Integrand size = 15, antiderivative size = 126 \[ \int x^6 \left (a+b x^4\right )^{3/4} \, dx=-\frac {3 a^2 x^3}{40 b \sqrt [4]{a+b x^4}}+\frac {a x^3 \left (a+b x^4\right )^{3/4}}{20 b}+\frac {1}{10} x^7 \left (a+b x^4\right )^{3/4}-\frac {3 a^{5/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 b^{3/2} \sqrt [4]{a+b x^4}} \]
-3/40*a^2*x^3/b/(b*x^4+a)^(1/4)+1/20*a*x^3*(b*x^4+a)^(3/4)/b+1/10*x^7*(b*x ^4+a)^(3/4)-3/40*a^(5/2)*(1+a/b/x^4)^(1/4)*x*(cos(1/2*arccot(x^2*b^(1/2)/a ^(1/2)))^2)^(1/2)/cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*a rccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/b^(3/2)/(b*x^4+a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.61 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.51 \[ \int x^6 \left (a+b x^4\right )^{3/4} \, dx=\frac {x^3 \left (a+b x^4\right )^{3/4} \left (a+b x^4-\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{\left (1+\frac {b x^4}{a}\right )^{3/4}}\right )}{10 b} \]
(x^3*(a + b*x^4)^(3/4)*(a + b*x^4 - (a*Hypergeometric2F1[-3/4, 3/4, 7/4, - ((b*x^4)/a)])/(1 + (b*x^4)/a)^(3/4)))/(10*b)
Time = 0.29 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {811, 843, 839, 813, 858, 807, 212}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^6 \left (a+b x^4\right )^{3/4} \, dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {3}{10} a \int \frac {x^6}{\sqrt [4]{b x^4+a}}dx+\frac {1}{10} x^7 \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {3}{10} a \left (\frac {x^3 \left (a+b x^4\right )^{3/4}}{6 b}-\frac {a \int \frac {x^2}{\sqrt [4]{b x^4+a}}dx}{2 b}\right )+\frac {1}{10} x^7 \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 839 |
\(\displaystyle \frac {3}{10} a \left (\frac {x^3 \left (a+b x^4\right )^{3/4}}{6 b}-\frac {a \left (\frac {x^3}{2 \sqrt [4]{a+b x^4}}-\frac {1}{2} a \int \frac {x^2}{\left (b x^4+a\right )^{5/4}}dx\right )}{2 b}\right )+\frac {1}{10} x^7 \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 813 |
\(\displaystyle \frac {3}{10} a \left (\frac {x^3 \left (a+b x^4\right )^{3/4}}{6 b}-\frac {a \left (\frac {x^3}{2 \sqrt [4]{a+b x^4}}-\frac {a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x^3}dx}{2 b \sqrt [4]{a+b x^4}}\right )}{2 b}\right )+\frac {1}{10} x^7 \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {3}{10} a \left (\frac {x^3 \left (a+b x^4\right )^{3/4}}{6 b}-\frac {a \left (\frac {a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x}d\frac {1}{x}}{2 b \sqrt [4]{a+b x^4}}+\frac {x^3}{2 \sqrt [4]{a+b x^4}}\right )}{2 b}\right )+\frac {1}{10} x^7 \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {3}{10} a \left (\frac {x^3 \left (a+b x^4\right )^{3/4}}{6 b}-\frac {a \left (\frac {a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x^2}}{4 b \sqrt [4]{a+b x^4}}+\frac {x^3}{2 \sqrt [4]{a+b x^4}}\right )}{2 b}\right )+\frac {1}{10} x^7 \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {3}{10} a \left (\frac {x^3 \left (a+b x^4\right )^{3/4}}{6 b}-\frac {a \left (\frac {\sqrt {a} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right )\right |2\right )}{2 \sqrt {b} \sqrt [4]{a+b x^4}}+\frac {x^3}{2 \sqrt [4]{a+b x^4}}\right )}{2 b}\right )+\frac {1}{10} x^7 \left (a+b x^4\right )^{3/4}\) |
(x^7*(a + b*x^4)^(3/4))/10 + (3*a*((x^3*(a + b*x^4)^(3/4))/(6*b) - (a*(x^3 /(2*(a + b*x^4)^(1/4)) + (Sqrt[a]*(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcTan [Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(2*Sqrt[b]*(a + b*x^4)^(1/4))))/(2*b)))/10
3.11.41.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
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] + Simp[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] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m , p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) )^(1/4)/(b*(a + b*x^4)^(1/4))) Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[x^3/(2*(a + b*x^4 )^(1/4)), x] - Simp[a/2 Int[x^2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b} , x] && PosQ[b/a]
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] - Simp[ 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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
\[\int x^{6} \left (b \,x^{4}+a \right )^{\frac {3}{4}}d x\]
\[ \int x^6 \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{6} \,d x } \]
Result contains complex when optimal does not.
Time = 0.69 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.31 \[ \int x^6 \left (a+b x^4\right )^{3/4} \, dx=\frac {a^{\frac {3}{4}} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} \]
a**(3/4)*x**7*gamma(7/4)*hyper((-3/4, 7/4), (11/4,), b*x**4*exp_polar(I*pi )/a)/(4*gamma(11/4))
\[ \int x^6 \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{6} \,d x } \]
\[ \int x^6 \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{6} \,d x } \]
Timed out. \[ \int x^6 \left (a+b x^4\right )^{3/4} \, dx=\int x^6\,{\left (b\,x^4+a\right )}^{3/4} \,d x \]