3.1063 \(\int x^6 (a+b x^4)^{5/4} \, dx\)

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

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.0504509, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {279, 321, 331, 298, 203, 206} \[ \frac{5 a^3 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{7/4}}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{7/4}}+\frac{5 a^2 x^3 \sqrt [4]{a+b x^4}}{384 b}+\frac{1}{12} x^7 \left (a+b x^4\right )^{5/4}+\frac{5}{96} a x^7 \sqrt [4]{a+b x^4} \]

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 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]

Rule 331

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]

Rule 298

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 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])

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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [C]  time = 0.063467, size = 69, normalized size = 0.56 \[ \frac{x^3 \sqrt [4]{a+b x^4} \left (\left (a+b x^4\right )^2-\frac{a^2 \, _2F_1\left (-\frac{5}{4},\frac{3}{4};\frac{7}{4};-\frac{b x^4}{a}\right )}{\sqrt [4]{\frac{b x^4}{a}+1}}\right )}{12 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6*(a + b*x^4)^(5/4),x]

[Out]

(x^3*(a + b*x^4)^(1/4)*((a + b*x^4)^2 - (a^2*Hypergeometric2F1[-5/4, 3/4, 7/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^
(1/4)))/(12*b)

________________________________________________________________________________________

Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int{x}^{6} \left ( b{x}^{4}+a \right ) ^{{\frac{5}{4}}}\, dx \end{align*}

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)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.65828, size = 545, normalized size = 4.4 \begin{align*} \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{align*}

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

________________________________________________________________________________________

Sympy [C]  time = 5.61634, size = 39, normalized size = 0.31 \begin{align*} \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{align*}

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))

________________________________________________________________________________________

Giac [B]  time = 1.17066, size = 397, normalized size = 3.2 \begin{align*} \frac{1}{3072} \,{\left (\frac{8 \,{\left (\frac{42 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}{\left (b + \frac{a}{x^{4}}\right )} b}{x} - \frac{15 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}} b^{2}}{x} + \frac{5 \,{\left (b^{2} x^{8} + 2 \, a b x^{4} + a^{2}\right )}{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x^{9}}\right )} x^{12}}{a^{3} b} - \frac{30 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} + \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right )}{b^{2}} - \frac{30 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} - \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right )}{b^{2}} - \frac{15 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \log \left (\sqrt{-b} + \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right )}{b^{2}} + \frac{15 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \log \left (\sqrt{-b} - \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right )}{b^{2}}\right )} a^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3072*(8*(42*(b*x^4 + a)^(1/4)*(b + a/x^4)*b/x - 15*(b*x^4 + a)^(1/4)*b^2/x + 5*(b^2*x^8 + 2*a*b*x^4 + a^2)*(
b*x^4 + a)^(1/4)/x^9)*x^12/(a^3*b) - 30*sqrt(2)*(-b)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2*(b*x^4 +
a)^(1/4)/x)/(-b)^(1/4))/b^2 - 30*sqrt(2)*(-b)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) - 2*(b*x^4 + a)^(
1/4)/x)/(-b)^(1/4))/b^2 - 15*sqrt(2)*(-b)^(1/4)*log(sqrt(-b) + sqrt(2)*(b*x^4 + a)^(1/4)*(-b)^(1/4)/x + sqrt(b
*x^4 + a)/x^2)/b^2 + 15*sqrt(2)*(-b)^(1/4)*log(sqrt(-b) - sqrt(2)*(b*x^4 + a)^(1/4)*(-b)^(1/4)/x + sqrt(b*x^4
+ a)/x^2)/b^2)*a^3