3.8.70 \(\int x^5 \sqrt {a+c x^4} \, dx\) [770]

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

[Out]

-1/16*a^2*arctanh(x^2*c^(1/2)/(c*x^4+a)^(1/2))/c^(3/2)+1/16*a*x^2*(c*x^4+a)^(1/2)/c+1/8*x^6*(c*x^4+a)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 74, normalized size of antiderivative = 1.00, 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 = {281, 285, 327, 223, 212} \begin {gather*} -\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} \end {gather*}

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 281

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 285

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 327

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]

Rubi steps

\begin {align*} \int x^5 \sqrt {a+c x^4} \, dx &=\frac {1}{2} \text {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 \text {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 \text {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 \text {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*}

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Mathematica [A]
time = 0.13, size = 63, normalized size = 0.85 \begin {gather*} \frac {x^2 \sqrt {a+c x^4} \left (a+2 c x^4\right )}{16 c}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {c} x^2}\right )}{16 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*Sqrt[a + c*x^4],x]

[Out]

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

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Maple [A]
time = 0.14, size = 63, normalized size = 0.85

method	result	size
risch	\(\frac {x^{2} \left (2 x^{4} c +a \right ) \sqrt {x^{4} c +a}}{16 c}-\frac {a^{2} \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{16 c^{\frac {3}{2}}}\)	\(53\)
default	\(\frac {x^{2} \left (x^{4} c +a \right )^{\frac {3}{2}}}{8 c}-\frac {a \,x^{2} \sqrt {x^{4} c +a}}{16 c}-\frac {a^{2} \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{16 c^{\frac {3}{2}}}\)	\(63\)
elliptic	\(\frac {x^{2} \left (x^{4} c +a \right )^{\frac {3}{2}}}{8 c}-\frac {a \,x^{2} \sqrt {x^{4} c +a}}{16 c}-\frac {a^{2} \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{16 c^{\frac {3}{2}}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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/c^(3/2)*a^2*ln(x^2*c^(1/2)+(c*x^4+a)^(1/2))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (58) = 116\).
time = 0.50, size = 120, normalized size = 1.62 \begin {gather*} \frac {a^{2} \log \left (-\frac {\sqrt {c} - \frac {\sqrt {c x^{4} + a}}{x^{2}}}{\sqrt {c} + \frac {\sqrt {c x^{4} + a}}{x^{2}}}\right )}{32 \, c^{\frac {3}{2}}} + \frac {\frac {\sqrt {c x^{4} + a} a^{2} c}{x^{2}} + \frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}} a^{2}}{x^{6}}}{16 \, {\left (c^{3} - \frac {2 \, {\left (c x^{4} + a\right )} c^{2}}{x^{4}} + \frac {{\left (c x^{4} + a\right )}^{2} c}{x^{8}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*a^2*log(-(sqrt(c) - sqrt(c*x^4 + a)/x^2)/(sqrt(c) + sqrt(c*x^4 + a)/x^2))/c^(3/2) + 1/16*(sqrt(c*x^4 + a)
*a^2*c/x^2 + (c*x^4 + a)^(3/2)*a^2/x^6)/(c^3 - 2*(c*x^4 + a)*c^2/x^4 + (c*x^4 + a)^2*c/x^8)

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Fricas [A]
time = 0.36, size = 127, normalized size = 1.72 \begin {gather*} \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 {gather*}

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 = 2.00, size = 95, normalized size = 1.28 \begin {gather*} \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 {gather*}

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 = 0.60, size = 54, normalized size = 0.73 \begin {gather*} \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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^5\,\sqrt {c\,x^4+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + c*x^4)^(1/2),x)

[Out]

int(x^5*(a + c*x^4)^(1/2), x)

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