Integrand size = 28, antiderivative size = 167 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=-\frac {2 (7 b B-A c) \sqrt {b x^2+c x^4}}{21 b x^{5/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}}+\frac {2 c^{3/4} (7 b B-A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{21 b^{5/4} \sqrt {b x^2+c x^4}} \]
-2/7*A*(c*x^4+b*x^2)^(3/2)/b/x^(13/2)-2/21*(-A*c+7*B*b)*(c*x^4+b*x^2)^(1/2 )/b/x^(5/2)+2/21*c^(3/4)*(-A*c+7*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*arct an(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)/b^(5/4)/(c*x^4+b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.59 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=-\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (3 A \left (b+c x^2\right ) \sqrt {1+\frac {c x^2}{b}}+(7 b B-A c) x^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{2},\frac {1}{4},-\frac {c x^2}{b}\right )\right )}{21 b x^{9/2} \sqrt {1+\frac {c x^2}{b}}} \]
(-2*Sqrt[x^2*(b + c*x^2)]*(3*A*(b + c*x^2)*Sqrt[1 + (c*x^2)/b] + (7*b*B - A*c)*x^2*Hypergeometric2F1[-3/4, -1/2, 1/4, -((c*x^2)/b)]))/(21*b*x^(9/2)* Sqrt[1 + (c*x^2)/b])
Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1944, 1425, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx\) |
\(\Big \downarrow \) 1944 |
\(\displaystyle \frac {(7 b B-A c) \int \frac {\sqrt {c x^4+b x^2}}{x^{7/2}}dx}{7 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}}\) |
\(\Big \downarrow \) 1425 |
\(\displaystyle \frac {(7 b B-A c) \left (\frac {2}{3} c \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx-\frac {2 \sqrt {b x^2+c x^4}}{3 x^{5/2}}\right )}{7 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}}\) |
\(\Big \downarrow \) 1431 |
\(\displaystyle \frac {(7 b B-A c) \left (\frac {2 c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 x^{5/2}}\right )}{7 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {(7 b B-A c) \left (\frac {4 c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 x^{5/2}}\right )}{7 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {(7 b B-A c) \left (\frac {2 c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 x^{5/2}}\right )}{7 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}}\) |
(-2*A*(b*x^2 + c*x^4)^(3/2))/(7*b*x^(13/2)) + ((7*b*B - A*c)*((-2*Sqrt[b*x ^2 + c*x^4])/(3*x^(5/2)) + (2*c^(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*b^(1/4)*Sqrt[b*x^2 + c*x^4])))/(7*b)
3.3.28.3.1 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 + 2*p + 1))), x] - Simp[2*c*(p/(d^4 *(m + 2*p + 1))) Int[(d*x)^(m + 4)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 0] && LtQ[m + 2*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[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b *c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)) Int[(e*x)^(m + n)*(a*x^ j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 ] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 , 0]
Time = 1.89 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.16
method | result | size |
risch | \(-\frac {2 \left (2 A c \,x^{2}+7 b B \,x^{2}+3 A b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{21 x^{\frac {9}{2}} b}-\frac {2 \left (A c -7 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{21 b \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\) | \(194\) |
default | \(-\frac {2 \sqrt {x^{4} c +b \,x^{2}}\, \left (A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c \,x^{3}-7 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,x^{3}+2 A \,c^{2} x^{4}+7 x^{4} B b c +5 A b c \,x^{2}+7 b^{2} B \,x^{2}+3 b^{2} A \right )}{21 x^{\frac {9}{2}} \left (c \,x^{2}+b \right ) b}\) | \(255\) |
-2/21*(2*A*c*x^2+7*B*b*x^2+3*A*b)/x^(9/2)/b*(x^2*(c*x^2+b))^(1/2)-2/21*(A* c-7*B*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 = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.43 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=\frac {2 \, {\left (2 \, {\left (7 \, B b - A c\right )} \sqrt {c} x^{5} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - \sqrt {c x^{4} + b x^{2}} {\left ({\left (7 \, B b + 2 \, A c\right )} x^{2} + 3 \, A b\right )} \sqrt {x}\right )}}{21 \, b x^{5}} \]
2/21*(2*(7*B*b - A*c)*sqrt(c)*x^5*weierstrassPInverse(-4*b/c, 0, x) - sqrt (c*x^4 + b*x^2)*((7*B*b + 2*A*c)*x^2 + 3*A*b)*sqrt(x))/(b*x^5)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{\frac {11}{2}}}\, dx \]
\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )}}{x^{\frac {11}{2}}} \,d x } \]
\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )}}{x^{\frac {11}{2}}} \,d x } \]
Timed out. \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=\int \frac {\left (B\,x^2+A\right )\,\sqrt {c\,x^4+b\,x^2}}{x^{11/2}} \,d x \]