3.219 \(\int \frac{x^{14}}{\sqrt{a+b x^3+c x^6}} \, dx\)

Optimal. Leaf size=171 \[ \frac{\left (48 a^2 c^2-120 a b^2 c+35 b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )}{384 c^{9/2}}-\frac{\left (5 b \left (21 b^2-44 a c\right )-2 c x^3 \left (35 b^2-36 a c\right )\right ) \sqrt{a+b x^3+c x^6}}{576 c^4}-\frac{7 b x^6 \sqrt{a+b x^3+c x^6}}{72 c^2}+\frac{x^9 \sqrt{a+b x^3+c x^6}}{12 c} \]

[Out]

(-7*b*x^6*Sqrt[a + b*x^3 + c*x^6])/(72*c^2) + (x^9*Sqrt[a + b*x^3 + c*x^6])/(12*c) - ((5*b*(21*b^2 - 44*a*c) -
2*c*(35*b^2 - 36*a*c)*x^3)*Sqrt[a + b*x^3 + c*x^6])/(576*c^4) + ((35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*ArcTanh[
(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(384*c^(9/2))

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Rubi [A]  time = 0.216241, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1357, 742, 832, 779, 621, 206} \[ \frac{\left (48 a^2 c^2-120 a b^2 c+35 b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )}{384 c^{9/2}}-\frac{\left (5 b \left (21 b^2-44 a c\right )-2 c x^3 \left (35 b^2-36 a c\right )\right ) \sqrt{a+b x^3+c x^6}}{576 c^4}-\frac{7 b x^6 \sqrt{a+b x^3+c x^6}}{72 c^2}+\frac{x^9 \sqrt{a+b x^3+c x^6}}{12 c} \]

Antiderivative was successfully verified.

[In]

Int[x^14/Sqrt[a + b*x^3 + c*x^6],x]

[Out]

(-7*b*x^6*Sqrt[a + b*x^3 + c*x^6])/(72*c^2) + (x^9*Sqrt[a + b*x^3 + c*x^6])/(12*c) - ((5*b*(21*b^2 - 44*a*c) -
2*c*(35*b^2 - 36*a*c)*x^3)*Sqrt[a + b*x^3 + c*x^6])/(576*c^4) + ((35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*ArcTanh[
(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(384*c^(9/2))

Rule 1357

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

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^{14}}{\sqrt{a+b x^3+c x^6}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{a+b x+c x^2}} \, dx,x,x^3\right )\\ &=\frac{x^9 \sqrt{a+b x^3+c x^6}}{12 c}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (-3 a-\frac{7 b x}{2}\right )}{\sqrt{a+b x+c x^2}} \, dx,x,x^3\right )}{12 c}\\ &=-\frac{7 b x^6 \sqrt{a+b x^3+c x^6}}{72 c^2}+\frac{x^9 \sqrt{a+b x^3+c x^6}}{12 c}+\frac{\operatorname{Subst}\left (\int \frac{x \left (7 a b+\frac{1}{4} \left (35 b^2-36 a c\right ) x\right )}{\sqrt{a+b x+c x^2}} \, dx,x,x^3\right )}{36 c^2}\\ &=-\frac{7 b x^6 \sqrt{a+b x^3+c x^6}}{72 c^2}+\frac{x^9 \sqrt{a+b x^3+c x^6}}{12 c}-\frac{\left (5 b \left (21 b^2-44 a c\right )-2 c \left (35 b^2-36 a c\right ) x^3\right ) \sqrt{a+b x^3+c x^6}}{576 c^4}+\frac{\left (35 b^4-120 a b^2 c+48 a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^3\right )}{384 c^4}\\ &=-\frac{7 b x^6 \sqrt{a+b x^3+c x^6}}{72 c^2}+\frac{x^9 \sqrt{a+b x^3+c x^6}}{12 c}-\frac{\left (5 b \left (21 b^2-44 a c\right )-2 c \left (35 b^2-36 a c\right ) x^3\right ) \sqrt{a+b x^3+c x^6}}{576 c^4}+\frac{\left (35 b^4-120 a b^2 c+48 a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x^3}{\sqrt{a+b x^3+c x^6}}\right )}{192 c^4}\\ &=-\frac{7 b x^6 \sqrt{a+b x^3+c x^6}}{72 c^2}+\frac{x^9 \sqrt{a+b x^3+c x^6}}{12 c}-\frac{\left (5 b \left (21 b^2-44 a c\right )-2 c \left (35 b^2-36 a c\right ) x^3\right ) \sqrt{a+b x^3+c x^6}}{576 c^4}+\frac{\left (35 b^4-120 a b^2 c+48 a^2 c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )}{384 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.113647, size = 137, normalized size = 0.8 \[ \frac{3 \left (48 a^2 c^2-120 a b^2 c+35 b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )+2 \sqrt{c} \sqrt{a+b x^3+c x^6} \left (4 b c \left (55 a-14 c x^6\right )+24 c^2 x^3 \left (2 c x^6-3 a\right )+70 b^2 c x^3-105 b^3\right )}{1152 c^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^14/Sqrt[a + b*x^3 + c*x^6],x]

[Out]

(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6]*(-105*b^3 + 70*b^2*c*x^3 + 4*b*c*(55*a - 14*c*x^6) + 24*c^2*x^3*(-3*a + 2*c
*x^6)) + 3*(35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(11
52*c^(9/2))

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Maple [F]  time = 0.023, size = 0, normalized size = 0. \begin{align*} \int{{x}^{14}{\frac{1}{\sqrt{c{x}^{6}+b{x}^{3}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/(c*x^6+b*x^3+a)^(1/2),x)

[Out]

int(x^14/(c*x^6+b*x^3+a)^(1/2),x)

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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^14/(c*x^6+b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.642, size = 717, normalized size = 4.19 \begin{align*} \left [\frac{3 \,{\left (35 \, b^{4} - 120 \, a b^{2} c + 48 \, a^{2} c^{2}\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} - 4 \, \sqrt{c x^{6} + b x^{3} + a}{\left (2 \, c x^{3} + b\right )} \sqrt{c} - 4 \, a c\right ) + 4 \,{\left (48 \, c^{4} x^{9} - 56 \, b c^{3} x^{6} - 105 \, b^{3} c + 220 \, a b c^{2} + 2 \,{\left (35 \, b^{2} c^{2} - 36 \, a c^{3}\right )} x^{3}\right )} \sqrt{c x^{6} + b x^{3} + a}}{2304 \, c^{5}}, -\frac{3 \,{\left (35 \, b^{4} - 120 \, a b^{2} c + 48 \, a^{2} c^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{6} + b x^{3} + a}{\left (2 \, c x^{3} + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) - 2 \,{\left (48 \, c^{4} x^{9} - 56 \, b c^{3} x^{6} - 105 \, b^{3} c + 220 \, a b c^{2} + 2 \,{\left (35 \, b^{2} c^{2} - 36 \, a c^{3}\right )} x^{3}\right )} \sqrt{c x^{6} + b x^{3} + a}}{1152 \, c^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(c*x^6+b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

[1/2304*(3*(35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*sqrt(c)*log(-8*c^2*x^6 - 8*b*c*x^3 - b^2 - 4*sqrt(c*x^6 + b*x^3
+ a)*(2*c*x^3 + b)*sqrt(c) - 4*a*c) + 4*(48*c^4*x^9 - 56*b*c^3*x^6 - 105*b^3*c + 220*a*b*c^2 + 2*(35*b^2*c^2
- 36*a*c^3)*x^3)*sqrt(c*x^6 + b*x^3 + a))/c^5, -1/1152*(3*(35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*sqrt(-c)*arctan(
1/2*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b)*sqrt(-c)/(c^2*x^6 + b*c*x^3 + a*c)) - 2*(48*c^4*x^9 - 56*b*c^3*x^6 -
105*b^3*c + 220*a*b*c^2 + 2*(35*b^2*c^2 - 36*a*c^3)*x^3)*sqrt(c*x^6 + b*x^3 + a))/c^5]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{14}}{\sqrt{a + b x^{3} + c x^{6}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**14/(c*x**6+b*x**3+a)**(1/2),x)

[Out]

Integral(x**14/sqrt(a + b*x**3 + c*x**6), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{14}}{\sqrt{c x^{6} + b x^{3} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(c*x^6+b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

integrate(x^14/sqrt(c*x^6 + b*x^3 + a), x)