Integrand size = 15, antiderivative size = 142 \[ \int x^4 \left (a+b x^2\right )^{5/4} \, dx=-\frac {4 a^3 x \sqrt [4]{a+b x^2}}{231 b^2}+\frac {2 a^2 x^3 \sqrt [4]{a+b x^2}}{231 b}+\frac {2}{33} a x^5 \sqrt [4]{a+b x^2}+\frac {2}{15} x^5 \left (a+b x^2\right )^{5/4}+\frac {8 a^{9/2} \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{231 b^{5/2} \left (a+b x^2\right )^{3/4}} \] Output:
-4/231*a^3*x*(b*x^2+a)^(1/4)/b^2+2/231*a^2*x^3*(b*x^2+a)^(1/4)/b+2/33*a*x^ 5*(b*x^2+a)^(1/4)+2/15*x^5*(b*x^2+a)^(5/4)+8/231*a^(9/2)*(1+b*x^2/a)^(3/4) *InverseJacobiAM(1/2*arctan(b^(1/2)*x/a^(1/2)),2^(1/2))/b^(5/2)/(b*x^2+a)^ (3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.57 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.56 \[ \int x^4 \left (a+b x^2\right )^{5/4} \, dx=\frac {2 x \sqrt [4]{a+b x^2} \left (-\left (\left (6 a-11 b x^2\right ) \left (a+b x^2\right )^2\right )+\frac {6 a^3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{\sqrt [4]{1+\frac {b x^2}{a}}}\right )}{165 b^2} \] Input:
Integrate[x^4*(a + b*x^2)^(5/4),x]
Output:
(2*x*(a + b*x^2)^(1/4)*(-((6*a - 11*b*x^2)*(a + b*x^2)^2) + (6*a^3*Hyperge ometric2F1[-5/4, 1/2, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(1/4)))/(165*b^2 )
Time = 0.22 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {248, 248, 262, 262, 231, 229}
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.
\(\displaystyle \int x^4 \left (a+b x^2\right )^{5/4} \, dx\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {1}{3} a \int x^4 \sqrt [4]{b x^2+a}dx+\frac {2}{15} x^5 \left (a+b x^2\right )^{5/4}\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {1}{3} a \left (\frac {1}{11} a \int \frac {x^4}{\left (b x^2+a\right )^{3/4}}dx+\frac {2}{11} x^5 \sqrt [4]{a+b x^2}\right )+\frac {2}{15} x^5 \left (a+b x^2\right )^{5/4}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{3} a \left (\frac {1}{11} a \left (\frac {2 x^3 \sqrt [4]{a+b x^2}}{7 b}-\frac {6 a \int \frac {x^2}{\left (b x^2+a\right )^{3/4}}dx}{7 b}\right )+\frac {2}{11} x^5 \sqrt [4]{a+b x^2}\right )+\frac {2}{15} x^5 \left (a+b x^2\right )^{5/4}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{3} a \left (\frac {1}{11} a \left (\frac {2 x^3 \sqrt [4]{a+b x^2}}{7 b}-\frac {6 a \left (\frac {2 x \sqrt [4]{a+b x^2}}{3 b}-\frac {2 a \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx}{3 b}\right )}{7 b}\right )+\frac {2}{11} x^5 \sqrt [4]{a+b x^2}\right )+\frac {2}{15} x^5 \left (a+b x^2\right )^{5/4}\) |
\(\Big \downarrow \) 231 |
\(\displaystyle \frac {1}{3} a \left (\frac {1}{11} a \left (\frac {2 x^3 \sqrt [4]{a+b x^2}}{7 b}-\frac {6 a \left (\frac {2 x \sqrt [4]{a+b x^2}}{3 b}-\frac {2 a \left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{3 b \left (a+b x^2\right )^{3/4}}\right )}{7 b}\right )+\frac {2}{11} x^5 \sqrt [4]{a+b x^2}\right )+\frac {2}{15} x^5 \left (a+b x^2\right )^{5/4}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {1}{3} a \left (\frac {1}{11} a \left (\frac {2 x^3 \sqrt [4]{a+b x^2}}{7 b}-\frac {6 a \left (\frac {2 x \sqrt [4]{a+b x^2}}{3 b}-\frac {4 a^{3/2} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 b^{3/2} \left (a+b x^2\right )^{3/4}}\right )}{7 b}\right )+\frac {2}{11} x^5 \sqrt [4]{a+b x^2}\right )+\frac {2}{15} x^5 \left (a+b x^2\right )^{5/4}\) |
Input:
Int[x^4*(a + b*x^2)^(5/4),x]
Output:
(2*x^5*(a + b*x^2)^(5/4))/15 + (a*((2*x^5*(a + b*x^2)^(1/4))/11 + (a*((2*x ^3*(a + b*x^2)^(1/4))/(7*b) - (6*a*((2*x*(a + b*x^2)^(1/4))/(3*b) - (4*a^( 3/2)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3 *b^(3/2)*(a + b*x^2)^(3/4))))/(7*b)))/11))/3
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) )*EllipticF[(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)^(-3/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(3/4)/( a + b*x^2)^(3/4) Int[1/(1 + b*(x^2/a))^(3/4), 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]
\[\int x^{4} \left (b \,x^{2}+a \right )^{\frac {5}{4}}d x\]
Input:
int(x^4*(b*x^2+a)^(5/4),x)
Output:
int(x^4*(b*x^2+a)^(5/4),x)
\[ \int x^4 \left (a+b x^2\right )^{5/4} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {5}{4}} x^{4} \,d x } \] Input:
integrate(x^4*(b*x^2+a)^(5/4),x, algorithm="fricas")
Output:
integral((b*x^6 + a*x^4)*(b*x^2 + a)^(1/4), x)
Result contains complex when optimal does not.
Time = 0.71 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.20 \[ \int x^4 \left (a+b x^2\right )^{5/4} \, dx=\frac {a^{\frac {5}{4}} x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} \] Input:
integrate(x**4*(b*x**2+a)**(5/4),x)
Output:
a**(5/4)*x**5*hyper((-5/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/5
\[ \int x^4 \left (a+b x^2\right )^{5/4} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {5}{4}} x^{4} \,d x } \] Input:
integrate(x^4*(b*x^2+a)^(5/4),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(5/4)*x^4, x)
\[ \int x^4 \left (a+b x^2\right )^{5/4} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {5}{4}} x^{4} \,d x } \] Input:
integrate(x^4*(b*x^2+a)^(5/4),x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(5/4)*x^4, x)
Timed out. \[ \int x^4 \left (a+b x^2\right )^{5/4} \, dx=\int x^4\,{\left (b\,x^2+a\right )}^{5/4} \,d x \] Input:
int(x^4*(a + b*x^2)^(5/4),x)
Output:
int(x^4*(a + b*x^2)^(5/4), x)
\[ \int x^4 \left (a+b x^2\right )^{5/4} \, dx=\frac {-\frac {4 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{3} x}{231}+\frac {2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} b \,x^{3}}{231}+\frac {32 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,b^{2} x^{5}}{165}+\frac {2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{3} x^{7}}{15}+\frac {4 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{4}}{231}}{b^{2}} \] Input:
int(x^4*(b*x^2+a)^(5/4),x)
Output:
(2*( - 10*(a + b*x**2)**(1/4)*a**3*x + 5*(a + b*x**2)**(1/4)*a**2*b*x**3 + 112*(a + b*x**2)**(1/4)*a*b**2*x**5 + 77*(a + b*x**2)**(1/4)*b**3*x**7 + 10*int((a + b*x**2)**(1/4)/(a + b*x**2),x)*a**4))/(1155*b**2)