3.770 \(\int x^5 \sqrt{a+c x^4} \, dx\)

Optimal. Leaf size=74 \[ -\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{16 c^{3/2}}+\frac{1}{8} x^6 \sqrt{a+c x^4}+\frac{a x^2 \sqrt{a+c x^4}}{16 c} \]

[Out]

(a*x^2*Sqrt[a + c*x^4])/(16*c) + (x^6*Sqrt[a + c*x^4])/8 - (a^2*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(16*c^
(3/2))

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Rubi [A]  time = 0.0461811, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {275, 279, 321, 217, 206} \[ -\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{16 c^{3/2}}+\frac{1}{8} x^6 \sqrt{a+c x^4}+\frac{a x^2 \sqrt{a+c x^4}}{16 c} \]

Antiderivative was successfully verified.

[In]

Int[x^5*Sqrt[a + c*x^4],x]

[Out]

(a*x^2*Sqrt[a + c*x^4])/(16*c) + (x^6*Sqrt[a + c*x^4])/8 - (a^2*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(16*c^
(3/2))

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 279

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

Rule 321

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 x^5 \sqrt{a+c x^4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 \sqrt{a+c x^2} \, dx,x,x^2\right )\\ &=\frac{1}{8} x^6 \sqrt{a+c x^4}+\frac{1}{8} a \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+c x^2}} \, dx,x,x^2\right )\\ &=\frac{a x^2 \sqrt{a+c x^4}}{16 c}+\frac{1}{8} x^6 \sqrt{a+c x^4}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^2}} \, dx,x,x^2\right )}{16 c}\\ &=\frac{a x^2 \sqrt{a+c x^4}}{16 c}+\frac{1}{8} x^6 \sqrt{a+c x^4}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x^2}{\sqrt{a+c x^4}}\right )}{16 c}\\ &=\frac{a x^2 \sqrt{a+c x^4}}{16 c}+\frac{1}{8} x^6 \sqrt{a+c x^4}-\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{16 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.102809, size = 74, normalized size = 1. \[ \frac{\sqrt{a+c x^4} \left (\sqrt{c} x^2 \left (a+2 c x^4\right )-\frac{a^{3/2} \sinh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{\sqrt{\frac{c x^4}{a}+1}}\right )}{16 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*Sqrt[a + c*x^4],x]

[Out]

(Sqrt[a + c*x^4]*(Sqrt[c]*x^2*(a + 2*c*x^4) - (a^(3/2)*ArcSinh[(Sqrt[c]*x^2)/Sqrt[a]])/Sqrt[1 + (c*x^4)/a]))/(
16*c^(3/2))

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Maple [A]  time = 0.011, size = 63, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}}{8\,c} \left ( c{x}^{4}+a \right ) ^{{\frac{3}{2}}}}-{\frac{a{x}^{2}}{16\,c}\sqrt{c{x}^{4}+a}}-{\frac{{a}^{2}}{16}\ln \left ({x}^{2}\sqrt{c}+\sqrt{c{x}^{4}+a} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x^2*(c*x^4+a)^(3/2)/c-1/16*a*x^2*(c*x^4+a)^(1/2)/c-1/16*a^2/c^(3/2)*ln(x^2*c^(1/2)+(c*x^4+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*(c*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55014, size = 300, normalized size = 4.05 \begin{align*} \left [\frac{a^{2} \sqrt{c} \log \left (-2 \, c x^{4} + 2 \, \sqrt{c x^{4} + a} \sqrt{c} x^{2} - a\right ) + 2 \,{\left (2 \, c^{2} x^{6} + a c x^{2}\right )} \sqrt{c x^{4} + a}}{32 \, c^{2}}, \frac{a^{2} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x^{2}}{\sqrt{c x^{4} + a}}\right ) +{\left (2 \, c^{2} x^{6} + a c x^{2}\right )} \sqrt{c x^{4} + a}}{16 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

[1/32*(a^2*sqrt(c)*log(-2*c*x^4 + 2*sqrt(c*x^4 + a)*sqrt(c)*x^2 - a) + 2*(2*c^2*x^6 + a*c*x^2)*sqrt(c*x^4 + a)
)/c^2, 1/16*(a^2*sqrt(-c)*arctan(sqrt(-c)*x^2/sqrt(c*x^4 + a)) + (2*c^2*x^6 + a*c*x^2)*sqrt(c*x^4 + a))/c^2]

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Sympy [A]  time = 4.6949, size = 95, normalized size = 1.28 \begin{align*} \frac{a^{\frac{3}{2}} x^{2}}{16 c \sqrt{1 + \frac{c x^{4}}{a}}} + \frac{3 \sqrt{a} x^{6}}{16 \sqrt{1 + \frac{c x^{4}}{a}}} - \frac{a^{2} \operatorname{asinh}{\left (\frac{\sqrt{c} x^{2}}{\sqrt{a}} \right )}}{16 c^{\frac{3}{2}}} + \frac{c x^{10}}{8 \sqrt{a} \sqrt{1 + \frac{c x^{4}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(c*x**4+a)**(1/2),x)

[Out]

a**(3/2)*x**2/(16*c*sqrt(1 + c*x**4/a)) + 3*sqrt(a)*x**6/(16*sqrt(1 + c*x**4/a)) - a**2*asinh(sqrt(c)*x**2/sqr
t(a))/(16*c**(3/2)) + c*x**10/(8*sqrt(a)*sqrt(1 + c*x**4/a))

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Giac [A]  time = 1.11629, size = 73, normalized size = 0.99 \begin{align*} \frac{1}{16} \, \sqrt{c x^{4} + a}{\left (2 \, x^{4} + \frac{a}{c}\right )} x^{2} + \frac{a^{2} \log \left ({\left | -\sqrt{c} x^{2} + \sqrt{c x^{4} + a} \right |}\right )}{16 \, c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^4+a)^(1/2),x, algorithm="giac")

[Out]

1/16*sqrt(c*x^4 + a)*(2*x^4 + a/c)*x^2 + 1/16*a^2*log(abs(-sqrt(c)*x^2 + sqrt(c*x^4 + a)))/c^(3/2)