Integrand size = 22, antiderivative size = 155 \[ \int \frac {(e x)^m (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=-\frac {d^2 (e x)^{1+m}}{b e (1-m) \left (a+b x^2\right )}+\frac {\left (b c^2 (1-m)+a d^2 (1+m)\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^2 b e (1-m) (1+m)}+\frac {2 c d (e x)^{2+m} \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{a^2 e^2 (2+m)} \] Output:
-d^2*(e*x)^(1+m)/b/e/(1-m)/(b*x^2+a)+(b*c^2*(1-m)+a*d^2*(1+m))*(e*x)^(1+m) *hypergeom([2, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)/a^2/b/e/(1-m)/(1+m)+2*c*d* (e*x)^(2+m)*hypergeom([2, 1+1/2*m],[2+1/2*m],-b*x^2/a)/a^2/e^2/(2+m)
Time = 0.04 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.85 \[ \int \frac {(e x)^m (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=\frac {x (e x)^m \left (a d^2 (2+m) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+\left (b c^2-a d^2\right ) (2+m) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+2 b c d (1+m) x \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )\right )}{a^2 b (1+m) (2+m)} \] Input:
Integrate[((e*x)^m*(c + d*x)^2)/(a + b*x^2)^2,x]
Output:
(x*(e*x)^m*(a*d^2*(2 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b* x^2)/a)] + (b*c^2 - a*d^2)*(2 + m)*Hypergeometric2F1[2, (1 + m)/2, (3 + m) /2, -((b*x^2)/a)] + 2*b*c*d*(1 + m)*x*Hypergeometric2F1[2, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)]))/(a^2*b*(1 + m)*(2 + m))
Time = 0.30 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {558, 25, 557, 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.
| \(\displaystyle \int \frac {(c+d x)^2 (e x)^m}{\left (a+b x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 558 |
| \(\displaystyle \frac {(e x)^{m+1} \left (-\frac {a d^2}{b}+c^2+2 c d x\right )}{2 a e \left (a+b x^2\right )}-\frac {\int -\frac {(e x)^m \left ((1-m) c^2-2 d m x c+\frac {a d^2 (m+1)}{b}\right )}{b x^2+a}dx}{2 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {(e x)^m \left ((1-m) c^2-2 d m x c+\frac {a d^2 (m+1)}{b}\right )}{b x^2+a}dx}{2 a}+\frac {(e x)^{m+1} \left (-\frac {a d^2}{b}+c^2+2 c d x\right )}{2 a e \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 557 |
| \(\displaystyle \frac {\left (\frac {a d^2 (m+1)}{b}+c^2 (1-m)\right ) \int \frac {(e x)^m}{b x^2+a}dx-\frac {2 c d m \int \frac {(e x)^{m+1}}{b x^2+a}dx}{e}}{2 a}+\frac {(e x)^{m+1} \left (-\frac {a d^2}{b}+c^2+2 c d x\right )}{2 a e \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {\frac {(e x)^{m+1} \left (\frac {a d^2 (m+1)}{b}+c^2 (1-m)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a e (m+1)}-\frac {2 c d m (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\frac {b x^2}{a}\right )}{a e^2 (m+2)}}{2 a}+\frac {(e x)^{m+1} \left (-\frac {a d^2}{b}+c^2+2 c d x\right )}{2 a e \left (a+b x^2\right )}\) |
Input:
Int[((e*x)^m*(c + d*x)^2)/(a + b*x^2)^2,x]
Output:
((e*x)^(1 + m)*(c^2 - (a*d^2)/b + 2*c*d*x))/(2*a*e*(a + b*x^2)) + (((c^2*( 1 - m) + (a*d^2*(1 + m))/b)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*e*(1 + m)) - (2*c*d*m*(e*x)^(2 + m)*Hypergeom etric2F1[1, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)])/(a*e^2*(2 + m)))/(2*a)
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[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym bol] :> Simp[c Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e Int[(e*x)^( m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, a + b*x^2, x], f = Coeff[PolynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 0], g = Coeff[Pol ynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(-(e*x)^(m + 1))* (f + g*x)*((a + b*x^2)^(p + 1)/(2*a*e*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + f*(m + 2*p + 3) + g*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, m}, x] && IGt Q[n, 1] && !IntegerQ[m] && LtQ[p, -1]
\[\int \frac {\left (e x \right )^{m} \left (d x +c \right )^{2}}{\left (b \,x^{2}+a \right )^{2}}d x\]
Input:
int((e*x)^m*(d*x+c)^2/(b*x^2+a)^2,x)
Output:
int((e*x)^m*(d*x+c)^2/(b*x^2+a)^2,x)
\[ \int \frac {(e x)^m (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x + c\right )}^{2} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \] Input:
integrate((e*x)^m*(d*x+c)^2/(b*x^2+a)^2,x, algorithm="fricas")
Output:
integral((d^2*x^2 + 2*c*d*x + c^2)*(e*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2), x)
\[ \int \frac {(e x)^m (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=\int \frac {\left (e x\right )^{m} \left (c + d x\right )^{2}}{\left (a + b x^{2}\right )^{2}}\, dx \] Input:
integrate((e*x)**m*(d*x+c)**2/(b*x**2+a)**2,x)
Output:
Integral((e*x)**m*(c + d*x)**2/(a + b*x**2)**2, x)
\[ \int \frac {(e x)^m (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x + c\right )}^{2} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \] Input:
integrate((e*x)^m*(d*x+c)^2/(b*x^2+a)^2,x, algorithm="maxima")
Output:
integrate((d*x + c)^2*(e*x)^m/(b*x^2 + a)^2, x)
\[ \int \frac {(e x)^m (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x + c\right )}^{2} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \] Input:
integrate((e*x)^m*(d*x+c)^2/(b*x^2+a)^2,x, algorithm="giac")
Output:
integrate((d*x + c)^2*(e*x)^m/(b*x^2 + a)^2, x)
Timed out. \[ \int \frac {(e x)^m (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=\int \frac {{\left (e\,x\right )}^m\,{\left (c+d\,x\right )}^2}{{\left (b\,x^2+a\right )}^2} \,d x \] Input:
int(((e*x)^m*(c + d*x)^2)/(a + b*x^2)^2,x)
Output:
int(((e*x)^m*(c + d*x)^2)/(a + b*x^2)^2, x)
\[ \int \frac {(e x)^m (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=\text {too large to display} \] Input:
int((e*x)^m*(d*x+c)^2/(b*x^2+a)^2,x)
Output:
(e**m*(2*x**m*c*d*m - 2*x**m*c*d + x**m*d**2*m*x - 2*x**m*d**2*x - 2*int(x **m/(a**2*m**2*x - 3*a**2*m*x + 2*a**2*x + 2*a*b*m**2*x**3 - 6*a*b*m*x**3 + 4*a*b*x**3 + b**2*m**2*x**5 - 3*b**2*m*x**5 + 2*b**2*x**5),x)*a**2*c*d*m **4 + 8*int(x**m/(a**2*m**2*x - 3*a**2*m*x + 2*a**2*x + 2*a*b*m**2*x**3 - 6*a*b*m*x**3 + 4*a*b*x**3 + b**2*m**2*x**5 - 3*b**2*m*x**5 + 2*b**2*x**5), x)*a**2*c*d*m**3 - 10*int(x**m/(a**2*m**2*x - 3*a**2*m*x + 2*a**2*x + 2*a* b*m**2*x**3 - 6*a*b*m*x**3 + 4*a*b*x**3 + b**2*m**2*x**5 - 3*b**2*m*x**5 + 2*b**2*x**5),x)*a**2*c*d*m**2 + 4*int(x**m/(a**2*m**2*x - 3*a**2*m*x + 2* a**2*x + 2*a*b*m**2*x**3 - 6*a*b*m*x**3 + 4*a*b*x**3 + b**2*m**2*x**5 - 3* b**2*m*x**5 + 2*b**2*x**5),x)*a**2*c*d*m - 2*int(x**m/(a**2*m**2*x - 3*a** 2*m*x + 2*a**2*x + 2*a*b*m**2*x**3 - 6*a*b*m*x**3 + 4*a*b*x**3 + b**2*m**2 *x**5 - 3*b**2*m*x**5 + 2*b**2*x**5),x)*a*b*c*d*m**4*x**2 + 8*int(x**m/(a* *2*m**2*x - 3*a**2*m*x + 2*a**2*x + 2*a*b*m**2*x**3 - 6*a*b*m*x**3 + 4*a*b *x**3 + b**2*m**2*x**5 - 3*b**2*m*x**5 + 2*b**2*x**5),x)*a*b*c*d*m**3*x**2 - 10*int(x**m/(a**2*m**2*x - 3*a**2*m*x + 2*a**2*x + 2*a*b*m**2*x**3 - 6* a*b*m*x**3 + 4*a*b*x**3 + b**2*m**2*x**5 - 3*b**2*m*x**5 + 2*b**2*x**5),x) *a*b*c*d*m**2*x**2 + 4*int(x**m/(a**2*m**2*x - 3*a**2*m*x + 2*a**2*x + 2*a *b*m**2*x**3 - 6*a*b*m*x**3 + 4*a*b*x**3 + b**2*m**2*x**5 - 3*b**2*m*x**5 + 2*b**2*x**5),x)*a*b*c*d*m*x**2 - int(x**m/(a**2*m**2 - 3*a**2*m + 2*a**2 + 2*a*b*m**2*x**2 - 6*a*b*m*x**2 + 4*a*b*x**2 + b**2*m**2*x**4 - 3*b**...