Integrand size = 26, antiderivative size = 218 \[ \int (e x)^{5/2} \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=-\frac {3 a^2 (2 b c-a d) e (e x)^{3/2}}{40 b^2 \sqrt [4]{a+b x^2}}+\frac {a (2 b c-a d) e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}{20 b^2}+\frac {(2 b c-a d) (e x)^{7/2} \left (a+b x^2\right )^{3/4}}{10 b e}+\frac {d (e x)^{7/2} \left (a+b x^2\right )^{7/4}}{7 b e}-\frac {3 a^{5/2} (2 b c-a d) e^2 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{40 b^{5/2} \sqrt [4]{a+b x^2}} \] Output:
-3/40*a^2*(-a*d+2*b*c)*e*(e*x)^(3/2)/b^2/(b*x^2+a)^(1/4)+1/20*a*(-a*d+2*b* c)*e*(e*x)^(3/2)*(b*x^2+a)^(3/4)/b^2+1/10*(-a*d+2*b*c)*(e*x)^(7/2)*(b*x^2+ a)^(3/4)/b/e+1/7*d*(e*x)^(7/2)*(b*x^2+a)^(7/4)/b/e-3/40*a^(5/2)*(-a*d+2*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 = 10.11 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.51 \[ \int (e x)^{5/2} \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\frac {e (e x)^{3/2} \left (a+b x^2\right )^{3/4} \left (-\left (\left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \left (7 a d-2 b \left (7 c+5 d x^2\right )\right )\right )+7 a (-2 b c+a d) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{70 b^2 \left (1+\frac {b x^2}{a}\right )^{3/4}} \] Input:
Integrate[(e*x)^(5/2)*(a + b*x^2)^(3/4)*(c + d*x^2),x]
Output:
(e*(e*x)^(3/2)*(a + b*x^2)^(3/4)*(-((a + b*x^2)*(1 + (b*x^2)/a)^(3/4)*(7*a *d - 2*b*(7*c + 5*d*x^2))) + 7*a*(-2*b*c + a*d)*Hypergeometric2F1[-3/4, 3/ 4, 7/4, -((b*x^2)/a)]))/(70*b^2*(1 + (b*x^2)/a)^(3/4))
Time = 0.30 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {363, 248, 262, 255, 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.
| \(\displaystyle \int (e x)^{5/2} \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {(2 b c-a d) \int (e x)^{5/2} \left (b x^2+a\right )^{3/4}dx}{2 b}+\frac {d (e x)^{7/2} \left (a+b x^2\right )^{7/4}}{7 b e}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {(2 b c-a d) \left (\frac {3}{10} a \int \frac {(e x)^{5/2}}{\sqrt [4]{b x^2+a}}dx+\frac {(e x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 e}\right )}{2 b}+\frac {d (e x)^{7/2} \left (a+b x^2\right )^{7/4}}{7 b e}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {(2 b c-a d) \left (\frac {3}{10} a \left (\frac {e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}{3 b}-\frac {a e^2 \int \frac {\sqrt {e x}}{\sqrt [4]{b x^2+a}}dx}{2 b}\right )+\frac {(e x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 e}\right )}{2 b}+\frac {d (e x)^{7/2} \left (a+b x^2\right )^{7/4}}{7 b e}\) |
| \(\Big \downarrow \) 255 |
| \(\displaystyle \frac {(2 b c-a d) \left (\frac {3}{10} a \left (\frac {e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}{3 b}-\frac {a e^2 \left (\frac {x \sqrt {e x}}{\sqrt [4]{a+b x^2}}-\frac {1}{2} a \int \frac {\sqrt {e x}}{\left (b x^2+a\right )^{5/4}}dx\right )}{2 b}\right )+\frac {(e x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 e}\right )}{2 b}+\frac {d (e x)^{7/2} \left (a+b x^2\right )^{7/4}}{7 b e}\) |
| \(\Big \downarrow \) 249 |
| \(\displaystyle \frac {(2 b c-a d) \left (\frac {3}{10} a \left (\frac {e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}{3 b}-\frac {a e^2 \left (\frac {x \sqrt {e x}}{\sqrt [4]{a+b x^2}}-\frac {a \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4} x^2}dx}{2 b \sqrt [4]{a+b x^2}}\right )}{2 b}\right )+\frac {(e x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 e}\right )}{2 b}+\frac {d (e x)^{7/2} \left (a+b x^2\right )^{7/4}}{7 b e}\) |
| \(\Big \downarrow \) 858 |
| \(\displaystyle \frac {(2 b c-a d) \left (\frac {3}{10} a \left (\frac {e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}{3 b}-\frac {a e^2 \left (\frac {a \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x}}{2 b \sqrt [4]{a+b x^2}}+\frac {x \sqrt {e x}}{\sqrt [4]{a+b x^2}}\right )}{2 b}\right )+\frac {(e x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 e}\right )}{2 b}+\frac {d (e x)^{7/2} \left (a+b x^2\right )^{7/4}}{7 b e}\) |
| \(\Big \downarrow \) 212 |
| \(\displaystyle \frac {(2 b c-a d) \left (\frac {3}{10} a \left (\frac {e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}{3 b}-\frac {a e^2 \left (\frac {\sqrt {a} \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x}\right )\right |2\right )}{\sqrt {b} \sqrt [4]{a+b x^2}}+\frac {x \sqrt {e x}}{\sqrt [4]{a+b x^2}}\right )}{2 b}\right )+\frac {(e x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 e}\right )}{2 b}+\frac {d (e x)^{7/2} \left (a+b x^2\right )^{7/4}}{7 b e}\) |
Input:
Int[(e*x)^(5/2)*(a + b*x^2)^(3/4)*(c + d*x^2),x]
Output:
(d*(e*x)^(7/2)*(a + b*x^2)^(7/4))/(7*b*e) + ((2*b*c - a*d)*(((e*x)^(7/2)*( a + b*x^2)^(3/4))/(5*e) + (3*a*((e*(e*x)^(3/2)*(a + b*x^2)^(3/4))/(3*b) - (a*e^2*((x*Sqrt[e*x])/(a + b*x^2)^(1/4) + (Sqrt[a]*(1 + a/(b*x^2))^(1/4)*S qrt[e*x]*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x)]/2, 2])/(Sqrt[b]*(a + b*x^2) ^(1/4))))/(2*b)))/10))/(2*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[((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[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[Sqrt[(c_)*(x_)]/((a_) + (b_.)*(x_)^2)^(1/4), x_Symbol] :> Simp[x*(Sqrt[ c*x]/(a + b*x^2)^(1/4)), x] - Simp[a/2 Int[Sqrt[c*x]/(a + b*x^2)^(5/4), x ], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]
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[((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]
\[\int \left (e x \right )^{\frac {5}{2}} \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (x^{2} d +c \right )d x\]
Input:
int((e*x)^(5/2)*(b*x^2+a)^(3/4)*(d*x^2+c),x)
Output:
int((e*x)^(5/2)*(b*x^2+a)^(3/4)*(d*x^2+c),x)
\[ \int (e x)^{5/2} \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}} \,d x } \] Input:
integrate((e*x)^(5/2)*(b*x^2+a)^(3/4)*(d*x^2+c),x, algorithm="fricas")
Output:
integral((d*e^2*x^4 + c*e^2*x^2)*(b*x^2 + a)^(3/4)*sqrt(e*x), x)
Result contains complex when optimal does not.
Time = 24.91 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.44 \[ \int (e x)^{5/2} \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\frac {a^{\frac {3}{4}} c e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {11}{4}\right )} + \frac {a^{\frac {3}{4}} d e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {15}{4}\right )} \] Input:
integrate((e*x)**(5/2)*(b*x**2+a)**(3/4)*(d*x**2+c),x)
Output:
a**(3/4)*c*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((-3/4, 7/4), (11/4,), b*x**2 *exp_polar(I*pi)/a)/(2*gamma(11/4)) + a**(3/4)*d*e**(5/2)*x**(11/2)*gamma( 11/4)*hyper((-3/4, 11/4), (15/4,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(15/4 ))
\[ \int (e x)^{5/2} \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}} \,d x } \] Input:
integrate((e*x)^(5/2)*(b*x^2+a)^(3/4)*(d*x^2+c),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(3/4)*(d*x^2 + c)*(e*x)^(5/2), x)
Timed out. \[ \int (e x)^{5/2} \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\text {Timed out} \] Input:
integrate((e*x)^(5/2)*(b*x^2+a)^(3/4)*(d*x^2+c),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int (e x)^{5/2} \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\int {\left (e\,x\right )}^{5/2}\,{\left (b\,x^2+a\right )}^{3/4}\,\left (d\,x^2+c\right ) \,d x \] Input:
int((e*x)^(5/2)*(a + b*x^2)^(3/4)*(c + d*x^2),x)
Output:
int((e*x)^(5/2)*(a + b*x^2)^(3/4)*(c + d*x^2), x)
\[ \int (e x)^{5/2} \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\frac {\sqrt {e}\, e^{2} \left (-14 \sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{2} d x +28 \sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}} a b c x +12 \sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}} a b d \,x^{3}+56 \sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}} b^{2} c \,x^{3}+40 \sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}} b^{2} d \,x^{5}+21 \left (\int \frac {\sqrt {x}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x \right ) a^{3} d -42 \left (\int \frac {\sqrt {x}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x \right ) a^{2} b c \right )}{280 b^{2}} \] Input:
int((e*x)^(5/2)*(b*x^2+a)^(3/4)*(d*x^2+c),x)
Output:
(sqrt(e)*e**2*( - 14*sqrt(x)*(a + b*x**2)**(3/4)*a**2*d*x + 28*sqrt(x)*(a + b*x**2)**(3/4)*a*b*c*x + 12*sqrt(x)*(a + b*x**2)**(3/4)*a*b*d*x**3 + 56* sqrt(x)*(a + b*x**2)**(3/4)*b**2*c*x**3 + 40*sqrt(x)*(a + b*x**2)**(3/4)*b **2*d*x**5 + 21*int((sqrt(x)*(a + b*x**2)**(3/4))/(a + b*x**2),x)*a**3*d - 42*int((sqrt(x)*(a + b*x**2)**(3/4))/(a + b*x**2),x)*a**2*b*c))/(280*b**2 )