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

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

[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))

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Rubi [A]  time = 0.202879, antiderivative size = 153, normalized size of antiderivative = 1., 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 = {1251, 832, 779, 621, 206} \[ \frac{\sqrt{a+b x^2+c x^4} \left (-16 a B c-2 c x^2 (5 b B-6 A c)-18 A b c+15 b^2 B\right )}{48 c^3}-\frac{\left (8 a A c^2-12 a b B c-6 A b^2 c+5 b^3 B\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{7/2}}+\frac{B x^4 \sqrt{a+b x^2+c x^4}}{6 c} \]

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 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]

Rule 832

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 779

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 621

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 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])

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} \operatorname{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{\operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.10728, size = 139, normalized size = 0.91 \[ \frac{2 \sqrt{c} \sqrt{a+b x^2+c x^4} \left (4 c \left (-4 a B+3 A c x^2+2 B c x^4\right )-2 b c \left (9 A+5 B x^2\right )+15 b^2 B\right )-3 \left (8 a A c^2-12 a b B c-6 A b^2 c+5 b^3 B\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{96 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]*(15*b^2*B - 2*b*c*(9*A + 5*B*x^2) + 4*c*(-4*a*B + 3*A*c*x^2 + 2*B*c*x^4)) -
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]*Sqrt[a + b*x^2 + c*x^4])])/
(96*c^(7/2))

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Maple [B]  time = 0.027, size = 286, normalized size = 1.9 \begin{align*}{\frac{B{x}^{4}}{6\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{5\,Bb{x}^{2}}{24\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{5\,{b}^{2}B}{16\,{c}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{5\,{b}^{3}B}{32}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{7}{2}}}}+{\frac{3\,abB}{8}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{aB}{3\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{A{x}^{2}}{4\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{3\,Ab}{8\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,A{b}^{2}}{16}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{Aa}{4}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

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)

[Out]

1/6*B*x^4*(c*x^4+b*x^2+a)^(1/2)/c-5/24*B*b/c^2*x^2*(c*x^4+b*x^2+a)^(1/2)+5/16*B*b^2/c^3*(c*x^4+b*x^2+a)^(1/2)-
5/32*B*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*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*B*a/c^2*(c*x^4+b*x^2+a)^(1/2)+1/4*A*x^2/c*(c*x^4+b*x^2+a)^(1/2)-3/8*A*b/c^2*(c*x^4+
b*x^2+a)^(1/2)+3/16*A*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*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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 1.55295, size = 725, normalized size = 4.74 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5} \left (A + B x^{2}\right )}{\sqrt{a + b x^{2} + c x^{4}}}\, dx \end{align*}

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 = 1.19961, size = 203, normalized size = 1.33 \begin{align*} \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^{2} - 6 \, A c^{3}}{c^{4}}\right )} x^{2} + \frac{15 \, B b^{2} c - 16 \, B a c^{2} - 18 \, A b c^{2}}{c^{4}}\right )} + \frac{{\left (5 \, B b^{3} c - 12 \, B a b c^{2} - 6 \, A b^{2} c^{2} + 8 \, A a c^{3}\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{9}{2}}} \end{align*}

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^2 - 6*A*c^3)/c^4)*x^2 + (15*B*b^2*c - 16*B*a*c^2 - 18*A*
b*c^2)/c^4) + 1/32*(5*B*b^3*c - 12*B*a*b*c^2 - 6*A*b^2*c^2 + 8*A*a*c^3)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 +
b*x^2 + a))*sqrt(c) - b))/c^(9/2)