Integrand size = 20, antiderivative size = 109 \[ \int \frac {(A+B x) (d+e x)}{\left (a+c x^2\right )^3} \, dx=-\frac {(A+B x) (a e-c d x)}{4 a c \left (a+c x^2\right )^2}-\frac {2 a B d-(3 A c d+a B e) x}{8 a^2 c \left (a+c x^2\right )}+\frac {(3 A c d+a B e) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{3/2}} \] Output:
-1/4*(B*x+A)*(-c*d*x+a*e)/a/c/(c*x^2+a)^2-1/8*(2*a*B*d-(3*A*c*d+B*a*e)*x)/ a^2/c/(c*x^2+a)+1/8*(3*A*c*d+B*a*e)*arctan(c^(1/2)*x/a^(1/2))/a^(5/2)/c^(3 /2)
Time = 0.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) (d+e x)}{\left (a+c x^2\right )^3} \, dx=\frac {3 A c^2 d x^3-a^2 (2 B d+2 A e+B e x)+a c x \left (5 A d+B e x^2\right )}{8 a^2 c \left (a+c x^2\right )^2}+\frac {(3 A c d+a B e) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{3/2}} \] Input:
Integrate[((A + B*x)*(d + e*x))/(a + c*x^2)^3,x]
Output:
(3*A*c^2*d*x^3 - a^2*(2*B*d + 2*A*e + B*e*x) + a*c*x*(5*A*d + B*e*x^2))/(8 *a^2*c*(a + c*x^2)^2) + ((3*A*c*d + a*B*e)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8 *a^(5/2)*c^(3/2))
Time = 0.22 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {675, 215, 218}
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.
\(\displaystyle \int \frac {(A+B x) (d+e x)}{\left (a+c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 675 |
\(\displaystyle \frac {(a B e+3 A c d) \int \frac {1}{\left (c x^2+a\right )^2}dx}{4 a c}-\frac {A e+B d}{4 c \left (a+c x^2\right )^2}+\frac {x (A c d-a B e)}{4 a c \left (a+c x^2\right )^2}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {(a B e+3 A c d) \left (\frac {\int \frac {1}{c x^2+a}dx}{2 a}+\frac {x}{2 a \left (a+c x^2\right )}\right )}{4 a c}-\frac {A e+B d}{4 c \left (a+c x^2\right )^2}+\frac {x (A c d-a B e)}{4 a c \left (a+c x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\left (\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {c}}+\frac {x}{2 a \left (a+c x^2\right )}\right ) (a B e+3 A c d)}{4 a c}-\frac {A e+B d}{4 c \left (a+c x^2\right )^2}+\frac {x (A c d-a B e)}{4 a c \left (a+c x^2\right )^2}\) |
Input:
Int[((A + B*x)*(d + e*x))/(a + c*x^2)^3,x]
Output:
-1/4*(B*d + A*e)/(c*(a + c*x^2)^2) + ((A*c*d - a*B*e)*x)/(4*a*c*(a + c*x^2 )^2) + ((3*A*c*d + a*B*e)*(x/(2*a*(a + c*x^2)) + ArcTan[(Sqrt[c]*x)/Sqrt[a ]]/(2*a^(3/2)*Sqrt[c])))/(4*a*c)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((d_) + (e_.)*(x_))*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[a*(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + (- Simp[(c*d*f - a*e*g)*x*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Simp[(a* e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)) Int[(a + c*x^2)^(p + 1), x], x]) / ; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1] && !(IntegerQ[p] && NiceSqrtQ [(-a)*c])
Time = 0.78 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.87
method | result | size |
default | \(\frac {\frac {\left (3 A c d +B a e \right ) x^{3}}{8 a^{2}}+\frac {\left (5 A c d -B a e \right ) x}{8 a c}-\frac {A e +B d}{4 c}}{\left (c \,x^{2}+a \right )^{2}}+\frac {\left (3 A c d +B a e \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 a^{2} c \sqrt {a c}}\) | \(95\) |
risch | \(\frac {\frac {\left (3 A c d +B a e \right ) x^{3}}{8 a^{2}}+\frac {\left (5 A c d -B a e \right ) x}{8 a c}-\frac {A e +B d}{4 c}}{\left (c \,x^{2}+a \right )^{2}}-\frac {3 \ln \left (c x +\sqrt {-a c}\right ) A d}{16 \sqrt {-a c}\, a^{2}}-\frac {\ln \left (c x +\sqrt {-a c}\right ) B e}{16 \sqrt {-a c}\, c a}+\frac {3 \ln \left (-c x +\sqrt {-a c}\right ) A d}{16 \sqrt {-a c}\, a^{2}}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) B e}{16 \sqrt {-a c}\, c a}\) | \(167\) |
Input:
int((B*x+A)*(e*x+d)/(c*x^2+a)^3,x,method=_RETURNVERBOSE)
Output:
(1/8*(3*A*c*d+B*a*e)/a^2*x^3+1/8*(5*A*c*d-B*a*e)/a/c*x-1/4*(A*e+B*d)/c)/(c *x^2+a)^2+1/8*(3*A*c*d+B*a*e)/a^2/c/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2))
Time = 0.09 (sec) , antiderivative size = 355, normalized size of antiderivative = 3.26 \[ \int \frac {(A+B x) (d+e x)}{\left (a+c x^2\right )^3} \, dx=\left [-\frac {4 \, B a^{3} c d + 4 \, A a^{3} c e - 2 \, {\left (3 \, A a c^{3} d + B a^{2} c^{2} e\right )} x^{3} + {\left (3 \, A a^{2} c d + B a^{3} e + {\left (3 \, A c^{3} d + B a c^{2} e\right )} x^{4} + 2 \, {\left (3 \, A a c^{2} d + B a^{2} c e\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 2 \, {\left (5 \, A a^{2} c^{2} d - B a^{3} c e\right )} x}{16 \, {\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}, -\frac {2 \, B a^{3} c d + 2 \, A a^{3} c e - {\left (3 \, A a c^{3} d + B a^{2} c^{2} e\right )} x^{3} - {\left (3 \, A a^{2} c d + B a^{3} e + {\left (3 \, A c^{3} d + B a c^{2} e\right )} x^{4} + 2 \, {\left (3 \, A a c^{2} d + B a^{2} c e\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (5 \, A a^{2} c^{2} d - B a^{3} c e\right )} x}{8 \, {\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}\right ] \] Input:
integrate((B*x+A)*(e*x+d)/(c*x^2+a)^3,x, algorithm="fricas")
Output:
[-1/16*(4*B*a^3*c*d + 4*A*a^3*c*e - 2*(3*A*a*c^3*d + B*a^2*c^2*e)*x^3 + (3 *A*a^2*c*d + B*a^3*e + (3*A*c^3*d + B*a*c^2*e)*x^4 + 2*(3*A*a*c^2*d + B*a^ 2*c*e)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) - 2*( 5*A*a^2*c^2*d - B*a^3*c*e)*x)/(a^3*c^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2), -1/ 8*(2*B*a^3*c*d + 2*A*a^3*c*e - (3*A*a*c^3*d + B*a^2*c^2*e)*x^3 - (3*A*a^2* c*d + B*a^3*e + (3*A*c^3*d + B*a*c^2*e)*x^4 + 2*(3*A*a*c^2*d + B*a^2*c*e)* x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - (5*A*a^2*c^2*d - B*a^3*c*e)*x)/(a^3 *c^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2)]
Time = 0.92 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.65 \[ \int \frac {(A+B x) (d+e x)}{\left (a+c x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{a^{5} c^{3}}} \cdot \left (3 A c d + B a e\right ) \log {\left (- a^{3} c \sqrt {- \frac {1}{a^{5} c^{3}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{5} c^{3}}} \cdot \left (3 A c d + B a e\right ) \log {\left (a^{3} c \sqrt {- \frac {1}{a^{5} c^{3}}} + x \right )}}{16} + \frac {- 2 A a^{2} e - 2 B a^{2} d + x^{3} \cdot \left (3 A c^{2} d + B a c e\right ) + x \left (5 A a c d - B a^{2} e\right )}{8 a^{4} c + 16 a^{3} c^{2} x^{2} + 8 a^{2} c^{3} x^{4}} \] Input:
integrate((B*x+A)*(e*x+d)/(c*x**2+a)**3,x)
Output:
-sqrt(-1/(a**5*c**3))*(3*A*c*d + B*a*e)*log(-a**3*c*sqrt(-1/(a**5*c**3)) + x)/16 + sqrt(-1/(a**5*c**3))*(3*A*c*d + B*a*e)*log(a**3*c*sqrt(-1/(a**5*c **3)) + x)/16 + (-2*A*a**2*e - 2*B*a**2*d + x**3*(3*A*c**2*d + B*a*c*e) + x*(5*A*a*c*d - B*a**2*e))/(8*a**4*c + 16*a**3*c**2*x**2 + 8*a**2*c**3*x**4 )
Time = 0.13 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) (d+e x)}{\left (a+c x^2\right )^3} \, dx=-\frac {2 \, B a^{2} d + 2 \, A a^{2} e - {\left (3 \, A c^{2} d + B a c e\right )} x^{3} - {\left (5 \, A a c d - B a^{2} e\right )} x}{8 \, {\left (a^{2} c^{3} x^{4} + 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )}} + \frac {{\left (3 \, A c d + B a e\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c} \] Input:
integrate((B*x+A)*(e*x+d)/(c*x^2+a)^3,x, algorithm="maxima")
Output:
-1/8*(2*B*a^2*d + 2*A*a^2*e - (3*A*c^2*d + B*a*c*e)*x^3 - (5*A*a*c*d - B*a ^2*e)*x)/(a^2*c^3*x^4 + 2*a^3*c^2*x^2 + a^4*c) + 1/8*(3*A*c*d + B*a*e)*arc tan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c)
Time = 0.12 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) (d+e x)}{\left (a+c x^2\right )^3} \, dx=\frac {{\left (3 \, A c d + B a e\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c} + \frac {3 \, A c^{2} d x^{3} + B a c e x^{3} + 5 \, A a c d x - B a^{2} e x - 2 \, B a^{2} d - 2 \, A a^{2} e}{8 \, {\left (c x^{2} + a\right )}^{2} a^{2} c} \] Input:
integrate((B*x+A)*(e*x+d)/(c*x^2+a)^3,x, algorithm="giac")
Output:
1/8*(3*A*c*d + B*a*e)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c) + 1/8*(3*A*c ^2*d*x^3 + B*a*c*e*x^3 + 5*A*a*c*d*x - B*a^2*e*x - 2*B*a^2*d - 2*A*a^2*e)/ ((c*x^2 + a)^2*a^2*c)
Time = 6.69 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) (d+e x)}{\left (a+c x^2\right )^3} \, dx=\frac {\frac {x^3\,\left (3\,A\,c\,d+B\,a\,e\right )}{8\,a^2}-\frac {A\,e+B\,d}{4\,c}+\frac {x\,\left (5\,A\,c\,d-B\,a\,e\right )}{8\,a\,c}}{a^2+2\,a\,c\,x^2+c^2\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (3\,A\,c\,d+B\,a\,e\right )}{8\,a^{5/2}\,c^{3/2}} \] Input:
int(((A + B*x)*(d + e*x))/(a + c*x^2)^3,x)
Output:
((x^3*(3*A*c*d + B*a*e))/(8*a^2) - (A*e + B*d)/(4*c) + (x*(5*A*c*d - B*a*e ))/(8*a*c))/(a^2 + c^2*x^4 + 2*a*c*x^2) + (atan((c^(1/2)*x)/a^(1/2))*(3*A* c*d + B*a*e))/(8*a^(5/2)*c^(3/2))
Time = 0.23 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.10 \[ \int \frac {(A+B x) (d+e x)}{\left (a+c x^2\right )^3} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a^{2} b e +3 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a^{2} c d +2 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a b c e \,x^{2}+6 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a \,c^{2} d \,x^{2}+\sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) b \,c^{2} e \,x^{4}+3 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) c^{3} d \,x^{4}-2 a^{3} c e -2 a^{2} b c d -a^{2} b c e x +5 a^{2} c^{2} d x +a b \,c^{2} e \,x^{3}+3 a \,c^{3} d \,x^{3}}{8 a^{2} c^{2} \left (c^{2} x^{4}+2 a c \,x^{2}+a^{2}\right )} \] Input:
int((B*x+A)*(e*x+d)/(c*x^2+a)^3,x)
Output:
(sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**2*b*e + 3*sqrt(c)*sqrt(a )*atan((c*x)/(sqrt(c)*sqrt(a)))*a**2*c*d + 2*sqrt(c)*sqrt(a)*atan((c*x)/(s qrt(c)*sqrt(a)))*a*b*c*e*x**2 + 6*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt (a)))*a*c**2*d*x**2 + sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*b*c**2 *e*x**4 + 3*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*c**3*d*x**4 - 2* a**3*c*e - 2*a**2*b*c*d - a**2*b*c*e*x + 5*a**2*c**2*d*x + a*b*c**2*e*x**3 + 3*a*c**3*d*x**3)/(8*a**2*c**2*(a**2 + 2*a*c*x**2 + c**2*x**4))