Integrand size = 28, antiderivative size = 243 \[ \int \frac {x^{13/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=-\frac {2 b^2 (13 b B-15 A c) \sqrt {b x^2+c x^4}}{77 c^4 \sqrt {x}}+\frac {6 b (13 b B-15 A c) x^{3/2} \sqrt {b x^2+c x^4}}{385 c^3}-\frac {2 (13 b B-15 A c) x^{7/2} \sqrt {b x^2+c x^4}}{165 c^2}+\frac {2 B x^{11/2} \sqrt {b x^2+c x^4}}{15 c}+\frac {b^{11/4} (13 b B-15 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{77 c^{17/4} \sqrt {b x^2+c x^4}} \]
6/385*b*(-15*A*c+13*B*b)*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c^3-2/165*(-15*A*c+13 *B*b)*x^(7/2)*(c*x^4+b*x^2)^(1/2)/c^2+2/15*B*x^(11/2)*(c*x^4+b*x^2)^(1/2)/ c-2/77*b^2*(-15*A*c+13*B*b)*(c*x^4+b*x^2)^(1/2)/c^4/x^(1/2)+1/77*b^(11/4)* (-15*A*c+13*B*b)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2* arctan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/b^ (1/4))),1/2*2^(1/2))*(b^(1/2)+x*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2) ^(1/2)/c^(17/4)/(c*x^4+b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.18 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.59 \[ \int \frac {x^{13/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 x^{3/2} \left (-\left (\left (b+c x^2\right ) \left (195 b^3 B-7 c^3 x^4 \left (15 A+11 B x^2\right )-9 b^2 c \left (25 A+13 B x^2\right )+b c^2 x^2 \left (135 A+91 B x^2\right )\right )\right )+15 b^3 (13 b B-15 A c) \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{1155 c^4 \sqrt {x^2 \left (b+c x^2\right )}} \]
(2*x^(3/2)*(-((b + c*x^2)*(195*b^3*B - 7*c^3*x^4*(15*A + 11*B*x^2) - 9*b^2 *c*(25*A + 13*B*x^2) + b*c^2*x^2*(135*A + 91*B*x^2))) + 15*b^3*(13*b*B - 1 5*A*c)*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]) )/(1155*c^4*Sqrt[x^2*(b + c*x^2)])
Time = 0.40 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1945, 1429, 1429, 1429, 1431, 266, 761}
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.
| \(\displaystyle \int \frac {x^{13/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx\) |
| \(\Big \downarrow \) 1945 |
| \(\displaystyle \frac {2 B x^{11/2} \sqrt {b x^2+c x^4}}{15 c}-\frac {(13 b B-15 A c) \int \frac {x^{13/2}}{\sqrt {c x^4+b x^2}}dx}{15 c}\) |
| \(\Big \downarrow \) 1429 |
| \(\displaystyle \frac {2 B x^{11/2} \sqrt {b x^2+c x^4}}{15 c}-\frac {(13 b B-15 A c) \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \int \frac {x^{9/2}}{\sqrt {c x^4+b x^2}}dx}{11 c}\right )}{15 c}\) |
| \(\Big \downarrow \) 1429 |
| \(\displaystyle \frac {2 B x^{11/2} \sqrt {b x^2+c x^4}}{15 c}-\frac {(13 b B-15 A c) \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \int \frac {x^{5/2}}{\sqrt {c x^4+b x^2}}dx}{7 c}\right )}{11 c}\right )}{15 c}\) |
| \(\Big \downarrow \) 1429 |
| \(\displaystyle \frac {2 B x^{11/2} \sqrt {b x^2+c x^4}}{15 c}-\frac {(13 b B-15 A c) \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )}{7 c}\right )}{11 c}\right )}{15 c}\) |
| \(\Big \downarrow \) 1431 |
| \(\displaystyle \frac {2 B x^{11/2} \sqrt {b x^2+c x^4}}{15 c}-\frac {(13 b B-15 A c) \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{11 c}\right )}{15 c}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2 B x^{11/2} \sqrt {b x^2+c x^4}}{15 c}-\frac {(13 b B-15 A c) \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{11 c}\right )}{15 c}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {2 B x^{11/2} \sqrt {b x^2+c x^4}}{15 c}-\frac {(13 b B-15 A c) \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 c^{5/4} \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{11 c}\right )}{15 c}\) |
(2*B*x^(11/2)*Sqrt[b*x^2 + c*x^4])/(15*c) - ((13*b*B - 15*A*c)*((2*x^(7/2) *Sqrt[b*x^2 + c*x^4])/(11*c) - (9*b*((2*x^(3/2)*Sqrt[b*x^2 + c*x^4])/(7*c) - (5*b*((2*Sqrt[b*x^2 + c*x^4])/(3*c*Sqrt[x]) - (b^(3/4)*x*(Sqrt[b] + Sqr t[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1 /4)*Sqrt[x])/b^(1/4)], 1/2])/(3*c^(5/4)*Sqrt[b*x^2 + c*x^4])))/(7*c)))/(11 *c)))/(15*c)
3.3.44.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b *d^2*((m + 2*p - 1)/(c*(m + 4*p + 1))) Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^ p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[m + 2*p - 1, 0] && NeQ[m + 4*p + 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p) Int[(d*x)^(m + 2*p)*(b + c *x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)) Int[(e* x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p *(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
Time = 1.97 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.98
| method | result | size |
| risch | \(\frac {2 \left (77 B \,c^{3} x^{6}+105 A \,c^{3} x^{4}-91 B b \,c^{2} x^{4}-135 A b \,c^{2} x^{2}+117 B \,b^{2} c \,x^{2}+225 b^{2} A c -195 B \,b^{3}\right ) x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}{1155 c^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {b^{3} \left (15 A c -13 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{77 c^{5} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(239\) |
| default | \(-\frac {\sqrt {x}\, \left (-154 B \,c^{5} x^{9}-210 A \,c^{5} x^{7}+28 B b \,c^{4} x^{7}+225 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{3} c +60 A b \,c^{4} x^{5}-195 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{4}-52 B \,b^{2} c^{3} x^{5}-180 A \,b^{2} c^{3} x^{3}+156 B \,b^{3} c^{2} x^{3}-450 A \,b^{3} c^{2} x +390 B \,b^{4} c x \right )}{1155 \sqrt {x^{4} c +b \,x^{2}}\, c^{5}}\) | \(298\) |
2/1155*(77*B*c^3*x^6+105*A*c^3*x^4-91*B*b*c^2*x^4-135*A*b*c^2*x^2+117*B*b^ 2*c*x^2+225*A*b^2*c-195*B*b^3)/c^4*x^(3/2)*(c*x^2+b)/(x^2*(c*x^2+b))^(1/2) -1/77*b^3*(15*A*c-13*B*b)/c^5*(-b*c)^(1/2)*((x+1/c*(-b*c)^(1/2))*c/(-b*c)^ (1/2))^(1/2)*(-2*(x-1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^( 1/2))^(1/2)/(c*x^3+b*x)^(1/2)*EllipticF(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/ 2))^(1/2),1/2*2^(1/2))*x^(1/2)/(x^2*(c*x^2+b))^(1/2)*(x*(c*x^2+b))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.50 \[ \int \frac {x^{13/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 \, {\left (15 \, {\left (13 \, B b^{4} - 15 \, A b^{3} c\right )} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (77 \, B c^{4} x^{6} - 195 \, B b^{3} c + 225 \, A b^{2} c^{2} - 7 \, {\left (13 \, B b c^{3} - 15 \, A c^{4}\right )} x^{4} + 9 \, {\left (13 \, B b^{2} c^{2} - 15 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{1155 \, c^{5} x} \]
2/1155*(15*(13*B*b^4 - 15*A*b^3*c)*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0 , x) + (77*B*c^4*x^6 - 195*B*b^3*c + 225*A*b^2*c^2 - 7*(13*B*b*c^3 - 15*A* c^4)*x^4 + 9*(13*B*b^2*c^2 - 15*A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2)*sqrt(x)) /(c^5*x)
Timed out. \[ \int \frac {x^{13/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\text {Timed out} \]
\[ \int \frac {x^{13/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{\frac {13}{2}}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \]
\[ \int \frac {x^{13/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{\frac {13}{2}}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \]
Timed out. \[ \int \frac {x^{13/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{13/2}\,\left (B\,x^2+A\right )}{\sqrt {c\,x^4+b\,x^2}} \,d x \]