3.8.90 \(\int x (a+c x^4)^{3/2} \, dx\) [790]

Optimal. Leaf size=71 \[ \frac {3}{16} a x^2 \sqrt {a+c x^4}+\frac {1}{8} x^2 \left (a+c x^4\right )^{3/2}+\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{16 \sqrt {c}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {281, 201, 223, 212} \begin {gather*} \frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{16 \sqrt {c}}+\frac {3}{16} a x^2 \sqrt {a+c x^4}+\frac {1}{8} x^2 \left (a+c x^4\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + c*x^4)^(3/2),x]

[Out]

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

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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]

Rubi steps

\begin {align*} \int x \left (a+c x^4\right )^{3/2} \, dx &=\frac {1}{2} \text {Subst}\left (\int \left (a+c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac {1}{8} x^2 \left (a+c x^4\right )^{3/2}+\frac {1}{8} (3 a) \text {Subst}\left (\int \sqrt {a+c x^2} \, dx,x,x^2\right )\\ &=\frac {3}{16} a x^2 \sqrt {a+c x^4}+\frac {1}{8} x^2 \left (a+c x^4\right )^{3/2}+\frac {1}{16} \left (3 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+c x^2}} \, dx,x,x^2\right )\\ &=\frac {3}{16} a x^2 \sqrt {a+c x^4}+\frac {1}{8} x^2 \left (a+c x^4\right )^{3/2}+\frac {1}{16} \left (3 a^2\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {a+c x^4}}\right )\\ &=\frac {3}{16} a x^2 \sqrt {a+c x^4}+\frac {1}{8} x^2 \left (a+c x^4\right )^{3/2}+\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{16 \sqrt {c}}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 62, normalized size = 0.87 \begin {gather*} \frac {1}{16} x^2 \sqrt {a+c x^4} \left (5 a+2 c x^4\right )+\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {c} x^2}\right )}{16 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(a + c*x^4)^(3/2),x]

[Out]

(x^2*Sqrt[a + c*x^4]*(5*a + 2*c*x^4))/16 + (3*a^2*ArcTanh[Sqrt[a + c*x^4]/(Sqrt[c]*x^2)])/(16*Sqrt[c])

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Maple [A]
time = 0.15, size = 58, normalized size = 0.82

method	result	size
risch	\(\frac {x^{2} \left (2 x^{4} c +5 a \right ) \sqrt {x^{4} c +a}}{16}+\frac {3 a^{2} \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{16 \sqrt {c}}\)	\(52\)
default	\(\frac {c \,x^{6} \sqrt {x^{4} c +a}}{8}+\frac {5 a \,x^{2} \sqrt {x^{4} c +a}}{16}+\frac {3 a^{2} \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{16 \sqrt {c}}\)	\(58\)
elliptic	\(\frac {c \,x^{6} \sqrt {x^{4} c +a}}{8}+\frac {5 a \,x^{2} \sqrt {x^{4} c +a}}{16}+\frac {3 a^{2} \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{16 \sqrt {c}}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*x^4+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/8*c*x^6*(c*x^4+a)^(1/2)+5/16*a*x^2*(c*x^4+a)^(1/2)+3/16*a^2*ln(x^2*c^(1/2)+(c*x^4+a)^(1/2))/c^(1/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (55) = 110\).
time = 0.50, size = 119, normalized size = 1.68 \begin {gather*} -\frac {3 \, 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 \, \sqrt {c}} - \frac {\frac {3 \, \sqrt {c x^{4} + a} a^{2} c}{x^{2}} - \frac {5 \, {\left (c x^{4} + a\right )}^{\frac {3}{2}} a^{2}}{x^{6}}}{16 \, {\left (c^{2} - \frac {2 \, {\left (c x^{4} + a\right )} c}{x^{4}} + \frac {{\left (c x^{4} + a\right )}^{2}}{x^{8}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

-3/32*a^2*log(-(sqrt(c) - sqrt(c*x^4 + a)/x^2)/(sqrt(c) + sqrt(c*x^4 + a)/x^2))/sqrt(c) - 1/16*(3*sqrt(c*x^4 +
a)*a^2*c/x^2 - 5*(c*x^4 + a)^(3/2)*a^2/x^6)/(c^2 - 2*(c*x^4 + a)*c/x^4 + (c*x^4 + a)^2/x^8)

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Fricas [A]
time = 0.38, size = 132, normalized size = 1.86 \begin {gather*} \left [\frac {3 \, 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} + 5 \, a c x^{2}\right )} \sqrt {c x^{4} + a}}{32 \, c}, -\frac {3 \, a^{2} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x^{2}}{\sqrt {c x^{4} + a}}\right ) - {\left (2 \, c^{2} x^{6} + 5 \, a c x^{2}\right )} \sqrt {c x^{4} + a}}{16 \, c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

[1/32*(3*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 + 5*a*c*x^2)*sqrt(c*x^4
+ a))/c, -1/16*(3*a^2*sqrt(-c)*arctan(sqrt(-c)*x^2/sqrt(c*x^4 + a)) - (2*c^2*x^6 + 5*a*c*x^2)*sqrt(c*x^4 + a))
/c]

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Sympy [A]
time = 1.51, size = 73, normalized size = 1.03 \begin {gather*} \frac {5 a^{\frac {3}{2}} x^{2} \sqrt {1 + \frac {c x^{4}}{a}}}{16} + \frac {\sqrt {a} c x^{6} \sqrt {1 + \frac {c x^{4}}{a}}}{8} + \frac {3 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x^{2}}{\sqrt {a}} \right )}}{16 \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x**4+a)**(3/2),x)

[Out]

5*a**(3/2)*x**2*sqrt(1 + c*x**4/a)/16 + sqrt(a)*c*x**6*sqrt(1 + c*x**4/a)/8 + 3*a**2*asinh(sqrt(c)*x**2/sqrt(a
))/(16*sqrt(c))

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Giac [A]
time = 1.33, size = 99, normalized size = 1.39 \begin {gather*} \frac {1}{4} \, {\left (\sqrt {c x^{4} + a} x^{2} - \frac {a \log \left ({\left | -\sqrt {c} x^{2} + \sqrt {c x^{4} + a} \right |}\right )}{\sqrt {c}}\right )} a + \frac {1}{16} \, {\left (\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 )}{c^{\frac {3}{2}}}\right )} c \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x^4+a)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,{\left (c\,x^4+a\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + c*x^4)^(3/2),x)

[Out]

int(x*(a + c*x^4)^(3/2), x)

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