Integrand size = 29, antiderivative size = 123 \[ \int \frac {(f+g x)^n \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx=-\frac {2^{-3+p} (d-e x) \left (\frac {d+e x}{d}\right )^{-p} (f+g x)^n \left (\frac {e (f+g x)}{e f+d g}\right )^{-n} \left (d^2-e^2 x^2\right )^p \operatorname {AppellF1}\left (1+p,3-p,-n,2+p,\frac {d-e x}{2 d},\frac {g (d-e x)}{e f+d g}\right )}{d^3 e (1+p)} \] Output:
-2^(-3+p)*(-e*x+d)*(g*x+f)^n*(-e^2*x^2+d^2)^p*AppellF1(p+1,-n,3-p,2+p,g*(- e*x+d)/(d*g+e*f),1/2*(-e*x+d)/d)/d^3/e/(p+1)/(((e*x+d)/d)^p)/((e*(g*x+f)/( d*g+e*f))^n)
\[ \int \frac {(f+g x)^n \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx=\int \frac {(f+g x)^n \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx \] Input:
Integrate[((f + g*x)^n*(d^2 - e^2*x^2)^p)/(d + e*x)^3,x]
Output:
Integrate[((f + g*x)^n*(d^2 - e^2*x^2)^p)/(d + e*x)^3, x]
Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {718, 157, 156, 155}
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.
\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^p (f+g x)^n}{(d+e x)^3} \, dx\) |
\(\Big \downarrow \) 718 |
\(\displaystyle (d-e x)^{-p} (d+e x)^{-p} \left (d^2-e^2 x^2\right )^p \int (d-e x)^p (d+e x)^{p-3} (f+g x)^ndx\) |
\(\Big \downarrow \) 157 |
\(\displaystyle 2^p \left (\frac {d-e x}{d}\right )^{-p} (d+e x)^{-p} \left (d^2-e^2 x^2\right )^p \int (d+e x)^{p-3} \left (\frac {1}{2}-\frac {e x}{2 d}\right )^p (f+g x)^ndx\) |
\(\Big \downarrow \) 156 |
\(\displaystyle 2^p \left (\frac {d-e x}{d}\right )^{-p} (d+e x)^{-p} \left (d^2-e^2 x^2\right )^p (f+g x)^n \left (\frac {e (f+g x)}{e f-d g}\right )^{-n} \int (d+e x)^{p-3} \left (\frac {1}{2}-\frac {e x}{2 d}\right )^p \left (\frac {e f}{e f-d g}+\frac {e g x}{e f-d g}\right )^ndx\) |
\(\Big \downarrow \) 155 |
\(\displaystyle -\frac {2^p \left (\frac {d-e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p (f+g x)^n \left (\frac {e (f+g x)}{e f-d g}\right )^{-n} \operatorname {AppellF1}\left (p-2,-p,-n,p-1,\frac {d+e x}{2 d},-\frac {g (d+e x)}{e f-d g}\right )}{e (2-p) (d+e x)^2}\) |
Input:
Int[((f + g*x)^n*(d^2 - e^2*x^2)^p)/(d + e*x)^3,x]
Output:
-((2^p*(f + g*x)^n*(d^2 - e^2*x^2)^p*AppellF1[-2 + p, -p, -n, -1 + p, (d + e*x)/(2*d), -((g*(d + e*x))/(e*f - d*g))])/(e*(2 - p)*((d - e*x)/d)^p*(d + e*x)^2*((e*(f + g*x))/(e*f - d*g))^n))
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ (b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p ]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]) Int[(a + b*x)^m*(c + d*x)^n*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n ]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & !GtQ[Simplify[b/(b*c - a*d)], 0] && !SimplerQ[c + d*x, a + b*x] && !Si mplerQ[e + f*x, a + b*x]
Int[((d_) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_) ^2)^(p_), x_Symbol] :> Simp[(a + c*x^2)^FracPart[p]/((d + e*x)^FracPart[p]* (a/d + (c*x)/e)^FracPart[p]) Int[(d + e*x)^(m + p)*(f + g*x)^n*(a/d + (c/ e)*x)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && EqQ[c*d^2 + a*e^2, 0]
\[\int \frac {\left (g x +f \right )^{n} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{3}}d x\]
Input:
int((g*x+f)^n*(-e^2*x^2+d^2)^p/(e*x+d)^3,x)
Output:
int((g*x+f)^n*(-e^2*x^2+d^2)^p/(e*x+d)^3,x)
\[ \int \frac {(f+g x)^n \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} {\left (g x + f\right )}^{n}}{{\left (e x + d\right )}^{3}} \,d x } \] Input:
integrate((g*x+f)^n*(-e^2*x^2+d^2)^p/(e*x+d)^3,x, algorithm="fricas")
Output:
integral((-e^2*x^2 + d^2)^p*(g*x + f)^n/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), x)
\[ \int \frac {(f+g x)^n \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p} \left (f + g x\right )^{n}}{\left (d + e x\right )^{3}}\, dx \] Input:
integrate((g*x+f)**n*(-e**2*x**2+d**2)**p/(e*x+d)**3,x)
Output:
Integral((-(-d + e*x)*(d + e*x))**p*(f + g*x)**n/(d + e*x)**3, x)
\[ \int \frac {(f+g x)^n \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} {\left (g x + f\right )}^{n}}{{\left (e x + d\right )}^{3}} \,d x } \] Input:
integrate((g*x+f)^n*(-e^2*x^2+d^2)^p/(e*x+d)^3,x, algorithm="maxima")
Output:
integrate((-e^2*x^2 + d^2)^p*(g*x + f)^n/(e*x + d)^3, x)
\[ \int \frac {(f+g x)^n \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} {\left (g x + f\right )}^{n}}{{\left (e x + d\right )}^{3}} \,d x } \] Input:
integrate((g*x+f)^n*(-e^2*x^2+d^2)^p/(e*x+d)^3,x, algorithm="giac")
Output:
integrate((-e^2*x^2 + d^2)^p*(g*x + f)^n/(e*x + d)^3, x)
Timed out. \[ \int \frac {(f+g x)^n \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx=\int \frac {{\left (f+g\,x\right )}^n\,{\left (d^2-e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^3} \,d x \] Input:
int(((f + g*x)^n*(d^2 - e^2*x^2)^p)/(d + e*x)^3,x)
Output:
int(((f + g*x)^n*(d^2 - e^2*x^2)^p)/(d + e*x)^3, x)
\[ \int \frac {(f+g x)^n \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx=\text {too large to display} \] Input:
int((g*x+f)^n*(-e^2*x^2+d^2)^p/(e*x+d)^3,x)
Output:
( - (f + g*x)**n*(d**2 - e**2*x**2)**p*d*g + (f + g*x)**n*(d**2 - e**2*x** 2)**p*e*f - int(((f + g*x)**n*(d**2 - e**2*x**2)**p*x**2)/(d**5*f*g + d**5 *g**2*x + d**4*e*f**2*p - d**4*e*f**2 + d**4*e*f*g*p*x + d**4*e*f*g*x + 2* d**4*e*g**2*x**2 + 2*d**3*e**2*f**2*p*x - 2*d**3*e**2*f**2*x + 2*d**3*e**2 *f*g*p*x**2 - 2*d**3*e**2*f*g*x**2 - 2*d**2*e**3*f*g*x**3 - 2*d**2*e**3*g* *2*x**4 - 2*d*e**4*f**2*p*x**3 + 2*d*e**4*f**2*x**3 - 2*d*e**4*f*g*p*x**4 + d*e**4*f*g*x**4 - d*e**4*g**2*x**5 - e**5*f**2*p*x**4 + e**5*f**2*x**4 - e**5*f*g*p*x**5 + e**5*f*g*x**5),x)*d**4*e**2*g**3*n - 2*int(((f + g*x)** n*(d**2 - e**2*x**2)**p*x**2)/(d**5*f*g + d**5*g**2*x + d**4*e*f**2*p - d* *4*e*f**2 + d**4*e*f*g*p*x + d**4*e*f*g*x + 2*d**4*e*g**2*x**2 + 2*d**3*e* *2*f**2*p*x - 2*d**3*e**2*f**2*x + 2*d**3*e**2*f*g*p*x**2 - 2*d**3*e**2*f* g*x**2 - 2*d**2*e**3*f*g*x**3 - 2*d**2*e**3*g**2*x**4 - 2*d*e**4*f**2*p*x* *3 + 2*d*e**4*f**2*x**3 - 2*d*e**4*f*g*p*x**4 + d*e**4*f*g*x**4 - d*e**4*g **2*x**5 - e**5*f**2*p*x**4 + e**5*f**2*x**4 - e**5*f*g*p*x**5 + e**5*f*g* x**5),x)*d**4*e**2*g**3*p - int(((f + g*x)**n*(d**2 - e**2*x**2)**p*x**2)/ (d**5*f*g + d**5*g**2*x + d**4*e*f**2*p - d**4*e*f**2 + d**4*e*f*g*p*x + d **4*e*f*g*x + 2*d**4*e*g**2*x**2 + 2*d**3*e**2*f**2*p*x - 2*d**3*e**2*f**2 *x + 2*d**3*e**2*f*g*p*x**2 - 2*d**3*e**2*f*g*x**2 - 2*d**2*e**3*f*g*x**3 - 2*d**2*e**3*g**2*x**4 - 2*d*e**4*f**2*p*x**3 + 2*d*e**4*f**2*x**3 - 2*d* e**4*f*g*p*x**4 + d*e**4*f*g*x**4 - d*e**4*g**2*x**5 - e**5*f**2*p*x**4...