Integrand size = 21, antiderivative size = 214 \[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2 \, dx=\frac {2 a \left (55 b^2 c^2-4 a d (5 b c-a d)\right ) x \sqrt [4]{a+b x^2}}{231 b^2}+\frac {2}{385} \left (55 c^2-\frac {4 a d (5 b c-a d)}{b^2}\right ) x \left (a+b x^2\right )^{5/4}+\frac {4 d (5 b c-a d) x \left (a+b x^2\right )^{9/4}}{55 b^2}+\frac {2 d^2 x^3 \left (a+b x^2\right )^{9/4}}{15 b}+\frac {2 a^{5/2} \left (55 b^2 c^2-4 a d (5 b c-a d)\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{231 b^{5/2} \left (a+b x^2\right )^{3/4}} \] Output:
2/231*a*(55*b^2*c^2-4*a*d*(-a*d+5*b*c))*x*(b*x^2+a)^(1/4)/b^2+2/385*(55*c^ 2-4*a*d*(-a*d+5*b*c)/b^2)*x*(b*x^2+a)^(5/4)+4/55*d*(-a*d+5*b*c)*x*(b*x^2+a )^(9/4)/b^2+2/15*d^2*x^3*(b*x^2+a)^(9/4)/b+2/231*a^(5/2)*(55*b^2*c^2-4*a*d *(-a*d+5*b*c))*(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 = 11.50 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.82 \[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2 \, dx=\frac {x \sqrt [4]{a+b x^2} \left (7 a \left (15 c^2+10 c d x^2+3 d^2 x^4\right ) \operatorname {Gamma}\left (-\frac {5}{4}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},\frac {7}{2},-\frac {b x^2}{a}\right )-8 b x^2 \left (2 c^2+3 c d x^2+d^2 x^4\right ) \operatorname {Gamma}\left (-\frac {1}{4}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {9}{2},-\frac {b x^2}{a}\right )-4 b x^2 \left (c+d x^2\right )^2 \operatorname {Gamma}\left (-\frac {1}{4}\right ) \, _3F_2\left (-\frac {1}{4},\frac {3}{2},2;1,\frac {9}{2};-\frac {b x^2}{a}\right )\right )}{105 \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Gamma}\left (-\frac {5}{4}\right )} \] Input:
Integrate[(a + b*x^2)^(5/4)*(c + d*x^2)^2,x]
Output:
(x*(a + b*x^2)^(1/4)*(7*a*(15*c^2 + 10*c*d*x^2 + 3*d^2*x^4)*Gamma[-5/4]*Hy pergeometric2F1[-5/4, 1/2, 7/2, -((b*x^2)/a)] - 8*b*x^2*(2*c^2 + 3*c*d*x^2 + d^2*x^4)*Gamma[-1/4]*Hypergeometric2F1[-1/4, 3/2, 9/2, -((b*x^2)/a)] - 4*b*x^2*(c + d*x^2)^2*Gamma[-1/4]*HypergeometricPFQ[{-1/4, 3/2, 2}, {1, 9/ 2}, -((b*x^2)/a)]))/(105*(1 + (b*x^2)/a)^(1/4)*Gamma[-5/4])
Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {318, 27, 299, 211, 211, 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.
| \(\displaystyle \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2 \, dx\) |
| \(\Big \downarrow \) 318 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \left (b x^2+a\right )^{5/4} \left (d (19 b c-6 a d) x^2+c (15 b c-2 a d)\right )dx}{15 b}+\frac {2 d x \left (a+b x^2\right )^{9/4} \left (c+d x^2\right )}{15 b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \left (b x^2+a\right )^{5/4} \left (d (19 b c-6 a d) x^2+c (15 b c-2 a d)\right )dx}{15 b}+\frac {2 d x \left (a+b x^2\right )^{9/4} \left (c+d x^2\right )}{15 b}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {3 \left (4 a^2 d^2-20 a b c d+55 b^2 c^2\right ) \int \left (b x^2+a\right )^{5/4}dx}{11 b}+\frac {2 d x \left (a+b x^2\right )^{9/4} (19 b c-6 a d)}{11 b}}{15 b}+\frac {2 d x \left (a+b x^2\right )^{9/4} \left (c+d x^2\right )}{15 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {3 \left (4 a^2 d^2-20 a b c d+55 b^2 c^2\right ) \left (\frac {5}{7} a \int \sqrt [4]{b x^2+a}dx+\frac {2}{7} x \left (a+b x^2\right )^{5/4}\right )}{11 b}+\frac {2 d x \left (a+b x^2\right )^{9/4} (19 b c-6 a d)}{11 b}}{15 b}+\frac {2 d x \left (a+b x^2\right )^{9/4} \left (c+d x^2\right )}{15 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {3 \left (4 a^2 d^2-20 a b c d+55 b^2 c^2\right ) \left (\frac {5}{7} a \left (\frac {1}{3} a \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )+\frac {2}{7} x \left (a+b x^2\right )^{5/4}\right )}{11 b}+\frac {2 d x \left (a+b x^2\right )^{9/4} (19 b c-6 a d)}{11 b}}{15 b}+\frac {2 d x \left (a+b x^2\right )^{9/4} \left (c+d x^2\right )}{15 b}\) |
| \(\Big \downarrow \) 231 |
| \(\displaystyle \frac {\frac {3 \left (4 a^2 d^2-20 a b c d+55 b^2 c^2\right ) \left (\frac {5}{7} a \left (\frac {a \left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{3 \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )+\frac {2}{7} x \left (a+b x^2\right )^{5/4}\right )}{11 b}+\frac {2 d x \left (a+b x^2\right )^{9/4} (19 b c-6 a d)}{11 b}}{15 b}+\frac {2 d x \left (a+b x^2\right )^{9/4} \left (c+d x^2\right )}{15 b}\) |
| \(\Big \downarrow \) 229 |
| \(\displaystyle \frac {\frac {3 \left (4 a^2 d^2-20 a b c d+55 b^2 c^2\right ) \left (\frac {5}{7} a \left (\frac {2 a^{3/2} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 \sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )+\frac {2}{7} x \left (a+b x^2\right )^{5/4}\right )}{11 b}+\frac {2 d x \left (a+b x^2\right )^{9/4} (19 b c-6 a d)}{11 b}}{15 b}+\frac {2 d x \left (a+b x^2\right )^{9/4} \left (c+d x^2\right )}{15 b}\) |
Input:
Int[(a + b*x^2)^(5/4)*(c + d*x^2)^2,x]
Output:
(2*d*x*(a + b*x^2)^(9/4)*(c + d*x^2))/(15*b) + ((2*d*(19*b*c - 6*a*d)*x*(a + b*x^2)^(9/4))/(11*b) + (3*(55*b^2*c^2 - 20*a*b*c*d + 4*a^2*d^2)*((2*x*( a + b*x^2)^(5/4))/7 + (5*a*((2*x*(a + b*x^2)^(1/4))/3 + (2*a^(3/2)*(1 + (b *x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3*Sqrt[b]*(a + b*x^2)^(3/4))))/7))/(11*b))/(15*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
\[\int \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (x^{2} d +c \right )^{2}d x\]
Input:
int((b*x^2+a)^(5/4)*(d*x^2+c)^2,x)
Output:
int((b*x^2+a)^(5/4)*(d*x^2+c)^2,x)
\[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2 \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{2} \,d x } \] Input:
integrate((b*x^2+a)^(5/4)*(d*x^2+c)^2,x, algorithm="fricas")
Output:
integral((b*d^2*x^6 + (2*b*c*d + a*d^2)*x^4 + a*c^2 + (b*c^2 + 2*a*c*d)*x^ 2)*(b*x^2 + a)^(1/4), x)
Result contains complex when optimal does not.
Time = 2.31 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.97 \[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2 \, dx=a^{\frac {5}{4}} c^{2} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {2 a^{\frac {5}{4}} c d x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {a^{\frac {5}{4}} d^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + \frac {\sqrt [4]{a} b c^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {2 \sqrt [4]{a} b c d x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + \frac {\sqrt [4]{a} b d^{2} x^{7} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7} \] Input:
integrate((b*x**2+a)**(5/4)*(d*x**2+c)**2,x)
Output:
a**(5/4)*c**2*x*hyper((-1/4, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a) + 2*a **(5/4)*c*d*x**3*hyper((-1/4, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + a**(5/4)*d**2*x**5*hyper((-1/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + a**(1/4)*b*c**2*x**3*hyper((-1/4, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a )/3 + 2*a**(1/4)*b*c*d*x**5*hyper((-1/4, 5/2), (7/2,), b*x**2*exp_polar(I* pi)/a)/5 + a**(1/4)*b*d**2*x**7*hyper((-1/4, 7/2), (9/2,), b*x**2*exp_pola r(I*pi)/a)/7
\[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2 \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{2} \,d x } \] Input:
integrate((b*x^2+a)^(5/4)*(d*x^2+c)^2,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(5/4)*(d*x^2 + c)^2, x)
\[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2 \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{2} \,d x } \] Input:
integrate((b*x^2+a)^(5/4)*(d*x^2+c)^2,x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(5/4)*(d*x^2 + c)^2, x)
Timed out. \[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2 \, dx=\int {\left (b\,x^2+a\right )}^{5/4}\,{\left (d\,x^2+c\right )}^2 \,d x \] Input:
int((a + b*x^2)^(5/4)*(c + d*x^2)^2,x)
Output:
int((a + b*x^2)^(5/4)*(c + d*x^2)^2, x)
\[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2 \, dx=\frac {-20 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{3} d^{2} x +100 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} b c d x +10 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} b \,d^{2} x^{3}+880 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,b^{2} c^{2} x +720 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,b^{2} c d \,x^{3}+224 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,b^{2} d^{2} x^{5}+330 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{3} c^{2} x^{3}+420 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{3} c d \,x^{5}+154 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{3} d^{2} x^{7}+20 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{4} d^{2}-100 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{3} b c d +275 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{2} b^{2} c^{2}}{1155 b^{2}} \] Input:
int((b*x^2+a)^(5/4)*(d*x^2+c)^2,x)
Output:
( - 20*(a + b*x**2)**(1/4)*a**3*d**2*x + 100*(a + b*x**2)**(1/4)*a**2*b*c* d*x + 10*(a + b*x**2)**(1/4)*a**2*b*d**2*x**3 + 880*(a + b*x**2)**(1/4)*a* b**2*c**2*x + 720*(a + b*x**2)**(1/4)*a*b**2*c*d*x**3 + 224*(a + b*x**2)** (1/4)*a*b**2*d**2*x**5 + 330*(a + b*x**2)**(1/4)*b**3*c**2*x**3 + 420*(a + b*x**2)**(1/4)*b**3*c*d*x**5 + 154*(a + b*x**2)**(1/4)*b**3*d**2*x**7 + 2 0*int((a + b*x**2)**(1/4)/(a + b*x**2),x)*a**4*d**2 - 100*int((a + b*x**2) **(1/4)/(a + b*x**2),x)*a**3*b*c*d + 275*int((a + b*x**2)**(1/4)/(a + b*x* *2),x)*a**2*b**2*c**2)/(1155*b**2)