3.204 \(\int x^5 (a+b x^3+c x^6)^{3/2} \, dx\)

Optimal. Leaf size=150 \[ -\frac {b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{256 c^{7/2}}+\frac {b \left (b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{128 c^3}-\frac {b \left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 c^2}+\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 c} \]

[Out]

-1/48*b*(2*c*x^3+b)*(c*x^6+b*x^3+a)^(3/2)/c^2+1/15*(c*x^6+b*x^3+a)^(5/2)/c-1/256*b*(-4*a*c+b^2)^2*arctanh(1/2*
(2*c*x^3+b)/c^(1/2)/(c*x^6+b*x^3+a)^(1/2))/c^(7/2)+1/128*b*(-4*a*c+b^2)*(2*c*x^3+b)*(c*x^6+b*x^3+a)^(1/2)/c^3

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Rubi [A]  time = 0.12, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1357, 640, 612, 621, 206} \[ \frac {b \left (b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{128 c^3}-\frac {b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{256 c^{7/2}}-\frac {b \left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 c^2}+\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 c} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^3 + c*x^6)^(3/2),x]

[Out]

(b*(b^2 - 4*a*c)*(b + 2*c*x^3)*Sqrt[a + b*x^3 + c*x^6])/(128*c^3) - (b*(b + 2*c*x^3)*(a + b*x^3 + c*x^6)^(3/2)
)/(48*c^2) + (a + b*x^3 + c*x^6)^(5/2)/(15*c) - (b*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b
*x^3 + c*x^6])])/(256*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 612

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

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 640

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

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 x^5 \left (a+b x^3+c x^6\right )^{3/2} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int x \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^3\right )\\ &=\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 c}-\frac {b \operatorname {Subst}\left (\int \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^3\right )}{6 c}\\ &=-\frac {b \left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 c^2}+\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 c}+\frac {\left (b \left (b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \sqrt {a+b x+c x^2} \, dx,x,x^3\right )}{32 c^2}\\ &=\frac {b \left (b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{128 c^3}-\frac {b \left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 c^2}+\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 c}-\frac {\left (b \left (b^2-4 a c\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{256 c^3}\\ &=\frac {b \left (b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{128 c^3}-\frac {b \left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 c^2}+\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 c}-\frac {\left (b \left (b^2-4 a c\right )^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 )}{128 c^3}\\ &=\frac {b \left (b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{128 c^3}-\frac {b \left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 c^2}+\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 c}-\frac {b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{256 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 149, normalized size = 0.99 \[ -\frac {b \left (b^2-4 a c\right ) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )-2 \sqrt {c} \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}\right )}{256 c^{7/2}}-\frac {b \left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 c^2}+\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^3 + c*x^6)^(3/2),x]

[Out]

-1/48*(b*(b + 2*c*x^3)*(a + b*x^3 + c*x^6)^(3/2))/c^2 + (a + b*x^3 + c*x^6)^(5/2)/(15*c) - (b*(b^2 - 4*a*c)*(-
2*Sqrt[c]*(b + 2*c*x^3)*Sqrt[a + b*x^3 + c*x^6] + (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^
3 + c*x^6])]))/(256*c^(7/2))

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fricas [A]  time = 1.11, size = 361, normalized size = 2.41 \[ \left [\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} + 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (128 \, c^{5} x^{12} + 176 \, b c^{4} x^{9} + 8 \, {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{6} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} - 2 \, {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{7680 \, c^{4}}, \frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) + 2 \, {\left (128 \, c^{5} x^{12} + 176 \, b c^{4} x^{9} + 8 \, {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{6} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} - 2 \, {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{3840 \, c^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^6+b*x^3+a)^(3/2),x, algorithm="fricas")

[Out]

[1/7680*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*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*(128*c^5*x^12 + 176*b*c^4*x^9 + 8*(b^2*c^3 + 32*a*c^4)*x^6 + 15*b^4*c -
100*a*b^2*c^2 + 128*a^2*c^3 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x^3)*sqrt(c*x^6 + b*x^3 + a))/c^4, 1/3840*(15*(b^5 -
8*a*b^3*c + 16*a^2*b*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*(128*c^5*x^12 + 176*b*c^4*x^9 + 8*(b^2*c^3 + 32*a*c^4)*x^6 + 15*b^4*c - 100*a*b^2*c^2 + 128*a^2
*c^3 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x^3)*sqrt(c*x^6 + b*x^3 + a))/c^4]

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giac [A]  time = 0.56, size = 172, normalized size = 1.15 \[ \frac {1}{1920} \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, c x^{3} + 11 \, b\right )} x^{3} + \frac {b^{2} c^{3} + 32 \, a c^{4}}{c^{4}}\right )} x^{3} - \frac {5 \, b^{3} c^{2} - 28 \, a b c^{3}}{c^{4}}\right )} x^{3} + \frac {15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3}}{c^{4}}\right )} + \frac {{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{3} - \sqrt {c x^{6} + b x^{3} + a}\right )} \sqrt {c} - b \right |}\right )}{256 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^6+b*x^3+a)^(3/2),x, algorithm="giac")

[Out]

1/1920*sqrt(c*x^6 + b*x^3 + a)*(2*(4*(2*(8*c*x^3 + 11*b)*x^3 + (b^2*c^3 + 32*a*c^4)/c^4)*x^3 - (5*b^3*c^2 - 28
*a*b*c^3)/c^4)*x^3 + (15*b^4*c - 100*a*b^2*c^2 + 128*a^2*c^3)/c^4) + 1/256*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*lo
g(abs(-2*(sqrt(c)*x^3 - sqrt(c*x^6 + b*x^3 + a))*sqrt(c) - b))/c^(7/2)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}} x^{5}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(c*x^6+b*x^3+a)^(3/2),x)

[Out]

int(x^5*(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^5*(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 positive, negative or zero?

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mupad [B]  time = 1.58, size = 223, normalized size = 1.49 \[ \frac {{\left (c\,x^6+b\,x^3+a\right )}^{5/2}}{15\,c}-\frac {b\,\left (\frac {3\,a\,\left (\ln \left (\sqrt {c\,x^6+b\,x^3+a}+\frac {c\,x^3+\frac {b}{2}}{\sqrt {c}}\right )\,\left (\frac {a}{2\,\sqrt {c}}-\frac {b^2}{8\,c^{3/2}}\right )+\frac {\left (2\,c\,x^3+b\right )\,\sqrt {c\,x^6+b\,x^3+a}}{4\,c}\right )}{4}+\frac {x^3\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2}}{4}-\frac {3\,b^2\,\left (\ln \left (\sqrt {c\,x^6+b\,x^3+a}+\frac {c\,x^3+\frac {b}{2}}{\sqrt {c}}\right )\,\left (\frac {a}{2\,\sqrt {c}}-\frac {b^2}{8\,c^{3/2}}\right )+\frac {\left (2\,c\,x^3+b\right )\,\sqrt {c\,x^6+b\,x^3+a}}{4\,c}\right )}{16\,c}+\frac {b\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2}}{8\,c}\right )}{6\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*x^3 + c*x^6)^(3/2),x)

[Out]

(a + b*x^3 + c*x^6)^(5/2)/(15*c) - (b*((3*a*(log((a + b*x^3 + c*x^6)^(1/2) + (b/2 + c*x^3)/c^(1/2))*(a/(2*c^(1
/2)) - b^2/(8*c^(3/2))) + ((b + 2*c*x^3)*(a + b*x^3 + c*x^6)^(1/2))/(4*c)))/4 + (x^3*(a + b*x^3 + c*x^6)^(3/2)
)/4 - (3*b^2*(log((a + b*x^3 + c*x^6)^(1/2) + (b/2 + c*x^3)/c^(1/2))*(a/(2*c^(1/2)) - b^2/(8*c^(3/2))) + ((b +
2*c*x^3)*(a + b*x^3 + c*x^6)^(1/2))/(4*c)))/(16*c) + (b*(a + b*x^3 + c*x^6)^(3/2))/(8*c)))/(6*c)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{5} \left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(c*x**6+b*x**3+a)**(3/2),x)

[Out]

Integral(x**5*(a + b*x**3 + c*x**6)**(3/2), x)

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