frac {2 A c}{3 x^{\frac {3}{2}}} - \frac {2 B b}{3 x^{\frac {3}{2}}} - \frac {2 B c}{\sqrt {x}} \]

[In]

integrate((B*x+A)*(c*x**2+b*x)/x**(9/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{x^{9/2}} \, dx=-\frac {2 \, {\left (15 \, B c x^{2} + 3 \, A b + 5 \, {\left (B b + A c\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x)/x^(9/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{x^{9/2}} \, dx=-\frac {2 \, {\left (15 \, B c x^{2} + 5 \, B b x + 5 \, A c x + 3 \, A b\right )}}{15 \, x^{\frac {5}{2}}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x)/x^(9/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{x^{9/2}} \, dx=-\frac {2\,B\,c\,x^2+\left (\frac {2\,A\,c}{3}+\frac {2\,B\,b}{3}\right )\,x+\frac {2\,A\,b}{5}}{x^{5/2}} \]

[In]

int(((b*x + c*x^2)*(A + B*x))/x^(9/2),x)

[Out]

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