3.2.69 \(\int \frac {x^5 (A+B x^2)}{\sqrt {a+b x^2+c x^4}} \, dx\) [169]

Optimal. Leaf size=153 \[ \frac {B x^4 \sqrt {a+b x^2+c x^4}}{6 c}+\frac {\left (15 b^2 B-18 A b c-16 a B c-2 c (5 b B-6 A c) x^2\right ) \sqrt {a+b x^2+c x^4}}{48 c^3}-\frac {\left (5 b^3 B-6 A b^2 c-12 a b B c+8 a A c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{32 c^{7/2}} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1265, 846, 793, 635, 212} \begin {gather*} \frac {\sqrt {a+b x^2+c x^4} \left (-16 a B c-2 c x^2 (5 b B-6 A c)-18 A b c+15 b^2 B\right )}{48 c^3}-\frac {\left (8 a A c^2-12 a b B c-6 A b^2 c+5 b^3 B\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{32 c^{7/2}}+\frac {B x^4 \sqrt {a+b x^2+c x^4}}{6 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 212

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

Rule 635

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

Rule 793

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

Rule 846

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

Rule 1265

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

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Mathematica [A]
time = 0.39, size = 135, normalized size = 0.88 \begin {gather*} \frac {\sqrt {a+b x^2+c x^4} \left (15 b^2 B-18 A b c-16 a B c-10 b B c x^2+12 A c^2 x^2+8 B c^2 x^4\right )}{48 c^3}+\frac {\left (5 b^3 B-6 A b^2 c-12 a b B c+8 a A c^2\right ) \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )}{32 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2 + c*x^4]*(15*b^2*B - 18*A*b*c - 16*a*B*c - 10*b*B*c*x^2 + 12*A*c^2*x^2 + 8*B*c^2*x^4))/(48*c^3
) + ((5*b^3*B - 6*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]])/(32*
c^(7/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(281\) vs. \(2(135)=270\).
time = 0.08, size = 282, normalized size = 1.84

method	result	size
risch	\(-\frac {\left (-8 c^{2} B \,x^{4}-12 A \,c^{2} x^{2}+10 B b c \,x^{2}+18 b c A +16 a c B -15 b^{2} B \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{48 c^{3}}-\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) a A}{4 c^{\frac {3}{2}}}+\frac {3 \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) A \,b^{2}}{16 c^{\frac {5}{2}}}+\frac {3 \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) a b B}{8 c^{\frac {5}{2}}}-\frac {5 \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) b^{3} B}{32 c^{\frac {7}{2}}}\)	\(213\)
default	\(B \left (\frac {x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{6 c}-\frac {5 b \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{24 c^{2}}+\frac {5 b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 c^{3}}-\frac {5 b^{3} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {7}{2}}}+\frac {3 b a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{8 c^{\frac {5}{2}}}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{3 c^{2}}\right )+A \left (\frac {x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}-\frac {3 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c^{2}}+\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {5}{2}}}-\frac {a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {3}{2}}}\right )\)	\(282\)
elliptic	\(\frac {B \,x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{6 c}-\frac {5 B b \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{24 c^{2}}+\frac {5 B \,b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 c^{3}}-\frac {5 \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) b^{3} B}{32 c^{\frac {7}{2}}}+\frac {3 \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) a b B}{8 c^{\frac {5}{2}}}-\frac {B a \sqrt {c \,x^{4}+b \,x^{2}+a}}{3 c^{2}}+\frac {A \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}-\frac {3 A b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c^{2}}+\frac {3 \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) A \,b^{2}}{16 c^{\frac {5}{2}}}-\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) a A}{4 c^{\frac {3}{2}}}\)	\(286\)

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)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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)^(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 deta

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Fricas [A]
time = 0.38, size = 315, normalized size = 2.06 \begin {gather*} \left [\frac {3 \, {\left (5 \, B b^{3} + 8 \, A a c^{2} - 6 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (8 \, B c^{3} x^{4} + 15 \, B b^{2} c - 2 \, {\left (8 \, B a + 9 \, A b\right )} c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{192 \, c^{4}}, \frac {3 \, {\left (5 \, B b^{3} + 8 \, A a c^{2} - 6 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \, {\left (8 \, B c^{3} x^{4} + 15 \, B b^{2} c - 2 \, {\left (8 \, B a + 9 \, A b\right )} c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{96 \, c^{4}}\right ] \end {gather*}

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)^(1/2),x, algorithm="fricas")

[Out]

[1/192*(3*(5*B*b^3 + 8*A*a*c^2 - 6*(2*B*a*b + A*b^2)*c)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 + 4*sqrt(c*x^
4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) + 4*(8*B*c^3*x^4 + 15*B*b^2*c - 2*(8*B*a + 9*A*b)*c^2 - 2*(5*B*b
*c^2 - 6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^4, 1/96*(3*(5*B*b^3 + 8*A*a*c^2 - 6*(2*B*a*b + A*b^2)*c)*sqrt(
-c)*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) + 2*(8*B*c^3*x^4 + 15
*B*b^2*c - 2*(8*B*a + 9*A*b)*c^2 - 2*(5*B*b*c^2 - 6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^4]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5} \left (A + B x^{2}\right )}{\sqrt {a + b x^{2} + c x^{4}}}\, dx \end {gather*}

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)**(1/2),x)

[Out]

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

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Giac [A]
time = 4.76, size = 138, normalized size = 0.90 \begin {gather*} \frac {1}{48} \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (\frac {4 \, B x^{2}}{c} - \frac {5 \, B b c - 6 \, A c^{2}}{c^{3}}\right )} x^{2} + \frac {15 \, B b^{2} - 16 \, B a c - 18 \, A b c}{c^{3}}\right )} + \frac {{\left (5 \, B b^{3} - 12 \, B a b c - 6 \, A b^{2} c + 8 \, A a c^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right )}{32 \, c^{\frac {7}{2}}} \end {gather*}

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)^(1/2),x, algorithm="giac")

[Out]

1/48*sqrt(c*x^4 + b*x^2 + a)*(2*(4*B*x^2/c - (5*B*b*c - 6*A*c^2)/c^3)*x^2 + (15*B*b^2 - 16*B*a*c - 18*A*b*c)/c
^3) + 1/32*(5*B*b^3 - 12*B*a*b*c - 6*A*b^2*c + 8*A*a*c^2)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*s
qrt(c) - b))/c^(7/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5\,\left (B\,x^2+A\right )}{\sqrt {c\,x^4+b\,x^2+a}} \,d x \end {gather*}

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)^(1/2),x)

[Out]

int((x^5*(A + B*x^2))/(a + b*x^2 + c*x^4)^(1/2), x)

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