3.11.64 \(\int x^2 (a+b x^4)^{5/4} \, dx\) [1064]

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

[Out]

5/32*a*x^3*(b*x^4+a)^(1/4)+1/8*x^3*(b*x^4+a)^(5/4)-5/64*a^2*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(3/4)+5/64*a^2
*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(3/4)

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Rubi [A]
time = 0.03, antiderivative size = 100, normalized size of antiderivative = 1.00, 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 = {285, 338, 304, 209, 212} \begin {gather*} -\frac {5 a^2 \text {ArcTan}\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} \end {gather*}

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 209

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

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 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 304

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 338

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]

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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.27, size = 91, normalized size = 0.91 \begin {gather*} \frac {1}{32} x^3 \sqrt [4]{a+b x^4} \left (9 a+4 b x^4\right )-\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 {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x^4)^(5/4),x]

[Out]

(x^3*(a + b*x^4)^(1/4)*(9*a + 4*b*x^4))/32 - (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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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int x^{2} \left (b \,x^{4}+a \right )^{\frac {5}{4}}\, dx\]

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 [A]
time = 0.50, size = 144, normalized size = 1.44 \begin {gather*} \frac {5 \, a^{2} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{64 \, b^{\frac {3}{4}}} - \frac {5 \, a^{2} \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{128 \, b^{\frac {3}{4}}} - \frac {\frac {5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} b}{x} - \frac {9 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2}}{x^{5}}}{32 \, {\left (b^{2} - \frac {2 \, {\left (b x^{4} + a\right )} b}{x^{4}} + \frac {{\left (b x^{4} + a\right )}^{2}}{x^{8}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/64*a^2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(3/4) - 5/128*a^2*log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/
4) + (b*x^4 + a)^(1/4)/x))/b^(3/4) - 1/32*(5*(b*x^4 + a)^(1/4)*a^2*b/x - 9*(b*x^4 + a)^(5/4)*a^2/x^5)/(b^2 - 2
*(b*x^4 + a)*b/x^4 + (b*x^4 + a)^2/x^8)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (76) = 152\).
time = 0.39, size = 209, normalized size = 2.09 \begin {gather*} \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 {gather*}

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] Result contains complex when optimal does not.
time = 1.00, size = 39, normalized size = 0.39 \begin {gather*} \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 {gather*}

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 [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^4 + a)^(5/4)*x^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\left (b\,x^4+a\right )}^{5/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x^4)^(5/4),x)

[Out]

int(x^2*(a + b*x^4)^(5/4), x)

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