Integrand size = 26, antiderivative size = 142 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {(6 b c-7 a d) e (e x)^{3/2}}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}}+\frac {\sqrt {a} (6 b c-7 a d) e^2 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 b^{5/2} \sqrt [4]{a+b x^2}} \] Output:
1/6*(-7*a*d+6*b*c)*e*(e*x)^(3/2)/b^2/(b*x^2+a)^(1/4)+1/3*d*(e*x)^(7/2)/b/e /(b*x^2+a)^(1/4)+1/2*a^(1/2)*(-7*a*d+6*b*c)*e^2*(1+a/b/x^2)^(1/4)*(e*x)^(1 /2)*EllipticE(sin(1/2*arccot(b^(1/2)*x/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 = 10.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.60 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {e (e x)^{3/2} \left (6 b c-7 a d+2 b d x^2+(-6 b c+7 a d) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{6 b^2 \sqrt [4]{a+b x^2}} \] Input:
Integrate[((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x]
Output:
(e*(e*x)^(3/2)*(6*b*c - 7*a*d + 2*b*d*x^2 + (-6*b*c + 7*a*d)*(1 + (b*x^2)/ a)^(1/4)*Hypergeometric2F1[3/4, 5/4, 7/4, -((b*x^2)/a)]))/(6*b^2*(a + b*x^ 2)^(1/4))
Time = 0.26 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {363, 250, 249, 858, 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.
\(\displaystyle \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 363 |
\(\displaystyle \frac {(6 b c-7 a d) \int \frac {(e x)^{5/2}}{\left (b x^2+a\right )^{5/4}}dx}{6 b}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 250 |
\(\displaystyle \frac {(6 b c-7 a d) \left (\frac {e (e x)^{3/2}}{b \sqrt [4]{a+b x^2}}-\frac {3 a e^2 \int \frac {\sqrt {e x}}{\left (b x^2+a\right )^{5/4}}dx}{2 b}\right )}{6 b}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 249 |
\(\displaystyle \frac {(6 b c-7 a d) \left (\frac {e (e x)^{3/2}}{b \sqrt [4]{a+b x^2}}-\frac {3 a e^2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4} x^2}dx}{2 b^2 \sqrt [4]{a+b x^2}}\right )}{6 b}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {(6 b c-7 a d) \left (\frac {3 a e^2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x}}{2 b^2 \sqrt [4]{a+b x^2}}+\frac {e (e x)^{3/2}}{b \sqrt [4]{a+b x^2}}\right )}{6 b}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {(6 b c-7 a d) \left (\frac {3 \sqrt {a} e^2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x}\right )\right |2\right )}{b^{3/2} \sqrt [4]{a+b x^2}}+\frac {e (e x)^{3/2}}{b \sqrt [4]{a+b x^2}}\right )}{6 b}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}}\) |
Input:
Int[((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x]
Output:
(d*(e*x)^(7/2))/(3*b*e*(a + b*x^2)^(1/4)) + ((6*b*c - 7*a*d)*((e*(e*x)^(3/ 2))/(b*(a + b*x^2)^(1/4)) + (3*Sqrt[a]*e^2*(1 + a/(b*x^2))^(1/4)*Sqrt[e*x] *EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x)]/2, 2])/(b^(3/2)*(a + b*x^2)^(1/4))) )/(6*b)
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[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[Sqrt[c* x]*((1 + a/(b*x^2))^(1/4)/(b*(a + b*x^2)^(1/4))) Int[1/(x^2*(1 + a/(b*x^2 ))^(5/4)), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]
Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[2*c*(( c*x)^(m - 1)/(b*(2*m - 3)*(a + b*x^2)^(1/4))), x] - Simp[2*a*c^2*((m - 1)/( b*(2*m - 3))) Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
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]
\[\int \frac {\left (e x \right )^{\frac {5}{2}} \left (x^{2} d +c \right )}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]
Input:
int((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)
Output:
int((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)
\[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="fricas")
Output:
integral((d*e^2*x^4 + c*e^2*x^2)*(b*x^2 + a)^(3/4)*sqrt(e*x)/(b^2*x^4 + 2* a*b*x^2 + a^2), x)
Result contains complex when optimal does not.
Time = 39.66 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.66 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {11}{4}\right )} + \frac {d e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {15}{4}\right )} \] Input:
integrate((e*x)**(5/2)*(d*x**2+c)/(b*x**2+a)**(5/4),x)
Output:
c*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((5/4, 7/4), (11/4,), b*x**2*exp_polar (I*pi)/a)/(2*a**(5/4)*gamma(11/4)) + d*e**(5/2)*x**(11/2)*gamma(11/4)*hype r((5/4, 11/4), (15/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(15/4))
\[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="maxima")
Output:
integrate((d*x^2 + c)*(e*x)^(5/2)/(b*x^2 + a)^(5/4), x)
\[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="giac")
Output:
integrate((d*x^2 + c)*(e*x)^(5/2)/(b*x^2 + a)^(5/4), x)
Timed out. \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int \frac {{\left (e\,x\right )}^{5/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \] Input:
int(((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x)
Output:
int(((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(5/4), x)
\[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\sqrt {e}\, e^{2} \left (\left (\int \frac {\sqrt {x}\, x^{4}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) d +\left (\int \frac {\sqrt {x}\, x^{2}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c \right ) \] Input:
int((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)
Output:
sqrt(e)*e**2*(int((sqrt(x)*x**4)/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1 /4)*b*x**2),x)*d + int((sqrt(x)*x**2)/((a + b*x**2)**(1/4)*a + (a + b*x**2 )**(1/4)*b*x**2),x)*c)