3.1.73 \(\int \frac {x^{14} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=140 \[ -\frac {7 b^{3/2} (9 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 c^{11/2}}-\frac {b^3 x (b B-A c)}{4 c^5 \left (b+c x^2\right )^2}+\frac {b^2 x (17 b B-13 A c)}{8 c^5 \left (b+c x^2\right )}+\frac {3 b x (2 b B-A c)}{c^5}-\frac {x^3 (3 b B-A c)}{3 c^4}+\frac {B x^5}{5 c^3} \]

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Rubi [A]  time = 0.23, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1584, 455, 1814, 1810, 205} \begin {gather*} \frac {b^2 x (17 b B-13 A c)}{8 c^5 \left (b+c x^2\right )}-\frac {b^3 x (b B-A c)}{4 c^5 \left (b+c x^2\right )^2}-\frac {7 b^{3/2} (9 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 c^{11/2}}-\frac {x^3 (3 b B-A c)}{3 c^4}+\frac {3 b x (2 b B-A c)}{c^5}+\frac {B x^5}{5 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^14*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(3*b*(2*b*B - A*c)*x)/c^5 - ((3*b*B - A*c)*x^3)/(3*c^4) + (B*x^5)/(5*c^3) - (b^3*(b*B - A*c)*x)/(4*c^5*(b + c*
x^2)^2) + (b^2*(17*b*B - 13*A*c)*x)/(8*c^5*(b + c*x^2)) - (7*b^(3/2)*(9*b*B - 5*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b
]])/(8*c^(11/2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1584

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]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^{14} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {x^8 \left (A+B x^2\right )}{\left (b+c x^2\right )^3} \, dx\\ &=-\frac {b^3 (b B-A c) x}{4 c^5 \left (b+c x^2\right )^2}-\frac {\int \frac {-b^3 (b B-A c)+4 b^2 c (b B-A c) x^2-4 b c^2 (b B-A c) x^4+4 c^3 (b B-A c) x^6-4 B c^4 x^8}{\left (b+c x^2\right )^2} \, dx}{4 c^5}\\ &=-\frac {b^3 (b B-A c) x}{4 c^5 \left (b+c x^2\right )^2}+\frac {b^2 (17 b B-13 A c) x}{8 c^5 \left (b+c x^2\right )}+\frac {\int \frac {-b^3 (15 b B-11 A c)+8 b^2 c (3 b B-2 A c) x^2-8 b c^2 (2 b B-A c) x^4+8 b B c^3 x^6}{b+c x^2} \, dx}{8 b c^5}\\ &=-\frac {b^3 (b B-A c) x}{4 c^5 \left (b+c x^2\right )^2}+\frac {b^2 (17 b B-13 A c) x}{8 c^5 \left (b+c x^2\right )}+\frac {\int \left (24 b^2 (2 b B-A c)-8 b c (3 b B-A c) x^2+8 b B c^2 x^4-\frac {7 \left (9 b^4 B-5 A b^3 c\right )}{b+c x^2}\right ) \, dx}{8 b c^5}\\ &=\frac {3 b (2 b B-A c) x}{c^5}-\frac {(3 b B-A c) x^3}{3 c^4}+\frac {B x^5}{5 c^3}-\frac {b^3 (b B-A c) x}{4 c^5 \left (b+c x^2\right )^2}+\frac {b^2 (17 b B-13 A c) x}{8 c^5 \left (b+c x^2\right )}-\frac {\left (7 b^2 (9 b B-5 A c)\right ) \int \frac {1}{b+c x^2} \, dx}{8 c^5}\\ &=\frac {3 b (2 b B-A c) x}{c^5}-\frac {(3 b B-A c) x^3}{3 c^4}+\frac {B x^5}{5 c^3}-\frac {b^3 (b B-A c) x}{4 c^5 \left (b+c x^2\right )^2}+\frac {b^2 (17 b B-13 A c) x}{8 c^5 \left (b+c x^2\right )}-\frac {7 b^{3/2} (9 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 c^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 133, normalized size = 0.95 \begin {gather*} \frac {x \left (-525 b^3 c \left (A-3 B x^2\right )+7 b^2 c^2 x^2 \left (72 B x^2-125 A\right )-8 b c^3 x^4 \left (35 A+9 B x^2\right )+8 c^4 x^6 \left (5 A+3 B x^2\right )+945 b^4 B\right )}{120 c^5 \left (b+c x^2\right )^2}-\frac {7 b^{3/2} (9 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 c^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^14*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(x*(945*b^4*B - 525*b^3*c*(A - 3*B*x^2) + 8*c^4*x^6*(5*A + 3*B*x^2) - 8*b*c^3*x^4*(35*A + 9*B*x^2) + 7*b^2*c^2
*x^2*(-125*A + 72*B*x^2)))/(120*c^5*(b + c*x^2)^2) - (7*b^(3/2)*(9*b*B - 5*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(
8*c^(11/2))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{14} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^14*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

IntegrateAlgebraic[(x^14*(A + B*x^2))/(b*x^2 + c*x^4)^3, x]

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fricas [A]  time = 0.40, size = 416, normalized size = 2.97 \begin {gather*} \left [\frac {48 \, B c^{4} x^{9} - 16 \, {\left (9 \, B b c^{3} - 5 \, A c^{4}\right )} x^{7} + 112 \, {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{5} + 350 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{3} - 105 \, {\left (9 \, B b^{4} - 5 \, A b^{3} c + {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{4} + 2 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x^{2} + 2 \, c x \sqrt {-\frac {b}{c}} - b}{c x^{2} + b}\right ) + 210 \, {\left (9 \, B b^{4} - 5 \, A b^{3} c\right )} x}{240 \, {\left (c^{7} x^{4} + 2 \, b c^{6} x^{2} + b^{2} c^{5}\right )}}, \frac {24 \, B c^{4} x^{9} - 8 \, {\left (9 \, B b c^{3} - 5 \, A c^{4}\right )} x^{7} + 56 \, {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{5} + 175 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{3} - 105 \, {\left (9 \, B b^{4} - 5 \, A b^{3} c + {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{4} + 2 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right ) + 105 \, {\left (9 \, B b^{4} - 5 \, A b^{3} c\right )} x}{120 \, {\left (c^{7} x^{4} + 2 \, b c^{6} x^{2} + b^{2} c^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

[1/240*(48*B*c^4*x^9 - 16*(9*B*b*c^3 - 5*A*c^4)*x^7 + 112*(9*B*b^2*c^2 - 5*A*b*c^3)*x^5 + 350*(9*B*b^3*c - 5*A
*b^2*c^2)*x^3 - 105*(9*B*b^4 - 5*A*b^3*c + (9*B*b^2*c^2 - 5*A*b*c^3)*x^4 + 2*(9*B*b^3*c - 5*A*b^2*c^2)*x^2)*sq
rt(-b/c)*log((c*x^2 + 2*c*x*sqrt(-b/c) - b)/(c*x^2 + b)) + 210*(9*B*b^4 - 5*A*b^3*c)*x)/(c^7*x^4 + 2*b*c^6*x^2
+ b^2*c^5), 1/120*(24*B*c^4*x^9 - 8*(9*B*b*c^3 - 5*A*c^4)*x^7 + 56*(9*B*b^2*c^2 - 5*A*b*c^3)*x^5 + 175*(9*B*b
^3*c - 5*A*b^2*c^2)*x^3 - 105*(9*B*b^4 - 5*A*b^3*c + (9*B*b^2*c^2 - 5*A*b*c^3)*x^4 + 2*(9*B*b^3*c - 5*A*b^2*c^
2)*x^2)*sqrt(b/c)*arctan(c*x*sqrt(b/c)/b) + 105*(9*B*b^4 - 5*A*b^3*c)*x)/(c^7*x^4 + 2*b*c^6*x^2 + b^2*c^5)]

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giac [A]  time = 0.18, size = 138, normalized size = 0.99 \begin {gather*} -\frac {7 \, {\left (9 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{5}} + \frac {17 \, B b^{3} c x^{3} - 13 \, A b^{2} c^{2} x^{3} + 15 \, B b^{4} x - 11 \, A b^{3} c x}{8 \, {\left (c x^{2} + b\right )}^{2} c^{5}} + \frac {3 \, B c^{12} x^{5} - 15 \, B b c^{11} x^{3} + 5 \, A c^{12} x^{3} + 90 \, B b^{2} c^{10} x - 45 \, A b c^{11} x}{15 \, c^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

-7/8*(9*B*b^3 - 5*A*b^2*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^5) + 1/8*(17*B*b^3*c*x^3 - 13*A*b^2*c^2*x^3 + 15
*B*b^4*x - 11*A*b^3*c*x)/((c*x^2 + b)^2*c^5) + 1/15*(3*B*c^12*x^5 - 15*B*b*c^11*x^3 + 5*A*c^12*x^3 + 90*B*b^2*
c^10*x - 45*A*b*c^11*x)/c^15

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maple [A]  time = 0.06, size = 174, normalized size = 1.24 \begin {gather*} -\frac {13 A \,b^{2} x^{3}}{8 \left (c \,x^{2}+b \right )^{2} c^{3}}+\frac {17 B \,b^{3} x^{3}}{8 \left (c \,x^{2}+b \right )^{2} c^{4}}+\frac {B \,x^{5}}{5 c^{3}}-\frac {11 A \,b^{3} x}{8 \left (c \,x^{2}+b \right )^{2} c^{4}}+\frac {A \,x^{3}}{3 c^{3}}+\frac {15 B \,b^{4} x}{8 \left (c \,x^{2}+b \right )^{2} c^{5}}-\frac {B b \,x^{3}}{c^{4}}+\frac {35 A \,b^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, c^{4}}-\frac {63 B \,b^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, c^{5}}-\frac {3 A b x}{c^{4}}+\frac {6 B \,b^{2} x}{c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

1/5*B*x^5/c^3+1/3/c^3*A*x^3-1/c^4*B*x^3*b-3/c^4*A*b*x+6/c^5*B*b^2*x-13/8*b^2/c^3/(c*x^2+b)^2*A*x^3+17/8*b^3/c^
4/(c*x^2+b)^2*B*x^3-11/8*b^3/c^4/(c*x^2+b)^2*A*x+15/8*b^4/c^5/(c*x^2+b)^2*B*x+35/8*b^2/c^4/(b*c)^(1/2)*arctan(
1/(b*c)^(1/2)*c*x)*A-63/8*b^3/c^5/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x)*B

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maxima [A]  time = 2.90, size = 147, normalized size = 1.05 \begin {gather*} \frac {{\left (17 \, B b^{3} c - 13 \, A b^{2} c^{2}\right )} x^{3} + {\left (15 \, B b^{4} - 11 \, A b^{3} c\right )} x}{8 \, {\left (c^{7} x^{4} + 2 \, b c^{6} x^{2} + b^{2} c^{5}\right )}} - \frac {7 \, {\left (9 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{5}} + \frac {3 \, B c^{2} x^{5} - 5 \, {\left (3 \, B b c - A c^{2}\right )} x^{3} + 45 \, {\left (2 \, B b^{2} - A b c\right )} x}{15 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

1/8*((17*B*b^3*c - 13*A*b^2*c^2)*x^3 + (15*B*b^4 - 11*A*b^3*c)*x)/(c^7*x^4 + 2*b*c^6*x^2 + b^2*c^5) - 7/8*(9*B
*b^3 - 5*A*b^2*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^5) + 1/15*(3*B*c^2*x^5 - 5*(3*B*b*c - A*c^2)*x^3 + 45*(2*
B*b^2 - A*b*c)*x)/c^5

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mupad [B]  time = 0.12, size = 177, normalized size = 1.26 \begin {gather*} \frac {x\,\left (\frac {15\,B\,b^4}{8}-\frac {11\,A\,b^3\,c}{8}\right )-x^3\,\left (\frac {13\,A\,b^2\,c^2}{8}-\frac {17\,B\,b^3\,c}{8}\right )}{b^2\,c^5+2\,b\,c^6\,x^2+c^7\,x^4}-x\,\left (\frac {3\,b\,\left (\frac {A}{c^3}-\frac {3\,B\,b}{c^4}\right )}{c}+\frac {3\,B\,b^2}{c^5}\right )+x^3\,\left (\frac {A}{3\,c^3}-\frac {B\,b}{c^4}\right )+\frac {B\,x^5}{5\,c^3}-\frac {7\,b^{3/2}\,\mathrm {atan}\left (\frac {b^{3/2}\,\sqrt {c}\,x\,\left (5\,A\,c-9\,B\,b\right )}{9\,B\,b^3-5\,A\,b^2\,c}\right )\,\left (5\,A\,c-9\,B\,b\right )}{8\,c^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^14*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)

[Out]

(x*((15*B*b^4)/8 - (11*A*b^3*c)/8) - x^3*((13*A*b^2*c^2)/8 - (17*B*b^3*c)/8))/(b^2*c^5 + c^7*x^4 + 2*b*c^6*x^2
) - x*((3*b*(A/c^3 - (3*B*b)/c^4))/c + (3*B*b^2)/c^5) + x^3*(A/(3*c^3) - (B*b)/c^4) + (B*x^5)/(5*c^3) - (7*b^(
3/2)*atan((b^(3/2)*c^(1/2)*x*(5*A*c - 9*B*b))/(9*B*b^3 - 5*A*b^2*c))*(5*A*c - 9*B*b))/(8*c^(11/2))

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sympy [A]  time = 1.40, size = 252, normalized size = 1.80 \begin {gather*} \frac {B x^{5}}{5 c^{3}} + x^{3} \left (\frac {A}{3 c^{3}} - \frac {B b}{c^{4}}\right ) + x \left (- \frac {3 A b}{c^{4}} + \frac {6 B b^{2}}{c^{5}}\right ) + \frac {7 \sqrt {- \frac {b^{3}}{c^{11}}} \left (- 5 A c + 9 B b\right ) \log {\left (- \frac {7 c^{5} \sqrt {- \frac {b^{3}}{c^{11}}} \left (- 5 A c + 9 B b\right )}{- 35 A b c + 63 B b^{2}} + x \right )}}{16} - \frac {7 \sqrt {- \frac {b^{3}}{c^{11}}} \left (- 5 A c + 9 B b\right ) \log {\left (\frac {7 c^{5} \sqrt {- \frac {b^{3}}{c^{11}}} \left (- 5 A c + 9 B b\right )}{- 35 A b c + 63 B b^{2}} + x \right )}}{16} + \frac {x^{3} \left (- 13 A b^{2} c^{2} + 17 B b^{3} c\right ) + x \left (- 11 A b^{3} c + 15 B b^{4}\right )}{8 b^{2} c^{5} + 16 b c^{6} x^{2} + 8 c^{7} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**14*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

B*x**5/(5*c**3) + x**3*(A/(3*c**3) - B*b/c**4) + x*(-3*A*b/c**4 + 6*B*b**2/c**5) + 7*sqrt(-b**3/c**11)*(-5*A*c
+ 9*B*b)*log(-7*c**5*sqrt(-b**3/c**11)*(-5*A*c + 9*B*b)/(-35*A*b*c + 63*B*b**2) + x)/16 - 7*sqrt(-b**3/c**11)
*(-5*A*c + 9*B*b)*log(7*c**5*sqrt(-b**3/c**11)*(-5*A*c + 9*B*b)/(-35*A*b*c + 63*B*b**2) + x)/16 + (x**3*(-13*A
*b**2*c**2 + 17*B*b**3*c) + x*(-11*A*b**3*c + 15*B*b**4))/(8*b**2*c**5 + 16*b*c**6*x**2 + 8*c**7*x**4)

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