Integrand size = 22, antiderivative size = 313 \[ \int \frac {(d x)^m}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {d (d x)^{-1+m} \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {c d \left (b^2 (3-m)+b \sqrt {b^2-4 a c} (3-m)-4 a c (5-m)\right ) (d x)^{-1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1+m),\frac {1+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) (1-m)}+\frac {c d \left (b^2 (3-m)-b \sqrt {b^2-4 a c} (3-m)-4 a c (5-m)\right ) (d x)^{-1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1+m),\frac {1+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) (1-m)} \] Output:
1/2*d*(d*x)^(-1+m)*(b*c*x^2-2*a*c+b^2)/a/(-4*a*c+b^2)/(c*x^4+b*x^2+a)-1/2* c*d*(b^2*(3-m)+b*(-4*a*c+b^2)^(1/2)*(3-m)-4*a*c*(5-m))*(d*x)^(-1+m)*hyperg eom([1, -1/2+1/2*m],[1/2+1/2*m],-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))/a/(-4*a*c +b^2)^(3/2)/(b-(-4*a*c+b^2)^(1/2))/(1-m)+1/2*c*d*(b^2*(3-m)-b*(-4*a*c+b^2) ^(1/2)*(3-m)-4*a*c*(5-m))*(d*x)^(-1+m)*hypergeom([1, -1/2+1/2*m],[1/2+1/2* m],-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/(b+(-4*a*c+b^2)^( 1/2))/(1-m)
Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
Time = 1.10 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.66 \[ \int \frac {(d x)^m}{\left (a x+b x^3+c x^5\right )^2} \, dx=-\frac {(d x)^m \left (2 b m \left (-6+m+4 m^2+m^3\right ) x^2 \operatorname {AppellF1}\left (\frac {1+m}{2},2,2,\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+2 c m \left (-2-m+2 m^2+m^3\right ) x^4 \operatorname {AppellF1}\left (\frac {3+m}{2},2,2,\frac {5+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+a (3+m) \left (-2 m \left (2+3 m+m^2\right )+b \left (-2-m+2 m^2+m^3\right ) x \text {RootSum}\left [a+b \text {$\#$1}^2+c \text {$\#$1}^4\&,\frac {\operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m}}{b \text {$\#$1}+2 c \text {$\#$1}^3}\&\right ]+c (-1+m) x \text {RootSum}\left [a+b \text {$\#$1}^2+c \text {$\#$1}^4\&,\frac {m x^2+m^2 x^2+2 m x \text {$\#$1}+m^2 x \text {$\#$1}+2 \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+3 m \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+m^2 \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+m \left (\frac {x}{\text {$\#$1}}\right )^{-m} \text {$\#$1}^2}{b \text {$\#$1}+2 c \text {$\#$1}^3}\&\right ]\right )\right )}{2 a^3 (-1+m) m (1+m) (2+m) (3+m) x} \] Input:
Integrate[(d*x)^m/(a*x + b*x^3 + c*x^5)^2,x]
Output:
-1/2*((d*x)^m*(2*b*m*(-6 + m + 4*m^2 + m^3)*x^2*AppellF1[(1 + m)/2, 2, 2, (3 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + 2*c*m*(-2 - m + 2*m^2 + m^3)*x^4*AppellF1[(3 + m)/2, 2, 2, (5 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a* c])] + a*(3 + m)*(-2*m*(2 + 3*m + m^2) + b*(-2 - m + 2*m^2 + m^3)*x*RootSu m[a + b*#1^2 + c*#1^4 & , Hypergeometric2F1[-m, -m, 1 - m, -(#1/(x - #1))] /((x/(x - #1))^m*(b*#1 + 2*c*#1^3)) & ] + c*(-1 + m)*x*RootSum[a + b*#1^2 + c*#1^4 & , (m*x^2 + m^2*x^2 + 2*m*x*#1 + m^2*x*#1 + (2*Hypergeometric2F1 [-m, -m, 1 - m, -(#1/(x - #1))]*#1^2)/(x/(x - #1))^m + (3*m*Hypergeometric 2F1[-m, -m, 1 - m, -(#1/(x - #1))]*#1^2)/(x/(x - #1))^m + (m^2*Hypergeomet ric2F1[-m, -m, 1 - m, -(#1/(x - #1))]*#1^2)/(x/(x - #1))^m + (m*#1^2)/(x/# 1)^m)/(b*#1 + 2*c*#1^3) & ])))/(a^3*(-1 + m)*m*(1 + m)*(2 + m)*(3 + m)*x)
Time = 0.65 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {9, 1441, 25, 1608, 27, 278}
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.
| \(\displaystyle \int \frac {(d x)^m}{\left (a x+b x^3+c x^5\right )^2} \, dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle d^2 \int \frac {(d x)^{m-2}}{\left (c x^4+b x^2+a\right )^2}dx\) |
| \(\Big \downarrow \) 1441 |
| \(\displaystyle d^2 \left (\frac {(d x)^{m-1} \left (-2 a c+b^2+b c x^2\right )}{2 a d \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(d x)^{m-2} \left ((3-m) b^2+c (3-m) x^2 b-2 a c (5-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle d^2 \left (\frac {\int \frac {(d x)^{m-2} \left ((3-m) b^2+c (3-m) x^2 b-2 a c (5-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(d x)^{m-1} \left (-2 a c+b^2+b c x^2\right )}{2 a d \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 1608 |
| \(\displaystyle d^2 \left (\frac {\frac {c \left (b (3-m) \sqrt {b^2-4 a c}-4 a c (5-m)+b^2 (3-m)\right ) \int \frac {2 (d x)^{m-2}}{2 c x^2+b-\sqrt {b^2-4 a c}}dx}{2 \sqrt {b^2-4 a c}}-\frac {c \left (-b (3-m) \sqrt {b^2-4 a c}-4 a c (5-m)+b^2 (3-m)\right ) \int \frac {2 (d x)^{m-2}}{2 c x^2+b+\sqrt {b^2-4 a c}}dx}{2 \sqrt {b^2-4 a c}}}{2 a \left (b^2-4 a c\right )}+\frac {(d x)^{m-1} \left (-2 a c+b^2+b c x^2\right )}{2 a d \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle d^2 \left (\frac {\frac {c \left (b (3-m) \sqrt {b^2-4 a c}-4 a c (5-m)+b^2 (3-m)\right ) \int \frac {(d x)^{m-2}}{2 c x^2+b-\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}-\frac {c \left (-b (3-m) \sqrt {b^2-4 a c}-4 a c (5-m)+b^2 (3-m)\right ) \int \frac {(d x)^{m-2}}{2 c x^2+b+\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}}{2 a \left (b^2-4 a c\right )}+\frac {(d x)^{m-1} \left (-2 a c+b^2+b c x^2\right )}{2 a d \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle d^2 \left (\frac {\frac {c (d x)^{m-1} \left (-b (3-m) \sqrt {b^2-4 a c}-4 a c (5-m)+b^2 (3-m)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m-1}{2},\frac {m+1}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (1-m) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {c (d x)^{m-1} \left (b (3-m) \sqrt {b^2-4 a c}-4 a c (5-m)+b^2 (3-m)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m-1}{2},\frac {m+1}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{d (1-m) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}}{2 a \left (b^2-4 a c\right )}+\frac {(d x)^{m-1} \left (-2 a c+b^2+b c x^2\right )}{2 a d \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
Input:
Int[(d*x)^m/(a*x + b*x^3 + c*x^5)^2,x]
Output:
d^2*(((d*x)^(-1 + m)*(b^2 - 2*a*c + b*c*x^2))/(2*a*(b^2 - 4*a*c)*d*(a + b* x^2 + c*x^4)) + (-((c*(b^2*(3 - m) + b*Sqrt[b^2 - 4*a*c]*(3 - m) - 4*a*c*( 5 - m))*(d*x)^(-1 + m)*Hypergeometric2F1[1, (-1 + m)/2, (1 + m)/2, (-2*c*x ^2)/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*d *(1 - m))) + (c*(b^2*(3 - m) - b*Sqrt[b^2 - 4*a*c]*(3 - m) - 4*a*c*(5 - m) )*(d*x)^(-1 + m)*Hypergeometric2F1[1, (-1 + m)/2, (1 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*d*(1 - m)))/(2*a*(b^2 - 4*a*c)))
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
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] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-(d*x)^(m + 1))*(b^2 - 2*a*c + b*c*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*a*d*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[(d*x)^m*(a + b*x^2 + c*x^4)^(p + 1)*Simp[b^2*(m + 2*p + 3) - 2*a*c*(m + 4*p + 5) + b*c*(m + 4*p + 7)*x^2, x], x], x] /; FreeQ[{a, b, c, d, m}, x ] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2))/((a_) + (b_.)*(x_)^2 + (c_.) *(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[(f*x)^m/(b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[(f*x)^m/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4*a*c, 0]
\[\int \frac {\left (d x \right )^{m}}{\left (c \,x^{5}+b \,x^{3}+x a \right )^{2}}d x\]
Input:
int((d*x)^m/(c*x^5+b*x^3+a*x)^2,x)
Output:
int((d*x)^m/(c*x^5+b*x^3+a*x)^2,x)
\[ \int \frac {(d x)^m}{\left (a x+b x^3+c x^5\right )^2} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{5} + b x^{3} + a x\right )}^{2}} \,d x } \] Input:
integrate((d*x)^m/(c*x^5+b*x^3+a*x)^2,x, algorithm="fricas")
Output:
integral((d*x)^m/(c^2*x^10 + 2*b*c*x^8 + (b^2 + 2*a*c)*x^6 + 2*a*b*x^4 + a ^2*x^2), x)
Timed out. \[ \int \frac {(d x)^m}{\left (a x+b x^3+c x^5\right )^2} \, dx=\text {Timed out} \] Input:
integrate((d*x)**m/(c*x**5+b*x**3+a*x)**2,x)
Output:
Timed out
\[ \int \frac {(d x)^m}{\left (a x+b x^3+c x^5\right )^2} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{5} + b x^{3} + a x\right )}^{2}} \,d x } \] Input:
integrate((d*x)^m/(c*x^5+b*x^3+a*x)^2,x, algorithm="maxima")
Output:
integrate((d*x)^m/(c*x^5 + b*x^3 + a*x)^2, x)
\[ \int \frac {(d x)^m}{\left (a x+b x^3+c x^5\right )^2} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{5} + b x^{3} + a x\right )}^{2}} \,d x } \] Input:
integrate((d*x)^m/(c*x^5+b*x^3+a*x)^2,x, algorithm="giac")
Output:
integrate((d*x)^m/(c*x^5 + b*x^3 + a*x)^2, x)
Timed out. \[ \int \frac {(d x)^m}{\left (a x+b x^3+c x^5\right )^2} \, dx=\int \frac {{\left (d\,x\right )}^m}{{\left (c\,x^5+b\,x^3+a\,x\right )}^2} \,d x \] Input:
int((d*x)^m/(a*x + b*x^3 + c*x^5)^2,x)
Output:
int((d*x)^m/(a*x + b*x^3 + c*x^5)^2, x)
\[ \int \frac {(d x)^m}{\left (a x+b x^3+c x^5\right )^2} \, dx=\text {too large to display} \] Input:
int((d*x)^m/(c*x^5+b*x^3+a*x)^2,x)
Output:
(d**m*(x**m - int(x**m/(a**2*m - a**2 + 2*a*b*m*x**2 - 2*a*b*x**2 + 2*a*c* m*x**4 - 2*a*c*x**4 + b**2*m*x**4 - b**2*x**4 + 2*b*c*m*x**6 - 2*b*c*x**6 + c**2*m*x**8 - c**2*x**8),x)*a*b*m**2*x + 4*int(x**m/(a**2*m - a**2 + 2*a *b*m*x**2 - 2*a*b*x**2 + 2*a*c*m*x**4 - 2*a*c*x**4 + b**2*m*x**4 - b**2*x* *4 + 2*b*c*m*x**6 - 2*b*c*x**6 + c**2*m*x**8 - c**2*x**8),x)*a*b*m*x - 3*i nt(x**m/(a**2*m - a**2 + 2*a*b*m*x**2 - 2*a*b*x**2 + 2*a*c*m*x**4 - 2*a*c* x**4 + b**2*m*x**4 - b**2*x**4 + 2*b*c*m*x**6 - 2*b*c*x**6 + c**2*m*x**8 - c**2*x**8),x)*a*b*x - int(x**m/(a**2*m - a**2 + 2*a*b*m*x**2 - 2*a*b*x**2 + 2*a*c*m*x**4 - 2*a*c*x**4 + b**2*m*x**4 - b**2*x**4 + 2*b*c*m*x**6 - 2* b*c*x**6 + c**2*m*x**8 - c**2*x**8),x)*b**2*m**2*x**3 + 4*int(x**m/(a**2*m - a**2 + 2*a*b*m*x**2 - 2*a*b*x**2 + 2*a*c*m*x**4 - 2*a*c*x**4 + b**2*m*x **4 - b**2*x**4 + 2*b*c*m*x**6 - 2*b*c*x**6 + c**2*m*x**8 - c**2*x**8),x)* b**2*m*x**3 - 3*int(x**m/(a**2*m - a**2 + 2*a*b*m*x**2 - 2*a*b*x**2 + 2*a* c*m*x**4 - 2*a*c*x**4 + b**2*m*x**4 - b**2*x**4 + 2*b*c*m*x**6 - 2*b*c*x** 6 + c**2*m*x**8 - c**2*x**8),x)*b**2*x**3 - int(x**m/(a**2*m - a**2 + 2*a* b*m*x**2 - 2*a*b*x**2 + 2*a*c*m*x**4 - 2*a*c*x**4 + b**2*m*x**4 - b**2*x** 4 + 2*b*c*m*x**6 - 2*b*c*x**6 + c**2*m*x**8 - c**2*x**8),x)*b*c*m**2*x**5 + 4*int(x**m/(a**2*m - a**2 + 2*a*b*m*x**2 - 2*a*b*x**2 + 2*a*c*m*x**4 - 2 *a*c*x**4 + b**2*m*x**4 - b**2*x**4 + 2*b*c*m*x**6 - 2*b*c*x**6 + c**2*m*x **8 - c**2*x**8),x)*b*c*m*x**5 - 3*int(x**m/(a**2*m - a**2 + 2*a*b*m*x*...