3.175 \(\int \frac {A+B x^2}{x^7 \sqrt {a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=177 \[ -\frac {\sqrt {a+b x^2+c x^4} \left (-16 a A c-18 a b B+15 A b^2\right )}{48 a^3 x^2}+\frac {(5 A b-6 a B) \sqrt {a+b x^2+c x^4}}{24 a^2 x^4}+\frac {\left (8 a^2 B c-12 a A b c-6 a b^2 B+5 A b^3\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{7/2}}-\frac {A \sqrt {a+b x^2+c x^4}}{6 a x^6} \]

[Out]

1/32*(-12*A*a*b*c+5*A*b^3+8*B*a^2*c-6*B*a*b^2)*arctanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^2+a)^(1/2))/a^(7/2)-
1/6*A*(c*x^4+b*x^2+a)^(1/2)/a/x^6+1/24*(5*A*b-6*B*a)*(c*x^4+b*x^2+a)^(1/2)/a^2/x^4-1/48*(-16*A*a*c+15*A*b^2-18
*B*a*b)*(c*x^4+b*x^2+a)^(1/2)/a^3/x^2

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1251, 834, 806, 724, 206} \[ -\frac {\sqrt {a+b x^2+c x^4} \left (-16 a A c-18 a b B+15 A b^2\right )}{48 a^3 x^2}+\frac {\left (8 a^2 B c-12 a A b c-6 a b^2 B+5 A b^3\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{7/2}}+\frac {(5 A b-6 a B) \sqrt {a+b x^2+c x^4}}{24 a^2 x^4}-\frac {A \sqrt {a+b x^2+c x^4}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*Sqrt[a + b*x^2 + c*x^4])/(6*a*x^6) + ((5*A*b - 6*a*B)*Sqrt[a + b*x^2 + c*x^4])/(24*a^2*x^4) - ((15*A*b^2 -
18*a*b*B - 16*a*A*c)*Sqrt[a + b*x^2 + c*x^4])/(48*a^3*x^2) + ((5*A*b^3 - 6*a*b^2*B - 12*a*A*b*c + 8*a^2*B*c)*
ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(32*a^(7/2))

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 724

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

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 834

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

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 {A+B x^2}{x^7 \sqrt {a+b x^2+c x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {A+B x}{x^4 \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac {A \sqrt {a+b x^2+c x^4}}{6 a x^6}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} (5 A b-6 a B)+2 A c x}{x^3 \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{6 a}\\ &=-\frac {A \sqrt {a+b x^2+c x^4}}{6 a x^6}+\frac {(5 A b-6 a B) \sqrt {a+b x^2+c x^4}}{24 a^2 x^4}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{4} \left (15 A b^2-18 a b B-16 a A c\right )+\frac {1}{2} (5 A b-6 a B) c x}{x^2 \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{12 a^2}\\ &=-\frac {A \sqrt {a+b x^2+c x^4}}{6 a x^6}+\frac {(5 A b-6 a B) \sqrt {a+b x^2+c x^4}}{24 a^2 x^4}-\frac {\left (15 A b^2-18 a b B-16 a A c\right ) \sqrt {a+b x^2+c x^4}}{48 a^3 x^2}-\frac {\left (5 A b^3-6 a b^2 B-12 a A b c+8 a^2 B c\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{32 a^3}\\ &=-\frac {A \sqrt {a+b x^2+c x^4}}{6 a x^6}+\frac {(5 A b-6 a B) \sqrt {a+b x^2+c x^4}}{24 a^2 x^4}-\frac {\left (15 A b^2-18 a b B-16 a A c\right ) \sqrt {a+b x^2+c x^4}}{48 a^3 x^2}+\frac {\left (5 A b^3-6 a b^2 B-12 a A b c+8 a^2 B c\right ) \operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x^2}{\sqrt {a+b x^2+c x^4}}\right )}{16 a^3}\\ &=-\frac {A \sqrt {a+b x^2+c x^4}}{6 a x^6}+\frac {(5 A b-6 a B) \sqrt {a+b x^2+c x^4}}{24 a^2 x^4}-\frac {\left (15 A b^2-18 a b B-16 a A c\right ) \sqrt {a+b x^2+c x^4}}{48 a^3 x^2}+\frac {\left (5 A b^3-6 a b^2 B-12 a A b c+8 a^2 B c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 148, normalized size = 0.84 \[ \frac {\left (8 a^2 B c-12 a A b c-6 a b^2 B+5 A b^3\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{7/2}}+\frac {\sqrt {a+b x^2+c x^4} \left (-4 a^2 \left (2 A+3 B x^2\right )+2 a \left (5 A b x^2+8 A c x^4+9 b B x^4\right )-15 A b^2 x^4\right )}{48 a^3 x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2 + c*x^4]*(-15*A*b^2*x^4 - 4*a^2*(2*A + 3*B*x^2) + 2*a*(5*A*b*x^2 + 9*b*B*x^4 + 8*A*c*x^4)))/(4
8*a^3*x^6) + ((5*A*b^3 - 6*a*b^2*B - 12*a*A*b*c + 8*a^2*B*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 +
c*x^4])])/(32*a^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.96, size = 339, normalized size = 1.92 \[ \left [\frac {3 \, {\left (6 \, B a b^{2} - 5 \, A b^{3} - 4 \, {\left (2 \, B a^{2} - 3 \, A a b\right )} c\right )} \sqrt {a} x^{6} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) + 4 \, {\left ({\left (18 \, B a^{2} b - 15 \, A a b^{2} + 16 \, A a^{2} c\right )} x^{4} - 8 \, A a^{3} - 2 \, {\left (6 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{192 \, a^{4} x^{6}}, \frac {3 \, {\left (6 \, B a b^{2} - 5 \, A b^{3} - 4 \, {\left (2 \, B a^{2} - 3 \, A a b\right )} c\right )} \sqrt {-a} x^{6} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (18 \, B a^{2} b - 15 \, A a b^{2} + 16 \, A a^{2} c\right )} x^{4} - 8 \, A a^{3} - 2 \, {\left (6 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{96 \, a^{4} x^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^7/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/192*(3*(6*B*a*b^2 - 5*A*b^3 - 4*(2*B*a^2 - 3*A*a*b)*c)*sqrt(a)*x^6*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 - 4*
sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8*a^2)/x^4) + 4*((18*B*a^2*b - 15*A*a*b^2 + 16*A*a^2*c)*x^4 -
8*A*a^3 - 2*(6*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(c*x^4 + b*x^2 + a))/(a^4*x^6), 1/96*(3*(6*B*a*b^2 - 5*A*b^3 - 4*(2
*B*a^2 - 3*A*a*b)*c)*sqrt(-a)*x^6*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2
+ a^2)) + 2*((18*B*a^2*b - 15*A*a*b^2 + 16*A*a^2*c)*x^4 - 8*A*a^3 - 2*(6*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(c*x^4 +
b*x^2 + a))/(a^4*x^6)]

________________________________________________________________________________________

giac [B]  time = 0.64, size = 571, normalized size = 3.23 \[ \frac {{\left (6 \, B a b^{2} - 5 \, A b^{3} - 8 \, B a^{2} c + 12 \, A a b c\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{16 \, \sqrt {-a} a^{3}} - \frac {18 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} B a b^{2} - 15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} A b^{3} - 24 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} B a^{2} c + 36 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} A a b c - 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} B a^{2} b^{2} + 40 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} A a b^{3} - 96 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} A a^{2} b c - 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} B a^{3} b \sqrt {c} - 96 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} A a^{3} c^{\frac {3}{2}} + 30 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} B a^{3} b^{2} - 33 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} A a^{2} b^{3} + 24 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} B a^{4} c - 36 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} A a^{3} b c + 48 \, B a^{4} b \sqrt {c} - 48 \, A a^{3} b^{2} \sqrt {c} + 32 \, A a^{4} c^{\frac {3}{2}}}{48 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{3} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^7/(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/16*(6*B*a*b^2 - 5*A*b^3 - 8*B*a^2*c + 12*A*a*b*c)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/
(sqrt(-a)*a^3) - 1/48*(18*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*B*a*b^2 - 15*(sqrt(c)*x^2 - sqrt(c*x^4 + b
*x^2 + a))^5*A*b^3 - 24*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*B*a^2*c + 36*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x
^2 + a))^5*A*a*b*c - 48*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*B*a^2*b^2 + 40*(sqrt(c)*x^2 - sqrt(c*x^4 + b
*x^2 + a))^3*A*a*b^3 - 96*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*A*a^2*b*c - 48*(sqrt(c)*x^2 - sqrt(c*x^4 +
b*x^2 + a))^2*B*a^3*b*sqrt(c) - 96*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2*A*a^3*c^(3/2) + 30*(sqrt(c)*x^2
- sqrt(c*x^4 + b*x^2 + a))*B*a^3*b^2 - 33*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*A*a^2*b^3 + 24*(sqrt(c)*x^2
- sqrt(c*x^4 + b*x^2 + a))*B*a^4*c - 36*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*A*a^3*b*c + 48*B*a^4*b*sqrt(c)
- 48*A*a^3*b^2*sqrt(c) + 32*A*a^4*c^(3/2))/(((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2 - a)^3*a^3)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 311, normalized size = 1.76 \[ -\frac {3 A b c \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{8 a^{\frac {5}{2}}}+\frac {5 A \,b^{3} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{32 a^{\frac {7}{2}}}+\frac {B c \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}-\frac {3 B \,b^{2} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{16 a^{\frac {5}{2}}}+\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, A c}{3 a^{2} x^{2}}-\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, A \,b^{2}}{16 a^{3} x^{2}}+\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, B b}{8 a^{2} x^{2}}+\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, A b}{24 a^{2} x^{4}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, B}{4 a \,x^{4}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, A}{6 a \,x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^7/(c*x^4+b*x^2+a)^(1/2),x)

[Out]

-1/6*A*(c*x^4+b*x^2+a)^(1/2)/a/x^6+5/24*A*b/a^2/x^4*(c*x^4+b*x^2+a)^(1/2)-5/16*A*b^2/a^3/x^2*(c*x^4+b*x^2+a)^(
1/2)+5/32*A*b^3/a^(7/2)*ln((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)-3/8*A*b/a^(5/2)*c*ln((b*x^2+2*a+2*
(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)+1/3*A/a^2*c/x^2*(c*x^4+b*x^2+a)^(1/2)-1/4*B/a/x^4*(c*x^4+b*x^2+a)^(1/2)+3/
8*B*b/a^2/x^2*(c*x^4+b*x^2+a)^(1/2)-3/16*B*b^2/a^(5/2)*ln((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)+1/4
*B*c/a^(3/2)*ln((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)

________________________________________________________________________________________

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((B*x^2+A)/x^7/(c*x^4+b*x^2+a)^(1/2),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, negative or zero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {B\,x^2+A}{x^7\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^7*(a + b*x^2 + c*x^4)^(1/2)),x)

[Out]

int((A + B*x^2)/(x^7*(a + b*x^2 + c*x^4)^(1/2)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x^{2}}{x^{7} \sqrt {a + b x^{2} + c x^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**7/(c*x**4+b*x**2+a)**(1/2),x)

[Out]

Integral((A + B*x**2)/(x**7*sqrt(a + b*x**2 + c*x**4)), x)

________________________________________________________________________________________