3.15.62 \(\int x^8 (b+a x^4)^{3/4} \, dx\)

Optimal. Leaf size=103 \[ \frac {5 b^3 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{256 a^{9/4}}+\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{256 a^{9/4}}+\frac {\left (a x^4+b\right )^{3/4} \left (32 a^2 x^9+12 a b x^5-15 b^2 x\right )}{384 a^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 125, normalized size of antiderivative = 1.21, 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 = {279, 321, 240, 212, 206, 203} \begin {gather*} \frac {5 b^3 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{256 a^{9/4}}+\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{256 a^{9/4}}-\frac {5 b^2 x \left (a x^4+b\right )^{3/4}}{128 a^2}+\frac {1}{12} x^9 \left (a x^4+b\right )^{3/4}+\frac {b x^5 \left (a x^4+b\right )^{3/4}}{32 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8*(b + a*x^4)^(3/4),x]

[Out]

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

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])

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 240

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.05, size = 94, normalized size = 0.91 \begin {gather*} \frac {x \left (a x^4+b\right )^{3/4} \left (\left (\frac {a x^4}{b}+1\right )^{3/4} \left (8 a^2 x^8+3 a b x^4-5 b^2\right )+5 b^2 \, _2F_1\left (-\frac {3}{4},\frac {1}{4};\frac {5}{4};-\frac {a x^4}{b}\right )\right )}{96 a^2 \left (\frac {a x^4}{b}+1\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8*(b + a*x^4)^(3/4),x]

[Out]

(x*(b + a*x^4)^(3/4)*((1 + (a*x^4)/b)^(3/4)*(-5*b^2 + 3*a*b*x^4 + 8*a^2*x^8) + 5*b^2*Hypergeometric2F1[-3/4, 1
/4, 5/4, -((a*x^4)/b)]))/(96*a^2*(1 + (a*x^4)/b)^(3/4))

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IntegrateAlgebraic [A]  time = 0.42, size = 103, normalized size = 1.00 \begin {gather*} \frac {\left (b+a x^4\right )^{3/4} \left (-15 b^2 x+12 a b x^5+32 a^2 x^9\right )}{384 a^2}+\frac {5 b^3 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{256 a^{9/4}}+\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{256 a^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^8*(b + a*x^4)^(3/4),x]

[Out]

((b + a*x^4)^(3/4)*(-15*b^2*x + 12*a*b*x^5 + 32*a^2*x^9))/(384*a^2) + (5*b^3*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1
/4)])/(256*a^(9/4)) + (5*b^3*ArcTanh[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(256*a^(9/4))

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fricas [B]  time = 0.54, size = 239, normalized size = 2.32 \begin {gather*} \frac {60 \, \left (\frac {b^{12}}{a^{9}}\right )^{\frac {1}{4}} a^{2} \arctan \left (-\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (\frac {b^{12}}{a^{9}}\right )^{\frac {1}{4}} a^{2} b^{9} - \left (\frac {b^{12}}{a^{9}}\right )^{\frac {1}{4}} a^{2} x \sqrt {\frac {\sqrt {\frac {b^{12}}{a^{9}}} a^{5} b^{12} x^{2} + \sqrt {a x^{4} + b} b^{18}}{x^{2}}}}{b^{12} x}\right ) + 15 \, \left (\frac {b^{12}}{a^{9}}\right )^{\frac {1}{4}} a^{2} \log \left (\frac {125 \, {\left ({\left (a x^{4} + b\right )}^{\frac {1}{4}} b^{9} + \left (\frac {b^{12}}{a^{9}}\right )^{\frac {3}{4}} a^{7} x\right )}}{x}\right ) - 15 \, \left (\frac {b^{12}}{a^{9}}\right )^{\frac {1}{4}} a^{2} \log \left (\frac {125 \, {\left ({\left (a x^{4} + b\right )}^{\frac {1}{4}} b^{9} - \left (\frac {b^{12}}{a^{9}}\right )^{\frac {3}{4}} a^{7} x\right )}}{x}\right ) + 4 \, {\left (32 \, a^{2} x^{9} + 12 \, a b x^{5} - 15 \, b^{2} x\right )} {\left (a x^{4} + b\right )}^{\frac {3}{4}}}{1536 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (a x^{4} + b\right )}^{\frac {3}{4}} x^{8}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^4 + b)^(3/4)*x^8, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.51, size = 189, normalized size = 1.83 \begin {gather*} -\frac {5 \, b^{3} {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {1}{4}}}\right )}}{512 \, a^{2}} - \frac {\frac {5 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} a^{2} b^{3}}{x^{3}} + \frac {42 \, {\left (a x^{4} + b\right )}^{\frac {7}{4}} a b^{3}}{x^{7}} - \frac {15 \, {\left (a x^{4} + b\right )}^{\frac {11}{4}} b^{3}}{x^{11}}}{384 \, {\left (a^{5} - \frac {3 \, {\left (a x^{4} + b\right )} a^{4}}{x^{4}} + \frac {3 \, {\left (a x^{4} + b\right )}^{2} a^{3}}{x^{8}} - \frac {{\left (a x^{4} + b\right )}^{3} a^{2}}{x^{12}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(b + a*x^4)^(3/4),x)

[Out]

int(x^8*(b + a*x^4)^(3/4), x)

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sympy [C]  time = 1.75, size = 39, normalized size = 0.38 \begin {gather*} \frac {b^{\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 {a x^{4} e^{i \pi }}{b}} \right )}}{4 \Gamma \left (\frac {13}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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