Integrand size = 19, antiderivative size = 145 \[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) \, dx=\frac {10 a (11 b c-2 a d) x \sqrt [4]{a+b x^2}}{231 b}+\frac {2 (11 b c-2 a d) x \left (a+b x^2\right )^{5/4}}{77 b}+\frac {2 d x \left (a+b x^2\right )^{9/4}}{11 b}+\frac {10 a^{5/2} (11 b c-2 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{231 b^{3/2} \left (a+b x^2\right )^{3/4}} \] Output:
10/231*a*(-2*a*d+11*b*c)*x*(b*x^2+a)^(1/4)/b+2/77*(-2*a*d+11*b*c)*x*(b*x^2 +a)^(5/4)/b+2/11*d*x*(b*x^2+a)^(9/4)/b+10/231*a^(5/2)*(-2*a*d+11*b*c)*(1+b *x^2/a)^(3/4)*InverseJacobiAM(1/2*arctan(b^(1/2)*x/a^(1/2)),2^(1/2))/b^(3/ 2)/(b*x^2+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.53 \[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) \, dx=\frac {2 x \sqrt [4]{a+b x^2} \left (d \left (a+b x^2\right )^2-\frac {a \left (-\frac {11 b c}{2}+a d\right ) \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{\sqrt [4]{1+\frac {b x^2}{a}}}\right )}{11 b} \] Input:
Integrate[(a + b*x^2)^(5/4)*(c + d*x^2),x]
Output:
(2*x*(a + b*x^2)^(1/4)*(d*(a + b*x^2)^2 - (a*((-11*b*c)/2 + a*d)*Hypergeom etric2F1[-5/4, 1/2, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(1/4)))/(11*b)
Time = 0.22 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {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.
| \(\displaystyle \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) \, dx\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {(11 b c-2 a d) \int \left (b x^2+a\right )^{5/4}dx}{11 b}+\frac {2 d x \left (a+b x^2\right )^{9/4}}{11 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {(11 b c-2 a d) \left (\frac {5}{7} a \int \sqrt [4]{b x^2+a}dx+\frac {2}{7} x \left (a+b x^2\right )^{5/4}\right )}{11 b}+\frac {2 d x \left (a+b x^2\right )^{9/4}}{11 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {(11 b c-2 a d) \left (\frac {5}{7} a \left (\frac {1}{3} a \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )+\frac {2}{7} x \left (a+b x^2\right )^{5/4}\right )}{11 b}+\frac {2 d x \left (a+b x^2\right )^{9/4}}{11 b}\) |
| \(\Big \downarrow \) 231 |
| \(\displaystyle \frac {(11 b c-2 a d) \left (\frac {5}{7} a \left (\frac {a \left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{3 \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )+\frac {2}{7} x \left (a+b x^2\right )^{5/4}\right )}{11 b}+\frac {2 d x \left (a+b x^2\right )^{9/4}}{11 b}\) |
| \(\Big \downarrow \) 229 |
| \(\displaystyle \frac {(11 b c-2 a d) \left (\frac {5}{7} a \left (\frac {2 a^{3/2} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 \sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )+\frac {2}{7} x \left (a+b x^2\right )^{5/4}\right )}{11 b}+\frac {2 d x \left (a+b x^2\right )^{9/4}}{11 b}\) |
Input:
Int[(a + b*x^2)^(5/4)*(c + d*x^2),x]
Output:
(2*d*x*(a + b*x^2)^(9/4))/(11*b) + ((11*b*c - 2*a*d)*((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)
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]
\[\int \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (x^{2} d +c \right )d x\]
Input:
int((b*x^2+a)^(5/4)*(d*x^2+c),x)
Output:
int((b*x^2+a)^(5/4)*(d*x^2+c),x)
\[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )} \,d x } \] Input:
integrate((b*x^2+a)^(5/4)*(d*x^2+c),x, algorithm="fricas")
Output:
integral((b*d*x^4 + (b*c + a*d)*x^2 + a*c)*(b*x^2 + a)^(1/4), x)
Result contains complex when optimal does not.
Time = 1.61 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.88 \[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) \, dx=a^{\frac {5}{4}} c x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {a^{\frac {5}{4}} d x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {\sqrt [4]{a} b c x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {\sqrt [4]{a} b d x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} \] Input:
integrate((b*x**2+a)**(5/4)*(d*x**2+c),x)
Output:
a**(5/4)*c*x*hyper((-1/4, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a) + a**(5/ 4)*d*x**3*hyper((-1/4, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + a**(1/4 )*b*c*x**3*hyper((-1/4, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + a**(1/ 4)*b*d*x**5*hyper((-1/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/5
\[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )} \,d x } \] Input:
integrate((b*x^2+a)^(5/4)*(d*x^2+c),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(5/4)*(d*x^2 + c), x)
\[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )} \,d x } \] Input:
integrate((b*x^2+a)^(5/4)*(d*x^2+c),x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(5/4)*(d*x^2 + c), x)
Timed out. \[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) \, dx=\int {\left (b\,x^2+a\right )}^{5/4}\,\left (d\,x^2+c\right ) \,d x \] Input:
int((a + b*x^2)^(5/4)*(c + d*x^2),x)
Output:
int((a + b*x^2)^(5/4)*(c + d*x^2), x)
\[ \int \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) \, dx=\frac {10 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} d x +176 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b c x +72 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b d \,x^{3}+66 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} c \,x^{3}+42 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} d \,x^{5}-10 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{3} d +55 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{2} b c}{231 b} \] Input:
int((b*x^2+a)^(5/4)*(d*x^2+c),x)
Output:
(10*(a + b*x**2)**(1/4)*a**2*d*x + 176*(a + b*x**2)**(1/4)*a*b*c*x + 72*(a + b*x**2)**(1/4)*a*b*d*x**3 + 66*(a + b*x**2)**(1/4)*b**2*c*x**3 + 42*(a + b*x**2)**(1/4)*b**2*d*x**5 - 10*int((a + b*x**2)**(1/4)/(a + b*x**2),x)* a**3*d + 55*int((a + b*x**2)**(1/4)/(a + b*x**2),x)*a**2*b*c)/(231*b)