3.1.41 \(\int \frac {(A+B x) (b x+c x^2)^3}{x^{10}} \, dx\) [41]

Optimal. Leaf size=75 \[ -\frac {A b^3}{6 x^6}-\frac {b^2 (b B+3 A c)}{5 x^5}-\frac {3 b c (b B+A c)}{4 x^4}-\frac {c^2 (3 b B+A c)}{3 x^3}-\frac {B c^3}{2 x^2} \]

[Out]

-1/6*A*b^3/x^6-1/5*b^2*(3*A*c+B*b)/x^5-3/4*b*c*(A*c+B*b)/x^4-1/3*c^2*(A*c+3*B*b)/x^3-1/2*B*c^3/x^2

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {779} \begin {gather*} -\frac {A b^3}{6 x^6}-\frac {b^2 (3 A c+b B)}{5 x^5}-\frac {c^2 (A c+3 b B)}{3 x^3}-\frac {3 b c (A c+b B)}{4 x^4}-\frac {B c^3}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^3)/x^10,x]

[Out]

-1/6*(A*b^3)/x^6 - (b^2*(b*B + 3*A*c))/(5*x^5) - (3*b*c*(b*B + A*c))/(4*x^4) - (c^2*(3*b*B + A*c))/(3*x^3) - (
B*c^3)/(2*x^2)

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*} \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x^{10}} \, dx &=\int \left (\frac {A b^3}{x^7}+\frac {b^2 (b B+3 A c)}{x^6}+\frac {3 b c (b B+A c)}{x^5}+\frac {c^2 (3 b B+A c)}{x^4}+\frac {B c^3}{x^3}\right ) \, dx\\ &=-\frac {A b^3}{6 x^6}-\frac {b^2 (b B+3 A c)}{5 x^5}-\frac {3 b c (b B+A c)}{4 x^4}-\frac {c^2 (3 b B+A c)}{3 x^3}-\frac {B c^3}{2 x^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 74, normalized size = 0.99 \begin {gather*} -\frac {3 B x \left (4 b^3+15 b^2 c x+20 b c^2 x^2+10 c^3 x^3\right )+A \left (10 b^3+36 b^2 c x+45 b c^2 x^2+20 c^3 x^3\right )}{60 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^3)/x^10,x]

[Out]

-1/60*(3*B*x*(4*b^3 + 15*b^2*c*x + 20*b*c^2*x^2 + 10*c^3*x^3) + A*(10*b^3 + 36*b^2*c*x + 45*b*c^2*x^2 + 20*c^3
*x^3))/x^6

________________________________________________________________________________________

Maple [A]
time = 0.54, size = 66, normalized size = 0.88

method	result	size
default	\(-\frac {A \,b^{3}}{6 x^{6}}-\frac {b^{2} \left (3 A c +B b \right )}{5 x^{5}}-\frac {3 b c \left (A c +B b \right )}{4 x^{4}}-\frac {c^{2} \left (A c +3 B b \right )}{3 x^{3}}-\frac {B \,c^{3}}{2 x^{2}}\)	\(66\)
risch	\(\frac {-\frac {B \,c^{3} x^{4}}{2}+\left (-\frac {1}{3} A \,c^{3}-B b \,c^{2}\right ) x^{3}+\left (-\frac {3}{4} A b \,c^{2}-\frac {3}{4} B \,b^{2} c \right ) x^{2}+\left (-\frac {3}{5} A \,b^{2} c -\frac {1}{5} B \,b^{3}\right ) x -\frac {A \,b^{3}}{6}}{x^{6}}\)	\(74\)
gosper	\(-\frac {30 B \,c^{3} x^{4}+20 A \,c^{3} x^{3}+60 B b \,c^{2} x^{3}+45 A b \,c^{2} x^{2}+45 B \,b^{2} c \,x^{2}+36 A \,b^{2} c x +12 B \,b^{3} x +10 A \,b^{3}}{60 x^{6}}\)	\(76\)
norman	\(\frac {\left (-\frac {1}{3} A \,c^{3}-B b \,c^{2}\right ) x^{6}+\left (-\frac {3}{4} A b \,c^{2}-\frac {3}{4} B \,b^{2} c \right ) x^{5}+\left (-\frac {3}{5} A \,b^{2} c -\frac {1}{5} B \,b^{3}\right ) x^{4}-\frac {A \,b^{3} x^{3}}{6}-\frac {B \,c^{3} x^{7}}{2}}{x^{9}}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^3/x^10,x,method=_RETURNVERBOSE)

[Out]

-1/6*A*b^3/x^6-1/5*b^2*(3*A*c+B*b)/x^5-3/4*b*c*(A*c+B*b)/x^4-1/3*c^2*(A*c+3*B*b)/x^3-1/2*B*c^3/x^2

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 73, normalized size = 0.97 \begin {gather*} -\frac {30 \, B c^{3} x^{4} + 10 \, A b^{3} + 20 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{3} + 45 \, {\left (B b^{2} c + A b c^{2}\right )} x^{2} + 12 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x}{60 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^10,x, algorithm="maxima")

[Out]

-1/60*(30*B*c^3*x^4 + 10*A*b^3 + 20*(3*B*b*c^2 + A*c^3)*x^3 + 45*(B*b^2*c + A*b*c^2)*x^2 + 12*(B*b^3 + 3*A*b^2
*c)*x)/x^6

________________________________________________________________________________________

Fricas [A]
time = 0.72, size = 73, normalized size = 0.97 \begin {gather*} -\frac {30 \, B c^{3} x^{4} + 10 \, A b^{3} + 20 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{3} + 45 \, {\left (B b^{2} c + A b c^{2}\right )} x^{2} + 12 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x}{60 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^10,x, algorithm="fricas")

[Out]

-1/60*(30*B*c^3*x^4 + 10*A*b^3 + 20*(3*B*b*c^2 + A*c^3)*x^3 + 45*(B*b^2*c + A*b*c^2)*x^2 + 12*(B*b^3 + 3*A*b^2
*c)*x)/x^6

________________________________________________________________________________________

Sympy [A]
time = 0.97, size = 82, normalized size = 1.09 \begin {gather*} \frac {- 10 A b^{3} - 30 B c^{3} x^{4} + x^{3} \left (- 20 A c^{3} - 60 B b c^{2}\right ) + x^{2} \left (- 45 A b c^{2} - 45 B b^{2} c\right ) + x \left (- 36 A b^{2} c - 12 B b^{3}\right )}{60 x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**3/x**10,x)

[Out]

(-10*A*b**3 - 30*B*c**3*x**4 + x**3*(-20*A*c**3 - 60*B*b*c**2) + x**2*(-45*A*b*c**2 - 45*B*b**2*c) + x*(-36*A*
b**2*c - 12*B*b**3))/(60*x**6)

________________________________________________________________________________________

Giac [A]
time = 0.81, size = 75, normalized size = 1.00 \begin {gather*} -\frac {30 \, B c^{3} x^{4} + 60 \, B b c^{2} x^{3} + 20 \, A c^{3} x^{3} + 45 \, B b^{2} c x^{2} + 45 \, A b c^{2} x^{2} + 12 \, B b^{3} x + 36 \, A b^{2} c x + 10 \, A b^{3}}{60 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^10,x, algorithm="giac")

[Out]

-1/60*(30*B*c^3*x^4 + 60*B*b*c^2*x^3 + 20*A*c^3*x^3 + 45*B*b^2*c*x^2 + 45*A*b*c^2*x^2 + 12*B*b^3*x + 36*A*b^2*
c*x + 10*A*b^3)/x^6

________________________________________________________________________________________

Mupad [B]
time = 0.04, size = 73, normalized size = 0.97 \begin {gather*} -\frac {x^2\,\left (\frac {3\,B\,b^2\,c}{4}+\frac {3\,A\,b\,c^2}{4}\right )+x\,\left (\frac {B\,b^3}{5}+\frac {3\,A\,c\,b^2}{5}\right )+\frac {A\,b^3}{6}+x^3\,\left (\frac {A\,c^3}{3}+B\,b\,c^2\right )+\frac {B\,c^3\,x^4}{2}}{x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^3*(A + B*x))/x^10,x)

[Out]

-(x^2*((3*A*b*c^2)/4 + (3*B*b^2*c)/4) + x*((B*b^3)/5 + (3*A*b^2*c)/5) + (A*b^3)/6 + x^3*((A*c^3)/3 + B*b*c^2)
+ (B*c^3*x^4)/2)/x^6

________________________________________________________________________________________