3.102 \(\int \frac {x^5 (A+B x^2)}{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=133 \[ \frac {\left (-a B c-A b c+b^2 B\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}+\frac {\left (2 a A c^2-3 a b B c-A b^2 c+b^3 B\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^3 \sqrt {b^2-4 a c}}-\frac {x^2 (b B-A c)}{2 c^2}+\frac {B x^4}{4 c} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1251, 800, 634, 618, 206, 628} \[ \frac {\left (-a B c-A b c+b^2 B\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}+\frac {\left (2 a A c^2-3 a b B c-A b^2 c+b^3 B\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^3 \sqrt {b^2-4 a c}}-\frac {x^2 (b B-A c)}{2 c^2}+\frac {B x^4}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(A + B*x^2))/(a + b*x^2 + c*x^4),x]

[Out]

-((b*B - A*c)*x^2)/(2*c^2) + (B*x^4)/(4*c) + ((b^3*B - A*b^2*c - 3*a*b*B*c + 2*a*A*c^2)*ArcTanh[(b + 2*c*x^2)/
Sqrt[b^2 - 4*a*c]])/(2*c^3*Sqrt[b^2 - 4*a*c]) + ((b^2*B - A*b*c - a*B*c)*Log[a + b*x^2 + c*x^4])/(4*c^3)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {x^5 \left (A+B x^2\right )}{a+b x^2+c x^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2 (A+B x)}{a+b x+c x^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {b B-A c}{c^2}+\frac {B x}{c}+\frac {a (b B-A c)+\left (b^2 B-A b c-a B c\right ) x}{c^2 \left (a+b x+c x^2\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac {(b B-A c) x^2}{2 c^2}+\frac {B x^4}{4 c}+\frac {\operatorname {Subst}\left (\int \frac {a (b B-A c)+\left (b^2 B-A b c-a B c\right ) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c^2}\\ &=-\frac {(b B-A c) x^2}{2 c^2}+\frac {B x^4}{4 c}+\frac {\left (b^2 B-A b c-a B c\right ) \operatorname {Subst}\left (\int \frac {b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3}-\frac {\left (b^3 B-A b^2 c-3 a b B c+2 a A c^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3}\\ &=-\frac {(b B-A c) x^2}{2 c^2}+\frac {B x^4}{4 c}+\frac {\left (b^2 B-A b c-a B c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}+\frac {\left (b^3 B-A b^2 c-3 a b B c+2 a A c^2\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 c^3}\\ &=-\frac {(b B-A c) x^2}{2 c^2}+\frac {B x^4}{4 c}+\frac {\left (b^3 B-A b^2 c-3 a b B c+2 a A c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^3 \sqrt {b^2-4 a c}}+\frac {\left (b^2 B-A b c-a B c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 126, normalized size = 0.95 \[ \frac {\left (-a B c-A b c+b^2 B\right ) \log \left (a+b x^2+c x^4\right )+\frac {2 \left (-2 a A c^2+3 a b B c+A b^2 c+b^3 (-B)\right ) \tan ^{-1}\left (\frac {b+2 c x^2}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+2 c x^2 (A c-b B)+B c^2 x^4}{4 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(A + B*x^2))/(a + b*x^2 + c*x^4),x]

[Out]

(2*c*(-(b*B) + A*c)*x^2 + B*c^2*x^4 + (2*(-(b^3*B) + A*b^2*c + 3*a*b*B*c - 2*a*A*c^2)*ArcTan[(b + 2*c*x^2)/Sqr
t[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + (b^2*B - A*b*c - a*B*c)*Log[a + b*x^2 + c*x^4])/(4*c^3)

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fricas [A]  time = 0.81, size = 421, normalized size = 3.17 \[ \left [\frac {{\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} x^{4} - 2 \, {\left (B b^{3} c + 4 \, A a c^{3} - {\left (4 \, B a b + A b^{2}\right )} c^{2}\right )} x^{2} + {\left (B b^{3} + 2 \, A a c^{2} - {\left (3 \, B a b + A b^{2}\right )} c\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c + {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) + {\left (B b^{4} + 4 \, {\left (B a^{2} + A a b\right )} c^{2} - {\left (5 \, B a b^{2} + A b^{3}\right )} c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}, \frac {{\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} x^{4} - 2 \, {\left (B b^{3} c + 4 \, A a c^{3} - {\left (4 \, B a b + A b^{2}\right )} c^{2}\right )} x^{2} + 2 \, {\left (B b^{3} + 2 \, A a c^{2} - {\left (3 \, B a b + A b^{2}\right )} c\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + {\left (B b^{4} + 4 \, {\left (B a^{2} + A a b\right )} c^{2} - {\left (5 \, B a b^{2} + A b^{3}\right )} c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

[1/4*((B*b^2*c^2 - 4*B*a*c^3)*x^4 - 2*(B*b^3*c + 4*A*a*c^3 - (4*B*a*b + A*b^2)*c^2)*x^2 + (B*b^3 + 2*A*a*c^2 -
(3*B*a*b + A*b^2)*c)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^4 + 2*b*c*x^2 + b^2 - 2*a*c + (2*c*x^2 + b)*sqrt(b^2 - 4*
a*c))/(c*x^4 + b*x^2 + a)) + (B*b^4 + 4*(B*a^2 + A*a*b)*c^2 - (5*B*a*b^2 + A*b^3)*c)*log(c*x^4 + b*x^2 + a))/(
b^2*c^3 - 4*a*c^4), 1/4*((B*b^2*c^2 - 4*B*a*c^3)*x^4 - 2*(B*b^3*c + 4*A*a*c^3 - (4*B*a*b + A*b^2)*c^2)*x^2 + 2
*(B*b^3 + 2*A*a*c^2 - (3*B*a*b + A*b^2)*c)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 -
4*a*c)) + (B*b^4 + 4*(B*a^2 + A*a*b)*c^2 - (5*B*a*b^2 + A*b^3)*c)*log(c*x^4 + b*x^2 + a))/(b^2*c^3 - 4*a*c^4)]

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giac [A]  time = 1.87, size = 126, normalized size = 0.95 \[ \frac {B c x^{4} - 2 \, B b x^{2} + 2 \, A c x^{2}}{4 \, c^{2}} + \frac {{\left (B b^{2} - B a c - A b c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{3}} - \frac {{\left (B b^{3} - 3 \, B a b c - A b^{2} c + 2 \, A a c^{2}\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.01, size = 261, normalized size = 1.96 \[ \frac {B \,x^{4}}{4 c}-\frac {A a \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {A \,b^{2} \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, c^{2}}+\frac {A \,x^{2}}{2 c}+\frac {3 B a b \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, c^{2}}-\frac {B \,b^{3} \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, c^{3}}-\frac {B b \,x^{2}}{2 c^{2}}-\frac {A b \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 c^{2}}-\frac {B a \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 c^{2}}+\frac {B \,b^{2} \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(B*x^2+A)/(c*x^4+b*x^2+a),x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 0.46, size = 1343, normalized size = 10.10 \[ x^2\,\left (\frac {A}{2\,c}-\frac {B\,b}{2\,c^2}\right )+\frac {B\,x^4}{4\,c}-\frac {\ln \left (c\,x^4+b\,x^2+a\right )\,\left (8\,B\,a^2\,c^2-10\,B\,a\,b^2\,c+8\,A\,a\,b\,c^2+2\,B\,b^4-2\,A\,b^3\,c\right )}{2\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}+\frac {\mathrm {atan}\left (\frac {2\,c^4\,\left (4\,a\,c-b^2\right )\,\left (x^2\,\left (\frac {\frac {\left (\frac {-6\,B\,b^3\,c^3+6\,A\,b^2\,c^4+10\,B\,a\,b\,c^4-4\,A\,a\,c^5}{c^4}-\frac {4\,b\,c^2\,\left (8\,B\,a^2\,c^2-10\,B\,a\,b^2\,c+8\,A\,a\,b\,c^2+2\,B\,b^4-2\,A\,b^3\,c\right )}{16\,a\,c^4-4\,b^2\,c^3}\right )\,\left (B\,b^3-A\,b^2\,c-3\,B\,a\,b\,c+2\,A\,a\,c^2\right )}{8\,c^3\,\sqrt {4\,a\,c-b^2}}-\frac {b\,\left (B\,b^3-A\,b^2\,c-3\,B\,a\,b\,c+2\,A\,a\,c^2\right )\,\left (8\,B\,a^2\,c^2-10\,B\,a\,b^2\,c+8\,A\,a\,b\,c^2+2\,B\,b^4-2\,A\,b^3\,c\right )}{2\,c\,\sqrt {4\,a\,c-b^2}\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}}{a}-\frac {b\,\left (\frac {\left (\frac {-6\,B\,b^3\,c^3+6\,A\,b^2\,c^4+10\,B\,a\,b\,c^4-4\,A\,a\,c^5}{c^4}-\frac {4\,b\,c^2\,\left (8\,B\,a^2\,c^2-10\,B\,a\,b^2\,c+8\,A\,a\,b\,c^2+2\,B\,b^4-2\,A\,b^3\,c\right )}{16\,a\,c^4-4\,b^2\,c^3}\right )\,\left (8\,B\,a^2\,c^2-10\,B\,a\,b^2\,c+8\,A\,a\,b\,c^2+2\,B\,b^4-2\,A\,b^3\,c\right )}{2\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}-\frac {-A^2\,a\,b\,c^3+A^2\,b^3\,c^2-A\,B\,a^2\,c^3+4\,A\,B\,a\,b^2\,c^2-2\,A\,B\,b^4\,c+2\,B^2\,a^2\,b\,c^2-3\,B^2\,a\,b^3\,c+B^2\,b^5}{c^4}+\frac {b\,{\left (B\,b^3-A\,b^2\,c-3\,B\,a\,b\,c+2\,A\,a\,c^2\right )}^2}{2\,c^4\,\left (4\,a\,c-b^2\right )}\right )}{2\,a\,\sqrt {4\,a\,c-b^2}}\right )+\frac {\frac {\left (\frac {8\,B\,a^2\,c^4-8\,B\,a\,b^2\,c^3+8\,A\,a\,b\,c^4}{c^4}-\frac {8\,a\,c^2\,\left (8\,B\,a^2\,c^2-10\,B\,a\,b^2\,c+8\,A\,a\,b\,c^2+2\,B\,b^4-2\,A\,b^3\,c\right )}{16\,a\,c^4-4\,b^2\,c^3}\right )\,\left (B\,b^3-A\,b^2\,c-3\,B\,a\,b\,c+2\,A\,a\,c^2\right )}{8\,c^3\,\sqrt {4\,a\,c-b^2}}-\frac {a\,\left (B\,b^3-A\,b^2\,c-3\,B\,a\,b\,c+2\,A\,a\,c^2\right )\,\left (8\,B\,a^2\,c^2-10\,B\,a\,b^2\,c+8\,A\,a\,b\,c^2+2\,B\,b^4-2\,A\,b^3\,c\right )}{c\,\sqrt {4\,a\,c-b^2}\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}}{a}-\frac {b\,\left (\frac {\left (\frac {8\,B\,a^2\,c^4-8\,B\,a\,b^2\,c^3+8\,A\,a\,b\,c^4}{c^4}-\frac {8\,a\,c^2\,\left (8\,B\,a^2\,c^2-10\,B\,a\,b^2\,c+8\,A\,a\,b\,c^2+2\,B\,b^4-2\,A\,b^3\,c\right )}{16\,a\,c^4-4\,b^2\,c^3}\right )\,\left (8\,B\,a^2\,c^2-10\,B\,a\,b^2\,c+8\,A\,a\,b\,c^2+2\,B\,b^4-2\,A\,b^3\,c\right )}{2\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}-\frac {A^2\,a\,b^2\,c^2+2\,A\,B\,a^2\,b\,c^2-2\,A\,B\,a\,b^3\,c+B^2\,a^3\,c^2-2\,B^2\,a^2\,b^2\,c+B^2\,a\,b^4}{c^4}+\frac {a\,{\left (B\,b^3-A\,b^2\,c-3\,B\,a\,b\,c+2\,A\,a\,c^2\right )}^2}{c^4\,\left (4\,a\,c-b^2\right )}\right )}{2\,a\,\sqrt {4\,a\,c-b^2}}\right )}{4\,A^2\,a^2\,c^4-4\,A^2\,a\,b^2\,c^3+A^2\,b^4\,c^2-12\,A\,B\,a^2\,b\,c^3+10\,A\,B\,a\,b^3\,c^2-2\,A\,B\,b^5\,c+9\,B^2\,a^2\,b^2\,c^2-6\,B^2\,a\,b^4\,c+B^2\,b^6}\right )\,\left (B\,b^3-A\,b^2\,c-3\,B\,a\,b\,c+2\,A\,a\,c^2\right )}{2\,c^3\,\sqrt {4\,a\,c-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(A + B*x^2))/(a + b*x^2 + c*x^4),x)

[Out]

x^2*(A/(2*c) - (B*b)/(2*c^2)) + (B*x^4)/(4*c) - (log(a + b*x^2 + c*x^4)*(2*B*b^4 + 8*B*a^2*c^2 - 2*A*b^3*c + 8
*A*a*b*c^2 - 10*B*a*b^2*c))/(2*(16*a*c^4 - 4*b^2*c^3)) + (atan((2*c^4*(4*a*c - b^2)*(x^2*(((((6*A*b^2*c^4 - 6*
B*b^3*c^3 - 4*A*a*c^5 + 10*B*a*b*c^4)/c^4 - (4*b*c^2*(2*B*b^4 + 8*B*a^2*c^2 - 2*A*b^3*c + 8*A*a*b*c^2 - 10*B*a
*b^2*c))/(16*a*c^4 - 4*b^2*c^3))*(B*b^3 + 2*A*a*c^2 - A*b^2*c - 3*B*a*b*c))/(8*c^3*(4*a*c - b^2)^(1/2)) - (b*(
B*b^3 + 2*A*a*c^2 - A*b^2*c - 3*B*a*b*c)*(2*B*b^4 + 8*B*a^2*c^2 - 2*A*b^3*c + 8*A*a*b*c^2 - 10*B*a*b^2*c))/(2*
c*(4*a*c - b^2)^(1/2)*(16*a*c^4 - 4*b^2*c^3)))/a - (b*((((6*A*b^2*c^4 - 6*B*b^3*c^3 - 4*A*a*c^5 + 10*B*a*b*c^4
)/c^4 - (4*b*c^2*(2*B*b^4 + 8*B*a^2*c^2 - 2*A*b^3*c + 8*A*a*b*c^2 - 10*B*a*b^2*c))/(16*a*c^4 - 4*b^2*c^3))*(2*
B*b^4 + 8*B*a^2*c^2 - 2*A*b^3*c + 8*A*a*b*c^2 - 10*B*a*b^2*c))/(2*(16*a*c^4 - 4*b^2*c^3)) - (B^2*b^5 + A^2*b^3
*c^2 - 2*A*B*b^4*c - A*B*a^2*c^3 - A^2*a*b*c^3 - 3*B^2*a*b^3*c + 2*B^2*a^2*b*c^2 + 4*A*B*a*b^2*c^2)/c^4 + (b*(
B*b^3 + 2*A*a*c^2 - A*b^2*c - 3*B*a*b*c)^2)/(2*c^4*(4*a*c - b^2))))/(2*a*(4*a*c - b^2)^(1/2))) + ((((8*B*a^2*c
^4 + 8*A*a*b*c^4 - 8*B*a*b^2*c^3)/c^4 - (8*a*c^2*(2*B*b^4 + 8*B*a^2*c^2 - 2*A*b^3*c + 8*A*a*b*c^2 - 10*B*a*b^2
*c))/(16*a*c^4 - 4*b^2*c^3))*(B*b^3 + 2*A*a*c^2 - A*b^2*c - 3*B*a*b*c))/(8*c^3*(4*a*c - b^2)^(1/2)) - (a*(B*b^
3 + 2*A*a*c^2 - A*b^2*c - 3*B*a*b*c)*(2*B*b^4 + 8*B*a^2*c^2 - 2*A*b^3*c + 8*A*a*b*c^2 - 10*B*a*b^2*c))/(c*(4*a
*c - b^2)^(1/2)*(16*a*c^4 - 4*b^2*c^3)))/a - (b*((((8*B*a^2*c^4 + 8*A*a*b*c^4 - 8*B*a*b^2*c^3)/c^4 - (8*a*c^2*
(2*B*b^4 + 8*B*a^2*c^2 - 2*A*b^3*c + 8*A*a*b*c^2 - 10*B*a*b^2*c))/(16*a*c^4 - 4*b^2*c^3))*(2*B*b^4 + 8*B*a^2*c
^2 - 2*A*b^3*c + 8*A*a*b*c^2 - 10*B*a*b^2*c))/(2*(16*a*c^4 - 4*b^2*c^3)) - (B^2*a*b^4 + B^2*a^3*c^2 + A^2*a*b^
2*c^2 - 2*B^2*a^2*b^2*c - 2*A*B*a*b^3*c + 2*A*B*a^2*b*c^2)/c^4 + (a*(B*b^3 + 2*A*a*c^2 - A*b^2*c - 3*B*a*b*c)^
2)/(c^4*(4*a*c - b^2))))/(2*a*(4*a*c - b^2)^(1/2))))/(B^2*b^6 + 4*A^2*a^2*c^4 + A^2*b^4*c^2 - 2*A*B*b^5*c + 9*
B^2*a^2*b^2*c^2 - 6*B^2*a*b^4*c - 4*A^2*a*b^2*c^3 + 10*A*B*a*b^3*c^2 - 12*A*B*a^2*b*c^3))*(B*b^3 + 2*A*a*c^2 -
A*b^2*c - 3*B*a*b*c))/(2*c^3*(4*a*c - b^2)^(1/2))

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sympy [B]  time = 43.89, size = 620, normalized size = 4.66 \[ \frac {B x^{4}}{4 c} + x^{2} \left (\frac {A}{2 c} - \frac {B b}{2 c^{2}}\right ) + \left (- \frac {\sqrt {- 4 a c + b^{2}} \left (- 2 A a c^{2} + A b^{2} c + 3 B a b c - B b^{3}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac {A b c + B a c - B b^{2}}{4 c^{3}}\right ) \log {\left (x^{2} + \frac {A a b c + 2 B a^{2} c - B a b^{2} + 8 a c^{3} \left (- \frac {\sqrt {- 4 a c + b^{2}} \left (- 2 A a c^{2} + A b^{2} c + 3 B a b c - B b^{3}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac {A b c + B a c - B b^{2}}{4 c^{3}}\right ) - 2 b^{2} c^{2} \left (- \frac {\sqrt {- 4 a c + b^{2}} \left (- 2 A a c^{2} + A b^{2} c + 3 B a b c - B b^{3}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac {A b c + B a c - B b^{2}}{4 c^{3}}\right )}{- 2 A a c^{2} + A b^{2} c + 3 B a b c - B b^{3}} \right )} + \left (\frac {\sqrt {- 4 a c + b^{2}} \left (- 2 A a c^{2} + A b^{2} c + 3 B a b c - B b^{3}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac {A b c + B a c - B b^{2}}{4 c^{3}}\right ) \log {\left (x^{2} + \frac {A a b c + 2 B a^{2} c - B a b^{2} + 8 a c^{3} \left (\frac {\sqrt {- 4 a c + b^{2}} \left (- 2 A a c^{2} + A b^{2} c + 3 B a b c - B b^{3}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac {A b c + B a c - B b^{2}}{4 c^{3}}\right ) - 2 b^{2} c^{2} \left (\frac {\sqrt {- 4 a c + b^{2}} \left (- 2 A a c^{2} + A b^{2} c + 3 B a b c - B b^{3}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac {A b c + B a c - B b^{2}}{4 c^{3}}\right )}{- 2 A a c^{2} + A b^{2} c + 3 B a b c - B b^{3}} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(B*x**2+A)/(c*x**4+b*x**2+a),x)

[Out]

B*x**4/(4*c) + x**2*(A/(2*c) - B*b/(2*c**2)) + (-sqrt(-4*a*c + b**2)*(-2*A*a*c**2 + A*b**2*c + 3*B*a*b*c - B*b
**3)/(4*c**3*(4*a*c - b**2)) - (A*b*c + B*a*c - B*b**2)/(4*c**3))*log(x**2 + (A*a*b*c + 2*B*a**2*c - B*a*b**2
+ 8*a*c**3*(-sqrt(-4*a*c + b**2)*(-2*A*a*c**2 + A*b**2*c + 3*B*a*b*c - B*b**3)/(4*c**3*(4*a*c - b**2)) - (A*b*
c + B*a*c - B*b**2)/(4*c**3)) - 2*b**2*c**2*(-sqrt(-4*a*c + b**2)*(-2*A*a*c**2 + A*b**2*c + 3*B*a*b*c - B*b**3
)/(4*c**3*(4*a*c - b**2)) - (A*b*c + B*a*c - B*b**2)/(4*c**3)))/(-2*A*a*c**2 + A*b**2*c + 3*B*a*b*c - B*b**3))
+ (sqrt(-4*a*c + b**2)*(-2*A*a*c**2 + A*b**2*c + 3*B*a*b*c - B*b**3)/(4*c**3*(4*a*c - b**2)) - (A*b*c + B*a*c
- B*b**2)/(4*c**3))*log(x**2 + (A*a*b*c + 2*B*a**2*c - B*a*b**2 + 8*a*c**3*(sqrt(-4*a*c + b**2)*(-2*A*a*c**2
+ A*b**2*c + 3*B*a*b*c - B*b**3)/(4*c**3*(4*a*c - b**2)) - (A*b*c + B*a*c - B*b**2)/(4*c**3)) - 2*b**2*c**2*(s
qrt(-4*a*c + b**2)*(-2*A*a*c**2 + A*b**2*c + 3*B*a*b*c - B*b**3)/(4*c**3*(4*a*c - b**2)) - (A*b*c + B*a*c - B*
b**2)/(4*c**3)))/(-2*A*a*c**2 + A*b**2*c + 3*B*a*b*c - B*b**3))

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