Integrand size = 22, antiderivative size = 108 \[ \int \frac {(A+B x) (d+e x)^2}{a+c x^2} \, dx=\frac {e (2 B d+A e) x}{c}+\frac {B e^2 x^2}{2 c}+\frac {\left (A c d^2-2 a B d e-a A e^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} c^{3/2}}+\frac {\left (B c d^2+2 A c d e-a B e^2\right ) \log \left (a+c x^2\right )}{2 c^2} \] Output:
e*(A*e+2*B*d)*x/c+1/2*B*e^2*x^2/c+(-A*a*e^2+A*c*d^2-2*B*a*d*e)*arctan(c^(1 /2)*x/a^(1/2))/a^(1/2)/c^(3/2)+1/2*(2*A*c*d*e-B*a*e^2+B*c*d^2)*ln(c*x^2+a) /c^2
Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) (d+e x)^2}{a+c x^2} \, dx=\frac {c e x (4 B d+2 A e+B e x)-\frac {2 \sqrt {c} \left (-A c d^2+2 a B d e+a A e^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a}}+\left (B c d^2+2 A c d e-a B e^2\right ) \log \left (a+c x^2\right )}{2 c^2} \] Input:
Integrate[((A + B*x)*(d + e*x)^2)/(a + c*x^2),x]
Output:
(c*e*x*(4*B*d + 2*A*e + B*e*x) - (2*Sqrt[c]*(-(A*c*d^2) + 2*a*B*d*e + a*A* e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/Sqrt[a] + (B*c*d^2 + 2*A*c*d*e - a*B*e^2 )*Log[a + c*x^2])/(2*c^2)
Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {657, 2009}
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.
\(\displaystyle \int \frac {(A+B x) (d+e x)^2}{a+c x^2} \, dx\) |
\(\Big \downarrow \) 657 |
\(\displaystyle \int \left (\frac {x \left (-a B e^2+2 A c d e+B c d^2\right )-a A e^2-2 a B d e+A c d^2}{c \left (a+c x^2\right )}+\frac {e (A e+2 B d)}{c}+\frac {B e^2 x}{c}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (-a A e^2-2 a B d e+A c d^2\right )}{\sqrt {a} c^{3/2}}+\frac {\log \left (a+c x^2\right ) \left (-a B e^2+2 A c d e+B c d^2\right )}{2 c^2}+\frac {e x (A e+2 B d)}{c}+\frac {B e^2 x^2}{2 c}\) |
Input:
Int[((A + B*x)*(d + e*x)^2)/(a + c*x^2),x]
Output:
(e*(2*B*d + A*e)*x)/c + (B*e^2*x^2)/(2*c) + ((A*c*d^2 - 2*a*B*d*e - a*A*e^ 2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(3/2)) + ((B*c*d^2 + 2*A*c*d*e - a*B*e^2)*Log[a + c*x^2])/(2*c^2)
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Time = 1.02 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.90
method | result | size |
default | \(\frac {e \left (\frac {1}{2} B e \,x^{2}+A e x +2 B d x \right )}{c}+\frac {\frac {\left (2 A c d e -B a \,e^{2}+B c \,d^{2}\right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (-A a \,e^{2}+A c \,d^{2}-2 B a d e \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}}{c}\) | \(97\) |
risch | \(\frac {B \,e^{2} x^{2}}{2 c}+\frac {e^{2} A x}{c}+\frac {2 e B d x}{c}+\frac {\ln \left (-A \,a^{2} e^{2}+A \,d^{2} a c -2 B \,a^{2} d e -\sqrt {-a c \left (A a \,e^{2}-A c \,d^{2}+2 B a d e \right )^{2}}\, x \right ) A d e}{c}-\frac {a \ln \left (-A \,a^{2} e^{2}+A \,d^{2} a c -2 B \,a^{2} d e -\sqrt {-a c \left (A a \,e^{2}-A c \,d^{2}+2 B a d e \right )^{2}}\, x \right ) B \,e^{2}}{2 c^{2}}+\frac {\ln \left (-A \,a^{2} e^{2}+A \,d^{2} a c -2 B \,a^{2} d e -\sqrt {-a c \left (A a \,e^{2}-A c \,d^{2}+2 B a d e \right )^{2}}\, x \right ) B \,d^{2}}{2 c}+\frac {\ln \left (-A \,a^{2} e^{2}+A \,d^{2} a c -2 B \,a^{2} d e -\sqrt {-a c \left (A a \,e^{2}-A c \,d^{2}+2 B a d e \right )^{2}}\, x \right ) \sqrt {-a c \left (A a \,e^{2}-A c \,d^{2}+2 B a d e \right )^{2}}}{2 c^{2} a}+\frac {\ln \left (-A \,a^{2} e^{2}+A \,d^{2} a c -2 B \,a^{2} d e +\sqrt {-a c \left (A a \,e^{2}-A c \,d^{2}+2 B a d e \right )^{2}}\, x \right ) A d e}{c}-\frac {a \ln \left (-A \,a^{2} e^{2}+A \,d^{2} a c -2 B \,a^{2} d e +\sqrt {-a c \left (A a \,e^{2}-A c \,d^{2}+2 B a d e \right )^{2}}\, x \right ) B \,e^{2}}{2 c^{2}}+\frac {\ln \left (-A \,a^{2} e^{2}+A \,d^{2} a c -2 B \,a^{2} d e +\sqrt {-a c \left (A a \,e^{2}-A c \,d^{2}+2 B a d e \right )^{2}}\, x \right ) B \,d^{2}}{2 c}-\frac {\ln \left (-A \,a^{2} e^{2}+A \,d^{2} a c -2 B \,a^{2} d e +\sqrt {-a c \left (A a \,e^{2}-A c \,d^{2}+2 B a d e \right )^{2}}\, x \right ) \sqrt {-a c \left (A a \,e^{2}-A c \,d^{2}+2 B a d e \right )^{2}}}{2 c^{2} a}\) | \(608\) |
Input:
int((B*x+A)*(e*x+d)^2/(c*x^2+a),x,method=_RETURNVERBOSE)
Output:
e/c*(1/2*B*e*x^2+A*e*x+2*B*d*x)+1/c*(1/2*(2*A*c*d*e-B*a*e^2+B*c*d^2)/c*ln( c*x^2+a)+(-A*a*e^2+A*c*d^2-2*B*a*d*e)/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2)))
Time = 0.08 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.18 \[ \int \frac {(A+B x) (d+e x)^2}{a+c x^2} \, dx=\left [\frac {B a c e^{2} x^{2} + {\left (A c d^{2} - 2 \, B a d e - A a e^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 2 \, {\left (2 \, B a c d e + A a c e^{2}\right )} x + {\left (B a c d^{2} + 2 \, A a c d e - B a^{2} e^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, a c^{2}}, \frac {B a c e^{2} x^{2} + 2 \, {\left (A c d^{2} - 2 \, B a d e - A a e^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + 2 \, {\left (2 \, B a c d e + A a c e^{2}\right )} x + {\left (B a c d^{2} + 2 \, A a c d e - B a^{2} e^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, a c^{2}}\right ] \] Input:
integrate((B*x+A)*(e*x+d)^2/(c*x^2+a),x, algorithm="fricas")
Output:
[1/2*(B*a*c*e^2*x^2 + (A*c*d^2 - 2*B*a*d*e - A*a*e^2)*sqrt(-a*c)*log((c*x^ 2 + 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 2*(2*B*a*c*d*e + A*a*c*e^2)*x + (B* a*c*d^2 + 2*A*a*c*d*e - B*a^2*e^2)*log(c*x^2 + a))/(a*c^2), 1/2*(B*a*c*e^2 *x^2 + 2*(A*c*d^2 - 2*B*a*d*e - A*a*e^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + 2*(2*B*a*c*d*e + A*a*c*e^2)*x + (B*a*c*d^2 + 2*A*a*c*d*e - B*a^2*e^2)*log (c*x^2 + a))/(a*c^2)]
Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (105) = 210\).
Time = 0.73 (sec) , antiderivative size = 425, normalized size of antiderivative = 3.94 \[ \int \frac {(A+B x) (d+e x)^2}{a+c x^2} \, dx=\frac {B e^{2} x^{2}}{2 c} + x \left (\frac {A e^{2}}{c} + \frac {2 B d e}{c}\right ) + \left (- \frac {- 2 A c d e + B a e^{2} - B c d^{2}}{2 c^{2}} - \frac {\sqrt {- a c^{5}} \left (A a e^{2} - A c d^{2} + 2 B a d e\right )}{2 a c^{4}}\right ) \log {\left (x + \frac {2 A a c d e - B a^{2} e^{2} + B a c d^{2} - 2 a c^{2} \left (- \frac {- 2 A c d e + B a e^{2} - B c d^{2}}{2 c^{2}} - \frac {\sqrt {- a c^{5}} \left (A a e^{2} - A c d^{2} + 2 B a d e\right )}{2 a c^{4}}\right )}{A a c e^{2} - A c^{2} d^{2} + 2 B a c d e} \right )} + \left (- \frac {- 2 A c d e + B a e^{2} - B c d^{2}}{2 c^{2}} + \frac {\sqrt {- a c^{5}} \left (A a e^{2} - A c d^{2} + 2 B a d e\right )}{2 a c^{4}}\right ) \log {\left (x + \frac {2 A a c d e - B a^{2} e^{2} + B a c d^{2} - 2 a c^{2} \left (- \frac {- 2 A c d e + B a e^{2} - B c d^{2}}{2 c^{2}} + \frac {\sqrt {- a c^{5}} \left (A a e^{2} - A c d^{2} + 2 B a d e\right )}{2 a c^{4}}\right )}{A a c e^{2} - A c^{2} d^{2} + 2 B a c d e} \right )} \] Input:
integrate((B*x+A)*(e*x+d)**2/(c*x**2+a),x)
Output:
B*e**2*x**2/(2*c) + x*(A*e**2/c + 2*B*d*e/c) + (-(-2*A*c*d*e + B*a*e**2 - B*c*d**2)/(2*c**2) - sqrt(-a*c**5)*(A*a*e**2 - A*c*d**2 + 2*B*a*d*e)/(2*a* c**4))*log(x + (2*A*a*c*d*e - B*a**2*e**2 + B*a*c*d**2 - 2*a*c**2*(-(-2*A* c*d*e + B*a*e**2 - B*c*d**2)/(2*c**2) - sqrt(-a*c**5)*(A*a*e**2 - A*c*d**2 + 2*B*a*d*e)/(2*a*c**4)))/(A*a*c*e**2 - A*c**2*d**2 + 2*B*a*c*d*e)) + (-( -2*A*c*d*e + B*a*e**2 - B*c*d**2)/(2*c**2) + sqrt(-a*c**5)*(A*a*e**2 - A*c *d**2 + 2*B*a*d*e)/(2*a*c**4))*log(x + (2*A*a*c*d*e - B*a**2*e**2 + B*a*c* d**2 - 2*a*c**2*(-(-2*A*c*d*e + B*a*e**2 - B*c*d**2)/(2*c**2) + sqrt(-a*c* *5)*(A*a*e**2 - A*c*d**2 + 2*B*a*d*e)/(2*a*c**4)))/(A*a*c*e**2 - A*c**2*d* *2 + 2*B*a*c*d*e))
Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) (d+e x)^2}{a+c x^2} \, dx=\frac {{\left (A c d^{2} - 2 \, B a d e - A a e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c} + \frac {B e^{2} x^{2} + 2 \, {\left (2 \, B d e + A e^{2}\right )} x}{2 \, c} + \frac {{\left (B c d^{2} + 2 \, A c d e - B a e^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} \] Input:
integrate((B*x+A)*(e*x+d)^2/(c*x^2+a),x, algorithm="maxima")
Output:
(A*c*d^2 - 2*B*a*d*e - A*a*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*c) + 1/2* (B*e^2*x^2 + 2*(2*B*d*e + A*e^2)*x)/c + 1/2*(B*c*d^2 + 2*A*c*d*e - B*a*e^2 )*log(c*x^2 + a)/c^2
Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) (d+e x)^2}{a+c x^2} \, dx=\frac {{\left (A c d^{2} - 2 \, B a d e - A a e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c} + \frac {{\left (B c d^{2} + 2 \, A c d e - B a e^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {B c e^{2} x^{2} + 4 \, B c d e x + 2 \, A c e^{2} x}{2 \, c^{2}} \] Input:
integrate((B*x+A)*(e*x+d)^2/(c*x^2+a),x, algorithm="giac")
Output:
(A*c*d^2 - 2*B*a*d*e - A*a*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*c) + 1/2* (B*c*d^2 + 2*A*c*d*e - B*a*e^2)*log(c*x^2 + a)/c^2 + 1/2*(B*c*e^2*x^2 + 4* B*c*d*e*x + 2*A*c*e^2*x)/c^2
Time = 6.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.06 \[ \int \frac {(A+B x) (d+e x)^2}{a+c x^2} \, dx=\frac {x\,\left (A\,e^2+2\,B\,d\,e\right )}{c}+\frac {\ln \left (c\,x^2+a\right )\,\left (-4\,B\,a^2\,c^2\,e^2+4\,B\,a\,c^3\,d^2+8\,A\,a\,c^3\,d\,e\right )}{8\,a\,c^4}-\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (-A\,c\,d^2+2\,B\,a\,d\,e+A\,a\,e^2\right )}{\sqrt {a}\,c^{3/2}}+\frac {B\,e^2\,x^2}{2\,c} \] Input:
int(((A + B*x)*(d + e*x)^2)/(a + c*x^2),x)
Output:
(x*(A*e^2 + 2*B*d*e))/c + (log(a + c*x^2)*(4*B*a*c^3*d^2 - 4*B*a^2*c^2*e^2 + 8*A*a*c^3*d*e))/(8*a*c^4) - (atan((c^(1/2)*x)/a^(1/2))*(A*a*e^2 - A*c*d ^2 + 2*B*a*d*e))/(a^(1/2)*c^(3/2)) + (B*e^2*x^2)/(2*c)
Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.28 \[ \int \frac {(A+B x) (d+e x)^2}{a+c x^2} \, dx=\frac {-2 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a \,e^{2}-4 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) b d e +2 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) c \,d^{2}-\mathrm {log}\left (c \,x^{2}+a \right ) a b \,e^{2}+2 \,\mathrm {log}\left (c \,x^{2}+a \right ) a c d e +\mathrm {log}\left (c \,x^{2}+a \right ) b c \,d^{2}+2 a c \,e^{2} x +4 b c d e x +b c \,e^{2} x^{2}}{2 c^{2}} \] Input:
int((B*x+A)*(e*x+d)^2/(c*x^2+a),x)
Output:
( - 2*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*e**2 - 4*sqrt(c)*sqr t(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*b*d*e + 2*sqrt(c)*sqrt(a)*atan((c*x)/(s qrt(c)*sqrt(a)))*c*d**2 - log(a + c*x**2)*a*b*e**2 + 2*log(a + c*x**2)*a*c *d*e + log(a + c*x**2)*b*c*d**2 + 2*a*c*e**2*x + 4*b*c*d*e*x + b*c*e**2*x* *2)/(2*c**2)