Integrand size = 28, antiderivative size = 167 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=-\frac {2 (5 b B-7 A c) \sqrt {b x^2+c x^4}}{21 c^2 \sqrt {x}}+\frac {2 B x^{3/2} \sqrt {b x^2+c x^4}}{7 c}+\frac {b^{3/4} (5 b B-7 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{21 c^{9/4} \sqrt {b x^2+c x^4}} \]
2/7*B*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c-2/21*(-7*A*c+5*B*b)*(c*x^4+b*x^2)^(1/2 )/c^2/x^(1/2)+1/21*b^(3/4)*(-7*A*c+5*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^(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 = 10.13 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.58 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 x^{3/2} \left (-\left (\left (b+c x^2\right ) \left (5 b B-7 A c-3 B c x^2\right )\right )+b (5 b B-7 A c) \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{21 c^2 \sqrt {x^2 \left (b+c x^2\right )}} \]
(2*x^(3/2)*(-((b + c*x^2)*(5*b*B - 7*A*c - 3*B*c*x^2)) + b*(5*b*B - 7*A*c) *Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(21* c^2*Sqrt[x^2*(b + c*x^2)])
Time = 0.32 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1945, 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.
\(\displaystyle \int \frac {x^{5/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle \frac {2 B x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {(5 b B-7 A c) \int \frac {x^{5/2}}{\sqrt {c x^4+b x^2}}dx}{7 c}\) |
\(\Big \downarrow \) 1429 |
\(\displaystyle \frac {2 B x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {(5 b B-7 A c) \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )}{7 c}\) |
\(\Big \downarrow \) 1431 |
\(\displaystyle \frac {2 B x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {(5 b B-7 A c) \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 B x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {(5 b B-7 A c) \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2 B x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {(5 b B-7 A c) \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 c^{5/4} \sqrt {b x^2+c x^4}}\right )}{7 c}\) |
(2*B*x^(3/2)*Sqrt[b*x^2 + c*x^4])/(7*c) - ((5*b*B - 7*A*c)*((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)*Sqrt[x])/b^(1/4)], 1/2])/(3*c^(5/4)*Sqrt[b*x^2 + c*x^4])))/(7*c)
3.3.48.3.1 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 = 2.31 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.14
method | result | size |
risch | \(\frac {2 \left (3 B c \,x^{2}+7 A c -5 B b \right ) x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}{21 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {b \left (7 A c -5 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{21 c^{3} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(191\) |
default | \(-\frac {\sqrt {x}\, \left (7 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b c -5 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2}-6 B \,c^{3} x^{5}-14 A \,c^{3} x^{3}+4 B b \,c^{2} x^{3}-14 A b \,c^{2} x +10 B \,b^{2} c x \right )}{21 \sqrt {x^{4} c +b \,x^{2}}\, c^{3}}\) | \(248\) |
2/21*(3*B*c*x^2+7*A*c-5*B*b)/c^2*x^(3/2)*(c*x^2+b)/(x^2*(c*x^2+b))^(1/2)-1 /21*b*(7*A*c-5*B*b)/c^3*(-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 = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.44 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 \, {\left ({\left (5 \, B b^{2} - 7 \, A b c\right )} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (3 \, B c^{2} x^{2} - 5 \, B b c + 7 \, A c^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{21 \, c^{3} x} \]
2/21*((5*B*b^2 - 7*A*b*c)*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, x) + (3 *B*c^2*x^2 - 5*B*b*c + 7*A*c^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(c^3*x)
\[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{\frac {5}{2}} \left (A + B x^{2}\right )}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]
\[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{\frac {5}{2}}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \]
\[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{\frac {5}{2}}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \]
Timed out. \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{5/2}\,\left (B\,x^2+A\right )}{\sqrt {c\,x^4+b\,x^2}} \,d x \]