3.11.32 \(\int x^8 (a+b x^4)^{3/4} \, dx\) [1032]

Optimal. Leaf size=125 \[ -\frac {5 a^2 x \left (a+b x^4\right )^{3/4}}{128 b^2}+\frac {a x^5 \left (a+b x^4\right )^{3/4}}{32 b}+\frac {1}{12} x^9 \left (a+b x^4\right )^{3/4}+\frac {5 a^3 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{9/4}}+\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{9/4}} \]

[Out]

-5/128*a^2*x*(b*x^4+a)^(3/4)/b^2+1/32*a*x^5*(b*x^4+a)^(3/4)/b+1/12*x^9*(b*x^4+a)^(3/4)+5/256*a^3*arctan(b^(1/4
)*x/(b*x^4+a)^(1/4))/b^(9/4)+5/256*a^3*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(9/4)

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Rubi [A]
time = 0.04, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {285, 327, 246, 218, 212, 209} \begin {gather*} \frac {5 a^3 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{9/4}}+\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{9/4}}-\frac {5 a^2 x \left (a+b x^4\right )^{3/4}}{128 b^2}+\frac {1}{12} x^9 \left (a+b x^4\right )^{3/4}+\frac {a x^5 \left (a+b x^4\right )^{3/4}}{32 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8*(a + b*x^4)^(3/4),x]

[Out]

(-5*a^2*x*(a + b*x^4)^(3/4))/(128*b^2) + (a*x^5*(a + b*x^4)^(3/4))/(32*b) + (x^9*(a + b*x^4)^(3/4))/12 + (5*a^
3*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(256*b^(9/4)) + (5*a^3*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(256*b
^(9/4))

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 218

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

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 327

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]

Rubi steps

\begin {align*} \int x^8 \left (a+b x^4\right )^{3/4} \, dx &=\frac {1}{12} x^9 \left (a+b x^4\right )^{3/4}+\frac {1}{4} a \int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx\\ &=\frac {a x^5 \left (a+b x^4\right )^{3/4}}{32 b}+\frac {1}{12} x^9 \left (a+b x^4\right )^{3/4}-\frac {\left (5 a^2\right ) \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx}{32 b}\\ &=-\frac {5 a^2 x \left (a+b x^4\right )^{3/4}}{128 b^2}+\frac {a x^5 \left (a+b x^4\right )^{3/4}}{32 b}+\frac {1}{12} x^9 \left (a+b x^4\right )^{3/4}+\frac {\left (5 a^3\right ) \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx}{128 b^2}\\ &=-\frac {5 a^2 x \left (a+b x^4\right )^{3/4}}{128 b^2}+\frac {a x^5 \left (a+b x^4\right )^{3/4}}{32 b}+\frac {1}{12} x^9 \left (a+b x^4\right )^{3/4}+\frac {\left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{128 b^2}\\ &=-\frac {5 a^2 x \left (a+b x^4\right )^{3/4}}{128 b^2}+\frac {a x^5 \left (a+b x^4\right )^{3/4}}{32 b}+\frac {1}{12} x^9 \left (a+b x^4\right )^{3/4}+\frac {\left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{256 b^2}+\frac {\left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{256 b^2}\\ &=-\frac {5 a^2 x \left (a+b x^4\right )^{3/4}}{128 b^2}+\frac {a x^5 \left (a+b x^4\right )^{3/4}}{32 b}+\frac {1}{12} x^9 \left (a+b x^4\right )^{3/4}+\frac {5 a^3 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{9/4}}+\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{9/4}}\\ \end {align*}

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Mathematica [A]
time = 0.36, size = 98, normalized size = 0.78 \begin {gather*} \frac {2 \sqrt [4]{b} x \left (a+b x^4\right )^{3/4} \left (-15 a^2+12 a b x^4+32 b^2 x^8\right )+15 a^3 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+15 a^3 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{768 b^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8*(a + b*x^4)^(3/4),x]

[Out]

(2*b^(1/4)*x*(a + b*x^4)^(3/4)*(-15*a^2 + 12*a*b*x^4 + 32*b^2*x^8) + 15*a^3*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/
4)] + 15*a^3*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(768*b^(9/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int x^{8} \left (b \,x^{4}+a \right )^{\frac {3}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(b*x^4+a)^(3/4),x)

[Out]

int(x^8*(b*x^4+a)^(3/4),x)

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Maxima [A]
time = 0.50, size = 189, normalized size = 1.51 \begin {gather*} -\frac {5 \, a^{3} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\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 )}{b^{\frac {1}{4}}}\right )}}{512 \, b^{2}} - \frac {\frac {5 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{3} b^{2}}{x^{3}} + \frac {42 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{3} b}{x^{7}} - \frac {15 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} a^{3}}{x^{11}}}{384 \, {\left (b^{5} - \frac {3 \, {\left (b x^{4} + a\right )} b^{4}}{x^{4}} + \frac {3 \, {\left (b x^{4} + a\right )}^{2} b^{3}}{x^{8}} - \frac {{\left (b x^{4} + a\right )}^{3} b^{2}}{x^{12}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

-5/512*a^3*(2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(1/4) + log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) +
(b*x^4 + a)^(1/4)/x))/b^(1/4))/b^2 - 1/384*(5*(b*x^4 + a)^(3/4)*a^3*b^2/x^3 + 42*(b*x^4 + a)^(7/4)*a^3*b/x^7 -
15*(b*x^4 + a)^(11/4)*a^3/x^11)/(b^5 - 3*(b*x^4 + a)*b^4/x^4 + 3*(b*x^4 + a)^2*b^3/x^8 - (b*x^4 + a)^3*b^2/x^
12)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (97) = 194\).
time = 0.41, size = 239, normalized size = 1.91 \begin {gather*} \frac {60 \, \left (\frac {a^{12}}{b^{9}}\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (\frac {a^{12}}{b^{9}}\right )^{\frac {1}{4}} a^{9} b^{2} - \left (\frac {a^{12}}{b^{9}}\right )^{\frac {1}{4}} b^{2} x \sqrt {\frac {\sqrt {\frac {a^{12}}{b^{9}}} a^{12} b^{5} x^{2} + \sqrt {b x^{4} + a} a^{18}}{x^{2}}}}{a^{12} x}\right ) + 15 \, \left (\frac {a^{12}}{b^{9}}\right )^{\frac {1}{4}} b^{2} \log \left (\frac {125 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{9} + \left (\frac {a^{12}}{b^{9}}\right )^{\frac {3}{4}} b^{7} x\right )}}{x}\right ) - 15 \, \left (\frac {a^{12}}{b^{9}}\right )^{\frac {1}{4}} b^{2} \log \left (\frac {125 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{9} - \left (\frac {a^{12}}{b^{9}}\right )^{\frac {3}{4}} b^{7} x\right )}}{x}\right ) + 4 \, {\left (32 \, b^{2} x^{9} + 12 \, a b x^{5} - 15 \, a^{2} x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{1536 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

1/1536*(60*(a^12/b^9)^(1/4)*b^2*arctan(-((b*x^4 + a)^(1/4)*(a^12/b^9)^(1/4)*a^9*b^2 - (a^12/b^9)^(1/4)*b^2*x*s
qrt((sqrt(a^12/b^9)*a^12*b^5*x^2 + sqrt(b*x^4 + a)*a^18)/x^2))/(a^12*x)) + 15*(a^12/b^9)^(1/4)*b^2*log(125*((b
*x^4 + a)^(1/4)*a^9 + (a^12/b^9)^(3/4)*b^7*x)/x) - 15*(a^12/b^9)^(1/4)*b^2*log(125*((b*x^4 + a)^(1/4)*a^9 - (a
^12/b^9)^(3/4)*b^7*x)/x) + 4*(32*b^2*x^9 + 12*a*b*x^5 - 15*a^2*x)*(b*x^4 + a)^(3/4))/b^2

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Sympy [C] Result contains complex when optimal does not.
time = 3.69, size = 39, normalized size = 0.31 \begin {gather*} \frac {a^{\frac {3}{4}} x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {13}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(b*x**4+a)**(3/4),x)

[Out]

a**(3/4)*x**9*gamma(9/4)*hyper((-3/4, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(13/4))

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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^8*(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/4)*x^8, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^8\,{\left (b\,x^4+a\right )}^{3/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(a + b*x^4)^(3/4),x)

[Out]

int(x^8*(a + b*x^4)^(3/4), x)

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