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

Optimal. Leaf size=98 \[ -\frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{64 a^{5/4}}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{64 a^{5/4}}+\frac {\left (a x^4-b\right )^{3/4} \left (4 a x^5-3 b x\right )}{32 a} \]

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Rubi [A]  time = 0.04, antiderivative size = 109, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {279, 321, 240, 212, 206, 203} \begin {gather*} -\frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{64 a^{5/4}}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{64 a^{5/4}}-\frac {3 b x \left (a x^4-b\right )^{3/4}}{32 a}+\frac {1}{8} x^5 \left (a x^4-b\right )^{3/4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(-b + a*x^4)^(3/4),x]

[Out]

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

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^4 \left (-b+a x^4\right )^{3/4} \, dx &=\frac {1}{8} x^5 \left (-b+a x^4\right )^{3/4}-\frac {1}{8} (3 b) \int \frac {x^4}{\sqrt [4]{-b+a x^4}} \, dx\\ &=-\frac {3 b x \left (-b+a x^4\right )^{3/4}}{32 a}+\frac {1}{8} x^5 \left (-b+a x^4\right )^{3/4}-\frac {\left (3 b^2\right ) \int \frac {1}{\sqrt [4]{-b+a x^4}} \, dx}{32 a}\\ &=-\frac {3 b x \left (-b+a x^4\right )^{3/4}}{32 a}+\frac {1}{8} x^5 \left (-b+a x^4\right )^{3/4}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{32 a}\\ &=-\frac {3 b x \left (-b+a x^4\right )^{3/4}}{32 a}+\frac {1}{8} x^5 \left (-b+a x^4\right )^{3/4}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{64 a}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{64 a}\\ &=-\frac {3 b x \left (-b+a x^4\right )^{3/4}}{32 a}+\frac {1}{8} x^5 \left (-b+a x^4\right )^{3/4}-\frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{64 a^{5/4}}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{64 a^{5/4}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 65, normalized size = 0.66 \begin {gather*} \frac {x \left (a x^4-b\right )^{3/4} \left (\frac {b \, _2F_1\left (-\frac {3}{4},\frac {1}{4};\frac {5}{4};\frac {a x^4}{b}\right )}{\left (1-\frac {a x^4}{b}\right )^{3/4}}+a x^4-b\right )}{8 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(-b + a*x^4)^(3/4),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.41, size = 98, normalized size = 1.00 \begin {gather*} \frac {\left (-b+a x^4\right )^{3/4} \left (-3 b x+4 a x^5\right )}{32 a}-\frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{64 a^{5/4}}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{64 a^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.67, size = 228, normalized size = 2.33 \begin {gather*} -\frac {12 \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a \arctan \left (-\frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}} \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a b^{6} - \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a x \sqrt {\frac {\sqrt {\frac {b^{8}}{a^{5}}} a^{3} b^{8} x^{2} + \sqrt {a x^{4} - b} b^{12}}{x^{2}}}}{b^{8} x}\right ) + 3 \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a \log \left (\frac {27 \, {\left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{6} + \left (\frac {b^{8}}{a^{5}}\right )^{\frac {3}{4}} a^{4} x\right )}}{x}\right ) - 3 \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a \log \left (\frac {27 \, {\left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{6} - \left (\frac {b^{8}}{a^{5}}\right )^{\frac {3}{4}} a^{4} x\right )}}{x}\right ) - 4 \, {\left (4 \, a x^{5} - 3 \, b x\right )} {\left (a x^{4} - b\right )}^{\frac {3}{4}}}{128 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(12*(b^8/a^5)^(1/4)*a*arctan(-((a*x^4 - b)^(1/4)*(b^8/a^5)^(1/4)*a*b^6 - (b^8/a^5)^(1/4)*a*x*sqrt((sqrt
(b^8/a^5)*a^3*b^8*x^2 + sqrt(a*x^4 - b)*b^12)/x^2))/(b^8*x)) + 3*(b^8/a^5)^(1/4)*a*log(27*((a*x^4 - b)^(1/4)*b
^6 + (b^8/a^5)^(3/4)*a^4*x)/x) - 3*(b^8/a^5)^(1/4)*a*log(27*((a*x^4 - b)^(1/4)*b^6 - (b^8/a^5)^(3/4)*a^4*x)/x)
- 4*(4*a*x^5 - 3*b*x)*(a*x^4 - b)^(3/4))/a

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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^{4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^4 - b)^(3/4)*x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.42, size = 162, normalized size = 1.65 \begin {gather*} \frac {3 \, b^{2} {\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 )}}{128 \, a} + \frac {\frac {{\left (a x^{4} - b\right )}^{\frac {3}{4}} a b^{2}}{x^{3}} + \frac {3 \, {\left (a x^{4} - b\right )}^{\frac {7}{4}} b^{2}}{x^{7}}}{32 \, {\left (a^{3} - \frac {2 \, {\left (a x^{4} - b\right )} a^{2}}{x^{4}} + \frac {{\left (a x^{4} - b\right )}^{2} a}{x^{8}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/128*b^2*(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 + 1/32*((a*x^4 - b)^(3/4)*a*b^2/x^3 + 3*(a*x^4 - b)^(7/4)*b^2/x^7)/(a^3 - 2*(a
*x^4 - b)*a^2/x^4 + (a*x^4 - b)^2*a/x^8)

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

[Out]

int(x^4*(a*x^4 - b)^(3/4), x)

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sympy [C]  time = 1.32, size = 42, normalized size = 0.43 \begin {gather*} \frac {b^{\frac {3}{4}} x^{5} e^{\frac {3 i \pi }{4}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {a x^{4}}{b}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a*x**4-b)**(3/4),x)

[Out]

b**(3/4)*x**5*exp(3*I*pi/4)*gamma(5/4)*hyper((-3/4, 5/4), (9/4,), a*x**4/b)/(4*gamma(9/4))

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