3.11.63 \(\int x^6 (a+b x^4)^{5/4} \, dx\) [1063]

Optimal. Leaf size=124 \[ \frac {5 a^2 x^3 \sqrt [4]{a+b x^4}}{384 b}+\frac {5}{96} a x^7 \sqrt [4]{a+b x^4}+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}+\frac {5 a^3 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{7/4}}-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{7/4}} \]

[Out]

5/384*a^2*x^3*(b*x^4+a)^(1/4)/b+5/96*a*x^7*(b*x^4+a)^(1/4)+1/12*x^7*(b*x^4+a)^(5/4)+5/256*a^3*arctan(b^(1/4)*x
/(b*x^4+a)^(1/4))/b^(7/4)-5/256*a^3*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(7/4)

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Rubi [A]
time = 0.03, antiderivative size = 124, 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, 338, 304, 209, 212} \begin {gather*} \frac {5 a^3 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{7/4}}-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{7/4}}+\frac {5 a^2 x^3 \sqrt [4]{a+b x^4}}{384 b}+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}+\frac {5}{96} a x^7 \sqrt [4]{a+b x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^6*(a + b*x^4)^(5/4),x]

[Out]

(5*a^2*x^3*(a + b*x^4)^(1/4))/(384*b) + (5*a*x^7*(a + b*x^4)^(1/4))/96 + (x^7*(a + b*x^4)^(5/4))/12 + (5*a^3*A
rcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(256*b^(7/4)) - (5*a^3*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(256*b^(7
/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 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]

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^6 \left (a+b x^4\right )^{5/4} \, dx &=\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}+\frac {1}{12} (5 a) \int x^6 \sqrt [4]{a+b x^4} \, dx\\ &=\frac {5}{96} a x^7 \sqrt [4]{a+b x^4}+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}+\frac {1}{96} \left (5 a^2\right ) \int \frac {x^6}{\left (a+b x^4\right )^{3/4}} \, dx\\ &=\frac {5 a^2 x^3 \sqrt [4]{a+b x^4}}{384 b}+\frac {5}{96} a x^7 \sqrt [4]{a+b x^4}+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}-\frac {\left (5 a^3\right ) \int \frac {x^2}{\left (a+b x^4\right )^{3/4}} \, dx}{128 b}\\ &=\frac {5 a^2 x^3 \sqrt [4]{a+b x^4}}{384 b}+\frac {5}{96} a x^7 \sqrt [4]{a+b x^4}+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}-\frac {\left (5 a^3\right ) \text {Subst}\left (\int \frac {x^2}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{128 b}\\ &=\frac {5 a^2 x^3 \sqrt [4]{a+b x^4}}{384 b}+\frac {5}{96} a x^7 \sqrt [4]{a+b x^4}+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}-\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^{3/2}}+\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^{3/2}}\\ &=\frac {5 a^2 x^3 \sqrt [4]{a+b x^4}}{384 b}+\frac {5}{96} a x^7 \sqrt [4]{a+b x^4}+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}+\frac {5 a^3 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{7/4}}-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{7/4}}\\ \end {align*}

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Mathematica [A]
time = 0.33, size = 100, normalized size = 0.81 \begin {gather*} \frac {2 b^{3/4} x^3 \sqrt [4]{a+b x^4} \left (5 a^2+52 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^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^6*(a + b*x^4)^(5/4),x]

[Out]

(2*b^(3/4)*x^3*(a + b*x^4)^(1/4)*(5*a^2 + 52*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^(7/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int x^{6} \left (b \,x^{4}+a \right )^{\frac {5}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(b*x^4+a)^(5/4),x)

[Out]

int(x^6*(b*x^4+a)^(5/4),x)

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Maxima [A]
time = 0.50, size = 191, normalized size = 1.54 \begin {gather*} \frac {\frac {15 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} b^{2}}{x} - \frac {42 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{3} b}{x^{5}} - \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{3}}{x^{9}}}{384 \, {\left (b^{4} - \frac {3 \, {\left (b x^{4} + a\right )} b^{3}}{x^{4}} + \frac {3 \, {\left (b x^{4} + a\right )}^{2} b^{2}}{x^{8}} - \frac {{\left (b x^{4} + a\right )}^{3} b}{x^{12}}\right )}} - \frac {5 \, {\left (\frac {2 \, a^{3} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {3}{4}}} - \frac {a^{3} \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 {3}{4}}}\right )}}{512 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

1/384*(15*(b*x^4 + a)^(1/4)*a^3*b^2/x - 42*(b*x^4 + a)^(5/4)*a^3*b/x^5 - 5*(b*x^4 + a)^(9/4)*a^3/x^9)/(b^4 - 3
*(b*x^4 + a)*b^3/x^4 + 3*(b*x^4 + a)^2*b^2/x^8 - (b*x^4 + a)^3*b/x^12) - 5/512*(2*a^3*arctan((b*x^4 + a)^(1/4)
/(b^(1/4)*x))/b^(3/4) - a^3*log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (b*x^4 + a)^(1/4)/x))/b^(3/4))/b

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (96) = 192\).
time = 0.41, size = 232, normalized size = 1.87 \begin {gather*} \frac {60 \, \left (\frac {a^{12}}{b^{7}}\right )^{\frac {1}{4}} b \arctan \left (-\frac {\left (\frac {a^{12}}{b^{7}}\right )^{\frac {3}{4}} {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} b^{5} - \left (\frac {a^{12}}{b^{7}}\right )^{\frac {3}{4}} b^{5} x \sqrt {\frac {\sqrt {b x^{4} + a} a^{6} + \sqrt {\frac {a^{12}}{b^{7}}} b^{4} x^{2}}{x^{2}}}}{a^{12} x}\right ) - 15 \, \left (\frac {a^{12}}{b^{7}}\right )^{\frac {1}{4}} b \log \left (\frac {5 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} + \left (\frac {a^{12}}{b^{7}}\right )^{\frac {1}{4}} b^{2} x\right )}}{x}\right ) + 15 \, \left (\frac {a^{12}}{b^{7}}\right )^{\frac {1}{4}} b \log \left (\frac {5 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} - \left (\frac {a^{12}}{b^{7}}\right )^{\frac {1}{4}} b^{2} x\right )}}{x}\right ) + 4 \, {\left (32 \, b^{2} x^{11} + 52 \, a b x^{7} + 5 \, a^{2} x^{3}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{1536 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

1/1536*(60*(a^12/b^7)^(1/4)*b*arctan(-((a^12/b^7)^(3/4)*(b*x^4 + a)^(1/4)*a^3*b^5 - (a^12/b^7)^(3/4)*b^5*x*sqr
t((sqrt(b*x^4 + a)*a^6 + sqrt(a^12/b^7)*b^4*x^2)/x^2))/(a^12*x)) - 15*(a^12/b^7)^(1/4)*b*log(5*((b*x^4 + a)^(1
/4)*a^3 + (a^12/b^7)^(1/4)*b^2*x)/x) + 15*(a^12/b^7)^(1/4)*b*log(5*((b*x^4 + a)^(1/4)*a^3 - (a^12/b^7)^(1/4)*b
^2*x)/x) + 4*(32*b^2*x^11 + 52*a*b*x^7 + 5*a^2*x^3)*(b*x^4 + a)^(1/4))/b

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Sympy [C] Result contains complex when optimal does not.
time = 2.45, size = 39, normalized size = 0.31 \begin {gather*} \frac {a^{\frac {5}{4}} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*(b*x**4+a)**(5/4),x)

[Out]

a**(5/4)*x**7*gamma(7/4)*hyper((-5/4, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(11/4))

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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^6*(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(5/4)*x^6, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^6\,{\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^6*(a + b*x^4)^(5/4),x)

[Out]

int(x^6*(a + b*x^4)^(5/4), x)

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