Integrand size = 21, antiderivative size = 174 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}+\frac {15 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{14 c^{13/4} \sqrt {b x^2+c x^4}} \]
-x^(11/2)/c/(c*x^4+b*x^2)^(1/2)+9/7*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c^2-15/7*b *(c*x^4+b*x^2)^(1/2)/c^3/x^(1/2)+15/14*b^(7/4)*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(s in(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^(13/4)/(c*x^4+b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.49 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^{3/2} \left (-15 b^2-6 b c x^2+2 c^2 x^4+15 b^2 \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{7 c^3 \sqrt {x^2 \left (b+c x^2\right )}} \]
(x^(3/2)*(-15*b^2 - 6*b*c*x^2 + 2*c^2*x^4 + 15*b^2*Sqrt[1 + (c*x^2)/b]*Hyp ergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(7*c^3*Sqrt[x^2*(b + c*x^2)] )
Time = 0.34 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1427, 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.
\(\displaystyle \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1427 |
\(\displaystyle \frac {9 \int \frac {x^{9/2}}{\sqrt {c x^4+b x^2}}dx}{2 c}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1429 |
\(\displaystyle \frac {9 \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \int \frac {x^{5/2}}{\sqrt {c x^4+b x^2}}dx}{7 c}\right )}{2 c}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1429 |
\(\displaystyle \frac {9 \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )}{7 c}\right )}{2 c}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1431 |
\(\displaystyle \frac {9 \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{2 c}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {9 \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{2 c}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {9 \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 c^{5/4} \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{2 c}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}\) |
-(x^(11/2)/(c*Sqrt[b*x^2 + c*x^4])) + (9*((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] + 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)) )/(2*c)
3.4.91.3.1 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)/(2*c*(p + 1))), x] - Simp[d^4*( (m + 2*p - 1)/(2*c*(p + 1))) Int[(d*x)^(m - 4)*(b*x^2 + c*x^4)^(p + 1), x ], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && LtQ[p, -1] && GtQ[m + 2*p + 1, 2]
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]
Time = 0.96 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.83
method | result | size |
default | \(\frac {x^{\frac {5}{2}} \left (c \,x^{2}+b \right ) \left (15 b^{2} \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+4 c^{3} x^{5}-12 b \,c^{2} x^{3}-30 b^{2} c x \right )}{14 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{4}}\) | \(144\) |
risch | \(-\frac {2 \left (-c \,x^{2}+4 b \right ) \left (c \,x^{2}+b \right ) x^{\frac {3}{2}}}{7 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {b^{2} \left (\frac {11 \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c \sqrt {c \,x^{3}+b x}}-7 b \left (\frac {x}{b \sqrt {\left (x^{2}+\frac {b}{c}\right ) c x}}+\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{2 b c \sqrt {c \,x^{3}+b x}}\right )\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{7 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(318\) |
1/14/(c*x^4+b*x^2)^(3/2)*x^(5/2)*(c*x^2+b)*(15*b^2*(-b*c)^(1/2)*((c*x+(-b* c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^( 1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2)) ^(1/2),1/2*2^(1/2))+4*c^3*x^5-12*b*c^2*x^3-30*b^2*c*x)/c^4
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.51 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {15 \, {\left (b^{2} c x^{3} + b^{3} x\right )} \sqrt {c} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (2 \, c^{3} x^{4} - 6 \, b c^{2} x^{2} - 15 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{7 \, {\left (c^{5} x^{3} + b c^{4} x\right )}} \]
1/7*(15*(b^2*c*x^3 + b^3*x)*sqrt(c)*weierstrassPInverse(-4*b/c, 0, x) + (2 *c^3*x^4 - 6*b*c^2*x^2 - 15*b^2*c)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(c^5*x^3 + b*c^4*x)
Timed out. \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{\frac {17}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{\frac {17}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{17/2}}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \]