Integrand size = 25, antiderivative size = 179 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^3} \, dx=-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{3 x^3}+\frac {3 e \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}-\frac {2 e^2 (8-p) \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{3 d^5 x}-\frac {e^3 (10-3 p) \left (d^2-e^2 x^2\right )^{-2+p} \operatorname {Hypergeometric2F1}\left (1,-2+p,-1+p,1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 (2-p)} \]
-1/3*d*(-e^2*x^2+d^2)^(-2+p)/x^3+3/2*e*(-e^2*x^2+d^2)^(-2+p)/x^2-2/3*e^2*( 8-p)*(-e^2*x^2+d^2)^p*hypergeom([-1/2, 3-p],[1/2],e^2*x^2/d^2)/d^5/x/((1-e ^2*x^2/d^2)^p)-1/2*e^3*(10-3*p)*(-e^2*x^2+d^2)^(-2+p)*hypergeom([1, -2+p], [-1+p],1-e^2*x^2/d^2)/d^2/(2-p)
Leaf count is larger than twice the leaf count of optimal. \(393\) vs. \(2(179)=358\).
Time = 0.63 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.20 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^3} \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (-\frac {8 d^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-p,-\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{x^3}-\frac {144 d^2 e^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{x}-\frac {36 d^3 e \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,\frac {d^2}{e^2 x^2}\right )}{(-1+p) x^2}+\frac {15\ 2^{3+p} e^3 (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {3\ 2^{3+p} e^3 (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {3\ 2^p e^3 (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}-\frac {120 d e^3 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )}{p}\right )}{24 d^7} \]
((d^2 - e^2*x^2)^p*((-8*d^4*Hypergeometric2F1[-3/2, -p, -1/2, (e^2*x^2)/d^ 2])/(x^3*(1 - (e^2*x^2)/d^2)^p) - (144*d^2*e^2*Hypergeometric2F1[-1/2, -p, 1/2, (e^2*x^2)/d^2])/(x*(1 - (e^2*x^2)/d^2)^p) - (36*d^3*e*Hypergeometric 2F1[1 - p, -p, 2 - p, d^2/(e^2*x^2)])/((-1 + p)*(1 - d^2/(e^2*x^2))^p*x^2) + (15*2^(3 + p)*e^3*(-d + e*x)*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (3*2^(3 + p)*e^3*(-d + e*x)*Hyp ergeometric2F1[2 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/ d)^p) + (3*2^p*e^3*(-d + e*x)*Hypergeometric2F1[3 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) - (120*d*e^3*Hypergeometric2F1[-p, -p, 1 - p, d^2/(e^2*x^2)])/(p*(1 - d^2/(e^2*x^2))^p)))/(24*d^7)
Time = 0.35 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {570, 543, 354, 25, 27, 87, 75, 359, 279, 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.
\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^3} \, dx\) |
\(\Big \downarrow \) 570 |
\(\displaystyle \int \frac {(d-e x)^3 \left (d^2-e^2 x^2\right )^{p-3}}{x^4}dx\) |
\(\Big \downarrow \) 543 |
\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{p-3} \left (-x^2 e^3-3 d^2 e\right )}{x^3}dx+\int \frac {\left (d^2-e^2 x^2\right )^{p-3} \left (d^3+3 e^2 x^2 d\right )}{x^4}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int -\frac {e \left (d^2-e^2 x^2\right )^{p-3} \left (3 d^2+e^2 x^2\right )}{x^4}dx^2+\int \frac {\left (d^2-e^2 x^2\right )^{p-3} \left (d^3+3 e^2 x^2 d\right )}{x^4}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{p-3} \left (d^3+3 e^2 x^2 d\right )}{x^4}dx-\frac {1}{2} \int \frac {e \left (d^2-e^2 x^2\right )^{p-3} \left (3 d^2+e^2 x^2\right )}{x^4}dx^2\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{p-3} \left (d^3+3 e^2 x^2 d\right )}{x^4}dx-\frac {1}{2} e \int \frac {\left (d^2-e^2 x^2\right )^{p-3} \left (3 d^2+e^2 x^2\right )}{x^4}dx^2\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{p-3} \left (d^3+3 e^2 x^2 d\right )}{x^4}dx-\frac {1}{2} e \left (e^2 (10-3 p) \int \frac {\left (d^2-e^2 x^2\right )^{p-3}}{x^2}dx^2-\frac {3 \left (d^2-e^2 x^2\right )^{p-2}}{x^2}\right )\) |
\(\Big \downarrow \) 75 |
\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{p-3} \left (d^3+3 e^2 x^2 d\right )}{x^4}dx-\frac {1}{2} e \left (\frac {e^2 (10-3 p) \left (d^2-e^2 x^2\right )^{p-2} \operatorname {Hypergeometric2F1}\left (1,p-2,p-1,1-\frac {e^2 x^2}{d^2}\right )}{d^2 (2-p)}-\frac {3 \left (d^2-e^2 x^2\right )^{p-2}}{x^2}\right )\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {2}{3} d e^2 (8-p) \int \frac {\left (d^2-e^2 x^2\right )^{p-3}}{x^2}dx-\frac {1}{2} e \left (\frac {e^2 (10-3 p) \left (d^2-e^2 x^2\right )^{p-2} \operatorname {Hypergeometric2F1}\left (1,p-2,p-1,1-\frac {e^2 x^2}{d^2}\right )}{d^2 (2-p)}-\frac {3 \left (d^2-e^2 x^2\right )^{p-2}}{x^2}\right )-\frac {d \left (d^2-e^2 x^2\right )^{p-2}}{3 x^3}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {2 e^2 (8-p) \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^{p-3}}{x^2}dx}{3 d^5}-\frac {1}{2} e \left (\frac {e^2 (10-3 p) \left (d^2-e^2 x^2\right )^{p-2} \operatorname {Hypergeometric2F1}\left (1,p-2,p-1,1-\frac {e^2 x^2}{d^2}\right )}{d^2 (2-p)}-\frac {3 \left (d^2-e^2 x^2\right )^{p-2}}{x^2}\right )-\frac {d \left (d^2-e^2 x^2\right )^{p-2}}{3 x^3}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {1}{2} e \left (\frac {e^2 (10-3 p) \left (d^2-e^2 x^2\right )^{p-2} \operatorname {Hypergeometric2F1}\left (1,p-2,p-1,1-\frac {e^2 x^2}{d^2}\right )}{d^2 (2-p)}-\frac {3 \left (d^2-e^2 x^2\right )^{p-2}}{x^2}\right )-\frac {d \left (d^2-e^2 x^2\right )^{p-2}}{3 x^3}-\frac {2 e^2 (8-p) \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{3 d^5 x}\) |
-1/3*(d*(d^2 - e^2*x^2)^(-2 + p))/x^3 - (2*e^2*(8 - p)*(d^2 - e^2*x^2)^p*H ypergeometric2F1[-1/2, 3 - p, 1/2, (e^2*x^2)/d^2])/(3*d^5*x*(1 - (e^2*x^2) /d^2)^p) - (e*((-3*(d^2 - e^2*x^2)^(-2 + p))/x^2 + (e^2*(10 - 3*p)*(d^2 - e^2*x^2)^(-2 + p)*Hypergeometric2F1[1, -2 + p, -1 + p, 1 - (e^2*x^2)/d^2]) /(d^2*(2 - p))))/2
3.3.94.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Module[{k}, Int[x^m*Sum[Binomial[n, 2*k]*c^(n - 2*k)*d^(2*k)*x^(2*k), {k, 0, n/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Binomial[n, 2*k + 1]*c^ (n - 2*k - 1)*d^(2*k + 1)*x^(2*k), {k, 0, (n - 1)/2}]*(a + b*x^2)^p, x]] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[n, 1] && IntegerQ[m] && !IntegerQ[2*p] && !(EqQ[m, 1] && EqQ[b*c^2 + a*d^2, 0])
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^(2*n)/a^n Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I LtQ[n, -1] && !(IGtQ[m, 0] && ILtQ[m + n, 0] && !GtQ[p, 1])
\[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{4} \left (e x +d \right )^{3}}d x\]
\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x^{4}} \,d x } \]
\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^3} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{4} \left (d + e x\right )^{3}}\, dx \]
\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x^{4}} \,d x } \]
\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^3} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^4\,{\left (d+e\,x\right )}^3} \,d x \]