Integrand size = 28, antiderivative size = 239 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx=-\frac {8 b^2 (b B-3 A c) \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}-\frac {4 b (b B-3 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c}-\frac {2 (b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{33 c \sqrt {x}}+\frac {2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}+\frac {4 b^{11/4} (b B-3 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{231 c^{9/4} \sqrt {b x^2+c x^4}} \]
2/15*B*(c*x^4+b*x^2)^(5/2)/c/x^(5/2)-2/33*(-3*A*c+B*b)*(c*x^4+b*x^2)^(3/2) /c/x^(1/2)-4/77*b*(-3*A*c+B*b)*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c-8/231*b^2*(-3 *A*c+B*b)*(c*x^4+b*x^2)^(1/2)/c^2/x^(1/2)+4/231*b^(11/4)*(-3*A*c+B*b)*x*(c os(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^(9/4)/(c*x^4 +b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.16 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.48 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (-\left (b+c x^2\right )^2 \sqrt {1+\frac {c x^2}{b}} \left (5 b B-15 A c-11 B c x^2\right )+5 b^2 (b B-3 A c) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{165 c^2 \sqrt {x} \sqrt {1+\frac {c x^2}{b}}} \]
(2*Sqrt[x^2*(b + c*x^2)]*(-((b + c*x^2)^2*Sqrt[1 + (c*x^2)/b]*(5*b*B - 15* A*c - 11*B*c*x^2)) + 5*b^2*(b*B - 3*A*c)*Hypergeometric2F1[-3/2, 1/4, 5/4, -((c*x^2)/b)]))/(165*c^2*Sqrt[x]*Sqrt[1 + (c*x^2)/b])
Time = 0.38 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1945, 1426, 1426, 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.
| \(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1945 |
| \(\displaystyle \frac {2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}-\frac {(b B-3 A c) \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{3/2}}dx}{3 c}\) |
| \(\Big \downarrow \) 1426 |
| \(\displaystyle \frac {2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}-\frac {(b B-3 A c) \left (\frac {6}{11} b \int \sqrt {x} \sqrt {c x^4+b x^2}dx+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}\right )}{3 c}\) |
| \(\Big \downarrow \) 1426 |
| \(\displaystyle \frac {2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}-\frac {(b B-3 A c) \left (\frac {6}{11} b \left (\frac {2}{7} b \int \frac {x^{5/2}}{\sqrt {c x^4+b x^2}}dx+\frac {2}{7} x^{3/2} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}\right )}{3 c}\) |
| \(\Big \downarrow \) 1429 |
| \(\displaystyle \frac {2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}-\frac {(b B-3 A c) \left (\frac {6}{11} b \left (\frac {2}{7} b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )+\frac {2}{7} x^{3/2} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}\right )}{3 c}\) |
| \(\Big \downarrow \) 1431 |
| \(\displaystyle \frac {2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}-\frac {(b B-3 A c) \left (\frac {6}{11} b \left (\frac {2}{7} b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{7} x^{3/2} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}\right )}{3 c}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}-\frac {(b B-3 A c) \left (\frac {6}{11} b \left (\frac {2}{7} b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{7} x^{3/2} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}\right )}{3 c}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {2 B \left (b x^2+c x^4\right )^{5/2}}{15 c x^{5/2}}-\frac {(b B-3 A c) \left (\frac {6}{11} b \left (\frac {2}{7} b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 c^{5/4} \sqrt {b x^2+c x^4}}\right )+\frac {2}{7} x^{3/2} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt {x}}\right )}{3 c}\) |
(2*B*(b*x^2 + c*x^4)^(5/2))/(15*c*x^(5/2)) - ((b*B - 3*A*c)*((2*(b*x^2 + c *x^4)^(3/2))/(11*Sqrt[x]) + (6*b*((2*x^(3/2)*Sqrt[b*x^2 + c*x^4])/7 + (2*b *((2*Sqrt[b*x^2 + c*x^4])/(3*c*Sqrt[x]) - (b^(3/4)*x*(Sqrt[b] + Sqrt[c]*x) *Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqr t[x])/b^(1/4)], 1/2])/(3*c^(5/4)*Sqrt[b*x^2 + c*x^4])))/7))/11))/(3*c)
3.3.36.3.1 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*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 4*p + 1))), x] + Simp[2*b*(p/(d^2 *(m + 4*p + 1))) Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 0] && NeQ[m + 4*p + 1, 0]
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 = 1.82 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.01
| method | result | size |
| risch | \(\frac {2 \left (77 B \,c^{3} x^{6}+105 A \,c^{3} x^{4}+119 B b \,c^{2} x^{4}+195 A b \,c^{2} x^{2}+12 B \,b^{2} c \,x^{2}+60 b^{2} A c -20 B \,b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{1155 c^{2} \sqrt {x}}-\frac {4 b^{3} \left (3 A c -B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{231 c^{3} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\) | \(241\) |
| default | \(-\frac {2 \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}} \left (-77 B \,c^{5} x^{9}-105 A \,c^{5} x^{7}-196 B b \,c^{4} x^{7}+30 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{3} c -300 A b \,c^{4} x^{5}-10 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{4}-131 B \,b^{2} c^{3} x^{5}-255 A \,b^{2} c^{3} x^{3}+8 B \,b^{3} c^{2} x^{3}-60 A \,b^{3} c^{2} x +20 B \,b^{4} c x \right )}{1155 x^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{2} c^{3}}\) | \(307\) |
2/1155/c^2*(77*B*c^3*x^6+105*A*c^3*x^4+119*B*b*c^2*x^4+195*A*b*c^2*x^2+12* B*b^2*c*x^2+60*A*b^2*c-20*B*b^3)/x^(1/2)*(x^2*(c*x^2+b))^(1/2)-4/231*b^3/c ^3*(3*A*c-B*b)*(-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^2*(c*x^2+b))^(1/2)/x^(3/2)/(c*x^2+b)*(x*(c*x^2+b))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.51 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx=\frac {2 \, {\left (20 \, {\left (B b^{4} - 3 \, A b^{3} c\right )} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (77 \, B c^{4} x^{6} - 20 \, B b^{3} c + 60 \, A b^{2} c^{2} + 7 \, {\left (17 \, B b c^{3} + 15 \, A c^{4}\right )} x^{4} + 3 \, {\left (4 \, B b^{2} c^{2} + 65 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{1155 \, c^{3} x} \]
2/1155*(20*(B*b^4 - 3*A*b^3*c)*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, x) + (77*B*c^4*x^6 - 20*B*b^3*c + 60*A*b^2*c^2 + 7*(17*B*b*c^3 + 15*A*c^4)*x ^4 + 3*(4*B*b^2*c^2 + 65*A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(c^3*x )
\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{3/2}} \,d x \]