3.1.25 \(\int \frac {(A+B x) (b x+c x^2)^2}{x^7} \, dx\)

Optimal. Leaf size=53 \[ -\frac {A b^2}{4 x^4}-\frac {b (2 A c+b B)}{3 x^3}-\frac {c (A c+2 b B)}{2 x^2}-\frac {B c^2}{x} \]

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Rubi [A]  time = 0.03, antiderivative size = 53, 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 = {765} \begin {gather*} -\frac {A b^2}{4 x^4}-\frac {b (2 A c+b B)}{3 x^3}-\frac {c (A c+2 b B)}{2 x^2}-\frac {B c^2}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^2)/x^7,x]

[Out]

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

Rule 765

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

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Mathematica [A]  time = 0.01, size = 50, normalized size = 0.94 \begin {gather*} -\frac {A \left (3 b^2+8 b c x+6 c^2 x^2\right )+4 B x \left (b^2+3 b c x+3 c^2 x^2\right )}{12 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(4*B*x*(b^2 + 3*b*c*x + 3*c^2*x^2) + A*(3*b^2 + 8*b*c*x + 6*c^2*x^2))/x^4

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) \left (b x+c x^2\right )^2}{x^7} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^2)/x^7,x]

[Out]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^2)/x^7, x]

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fricas [A]  time = 0.39, size = 51, normalized size = 0.96 \begin {gather*} -\frac {12 \, B c^{2} x^{3} + 3 \, A b^{2} + 6 \, {\left (2 \, B b c + A c^{2}\right )} x^{2} + 4 \, {\left (B b^{2} + 2 \, A b c\right )} x}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*B*c^2*x^3 + 3*A*b^2 + 6*(2*B*b*c + A*c^2)*x^2 + 4*(B*b^2 + 2*A*b*c)*x)/x^4

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giac [A]  time = 0.15, size = 51, normalized size = 0.96 \begin {gather*} -\frac {12 \, B c^{2} x^{3} + 12 \, B b c x^{2} + 6 \, A c^{2} x^{2} + 4 \, B b^{2} x + 8 \, A b c x + 3 \, A b^{2}}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*B*c^2*x^3 + 12*B*b*c*x^2 + 6*A*c^2*x^2 + 4*B*b^2*x + 8*A*b*c*x + 3*A*b^2)/x^4

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maple [A]  time = 0.05, size = 48, normalized size = 0.91 \begin {gather*} -\frac {B \,c^{2}}{x}-\frac {A \,b^{2}}{4 x^{4}}-\frac {\left (A c +2 b B \right ) c}{2 x^{2}}-\frac {\left (2 A c +b B \right ) b}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^2/x^7,x)

[Out]

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

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maxima [A]  time = 0.92, size = 51, normalized size = 0.96 \begin {gather*} -\frac {12 \, B c^{2} x^{3} + 3 \, A b^{2} + 6 \, {\left (2 \, B b c + A c^{2}\right )} x^{2} + 4 \, {\left (B b^{2} + 2 \, A b c\right )} x}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*B*c^2*x^3 + 3*A*b^2 + 6*(2*B*b*c + A*c^2)*x^2 + 4*(B*b^2 + 2*A*b*c)*x)/x^4

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mupad [B]  time = 0.04, size = 49, normalized size = 0.92 \begin {gather*} -\frac {x^2\,\left (\frac {A\,c^2}{2}+B\,b\,c\right )+\frac {A\,b^2}{4}+x\,\left (\frac {B\,b^2}{3}+\frac {2\,A\,c\,b}{3}\right )+B\,c^2\,x^3}{x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^2*(A + B*x))/x^7,x)

[Out]

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

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sympy [A]  time = 0.68, size = 56, normalized size = 1.06 \begin {gather*} \frac {- 3 A b^{2} - 12 B c^{2} x^{3} + x^{2} \left (- 6 A c^{2} - 12 B b c\right ) + x \left (- 8 A b c - 4 B b^{2}\right )}{12 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*A*b**2 - 12*B*c**2*x**3 + x**2*(-6*A*c**2 - 12*B*b*c) + x*(-8*A*b*c - 4*B*b**2))/(12*x**4)

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