Integrand size = 15, antiderivative size = 143 \[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\frac {8 a^3 x}{65 b^2 \sqrt [4]{a+b x^2}}-\frac {4 a^2 x \left (a+b x^2\right )^{3/4}}{65 b^2}+\frac {2 a x^3 \left (a+b x^2\right )^{3/4}}{39 b}+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}-\frac {8 a^{7/2} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{65 b^{5/2} \sqrt [4]{a+b x^2}} \]
8/65*a^3*x/b^2/(b*x^2+a)^(1/4)-4/65*a^2*x*(b*x^2+a)^(3/4)/b^2+2/39*a*x^3*( b*x^2+a)^(3/4)/b+2/13*x^5*(b*x^2+a)^(3/4)-8/65*a^(7/2)*(1+b*x^2/a)^(1/4)*( cos(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/ 2)))*EllipticE(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))/b^(5/2)/(b*x^2+ a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.65 \[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\frac {2 x \left (a+b x^2\right )^{3/4} \left (\left (1+\frac {b x^2}{a}\right )^{3/4} \left (-2 a^2+a b x^2+3 b^2 x^4\right )+2 a^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{39 b^2 \left (1+\frac {b x^2}{a}\right )^{3/4}} \]
(2*x*(a + b*x^2)^(3/4)*((1 + (b*x^2)/a)^(3/4)*(-2*a^2 + a*b*x^2 + 3*b^2*x^ 4) + 2*a^2*Hypergeometric2F1[-3/4, 1/2, 3/2, -((b*x^2)/a)]))/(39*b^2*(1 + (b*x^2)/a)^(3/4))
Time = 0.23 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {248, 262, 262, 227, 225, 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.
\(\displaystyle \int x^4 \left (a+b x^2\right )^{3/4} \, dx\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {3}{13} a \int \frac {x^4}{\sqrt [4]{b x^2+a}}dx+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {3}{13} a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \int \frac {x^2}{\sqrt [4]{b x^2+a}}dx}{3 b}\right )+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {3}{13} a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \int \frac {1}{\sqrt [4]{b x^2+a}}dx}{5 b}\right )}{3 b}\right )+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}\) |
\(\Big \downarrow \) 227 |
\(\displaystyle \frac {3}{13} a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{5 b \sqrt [4]{a+b x^2}}\right )}{3 b}\right )+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}\) |
\(\Big \downarrow \) 225 |
\(\displaystyle \frac {3}{13} a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{5 b \sqrt [4]{a+b x^2}}\right )}{3 b}\right )+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {3}{13} a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{5 b \sqrt [4]{a+b x^2}}\right )}{3 b}\right )+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}\) |
(2*x^5*(a + b*x^2)^(3/4))/13 + (3*a*((2*x^3*(a + b*x^2)^(3/4))/(9*b) - (2* a*((2*x*(a + b*x^2)^(3/4))/(5*b) - (2*a*(1 + (b*x^2)/a)^(1/4)*((2*x)/(1 + (b*x^2)/a)^(1/4) - (2*Sqrt[a]*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2]) /Sqrt[b]))/(5*b*(a + b*x^2)^(1/4))))/(3*b)))/13
3.8.99.3.1 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[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) , x] - Simp[a Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ a, 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( a + b*x^2)^(1/4) Int[1/(1 + b*(x^2/a))^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
\[\int x^{4} \left (b \,x^{2}+a \right )^{\frac {3}{4}}d x\]
\[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {3}{4}} x^{4} \,d x } \]
Result contains complex when optimal does not.
Time = 0.67 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.20 \[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\frac {a^{\frac {3}{4}} x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} \]
\[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {3}{4}} x^{4} \,d x } \]
\[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {3}{4}} x^{4} \,d x } \]
Timed out. \[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\int x^4\,{\left (b\,x^2+a\right )}^{3/4} \,d x \]