Integrand size = 25, antiderivative size = 166 \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=-\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{1+p}}{x}-2 e^3 (1+3 p) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )-\frac {e^2 (3-p) \left (d^2-e^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1-\frac {e^2 x^2}{d^2}\right )}{2 d (1+p)} \]
-1/2*d*(-e^2*x^2+d^2)^(p+1)/x^2-3*e*(-e^2*x^2+d^2)^(p+1)/x-2*e^3*(1+3*p)*x *(-e^2*x^2+d^2)^p*hypergeom([1/2, -p],[3/2],e^2*x^2/d^2)/((1-e^2*x^2/d^2)^ p)-1/2*e^2*(3-p)*(-e^2*x^2+d^2)^(p+1)*hypergeom([1, p+1],[2+p],1-e^2*x^2/d ^2)/d/(p+1)
Time = 0.28 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.10 \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=\frac {e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (-6 d^3 (1+p) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )+e x \left (2 d e (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )-\left (d^2-e^2 x^2\right ) \left (1-\frac {e^2 x^2}{d^2}\right )^p \left (3 \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1-\frac {e^2 x^2}{d^2}\right )+\operatorname {Hypergeometric2F1}\left (2,1+p,2+p,1-\frac {e^2 x^2}{d^2}\right )\right )\right )\right )}{2 d (1+p) x} \]
(e*(d^2 - e^2*x^2)^p*(-6*d^3*(1 + p)*Hypergeometric2F1[-1/2, -p, 1/2, (e^2 *x^2)/d^2] + e*x*(2*d*e*(1 + p)*x*Hypergeometric2F1[1/2, -p, 3/2, (e^2*x^2 )/d^2] - (d^2 - e^2*x^2)*(1 - (e^2*x^2)/d^2)^p*(3*Hypergeometric2F1[1, 1 + p, 2 + p, 1 - (e^2*x^2)/d^2] + Hypergeometric2F1[2, 1 + p, 2 + p, 1 - (e^ 2*x^2)/d^2]))))/(2*d*(1 + p)*x*(1 - (e^2*x^2)/d^2)^p)
Time = 0.29 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {543, 354, 27, 87, 75, 359, 238, 237}
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.
| \(\displaystyle \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx\) |
| \(\Big \downarrow \) 543 |
| \(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^p \left (x^2 e^3+3 d^2 e\right )}{x^2}dx+\int \frac {\left (d^2-e^2 x^2\right )^p \left (d^3+3 e^2 x^2 d\right )}{x^3}dx\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{2} \int \frac {d \left (d^2-e^2 x^2\right )^p \left (d^2+3 e^2 x^2\right )}{x^4}dx^2+\int \frac {\left (d^2-e^2 x^2\right )^p \left (x^2 e^3+3 d^2 e\right )}{x^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} d \int \frac {\left (d^2-e^2 x^2\right )^p \left (d^2+3 e^2 x^2\right )}{x^4}dx^2+\int \frac {\left (d^2-e^2 x^2\right )^p \left (x^2 e^3+3 d^2 e\right )}{x^2}dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {1}{2} d \left (e^2 (3-p) \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2}dx^2-\frac {\left (d^2-e^2 x^2\right )^{p+1}}{x^2}\right )+\int \frac {\left (d^2-e^2 x^2\right )^p \left (x^2 e^3+3 d^2 e\right )}{x^2}dx\) |
| \(\Big \downarrow \) 75 |
| \(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^p \left (x^2 e^3+3 d^2 e\right )}{x^2}dx+\frac {1}{2} d \left (-\frac {e^2 (3-p) \left (d^2-e^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,1-\frac {e^2 x^2}{d^2}\right )}{d^2 (p+1)}-\frac {\left (d^2-e^2 x^2\right )^{p+1}}{x^2}\right )\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle -2 e^3 (3 p+1) \int \left (d^2-e^2 x^2\right )^pdx+\frac {1}{2} d \left (-\frac {e^2 (3-p) \left (d^2-e^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,1-\frac {e^2 x^2}{d^2}\right )}{d^2 (p+1)}-\frac {\left (d^2-e^2 x^2\right )^{p+1}}{x^2}\right )-\frac {3 e \left (d^2-e^2 x^2\right )^{p+1}}{x}\) |
| \(\Big \downarrow \) 238 |
| \(\displaystyle -2 e^3 (3 p+1) \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \int \left (1-\frac {e^2 x^2}{d^2}\right )^pdx+\frac {1}{2} d \left (-\frac {e^2 (3-p) \left (d^2-e^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,1-\frac {e^2 x^2}{d^2}\right )}{d^2 (p+1)}-\frac {\left (d^2-e^2 x^2\right )^{p+1}}{x^2}\right )-\frac {3 e \left (d^2-e^2 x^2\right )^{p+1}}{x}\) |
| \(\Big \downarrow \) 237 |
| \(\displaystyle \frac {1}{2} d \left (-\frac {e^2 (3-p) \left (d^2-e^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,1-\frac {e^2 x^2}{d^2}\right )}{d^2 (p+1)}-\frac {\left (d^2-e^2 x^2\right )^{p+1}}{x^2}\right )-\frac {3 e \left (d^2-e^2 x^2\right )^{p+1}}{x}-2 e^3 (3 p+1) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )\) |
(-3*e*(d^2 - e^2*x^2)^(1 + p))/x - (2*e^3*(1 + 3*p)*x*(d^2 - e^2*x^2)^p*Hy pergeometric2F1[1/2, -p, 3/2, (e^2*x^2)/d^2])/(1 - (e^2*x^2)/d^2)^p + (d*( -((d^2 - e^2*x^2)^(1 + p)/x^2) - (e^2*(3 - p)*(d^2 - e^2*x^2)^(1 + p)*Hype rgeometric2F1[1, 1 + p, 2 + p, 1 - (e^2*x^2)/d^2])/(d^2*(1 + p))))/2
3.3.66.3.1 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[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] && !IntegerQ[2*p ] && GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) ^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(1 + b*(x^2/a))^p, x], x] / ; FreeQ[{a, b, p}, x] && !IntegerQ[2*p] && !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])
\[\int \frac {\left (e x +d \right )^{3} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{3}}d x\]
\[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{x^{3}} \,d x } \]
Result contains complex when optimal does not.
Time = 3.17 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.05 \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=- \frac {d^{3} e^{2 p} x^{2 p - 2} e^{i \pi p} \Gamma \left (1 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 - p \\ 2 - p \end {matrix}\middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{2 \Gamma \left (2 - p\right )} - \frac {3 d^{2} d^{2 p} e {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{x} - \frac {3 d e^{2} e^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} + d^{2 p} e^{3} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} \]
-d**3*e**(2*p)*x**(2*p - 2)*exp(I*pi*p)*gamma(1 - p)*hyper((-p, 1 - p), (2 - p,), d**2/(e**2*x**2))/(2*gamma(2 - p)) - 3*d**2*d**(2*p)*e*hyper((-1/2 , -p), (1/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)/x - 3*d*e**2*e**(2*p)*x** (2*p)*exp(I*pi*p)*gamma(-p)*hyper((-p, -p), (1 - p,), d**2/(e**2*x**2))/(2 *gamma(1 - p)) + d**(2*p)*e**3*x*hyper((1/2, -p), (3/2,), e**2*x**2*exp_po lar(2*I*pi)/d**2)
\[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{x^{3}} \,d x } \]
\[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p\,{\left (d+e\,x\right )}^3}{x^3} \,d x \]