3.1.10 \(\int \frac {(A+B x+C x^2) (a+b x^2+c x^4)}{x^7} \, dx\) [10]

Optimal. Leaf size=68 \[ -\frac {a A}{6 x^6}-\frac {a B}{5 x^5}-\frac {A b+a C}{4 x^4}-\frac {b B}{3 x^3}-\frac {A c+b C}{2 x^2}-\frac {B c}{x}+c C \log (x) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {1642} \begin {gather*} -\frac {a C+A b}{4 x^4}-\frac {a A}{6 x^6}-\frac {a B}{5 x^5}-\frac {A c+b C}{2 x^2}-\frac {b B}{3 x^3}-\frac {B c}{x}+c C \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x + C*x^2)*(a + b*x^2 + c*x^4))/x^7,x]

[Out]

-1/6*(a*A)/x^6 - (a*B)/(5*x^5) - (A*b + a*C)/(4*x^4) - (b*B)/(3*x^3) - (A*c + b*C)/(2*x^2) - (B*c)/x + c*C*Log
[x]

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 68, normalized size = 1.00 \begin {gather*} -\frac {a (10 A+3 x (4 B+5 C x))+5 x^2 \left (3 A \left (b+2 c x^2\right )+2 x \left (2 b B+3 b C x+6 B c x^2\right )\right )}{60 x^6}+c C \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x + C*x^2)*(a + b*x^2 + c*x^4))/x^7,x]

[Out]

-1/60*(a*(10*A + 3*x*(4*B + 5*C*x)) + 5*x^2*(3*A*(b + 2*c*x^2) + 2*x*(2*b*B + 3*b*C*x + 6*B*c*x^2)))/x^6 + c*C
*Log[x]

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Maple [A]
time = 0.01, size = 59, normalized size = 0.87

method result size
default \(-\frac {a B}{5 x^{5}}-\frac {A b +a C}{4 x^{4}}-\frac {A c +b C}{2 x^{2}}-\frac {B c}{x}-\frac {b B}{3 x^{3}}+c C \ln \left (x \right )-\frac {a A}{6 x^{6}}\) \(59\)
norman \(\frac {\left (-\frac {A b}{4}-\frac {a C}{4}\right ) x^{2}+\left (-\frac {A c}{2}-\frac {b C}{2}\right ) x^{4}-\frac {a A}{6}-\frac {B b \,x^{3}}{3}-B c \,x^{5}-\frac {a B x}{5}}{x^{6}}+c C \ln \left (x \right )\) \(61\)
risch \(\frac {\left (-\frac {A b}{4}-\frac {a C}{4}\right ) x^{2}+\left (-\frac {A c}{2}-\frac {b C}{2}\right ) x^{4}-\frac {a A}{6}-\frac {B b \,x^{3}}{3}-B c \,x^{5}-\frac {a B x}{5}}{x^{6}}+c C \ln \left (x \right )\) \(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)*(c*x^4+b*x^2+a)/x^7,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 59, normalized size = 0.87 \begin {gather*} C c \log \left (x\right ) - \frac {60 \, B c x^{5} + 20 \, B b x^{3} + 30 \, {\left (C b + A c\right )} x^{4} + 12 \, B a x + 15 \, {\left (C a + A b\right )} x^{2} + 10 \, A a}{60 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(c*x^4+b*x^2+a)/x^7,x, algorithm="maxima")

[Out]

C*c*log(x) - 1/60*(60*B*c*x^5 + 20*B*b*x^3 + 30*(C*b + A*c)*x^4 + 12*B*a*x + 15*(C*a + A*b)*x^2 + 10*A*a)/x^6

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Fricas [A]
time = 0.38, size = 62, normalized size = 0.91 \begin {gather*} \frac {60 \, C c x^{6} \log \left (x\right ) - 60 \, B c x^{5} - 20 \, B b x^{3} - 30 \, {\left (C b + A c\right )} x^{4} - 12 \, B a x - 15 \, {\left (C a + A b\right )} x^{2} - 10 \, A a}{60 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(c*x^4+b*x^2+a)/x^7,x, algorithm="fricas")

[Out]

1/60*(60*C*c*x^6*log(x) - 60*B*c*x^5 - 20*B*b*x^3 - 30*(C*b + A*c)*x^4 - 12*B*a*x - 15*(C*a + A*b)*x^2 - 10*A*
a)/x^6

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Sympy [A]
time = 9.68, size = 70, normalized size = 1.03 \begin {gather*} C c \log {\left (x \right )} + \frac {- 10 A a - 12 B a x - 20 B b x^{3} - 60 B c x^{5} + x^{4} \left (- 30 A c - 30 C b\right ) + x^{2} \left (- 15 A b - 15 C a\right )}{60 x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)*(c*x**4+b*x**2+a)/x**7,x)

[Out]

C*c*log(x) + (-10*A*a - 12*B*a*x - 20*B*b*x**3 - 60*B*c*x**5 + x**4*(-30*A*c - 30*C*b) + x**2*(-15*A*b - 15*C*
a))/(60*x**6)

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Giac [A]
time = 4.20, size = 60, normalized size = 0.88 \begin {gather*} C c \log \left ({\left | x \right |}\right ) - \frac {60 \, B c x^{5} + 20 \, B b x^{3} + 30 \, {\left (C b + A c\right )} x^{4} + 12 \, B a x + 15 \, {\left (C a + A b\right )} x^{2} + 10 \, A a}{60 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(c*x^4+b*x^2+a)/x^7,x, algorithm="giac")

[Out]

C*c*log(abs(x)) - 1/60*(60*B*c*x^5 + 20*B*b*x^3 + 30*(C*b + A*c)*x^4 + 12*B*a*x + 15*(C*a + A*b)*x^2 + 10*A*a)
/x^6

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Mupad [B]
time = 0.79, size = 60, normalized size = 0.88 \begin {gather*} C\,c\,\ln \left (x\right )-\frac {B\,c\,x^5+\left (\frac {A\,c}{2}+\frac {C\,b}{2}\right )\,x^4+\frac {B\,b\,x^3}{3}+\left (\frac {A\,b}{4}+\frac {C\,a}{4}\right )\,x^2+\frac {B\,a\,x}{5}+\frac {A\,a}{6}}{x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x + C*x^2)*(a + b*x^2 + c*x^4))/x^7,x)

[Out]

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

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