3.234 \(\int \frac {x^{14}}{(a+b x^3+c x^6)^{3/2}} \, dx\)

Optimal. Leaf size=195 \[ \frac {\left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{8 c^{7/2}}-\frac {\left (b \left (15 b^2-52 a c\right )-2 c x^3 \left (5 b^2-12 a c\right )\right ) \sqrt {a+b x^3+c x^6}}{12 c^3 \left (b^2-4 a c\right )}-\frac {2 b x^6 \sqrt {a+b x^3+c x^6}}{3 c \left (b^2-4 a c\right )}+\frac {2 x^9 \left (2 a+b x^3\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}} \]

[Out]

1/8*(-4*a*c+5*b^2)*arctanh(1/2*(2*c*x^3+b)/c^(1/2)/(c*x^6+b*x^3+a)^(1/2))/c^(7/2)+2/3*x^9*(b*x^3+2*a)/(-4*a*c+
b^2)/(c*x^6+b*x^3+a)^(1/2)-2/3*b*x^6*(c*x^6+b*x^3+a)^(1/2)/c/(-4*a*c+b^2)-1/12*(b*(-52*a*c+15*b^2)-2*c*(-12*a*
c+5*b^2)*x^3)*(c*x^6+b*x^3+a)^(1/2)/c^3/(-4*a*c+b^2)

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Rubi [A]  time = 0.23, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1357, 738, 832, 779, 621, 206} \[ -\frac {\left (b \left (15 b^2-52 a c\right )-2 c x^3 \left (5 b^2-12 a c\right )\right ) \sqrt {a+b x^3+c x^6}}{12 c^3 \left (b^2-4 a c\right )}+\frac {\left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{8 c^{7/2}}+\frac {2 x^9 \left (2 a+b x^3\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 b x^6 \sqrt {a+b x^3+c x^6}}{3 c \left (b^2-4 a c\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^14/(a + b*x^3 + c*x^6)^(3/2),x]

[Out]

(2*x^9*(2*a + b*x^3))/(3*(b^2 - 4*a*c)*Sqrt[a + b*x^3 + c*x^6]) - (2*b*x^6*Sqrt[a + b*x^3 + c*x^6])/(3*c*(b^2
- 4*a*c)) - ((b*(15*b^2 - 52*a*c) - 2*c*(5*b^2 - 12*a*c)*x^3)*Sqrt[a + b*x^3 + c*x^6])/(12*c^3*(b^2 - 4*a*c))
+ ((5*b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(8*c^(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 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 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, 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] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

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 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 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]]

Rubi steps

\begin {align*} \int \frac {x^{14}}{\left (a+b x^3+c x^6\right )^{3/2}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx,x,x^3\right )\\ &=\frac {2 x^9 \left (2 a+b x^3\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 \operatorname {Subst}\left (\int \frac {x^2 (6 a+3 b x)}{\sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{3 \left (b^2-4 a c\right )}\\ &=\frac {2 x^9 \left (2 a+b x^3\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 b x^6 \sqrt {a+b x^3+c x^6}}{3 c \left (b^2-4 a c\right )}-\frac {2 \operatorname {Subst}\left (\int \frac {x \left (-6 a b-\frac {3}{2} \left (5 b^2-12 a c\right ) x\right )}{\sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{9 c \left (b^2-4 a c\right )}\\ &=\frac {2 x^9 \left (2 a+b x^3\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 b x^6 \sqrt {a+b x^3+c x^6}}{3 c \left (b^2-4 a c\right )}-\frac {\left (b \left (15 b^2-52 a c\right )-2 c \left (5 b^2-12 a c\right ) x^3\right ) \sqrt {a+b x^3+c x^6}}{12 c^3 \left (b^2-4 a c\right )}+\frac {\left (5 b^2-4 a c\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{8 c^3}\\ &=\frac {2 x^9 \left (2 a+b x^3\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 b x^6 \sqrt {a+b x^3+c x^6}}{3 c \left (b^2-4 a c\right )}-\frac {\left (b \left (15 b^2-52 a c\right )-2 c \left (5 b^2-12 a c\right ) x^3\right ) \sqrt {a+b x^3+c x^6}}{12 c^3 \left (b^2-4 a c\right )}+\frac {\left (5 b^2-4 a c\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 )}{4 c^3}\\ &=\frac {2 x^9 \left (2 a+b x^3\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 b x^6 \sqrt {a+b x^3+c x^6}}{3 c \left (b^2-4 a c\right )}-\frac {\left (b \left (15 b^2-52 a c\right )-2 c \left (5 b^2-12 a c\right ) x^3\right ) \sqrt {a+b x^3+c x^6}}{12 c^3 \left (b^2-4 a c\right )}+\frac {\left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{8 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 181, normalized size = 0.93 \[ \frac {\frac {2 \sqrt {c} \left (4 a^2 c \left (6 c x^3-13 b\right )+a \left (15 b^3-62 b^2 c x^3-20 b c^2 x^6+8 c^3 x^9\right )+b^2 x^3 \left (15 b^2+5 b c x^3-2 c^2 x^6\right )\right )}{\sqrt {a+b x^3+c x^6}}-3 \left (16 a^2 c^2-24 a b^2 c+5 b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{24 c^{7/2} \left (4 a c-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^14/(a + b*x^3 + c*x^6)^(3/2),x]

[Out]

((2*Sqrt[c]*(4*a^2*c*(-13*b + 6*c*x^3) + b^2*x^3*(15*b^2 + 5*b*c*x^3 - 2*c^2*x^6) + a*(15*b^3 - 62*b^2*c*x^3 -
20*b*c^2*x^6 + 8*c^3*x^9)))/Sqrt[a + b*x^3 + c*x^6] - 3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*ArcTanh[(b + 2*c*x^
3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(24*c^(7/2)*(-b^2 + 4*a*c))

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fricas [A]  time = 1.41, size = 591, normalized size = 3.03 \[ \left [-\frac {3 \, {\left ({\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{6} + 5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{3}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} + 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{9} - 5 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{6} - 15 \, a b^{3} c + 52 \, a^{2} b c^{2} - {\left (15 \, b^{4} c - 62 \, a b^{2} c^{2} + 24 \, a^{2} c^{3}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{48 \, {\left (a b^{2} c^{4} - 4 \, a^{2} c^{5} + {\left (b^{2} c^{5} - 4 \, a c^{6}\right )} x^{6} + {\left (b^{3} c^{4} - 4 \, a b c^{5}\right )} x^{3}\right )}}, -\frac {3 \, {\left ({\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{6} + 5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) - 2 \, {\left (2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{9} - 5 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{6} - 15 \, a b^{3} c + 52 \, a^{2} b c^{2} - {\left (15 \, b^{4} c - 62 \, a b^{2} c^{2} + 24 \, a^{2} c^{3}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{24 \, {\left (a b^{2} c^{4} - 4 \, a^{2} c^{5} + {\left (b^{2} c^{5} - 4 \, a c^{6}\right )} x^{6} + {\left (b^{3} c^{4} - 4 \, a b c^{5}\right )} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(3*((5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*x^6 + 5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2 + (5*b^5 - 24*a*b^3
*c + 16*a^2*b*c^2)*x^3)*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)*sqr
t(c) - 4*a*c) - 4*(2*(b^2*c^3 - 4*a*c^4)*x^9 - 5*(b^3*c^2 - 4*a*b*c^3)*x^6 - 15*a*b^3*c + 52*a^2*b*c^2 - (15*b
^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*x^3)*sqrt(c*x^6 + b*x^3 + a))/(a*b^2*c^4 - 4*a^2*c^5 + (b^2*c^5 - 4*a*c^6)*x
^6 + (b^3*c^4 - 4*a*b*c^5)*x^3), -1/24*(3*((5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*x^6 + 5*a*b^4 - 24*a^2*b^2*c
+ 16*a^3*c^2 + (5*b^5 - 24*a*b^3*c + 16*a^2*b*c^2)*x^3)*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*(2*(b^2*c^3 - 4*a*c^4)*x^9 - 5*(b^3*c^2 - 4*a*b*c^3)*x^6 - 15*a*b^
3*c + 52*a^2*b*c^2 - (15*b^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*x^3)*sqrt(c*x^6 + b*x^3 + a))/(a*b^2*c^4 - 4*a^2*c
^5 + (b^2*c^5 - 4*a*c^6)*x^6 + (b^3*c^4 - 4*a*b*c^5)*x^3)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{14}}{{\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {x^{14}}{\left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^14/(c*x^6+b*x^3+a)^(3/2),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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{14}}{{\left (c\,x^6+b\,x^3+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/(a + b*x^3 + c*x^6)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{14}}{\left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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