3.789 \(\int x^5 (a+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=95 \[ -\frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{32 c^{3/2}}+\frac{a^2 x^2 \sqrt{a+c x^4}}{32 c}+\frac{1}{12} x^6 \left (a+c x^4\right )^{3/2}+\frac{1}{16} a x^6 \sqrt{a+c x^4} \]

[Out]

(a^2*x^2*Sqrt[a + c*x^4])/(32*c) + (a*x^6*Sqrt[a + c*x^4])/16 + (x^6*(a + c*x^4)^(3/2))/12 - (a^3*ArcTanh[(Sqr
t[c]*x^2)/Sqrt[a + c*x^4]])/(32*c^(3/2))

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Rubi [A]  time = 0.0591329, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{32 c^{3/2}}+\frac{a^2 x^2 \sqrt{a+c x^4}}{32 c}+\frac{1}{12} x^6 \left (a+c x^4\right )^{3/2}+\frac{1}{16} a x^6 \sqrt{a+c x^4} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + c*x^4)^(3/2),x]

[Out]

(a^2*x^2*Sqrt[a + c*x^4])/(32*c) + (a*x^6*Sqrt[a + c*x^4])/16 + (x^6*(a + c*x^4)^(3/2))/12 - (a^3*ArcTanh[(Sqr
t[c]*x^2)/Sqrt[a + c*x^4]])/(32*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 \left (a+c x^4\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 \left (a+c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{1}{12} x^6 \left (a+c x^4\right )^{3/2}+\frac{1}{4} a \operatorname{Subst}\left (\int x^2 \sqrt{a+c x^2} \, dx,x,x^2\right )\\ &=\frac{1}{16} a x^6 \sqrt{a+c x^4}+\frac{1}{12} x^6 \left (a+c x^4\right )^{3/2}+\frac{1}{16} a^2 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+c x^2}} \, dx,x,x^2\right )\\ &=\frac{a^2 x^2 \sqrt{a+c x^4}}{32 c}+\frac{1}{16} a x^6 \sqrt{a+c x^4}+\frac{1}{12} x^6 \left (a+c x^4\right )^{3/2}-\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^2}} \, dx,x,x^2\right )}{32 c}\\ &=\frac{a^2 x^2 \sqrt{a+c x^4}}{32 c}+\frac{1}{16} a x^6 \sqrt{a+c x^4}+\frac{1}{12} x^6 \left (a+c x^4\right )^{3/2}-\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x^2}{\sqrt{a+c x^4}}\right )}{32 c}\\ &=\frac{a^2 x^2 \sqrt{a+c x^4}}{32 c}+\frac{1}{16} a x^6 \sqrt{a+c x^4}+\frac{1}{12} x^6 \left (a+c x^4\right )^{3/2}-\frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a+c x^4}}\right )}{32 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.12345, size = 87, normalized size = 0.92 \[ \frac{\sqrt{a+c x^4} \left (\sqrt{c} x^2 \left (3 a^2+14 a c x^4+8 c^2 x^8\right )-\frac{3 a^{5/2} \sinh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{\sqrt{\frac{c x^4}{a}+1}}\right )}{96 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + c*x^4)^(3/2),x]

[Out]

(Sqrt[a + c*x^4]*(Sqrt[c]*x^2*(3*a^2 + 14*a*c*x^4 + 8*c^2*x^8) - (3*a^(5/2)*ArcSinh[(Sqrt[c]*x^2)/Sqrt[a]])/Sq
rt[1 + (c*x^4)/a]))/(96*c^(3/2))

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Maple [A]  time = 0.016, size = 78, normalized size = 0.8 \begin{align*}{\frac{c{x}^{10}}{12}\sqrt{c{x}^{4}+a}}+{\frac{7\,{x}^{6}a}{48}\sqrt{c{x}^{4}+a}}+{\frac{{a}^{2}{x}^{2}}{32\,c}\sqrt{c{x}^{4}+a}}-{\frac{{a}^{3}}{32}\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)^(3/2),x)

[Out]

1/12*c*x^10*(c*x^4+a)^(1/2)+7/48*a*x^6*(c*x^4+a)^(1/2)+1/32*a^2*x^2*(c*x^4+a)^(1/2)/c-1/32/c^(3/2)*a^3*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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.63983, size = 360, normalized size = 3.79 \begin{align*} \left [\frac{3 \, a^{3} \sqrt{c} \log \left (-2 \, c x^{4} + 2 \, \sqrt{c x^{4} + a} \sqrt{c} x^{2} - a\right ) + 2 \,{\left (8 \, c^{3} x^{10} + 14 \, a c^{2} x^{6} + 3 \, a^{2} c x^{2}\right )} \sqrt{c x^{4} + a}}{192 \, c^{2}}, \frac{3 \, a^{3} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x^{2}}{\sqrt{c x^{4} + a}}\right ) +{\left (8 \, c^{3} x^{10} + 14 \, a c^{2} x^{6} + 3 \, a^{2} c x^{2}\right )} \sqrt{c x^{4} + a}}{96 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

[1/192*(3*a^3*sqrt(c)*log(-2*c*x^4 + 2*sqrt(c*x^4 + a)*sqrt(c)*x^2 - a) + 2*(8*c^3*x^10 + 14*a*c^2*x^6 + 3*a^2
*c*x^2)*sqrt(c*x^4 + a))/c^2, 1/96*(3*a^3*sqrt(-c)*arctan(sqrt(-c)*x^2/sqrt(c*x^4 + a)) + (8*c^3*x^10 + 14*a*c
^2*x^6 + 3*a^2*c*x^2)*sqrt(c*x^4 + a))/c^2]

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Sympy [A]  time = 7.24864, size = 122, normalized size = 1.28 \begin{align*} \frac{a^{\frac{5}{2}} x^{2}}{32 c \sqrt{1 + \frac{c x^{4}}{a}}} + \frac{17 a^{\frac{3}{2}} x^{6}}{96 \sqrt{1 + \frac{c x^{4}}{a}}} + \frac{11 \sqrt{a} c x^{10}}{48 \sqrt{1 + \frac{c x^{4}}{a}}} - \frac{a^{3} \operatorname{asinh}{\left (\frac{\sqrt{c} x^{2}}{\sqrt{a}} \right )}}{32 c^{\frac{3}{2}}} + \frac{c^{2} x^{14}}{12 \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)**(3/2),x)

[Out]

a**(5/2)*x**2/(32*c*sqrt(1 + c*x**4/a)) + 17*a**(3/2)*x**6/(96*sqrt(1 + c*x**4/a)) + 11*sqrt(a)*c*x**10/(48*sq
rt(1 + c*x**4/a)) - a**3*asinh(sqrt(c)*x**2/sqrt(a))/(32*c**(3/2)) + c**2*x**14/(12*sqrt(a)*sqrt(1 + c*x**4/a)
)

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Giac [A]  time = 1.1534, size = 90, normalized size = 0.95 \begin{align*} \frac{1}{96} \,{\left (2 \,{\left (4 \, c x^{4} + 7 \, a\right )} x^{4} + \frac{3 \, a^{2}}{c}\right )} \sqrt{c x^{4} + a} x^{2} + \frac{a^{3} \log \left ({\left | -\sqrt{c} x^{2} + \sqrt{c x^{4} + a} \right |}\right )}{32 \, c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^4+a)^(3/2),x, algorithm="giac")

[Out]

1/96*(2*(4*c*x^4 + 7*a)*x^4 + 3*a^2/c)*sqrt(c*x^4 + a)*x^2 + 1/32*a^3*log(abs(-sqrt(c)*x^2 + sqrt(c*x^4 + a)))
/c^(3/2)