Integrand size = 21, antiderivative size = 279 \[ \int \frac {\left (a+b x^2\right )^{5/4}}{\left (c+d x^2\right )^2} \, dx=-\frac {(b c-a d) x \sqrt [4]{a+b x^2}}{2 c d \left (c+d x^2\right )}+\frac {\sqrt {a} \sqrt {b} (3 b c+a d) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{2 c d^2 \left (a+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a} (3 b c+2 a d) \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 c d^2 x}-\frac {\sqrt [4]{a} (3 b c+2 a d) \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 c d^2 x} \]
-1/2*(-a*d+b*c)*x*(b*x^2+a)^(1/4)/c/d/(d*x^2+c)+1/2*(a*d+3*b*c)*(1+b*x^2/a )^(3/4)*(cos(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1 /2)/a^(1/2)))*EllipticF(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))*a^(1/2 )*b^(1/2)/c/d^2/(b*x^2+a)^(3/4)-1/4*a^(1/4)*(2*a*d+3*b*c)*EllipticPi((b*x^ 2+a)^(1/4)/a^(1/4),-a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)*(-b*x^2/a)^(1/2)/c/ d^2/x-1/4*a^(1/4)*(2*a*d+3*b*c)*EllipticPi((b*x^2+a)^(1/4)/a^(1/4),a^(1/2) *d^(1/2)/(a*d-b*c)^(1/2),I)*(-b*x^2/a)^(1/2)/c/d^2/x
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.36 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+b x^2\right )^{5/4}}{\left (c+d x^2\right )^2} \, dx=\frac {x \left (b (3 b c+a d) x^2 \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {6 c \left (-6 a c \left (2 a^2 d-b^2 c x^2+a b d x^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+(-b c+a d) x^2 \left (a+b x^2\right ) \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}{\left (c+d x^2\right ) \left (-6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}\right )}{12 c^2 d \left (a+b x^2\right )^{3/4}} \]
(x*(b*(3*b*c + a*d)*x^2*(1 + (b*x^2)/a)^(3/4)*AppellF1[3/2, 3/4, 1, 5/2, - ((b*x^2)/a), -((d*x^2)/c)] + (6*c*(-6*a*c*(2*a^2*d - b^2*c*x^2 + a*b*d*x^2 )*AppellF1[1/2, 3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + (-(b*c) + a*d)* x^2*(a + b*x^2)*(4*a*d*AppellF1[3/2, 3/4, 2, 5/2, -((b*x^2)/a), -((d*x^2)/ c)] + 3*b*c*AppellF1[3/2, 7/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))/((c + d*x^2)*(-6*a*c*AppellF1[1/2, 3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + x^2*(4*a*d*AppellF1[3/2, 3/4, 2, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + 3*b*c* AppellF1[3/2, 7/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))))/(12*c^2*d*(a + b*x^2)^(3/4))
Time = 0.46 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {315, 27, 405, 231, 229, 312, 118, 25, 925, 1542}
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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^{5/4}}{\left (c+d x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle \frac {\int \frac {b (3 b c+a d) x^2+2 a (b c+a d)}{2 \left (b x^2+a\right )^{3/4} \left (d x^2+c\right )}dx}{2 c d}-\frac {x \sqrt [4]{a+b x^2} (b c-a d)}{2 c d \left (c+d x^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {b (3 b c+a d) x^2+2 a (b c+a d)}{\left (b x^2+a\right )^{3/4} \left (d x^2+c\right )}dx}{4 c d}-\frac {x \sqrt [4]{a+b x^2} (b c-a d)}{2 c d \left (c+d x^2\right )}\) |
| \(\Big \downarrow \) 405 |
| \(\displaystyle \frac {\frac {b (a d+3 b c) \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx}{d}-\frac {(b c-a d) (2 a d+3 b c) \int \frac {1}{\left (b x^2+a\right )^{3/4} \left (d x^2+c\right )}dx}{d}}{4 c d}-\frac {x \sqrt [4]{a+b x^2} (b c-a d)}{2 c d \left (c+d x^2\right )}\) |
| \(\Big \downarrow \) 231 |
| \(\displaystyle \frac {\frac {b \left (\frac {b x^2}{a}+1\right )^{3/4} (a d+3 b c) \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{d \left (a+b x^2\right )^{3/4}}-\frac {(b c-a d) (2 a d+3 b c) \int \frac {1}{\left (b x^2+a\right )^{3/4} \left (d x^2+c\right )}dx}{d}}{4 c d}-\frac {x \sqrt [4]{a+b x^2} (b c-a d)}{2 c d \left (c+d x^2\right )}\) |
| \(\Big \downarrow \) 229 |
| \(\displaystyle \frac {\frac {2 \sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (a d+3 b c) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{d \left (a+b x^2\right )^{3/4}}-\frac {(b c-a d) (2 a d+3 b c) \int \frac {1}{\left (b x^2+a\right )^{3/4} \left (d x^2+c\right )}dx}{d}}{4 c d}-\frac {x \sqrt [4]{a+b x^2} (b c-a d)}{2 c d \left (c+d x^2\right )}\) |
| \(\Big \downarrow \) 312 |
| \(\displaystyle \frac {\frac {2 \sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (a d+3 b c) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{d \left (a+b x^2\right )^{3/4}}-\frac {\sqrt {-\frac {b x^2}{a}} (b c-a d) (2 a d+3 b c) \int \frac {1}{\sqrt {-\frac {b x^2}{a}} \left (b x^2+a\right )^{3/4} \left (d x^2+c\right )}dx^2}{2 d x}}{4 c d}-\frac {x \sqrt [4]{a+b x^2} (b c-a d)}{2 c d \left (c+d x^2\right )}\) |
| \(\Big \downarrow \) 118 |
| \(\displaystyle \frac {\frac {2 \sqrt {-\frac {b x^2}{a}} (b c-a d) (2 a d+3 b c) \int -\frac {1}{\sqrt {1-\frac {x^8}{a}} \left (d x^8+b c-a d\right )}d\sqrt [4]{b x^2+a}}{d x}+\frac {2 \sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (a d+3 b c) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{d \left (a+b x^2\right )^{3/4}}}{4 c d}-\frac {x \sqrt [4]{a+b x^2} (b c-a d)}{2 c d \left (c+d x^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {2 \sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (a d+3 b c) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{d \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt {-\frac {b x^2}{a}} (b c-a d) (2 a d+3 b c) \int \frac {1}{\sqrt {1-\frac {x^8}{a}} \left (d x^8+b c-a d\right )}d\sqrt [4]{b x^2+a}}{d x}}{4 c d}-\frac {x \sqrt [4]{a+b x^2} (b c-a d)}{2 c d \left (c+d x^2\right )}\) |
| \(\Big \downarrow \) 925 |
| \(\displaystyle \frac {\frac {2 \sqrt {-\frac {b x^2}{a}} (b c-a d) (2 a d+3 b c) \left (-\frac {\int \frac {1}{\left (1-\frac {\sqrt {d} x^4}{\sqrt {a d-b c}}\right ) \sqrt {1-\frac {x^8}{a}}}d\sqrt [4]{b x^2+a}}{2 (b c-a d)}-\frac {\int \frac {1}{\left (\frac {\sqrt {d} x^4}{\sqrt {a d-b c}}+1\right ) \sqrt {1-\frac {x^8}{a}}}d\sqrt [4]{b x^2+a}}{2 (b c-a d)}\right )}{d x}+\frac {2 \sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (a d+3 b c) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{d \left (a+b x^2\right )^{3/4}}}{4 c d}-\frac {x \sqrt [4]{a+b x^2} (b c-a d)}{2 c d \left (c+d x^2\right )}\) |
| \(\Big \downarrow \) 1542 |
| \(\displaystyle \frac {\frac {2 \sqrt {-\frac {b x^2}{a}} (b c-a d) (2 a d+3 b c) \left (-\frac {\sqrt [4]{a} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}},\arcsin \left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right ),-1\right )}{2 (b c-a d)}-\frac {\sqrt [4]{a} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}},\arcsin \left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right ),-1\right )}{2 (b c-a d)}\right )}{d x}+\frac {2 \sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (a d+3 b c) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{d \left (a+b x^2\right )^{3/4}}}{4 c d}-\frac {x \sqrt [4]{a+b x^2} (b c-a d)}{2 c d \left (c+d x^2\right )}\) |
-1/2*((b*c - a*d)*x*(a + b*x^2)^(1/4))/(c*d*(c + d*x^2)) + ((2*Sqrt[a]*Sqr t[b]*(3*b*c + a*d)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt [a]]/2, 2])/(d*(a + b*x^2)^(3/4)) + (2*(b*c - a*d)*(3*b*c + 2*a*d)*Sqrt[-( (b*x^2)/a)]*(-1/2*(a^(1/4)*EllipticPi[-((Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a* d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(b*c - a*d) - (a^(1/4)*Ellipt icPi[(Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d], ArcSin[(a + b*x^2)^(1/4)/a^(1/4 )], -1])/(2*(b*c - a*d))))/(d*x))/(4*c*d)
3.4.32.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^( 3/4)), x_] :> Simp[-4 Subst[Int[1/((b*e - a*f - b*x^4)*Sqrt[c - d*(e/f) + d*(x^4/f)]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d, e, f}, x] & & GtQ[-f/(d*e - c*f), 0]
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[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Sim p[Sqrt[(-b)*(x^2/a)]/(2*x) Subst[Int[1/(Sqrt[(-b)*(x/a)]*(a + b*x)^(3/4)* (c + d*x)), x], x, x^2], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[(((a_) + (b_.)*(x_)^2)^(p_)*((e_) + (f_.)*(x_)^2))/((c_) + (d_.)*(x_)^2 ), x_Symbol] :> Simp[f/d Int[(a + b*x^2)^p, x], x] + Simp[(d*e - c*f)/d Int[(a + b*x^2)^p/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, p}, x]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Simp[ 1/(2*c) Int[1/(Sqrt[a + b*x^4]*(1 - Rt[-d/c, 2]*x^2)), x], x] + Simp[1/(2 *c) Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-d/c, 2]*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[ {q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x ], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]
\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {5}{4}}}{\left (d \,x^{2}+c \right )^{2}}d x\]
Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/4}}{\left (c+d x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b x^2\right )^{5/4}}{\left (c+d x^2\right )^2} \, dx=\int \frac {\left (a + b x^{2}\right )^{\frac {5}{4}}}{\left (c + d x^{2}\right )^{2}}\, dx \]
\[ \int \frac {\left (a+b x^2\right )^{5/4}}{\left (c+d x^2\right )^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {5}{4}}}{{\left (d x^{2} + c\right )}^{2}} \,d x } \]
\[ \int \frac {\left (a+b x^2\right )^{5/4}}{\left (c+d x^2\right )^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {5}{4}}}{{\left (d x^{2} + c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/4}}{\left (c+d x^2\right )^2} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{5/4}}{{\left (d\,x^2+c\right )}^2} \,d x \]