Integrand size = 21, antiderivative size = 287 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13/2}} \, dx=\frac {24 c^{3/2} x^{3/2} \left (b+c x^2\right )}{5 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {12 c \sqrt {b x^2+c x^4}}{5 x^{3/2}}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{5 x^{11/2}}-\frac {24 \sqrt [4]{b} c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {12 \sqrt [4]{b} c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{5 \sqrt {b x^2+c x^4}} \] Output:
24/5*c^(3/2)*x^(3/2)*(c*x^2+b)/(b^(1/2)+c^(1/2)*x)/(c*x^4+b*x^2)^(1/2)-12/ 5*c*(c*x^4+b*x^2)^(1/2)/x^(3/2)-2/5*(c*x^4+b*x^2)^(3/2)/x^(11/2)-24/5*b^(1 /4)*c^(5/4)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)* EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))/(c*x^4+b*x^2 )^(1/2)+12/5*b^(1/4)*c^(5/4)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^( 1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)),1/2*2^( 1/2))/(c*x^4+b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.20 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13/2}} \, dx=-\frac {2 b \sqrt {x^2 \left (b+c x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {5}{4},-\frac {1}{4},-\frac {c x^2}{b}\right )}{5 x^{7/2} \sqrt {1+\frac {c x^2}{b}}} \] Input:
Integrate[(b*x^2 + c*x^4)^(3/2)/x^(13/2),x]
Output:
(-2*b*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-3/2, -5/4, -1/4, -((c*x^2)/ b)])/(5*x^(7/2)*Sqrt[1 + (c*x^2)/b])
Time = 0.68 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {1425, 1425, 1431, 266, 834, 27, 761, 1510}
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.
\(\displaystyle \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13/2}} \, dx\) |
\(\Big \downarrow \) 1425 |
\(\displaystyle \frac {6}{5} c \int \frac {\sqrt {c x^4+b x^2}}{x^{5/2}}dx-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{5 x^{11/2}}\) |
\(\Big \downarrow \) 1425 |
\(\displaystyle \frac {6}{5} c \left (2 c \int \frac {x^{3/2}}{\sqrt {c x^4+b x^2}}dx-\frac {2 \sqrt {b x^2+c x^4}}{x^{3/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{5 x^{11/2}}\) |
\(\Big \downarrow \) 1431 |
\(\displaystyle \frac {6}{5} c \left (\frac {2 c x \sqrt {b+c x^2} \int \frac {\sqrt {x}}{\sqrt {c x^2+b}}dx}{\sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{x^{3/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{5 x^{11/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {6}{5} c \left (\frac {4 c x \sqrt {b+c x^2} \int \frac {x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{x^{3/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{5 x^{11/2}}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {6}{5} c \left (\frac {4 c x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\sqrt {b} \int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {b} \sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{\sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{x^{3/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{5 x^{11/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {6}{5} c \left (\frac {4 c x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{\sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{x^{3/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{5 x^{11/2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {6}{5} c \left (\frac {4 c x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{\sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{x^{3/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{5 x^{11/2}}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {6}{5} c \left (\frac {4 c x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {b+c x^2}}-\frac {\sqrt {x} \sqrt {b+c x^2}}{\sqrt {b}+\sqrt {c} x}}{\sqrt {c}}\right )}{\sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{x^{3/2}}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{5 x^{11/2}}\) |
Input:
Int[(b*x^2 + c*x^4)^(3/2)/x^(13/2),x]
Output:
(-2*(b*x^2 + c*x^4)^(3/2))/(5*x^(11/2)) + (6*c*((-2*Sqrt[b*x^2 + c*x^4])/x ^(3/2) + (4*c*x*Sqrt[b + c*x^2]*(-((-((Sqrt[x]*Sqrt[b + c*x^2])/(Sqrt[b] + Sqrt[c]*x)) + (b^(1/4)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(c^(1/4 )*Sqrt[b + c*x^2]))/Sqrt[c]) + (b^(1/4)*(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])/(2*c^(3/4)*Sqrt[b + c*x^2])))/Sqrt[b*x^2 + c*x^4]))/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], 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[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Time = 1.04 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.77
method | result | size |
risch | \(-\frac {2 \left (7 c \,x^{2}+b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{5 x^{\frac {7}{2}}}+\frac {12 c \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{5 \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\) | \(220\) |
default | \(\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (12 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b c \,x^{2}-6 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b c \,x^{2}-7 c^{2} x^{4}-8 b c \,x^{2}-b^{2}\right )}{5 x^{\frac {11}{2}} \left (c \,x^{2}+b \right )^{2}}\) | \(221\) |
Input:
int((c*x^4+b*x^2)^(3/2)/x^(13/2),x,method=_RETURNVERBOSE)
Output:
-2/5*(7*c*x^2+b)/x^(7/2)*(x^2*(c*x^2+b))^(1/2)+12/5*c*(-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)*(-c/(-b*c)^(1/2)*x)^(1/2)/(c*x^3+b*x)^(1/2)*(-2/c*(-b*c)^(1/2)*E llipticE(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))+1/c*(-b* c)^(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)
Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.20 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13/2}} \, dx=-\frac {2 \, {\left (12 \, c^{\frac {3}{2}} x^{4} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) + \sqrt {c x^{4} + b x^{2}} {\left (7 \, c x^{2} + b\right )} \sqrt {x}\right )}}{5 \, x^{4}} \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^(13/2),x, algorithm="fricas")
Output:
-2/5*(12*c^(3/2)*x^4*weierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4*b/c , 0, x)) + sqrt(c*x^4 + b*x^2)*(7*c*x^2 + b)*sqrt(x))/x^4
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13/2}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{\frac {13}{2}}}\, dx \] Input:
integrate((c*x**4+b*x**2)**(3/2)/x**(13/2),x)
Output:
Integral((x**2*(b + c*x**2))**(3/2)/x**(13/2), x)
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {13}{2}}} \,d x } \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^(13/2),x, algorithm="maxima")
Output:
integrate((c*x^4 + b*x^2)^(3/2)/x^(13/2), x)
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {13}{2}}} \,d x } \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^(13/2),x, algorithm="giac")
Output:
integrate((c*x^4 + b*x^2)^(3/2)/x^(13/2), x)
Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13/2}} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{13/2}} \,d x \] Input:
int((b*x^2 + c*x^4)^(3/2)/x^(13/2),x)
Output:
int((b*x^2 + c*x^4)^(3/2)/x^(13/2), x)
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13/2}} \, dx=\frac {-2 \sqrt {c \,x^{2}+b}\, b +2 \sqrt {c \,x^{2}+b}\, c \,x^{2}-4 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{6}+b \,x^{4}}d x \right ) b^{2} x^{2}}{\sqrt {x}\, x^{2}} \] Input:
int((c*x^4+b*x^2)^(3/2)/x^(13/2),x)
Output:
(2*( - sqrt(b + c*x**2)*b + sqrt(b + c*x**2)*c*x**2 - 2*sqrt(x)*int((sqrt( x)*sqrt(b + c*x**2))/(b*x**4 + c*x**6),x)*b**2*x**2))/(sqrt(x)*x**2)