3.790 \(\int x (a+c x^4)^{3/2} \, dx\)

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

[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])

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Rubi [A]  time = 0.0359085, antiderivative size = 71, normalized size of antiderivative = 1., 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 = {275, 195, 217, 206} \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{16 \sqrt{c}}+\frac{3}{16} a x^2 \sqrt{a+c x^4}+\frac{1}{8} x^2 \left (a+c x^4\right )^{3/2} \]

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 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 195

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 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 \left (a+c x^4\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{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) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.0969913, size = 69, normalized size = 0.97 \[ \frac{1}{16} \sqrt{a+c x^4} \left (\frac{3 a^{3/2} \sinh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{\sqrt{c} \sqrt{\frac{c x^4}{a}+1}}+5 a x^2+2 c x^6\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + c*x^4)^(3/2),x]

[Out]

(Sqrt[a + c*x^4]*(5*a*x^2 + 2*c*x^6 + (3*a^(3/2)*ArcSinh[(Sqrt[c]*x^2)/Sqrt[a]])/(Sqrt[c]*Sqrt[1 + (c*x^4)/a])
))/16

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Maple [A]  time = 0.01, size = 58, normalized size = 0.8 \begin{align*}{\frac{c{x}^{6}}{8}\sqrt{c{x}^{4}+a}}+{\frac{5\,a{x}^{2}}{16}\sqrt{c{x}^{4}+a}}+{\frac{3\,{a}^{2}}{16}\ln \left ({x}^{2}\sqrt{c}+\sqrt{c{x}^{4}+a} \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58657, size = 306, normalized size = 4.31 \begin{align*} \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{align*}

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 = 3.88753, size = 73, normalized size = 1.03 \begin{align*} \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{align*}

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 = 1.1194, size = 72, normalized size = 1.01 \begin{align*} \frac{1}{16} \,{\left (2 \, c x^{4} + 5 \, a\right )} \sqrt{c x^{4} + a} x^{2} - \frac{3 \, a^{2} \log \left ({\left | -\sqrt{c} x^{2} + \sqrt{c x^{4} + a} \right |}\right )}{16 \, \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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