Integrand size = 21, antiderivative size = 304 \[ \int \frac {1}{\left (a+b x^2\right )^{11/4} \left (c+d x^2\right )} \, dx=\frac {2 b x}{7 a (b c-a d) \left (a+b x^2\right )^{7/4}}+\frac {2 b (5 b c-12 a d) x}{21 a^2 (b c-a d)^2 \left (a+b x^2\right )^{3/4}}+\frac {2 \sqrt {b} (5 b c-12 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{21 a^{3/2} (b c-a d)^2 \left (a+b x^2\right )^{3/4}}+\frac {\sqrt [4]{a} d^2 \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{(b c-a d)^3 x}+\frac {\sqrt [4]{a} d^2 \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{(b c-a d)^3 x} \]
2/7*b*x/a/(-a*d+b*c)/(b*x^2+a)^(7/4)+2/21*b*(-12*a*d+5*b*c)*x/a^2/(-a*d+b* c)^2/(b*x^2+a)^(3/4)+2/21*(-12*a*d+5*b*c)*(1+b*x^2/a)^(3/4)*(cos(1/2*arcta n(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/2)))*Elliptic F(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))*b^(1/2)/a^(3/2)/(-a*d+b*c)^2 /(b*x^2+a)^(3/4)+a^(1/4)*d^2*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)/(-a*d+b*c)^3/x+a^(1/4)*d^2*Elli pticPi((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)/(-a*d+b*c)^3/x
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.83 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.42 \[ \int \frac {1}{\left (a+b x^2\right )^{11/4} \left (c+d x^2\right )} \, dx=-\frac {x \left (\frac {b d (-5 b c+12 a d) x^2 \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c}+\frac {6 \left (3 a c \left (21 a^3 d^2+5 b^3 c x^2 \left (3 c+2 d x^2\right )-3 a^2 b d \left (14 c+3 d x^2\right )+a b^2 \left (21 c^2-20 c d x^2-24 d^2 x^4\right )\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b x^2 \left (c+d x^2\right ) \left (15 a^2 d-5 b^2 c x^2+a b \left (-8 c+12 d x^2\right )\right ) \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}{\left (a+b x^2\right ) \left (c+d x^2\right ) \left (-6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}\right )}{63 a^2 (b c-a d)^2 \left (a+b x^2\right )^{3/4}} \]
-1/63*(x*((b*d*(-5*b*c + 12*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)])/c + (6*(3*a*c*(21*a^3*d^2 + 5*b^3 *c*x^2*(3*c + 2*d*x^2) - 3*a^2*b*d*(14*c + 3*d*x^2) + a*b^2*(21*c^2 - 20*c *d*x^2 - 24*d^2*x^4))*AppellF1[1/2, 3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c )] + b*x^2*(c + d*x^2)*(15*a^2*d - 5*b^2*c*x^2 + a*b*(-8*c + 12*d*x^2))*(4 *a*d*AppellF1[3/2, 3/4, 2, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + 3*b*c*Appell F1[3/2, 7/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))/((a + b*x^2)*(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*Appell F1[3/2, 7/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))))/(a^2*(b*c - a*d)^2*( a + b*x^2)^(3/4))
Time = 0.52 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {316, 27, 402, 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.
| \(\displaystyle \int \frac {1}{\left (a+b x^2\right )^{11/4} \left (c+d x^2\right )} \, dx\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)}-\frac {2 \int -\frac {5 b d x^2+5 b c-7 a d}{2 \left (b x^2+a\right )^{7/4} \left (d x^2+c\right )}dx}{7 a (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {5 b d x^2+5 b c-7 a d}{\left (b x^2+a\right )^{7/4} \left (d x^2+c\right )}dx}{7 a (b c-a d)}+\frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {2 b x (5 b c-12 a d)}{3 a \left (a+b x^2\right )^{3/4} (b c-a d)}-\frac {2 \int -\frac {5 b^2 c^2-12 a b d c+21 a^2 d^2+b d (5 b c-12 a d) x^2}{2 \left (b x^2+a\right )^{3/4} \left (d x^2+c\right )}dx}{3 a (b c-a d)}}{7 a (b c-a d)}+\frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {5 b^2 c^2-12 a b d c+21 a^2 d^2+b d (5 b c-12 a d) x^2}{\left (b x^2+a\right )^{3/4} \left (d x^2+c\right )}dx}{3 a (b c-a d)}+\frac {2 b x (5 b c-12 a d)}{3 a \left (a+b x^2\right )^{3/4} (b c-a d)}}{7 a (b c-a d)}+\frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)}\) |
| \(\Big \downarrow \) 405 |
| \(\displaystyle \frac {\frac {21 a^2 d^2 \int \frac {1}{\left (b x^2+a\right )^{3/4} \left (d x^2+c\right )}dx+b (5 b c-12 a d) \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx}{3 a (b c-a d)}+\frac {2 b x (5 b c-12 a d)}{3 a \left (a+b x^2\right )^{3/4} (b c-a d)}}{7 a (b c-a d)}+\frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)}\) |
| \(\Big \downarrow \) 231 |
| \(\displaystyle \frac {\frac {21 a^2 d^2 \int \frac {1}{\left (b x^2+a\right )^{3/4} \left (d x^2+c\right )}dx+\frac {b \left (\frac {b x^2}{a}+1\right )^{3/4} (5 b c-12 a d) \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{\left (a+b x^2\right )^{3/4}}}{3 a (b c-a d)}+\frac {2 b x (5 b c-12 a d)}{3 a \left (a+b x^2\right )^{3/4} (b c-a d)}}{7 a (b c-a d)}+\frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)}\) |
| \(\Big \downarrow \) 229 |
| \(\displaystyle \frac {\frac {21 a^2 d^2 \int \frac {1}{\left (b x^2+a\right )^{3/4} \left (d x^2+c\right )}dx+\frac {2 \sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (5 b c-12 a d) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\left (a+b x^2\right )^{3/4}}}{3 a (b c-a d)}+\frac {2 b x (5 b c-12 a d)}{3 a \left (a+b x^2\right )^{3/4} (b c-a d)}}{7 a (b c-a d)}+\frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)}\) |
| \(\Big \downarrow \) 312 |
| \(\displaystyle \frac {\frac {\frac {21 a^2 d^2 \sqrt {-\frac {b x^2}{a}} \int \frac {1}{\sqrt {-\frac {b x^2}{a}} \left (b x^2+a\right )^{3/4} \left (d x^2+c\right )}dx^2}{2 x}+\frac {2 \sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (5 b c-12 a d) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\left (a+b x^2\right )^{3/4}}}{3 a (b c-a d)}+\frac {2 b x (5 b c-12 a d)}{3 a \left (a+b x^2\right )^{3/4} (b c-a d)}}{7 a (b c-a d)}+\frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)}\) |
| \(\Big \downarrow \) 118 |
| \(\displaystyle \frac {\frac {\frac {2 \sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (5 b c-12 a d) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\left (a+b x^2\right )^{3/4}}-\frac {42 a^2 d^2 \sqrt {-\frac {b x^2}{a}} \int -\frac {1}{\sqrt {1-\frac {x^8}{a}} \left (d x^8+b c-a d\right )}d\sqrt [4]{b x^2+a}}{x}}{3 a (b c-a d)}+\frac {2 b x (5 b c-12 a d)}{3 a \left (a+b x^2\right )^{3/4} (b c-a d)}}{7 a (b c-a d)}+\frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {42 a^2 d^2 \sqrt {-\frac {b x^2}{a}} \int \frac {1}{\sqrt {1-\frac {x^8}{a}} \left (d x^8+b c-a d\right )}d\sqrt [4]{b x^2+a}}{x}+\frac {2 \sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (5 b c-12 a d) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\left (a+b x^2\right )^{3/4}}}{3 a (b c-a d)}+\frac {2 b x (5 b c-12 a d)}{3 a \left (a+b x^2\right )^{3/4} (b c-a d)}}{7 a (b c-a d)}+\frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)}\) |
| \(\Big \downarrow \) 925 |
| \(\displaystyle \frac {\frac {\frac {2 \sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (5 b c-12 a d) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\left (a+b x^2\right )^{3/4}}-\frac {42 a^2 d^2 \sqrt {-\frac {b x^2}{a}} \left (-\frac {\int \frac {1}{\left (1-\frac {\sqrt {d} x^4}{\sqrt {a d-b c}}\right ) \sqrt {1-\frac {x^8}{a}}}d\sqrt [4]{b x^2+a}}{2 (b c-a d)}-\frac {\int \frac {1}{\left (\frac {\sqrt {d} x^4}{\sqrt {a d-b c}}+1\right ) \sqrt {1-\frac {x^8}{a}}}d\sqrt [4]{b x^2+a}}{2 (b c-a d)}\right )}{x}}{3 a (b c-a d)}+\frac {2 b x (5 b c-12 a d)}{3 a \left (a+b x^2\right )^{3/4} (b c-a d)}}{7 a (b c-a d)}+\frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)}\) |
| \(\Big \downarrow \) 1542 |
| \(\displaystyle \frac {\frac {\frac {2 \sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (5 b c-12 a d) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\left (a+b x^2\right )^{3/4}}-\frac {42 a^2 d^2 \sqrt {-\frac {b x^2}{a}} \left (-\frac {\sqrt [4]{a} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}},\arcsin \left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right ),-1\right )}{2 (b c-a d)}-\frac {\sqrt [4]{a} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}},\arcsin \left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right ),-1\right )}{2 (b c-a d)}\right )}{x}}{3 a (b c-a d)}+\frac {2 b x (5 b c-12 a d)}{3 a \left (a+b x^2\right )^{3/4} (b c-a d)}}{7 a (b c-a d)}+\frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)}\) |
(2*b*x)/(7*a*(b*c - a*d)*(a + b*x^2)^(7/4)) + ((2*b*(5*b*c - 12*a*d)*x)/(3 *a*(b*c - a*d)*(a + b*x^2)^(3/4)) + ((2*Sqrt[a]*Sqrt[b]*(5*b*c - 12*a*d)*( 1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(a + b*x ^2)^(3/4) - (42*a^2*d^2*Sqrt[-((b*x^2)/a)]*(-1/2*(a^(1/4)*EllipticPi[-((Sq rt[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)*EllipticPi[(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))))/x)/(3*a*(b*c - a *d)))/(7*a*(b*c - a*d))
3.4.30.3.1 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[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
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]
\[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {11}{4}} \left (d \,x^{2}+c \right )}d x\]
Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{11/4} \left (c+d x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\left (a+b x^2\right )^{11/4} \left (c+d x^2\right )} \, dx=\int \frac {1}{\left (a + b x^{2}\right )^{\frac {11}{4}} \left (c + d x^{2}\right )}\, dx \]
\[ \int \frac {1}{\left (a+b x^2\right )^{11/4} \left (c+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {11}{4}} {\left (d x^{2} + c\right )}} \,d x } \]
\[ \int \frac {1}{\left (a+b x^2\right )^{11/4} \left (c+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {11}{4}} {\left (d x^{2} + c\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{11/4} \left (c+d x^2\right )} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{11/4}\,\left (d\,x^2+c\right )} \,d x \]