3.1.40 \(\int \frac {(A+B x^2) (b x^2+c x^4)^3}{x^{17}} \, dx\) [40]

Optimal. Leaf size=49 \[ -\frac {A \left (b+c x^2\right )^4}{10 b x^{10}}-\frac {(5 b B-A c) \left (b+c x^2\right )^4}{40 b^2 x^8} \]

[Out]

-1/10*A*(c*x^2+b)^4/b/x^10-1/40*(-A*c+5*B*b)*(c*x^2+b)^4/b^2/x^8

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Rubi [A]
time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1598, 457, 79, 37} \begin {gather*} -\frac {\left (b+c x^2\right )^4 (5 b B-A c)}{40 b^2 x^8}-\frac {A \left (b+c x^2\right )^4}{10 b x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^17,x]

[Out]

-1/10*(A*(b + c*x^2)^4)/(b*x^10) - ((5*b*B - A*c)*(b + c*x^2)^4)/(40*b^2*x^8)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{17}} \, dx &=\int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^{11}} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {(A+B x) (b+c x)^3}{x^6} \, dx,x,x^2\right )\\ &=-\frac {A \left (b+c x^2\right )^4}{10 b x^{10}}+\frac {(5 b B-A c) \text {Subst}\left (\int \frac {(b+c x)^3}{x^5} \, dx,x,x^2\right )}{10 b}\\ &=-\frac {A \left (b+c x^2\right )^4}{10 b x^{10}}-\frac {(5 b B-A c) \left (b+c x^2\right )^4}{40 b^2 x^8}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 78, normalized size = 1.59 \begin {gather*} -\frac {5 B x^2 \left (b^3+4 b^2 c x^2+6 b c^2 x^4+4 c^3 x^6\right )+A \left (4 b^3+15 b^2 c x^2+20 b c^2 x^4+10 c^3 x^6\right )}{40 x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^17,x]

[Out]

-1/40*(5*B*x^2*(b^3 + 4*b^2*c*x^2 + 6*b*c^2*x^4 + 4*c^3*x^6) + A*(4*b^3 + 15*b^2*c*x^2 + 20*b*c^2*x^4 + 10*c^3
*x^6))/x^10

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Maple [A]
time = 0.41, size = 66, normalized size = 1.35

method	result	size
default	\(-\frac {c^{2} \left (A c +3 B b \right )}{4 x^{4}}-\frac {b c \left (A c +B b \right )}{2 x^{6}}-\frac {B \,c^{3}}{2 x^{2}}-\frac {b^{2} \left (3 A c +B b \right )}{8 x^{8}}-\frac {A \,b^{3}}{10 x^{10}}\)	\(66\)
risch	\(\frac {-\frac {B \,c^{3} x^{8}}{2}+\left (-\frac {1}{4} A \,c^{3}-\frac {3}{4} B b \,c^{2}\right ) x^{6}+\left (-\frac {1}{2} A b \,c^{2}-\frac {1}{2} B \,b^{2} c \right ) x^{4}+\left (-\frac {3}{8} A \,b^{2} c -\frac {1}{8} B \,b^{3}\right ) x^{2}-\frac {A \,b^{3}}{10}}{x^{10}}\)	\(76\)
norman	\(\frac {\left (-\frac {1}{4} A \,c^{3}-\frac {3}{4} B b \,c^{2}\right ) x^{12}+\left (-\frac {1}{2} A b \,c^{2}-\frac {1}{2} B \,b^{2} c \right ) x^{10}+\left (-\frac {3}{8} A \,b^{2} c -\frac {1}{8} B \,b^{3}\right ) x^{8}-\frac {A \,b^{3} x^{6}}{10}-\frac {B \,c^{3} x^{14}}{2}}{x^{16}}\)	\(79\)
gosper	\(-\frac {20 B \,c^{3} x^{8}+10 A \,c^{3} x^{6}+30 x^{6} B b \,c^{2}+20 A b \,c^{2} x^{4}+20 x^{4} B \,b^{2} c +15 A \,b^{2} c \,x^{2}+5 x^{2} B \,b^{3}+4 A \,b^{3}}{40 x^{10}}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^3/x^17,x,method=_RETURNVERBOSE)

[Out]

-1/4*c^2*(A*c+3*B*b)/x^4-1/2*b*c*(A*c+B*b)/x^6-1/2*B*c^3/x^2-1/8*b^2*(3*A*c+B*b)/x^8-1/10*A*b^3/x^10

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Maxima [A]
time = 0.28, size = 75, normalized size = 1.53 \begin {gather*} -\frac {20 \, B c^{3} x^{8} + 10 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 20 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + 4 \, A b^{3} + 5 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{40 \, x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^17,x, algorithm="maxima")

[Out]

-1/40*(20*B*c^3*x^8 + 10*(3*B*b*c^2 + A*c^3)*x^6 + 20*(B*b^2*c + A*b*c^2)*x^4 + 4*A*b^3 + 5*(B*b^3 + 3*A*b^2*c
)*x^2)/x^10

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Fricas [A]
time = 1.46, size = 75, normalized size = 1.53 \begin {gather*} -\frac {20 \, B c^{3} x^{8} + 10 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 20 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + 4 \, A b^{3} + 5 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{40 \, x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^17,x, algorithm="fricas")

[Out]

-1/40*(20*B*c^3*x^8 + 10*(3*B*b*c^2 + A*c^3)*x^6 + 20*(B*b^2*c + A*b*c^2)*x^4 + 4*A*b^3 + 5*(B*b^3 + 3*A*b^2*c
)*x^2)/x^10

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Sympy [A]
time = 3.55, size = 83, normalized size = 1.69 \begin {gather*} \frac {- 4 A b^{3} - 20 B c^{3} x^{8} + x^{6} \left (- 10 A c^{3} - 30 B b c^{2}\right ) + x^{4} \left (- 20 A b c^{2} - 20 B b^{2} c\right ) + x^{2} \left (- 15 A b^{2} c - 5 B b^{3}\right )}{40 x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**17,x)

[Out]

(-4*A*b**3 - 20*B*c**3*x**8 + x**6*(-10*A*c**3 - 30*B*b*c**2) + x**4*(-20*A*b*c**2 - 20*B*b**2*c) + x**2*(-15*
A*b**2*c - 5*B*b**3))/(40*x**10)

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Giac [A]
time = 0.64, size = 79, normalized size = 1.61 \begin {gather*} -\frac {20 \, B c^{3} x^{8} + 30 \, B b c^{2} x^{6} + 10 \, A c^{3} x^{6} + 20 \, B b^{2} c x^{4} + 20 \, A b c^{2} x^{4} + 5 \, B b^{3} x^{2} + 15 \, A b^{2} c x^{2} + 4 \, A b^{3}}{40 \, x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^17,x, algorithm="giac")

[Out]

-1/40*(20*B*c^3*x^8 + 30*B*b*c^2*x^6 + 10*A*c^3*x^6 + 20*B*b^2*c*x^4 + 20*A*b*c^2*x^4 + 5*B*b^3*x^2 + 15*A*b^2
*c*x^2 + 4*A*b^3)/x^10

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Mupad [B]
time = 0.07, size = 76, normalized size = 1.55 \begin {gather*} -\frac {x^4\,\left (\frac {B\,b^2\,c}{2}+\frac {A\,b\,c^2}{2}\right )+\frac {A\,b^3}{10}+x^2\,\left (\frac {B\,b^3}{8}+\frac {3\,A\,c\,b^2}{8}\right )+x^6\,\left (\frac {A\,c^3}{4}+\frac {3\,B\,b\,c^2}{4}\right )+\frac {B\,c^3\,x^8}{2}}{x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^17,x)

[Out]

-(x^4*((A*b*c^2)/2 + (B*b^2*c)/2) + (A*b^3)/10 + x^2*((B*b^3)/8 + (3*A*b^2*c)/8) + x^6*((A*c^3)/4 + (3*B*b*c^2
)/4) + (B*c^3*x^8)/2)/x^10

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