3.1064 \(\int x^2 (a+b x^4)^{5/4} \, dx\)

Optimal. Leaf size=100 \[ -\frac{5 a^2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{3/4}}+\frac{5 a^2 \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{3/4}}+\frac{1}{8} x^3 \left (a+b x^4\right )^{5/4}+\frac{5}{32} a x^3 \sqrt [4]{a+b x^4} \]

[Out]

(5*a*x^3*(a + b*x^4)^(1/4))/32 + (x^3*(a + b*x^4)^(5/4))/8 - (5*a^2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64
*b^(3/4)) + (5*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^(3/4))

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Rubi [A]  time = 0.0370888, antiderivative size = 100, 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 = {279, 331, 298, 203, 206} \[ -\frac{5 a^2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{3/4}}+\frac{5 a^2 \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{3/4}}+\frac{1}{8} x^3 \left (a+b x^4\right )^{5/4}+\frac{5}{32} a x^3 \sqrt [4]{a+b x^4} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^4)^(5/4),x]

[Out]

(5*a*x^3*(a + b*x^4)^(1/4))/32 + (x^3*(a + b*x^4)^(5/4))/8 - (5*a^2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64
*b^(3/4)) + (5*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^(3/4))

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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^2 \left (a+b x^4\right )^{5/4} \, dx &=\frac{1}{8} x^3 \left (a+b x^4\right )^{5/4}+\frac{1}{8} (5 a) \int x^2 \sqrt [4]{a+b x^4} \, dx\\ &=\frac{5}{32} a x^3 \sqrt [4]{a+b x^4}+\frac{1}{8} x^3 \left (a+b x^4\right )^{5/4}+\frac{1}{32} \left (5 a^2\right ) \int \frac{x^2}{\left (a+b x^4\right )^{3/4}} \, dx\\ &=\frac{5}{32} a x^3 \sqrt [4]{a+b x^4}+\frac{1}{8} x^3 \left (a+b x^4\right )^{5/4}+\frac{1}{32} \left (5 a^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-b x^4} \, dx,x,\frac{x}{\sqrt [4]{a+b x^4}}\right )\\ &=\frac{5}{32} a x^3 \sqrt [4]{a+b x^4}+\frac{1}{8} x^3 \left (a+b x^4\right )^{5/4}+\frac{\left (5 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{b} x^2} \, dx,x,\frac{x}{\sqrt [4]{a+b x^4}}\right )}{64 \sqrt{b}}-\frac{\left (5 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{b} x^2} \, dx,x,\frac{x}{\sqrt [4]{a+b x^4}}\right )}{64 \sqrt{b}}\\ &=\frac{5}{32} a x^3 \sqrt [4]{a+b x^4}+\frac{1}{8} x^3 \left (a+b x^4\right )^{5/4}-\frac{5 a^2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{3/4}}+\frac{5 a^2 \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0083934, size = 52, normalized size = 0.52 \[ \frac{a x^3 \sqrt [4]{a+b x^4} \, _2F_1\left (-\frac{5}{4},\frac{3}{4};\frac{7}{4};-\frac{b x^4}{a}\right )}{3 \sqrt [4]{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x^4)^(5/4),x]

[Out]

(a*x^3*(a + b*x^4)^(1/4)*Hypergeometric2F1[-5/4, 3/4, 7/4, -((b*x^4)/a)])/(3*(1 + (b*x^4)/a)^(1/4))

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Maple [F]  time = 0.027, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( b{x}^{4}+a \right ) ^{{\frac{5}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.67821, size = 494, normalized size = 4.94 \begin{align*} \frac{1}{32} \,{\left (4 \, b x^{7} + 9 \, a x^{3}\right )}{\left (b x^{4} + a\right )}^{\frac{1}{4}} - \frac{5}{32} \, \left (\frac{a^{8}}{b^{3}}\right )^{\frac{1}{4}} \arctan \left (-\frac{\left (\frac{a^{8}}{b^{3}}\right )^{\frac{3}{4}}{\left (b x^{4} + a\right )}^{\frac{1}{4}} a^{2} b^{2} - \left (\frac{a^{8}}{b^{3}}\right )^{\frac{3}{4}} b^{2} x \sqrt{\frac{\sqrt{b x^{4} + a} a^{4} + \sqrt{\frac{a^{8}}{b^{3}}} b^{2} x^{2}}{x^{2}}}}{a^{8} x}\right ) + \frac{5}{128} \, \left (\frac{a^{8}}{b^{3}}\right )^{\frac{1}{4}} \log \left (\frac{5 \,{\left ({\left (b x^{4} + a\right )}^{\frac{1}{4}} a^{2} + \left (\frac{a^{8}}{b^{3}}\right )^{\frac{1}{4}} b x\right )}}{x}\right ) - \frac{5}{128} \, \left (\frac{a^{8}}{b^{3}}\right )^{\frac{1}{4}} \log \left (\frac{5 \,{\left ({\left (b x^{4} + a\right )}^{\frac{1}{4}} a^{2} - \left (\frac{a^{8}}{b^{3}}\right )^{\frac{1}{4}} b x\right )}}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(4*b*x^7 + 9*a*x^3)*(b*x^4 + a)^(1/4) - 5/32*(a^8/b^3)^(1/4)*arctan(-((a^8/b^3)^(3/4)*(b*x^4 + a)^(1/4)*a
^2*b^2 - (a^8/b^3)^(3/4)*b^2*x*sqrt((sqrt(b*x^4 + a)*a^4 + sqrt(a^8/b^3)*b^2*x^2)/x^2))/(a^8*x)) + 5/128*(a^8/
b^3)^(1/4)*log(5*((b*x^4 + a)^(1/4)*a^2 + (a^8/b^3)^(1/4)*b*x)/x) - 5/128*(a^8/b^3)^(1/4)*log(5*((b*x^4 + a)^(
1/4)*a^2 - (a^8/b^3)^(1/4)*b*x)/x)

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Sympy [C]  time = 3.0933, size = 39, normalized size = 0.39 \begin{align*} \frac{a^{\frac{5}{4}} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(5/4)*x**3*gamma(3/4)*hyper((-5/4, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(7/4))

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Giac [B]  time = 1.18454, size = 346, normalized size = 3.46 \begin{align*} \frac{1}{256} \,{\left (\frac{8 \, x^{8}{\left (\frac{9 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}{\left (b + \frac{a}{x^{4}}\right )}}{x} - \frac{5 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}} b}{x}\right )}}{a^{2}} + \frac{10 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} + \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right )}{b} + \frac{10 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} - \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right )}{b} + \frac{5 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \log \left (\sqrt{-b} + \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right )}{b} - \frac{5 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \log \left (\sqrt{-b} - \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right )}{b}\right )} a^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(8*x^8*(9*(b*x^4 + a)^(1/4)*(b + a/x^4)/x - 5*(b*x^4 + a)^(1/4)*b/x)/a^2 + 10*sqrt(2)*(-b)^(1/4)*arctan(
1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2*(b*x^4 + a)^(1/4)/x)/(-b)^(1/4))/b + 10*sqrt(2)*(-b)^(1/4)*arctan(-1/2*sqr
t(2)*(sqrt(2)*(-b)^(1/4) - 2*(b*x^4 + a)^(1/4)/x)/(-b)^(1/4))/b + 5*sqrt(2)*(-b)^(1/4)*log(sqrt(-b) + sqrt(2)*
(b*x^4 + a)^(1/4)*(-b)^(1/4)/x + sqrt(b*x^4 + a)/x^2)/b - 5*sqrt(2)*(-b)^(1/4)*log(sqrt(-b) - sqrt(2)*(b*x^4 +
a)^(1/4)*(-b)^(1/4)/x + sqrt(b*x^4 + a)/x^2)/b)*a^2