Integrand size = 22, antiderivative size = 167 \[ \int \frac {(A+B x) (d+e x)^3}{a+c x^2} \, dx=\frac {e \left (3 B c d^2+3 A c d e-a B e^2\right ) x}{c^2}+\frac {e^2 (3 B d+A e) x^2}{2 c}+\frac {B e^3 x^3}{3 c}+\frac {\left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} c^{5/2}}+\frac {\left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) \log \left (a+c x^2\right )}{2 c^2} \] Output:
e*(3*A*c*d*e-B*a*e^2+3*B*c*d^2)*x/c^2+1/2*e^2*(A*e+3*B*d)*x^2/c+1/3*B*e^3* x^3/c+(A*c*d*(-3*a*e^2+c*d^2)-a*B*e*(-a*e^2+3*c*d^2))*arctan(c^(1/2)*x/a^( 1/2))/a^(1/2)/c^(5/2)+1/2*(-A*a*e^3+3*A*c*d^2*e-3*B*a*d*e^2+B*c*d^3)*ln(c* x^2+a)/c^2
Time = 0.14 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) (d+e x)^3}{a+c x^2} \, dx=\frac {\left (A c d \left (c d^2-3 a e^2\right )+a B e \left (-3 c d^2+a e^2\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} c^{5/2}}+\frac {e x \left (-6 a B e^2+3 A c e (6 d+e x)+B c \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+3 \left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) \log \left (a+c x^2\right )}{6 c^2} \] Input:
Integrate[((A + B*x)*(d + e*x)^3)/(a + c*x^2),x]
Output:
((A*c*d*(c*d^2 - 3*a*e^2) + a*B*e*(-3*c*d^2 + a*e^2))*ArcTan[(Sqrt[c]*x)/S qrt[a]])/(Sqrt[a]*c^(5/2)) + (e*x*(-6*a*B*e^2 + 3*A*c*e*(6*d + e*x) + B*c* (18*d^2 + 9*d*e*x + 2*e^2*x^2)) + 3*(B*c*d^3 + 3*A*c*d^2*e - 3*a*B*d*e^2 - a*A*e^3)*Log[a + c*x^2])/(6*c^2)
Time = 0.35 (sec) , antiderivative size = 167, 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.
| \(\displaystyle \int \frac {(A+B x) (d+e x)^3}{a+c x^2} \, dx\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \int \left (\frac {e \left (-a B e^2+3 A c d e+3 B c d^2\right )}{c^2}+\frac {c x \left (-a A e^3-3 a B d e^2+3 A c d^2 e+B c d^3\right )+A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )}{c^2 \left (a+c x^2\right )}+\frac {e^2 x (A e+3 B d)}{c}+\frac {B e^3 x^2}{c}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt {a} c^{5/2}}+\frac {e x \left (-a B e^2+3 A c d e+3 B c d^2\right )}{c^2}+\frac {\log \left (a+c x^2\right ) \left (-a A e^3-3 a B d e^2+3 A c d^2 e+B c d^3\right )}{2 c^2}+\frac {e^2 x^2 (A e+3 B d)}{2 c}+\frac {B e^3 x^3}{3 c}\) |
Input:
Int[((A + B*x)*(d + e*x)^3)/(a + c*x^2),x]
Output:
(e*(3*B*c*d^2 + 3*A*c*d*e - a*B*e^2)*x)/c^2 + (e^2*(3*B*d + A*e)*x^2)/(2*c ) + (B*e^3*x^3)/(3*c) + ((A*c*d*(c*d^2 - 3*a*e^2) - a*B*e*(3*c*d^2 - a*e^2 ))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(5/2)) + ((B*c*d^3 + 3*A*c*d^2* e - 3*a*B*d*e^2 - a*A*e^3)*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 = 1.07 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.98
| method | result | size |
| default | \(\frac {e \left (\frac {1}{3} B c \,x^{3} e^{2}+\frac {1}{2} A c \,e^{2} x^{2}+\frac {3}{2} B c d e \,x^{2}+3 A c d e x -B a \,e^{2} x +3 B c \,d^{2} x \right )}{c^{2}}+\frac {\frac {\left (-A a c \,e^{3}+3 A \,c^{2} d^{2} e -3 B a c d \,e^{2}+B \,c^{2} d^{3}\right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (-3 A a c d \,e^{2}+A \,c^{2} d^{3}+B \,e^{3} a^{2}-3 B a c \,d^{2} e \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}}{c^{2}}\) | \(164\) |
| risch | \(\text {Expression too large to display}\) | \(1143\) |
Input:
int((B*x+A)*(e*x+d)^3/(c*x^2+a),x,method=_RETURNVERBOSE)
Output:
e/c^2*(1/3*B*c*x^3*e^2+1/2*A*c*e^2*x^2+3/2*B*c*d*e*x^2+3*A*c*d*e*x-B*a*e^2 *x+3*B*c*d^2*x)+1/c^2*(1/2*(-A*a*c*e^3+3*A*c^2*d^2*e-3*B*a*c*d*e^2+B*c^2*d ^3)/c*ln(c*x^2+a)+(-3*A*a*c*d*e^2+A*c^2*d^3+B*a^2*e^3-3*B*a*c*d^2*e)/(a*c) ^(1/2)*arctan(c*x/(a*c)^(1/2)))
Time = 0.13 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.38 \[ \int \frac {(A+B x) (d+e x)^3}{a+c x^2} \, dx=\left [\frac {2 \, B a c^{2} e^{3} x^{3} + 3 \, {\left (3 \, B a c^{2} d e^{2} + A a c^{2} e^{3}\right )} x^{2} - 3 \, {\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 6 \, {\left (3 \, B a c^{2} d^{2} e + 3 \, A a c^{2} d e^{2} - B a^{2} c e^{3}\right )} x + 3 \, {\left (B a c^{2} d^{3} + 3 \, A a c^{2} d^{2} e - 3 \, B a^{2} c d e^{2} - A a^{2} c e^{3}\right )} \log \left (c x^{2} + a\right )}{6 \, a c^{3}}, \frac {2 \, B a c^{2} e^{3} x^{3} + 3 \, {\left (3 \, B a c^{2} d e^{2} + A a c^{2} e^{3}\right )} x^{2} + 6 \, {\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + 6 \, {\left (3 \, B a c^{2} d^{2} e + 3 \, A a c^{2} d e^{2} - B a^{2} c e^{3}\right )} x + 3 \, {\left (B a c^{2} d^{3} + 3 \, A a c^{2} d^{2} e - 3 \, B a^{2} c d e^{2} - A a^{2} c e^{3}\right )} \log \left (c x^{2} + a\right )}{6 \, a c^{3}}\right ] \] Input:
integrate((B*x+A)*(e*x+d)^3/(c*x^2+a),x, algorithm="fricas")
Output:
[1/6*(2*B*a*c^2*e^3*x^3 + 3*(3*B*a*c^2*d*e^2 + A*a*c^2*e^3)*x^2 - 3*(A*c^2 *d^3 - 3*B*a*c*d^2*e - 3*A*a*c*d*e^2 + B*a^2*e^3)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 6*(3*B*a*c^2*d^2*e + 3*A*a*c^2*d*e^2 - B*a^2*c*e^3)*x + 3*(B*a*c^2*d^3 + 3*A*a*c^2*d^2*e - 3*B*a^2*c*d*e^2 - A*a^ 2*c*e^3)*log(c*x^2 + a))/(a*c^3), 1/6*(2*B*a*c^2*e^3*x^3 + 3*(3*B*a*c^2*d* e^2 + A*a*c^2*e^3)*x^2 + 6*(A*c^2*d^3 - 3*B*a*c*d^2*e - 3*A*a*c*d*e^2 + B* a^2*e^3)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + 6*(3*B*a*c^2*d^2*e + 3*A*a*c^2* d*e^2 - B*a^2*c*e^3)*x + 3*(B*a*c^2*d^3 + 3*A*a*c^2*d^2*e - 3*B*a^2*c*d*e^ 2 - A*a^2*c*e^3)*log(c*x^2 + a))/(a*c^3)]
Leaf count of result is larger than twice the leaf count of optimal. 641 vs. \(2 (170) = 340\).
Time = 1.08 (sec) , antiderivative size = 641, normalized size of antiderivative = 3.84 \[ \int \frac {(A+B x) (d+e x)^3}{a+c x^2} \, dx=\frac {B e^{3} x^{3}}{3 c} + x^{2} \left (\frac {A e^{3}}{2 c} + \frac {3 B d e^{2}}{2 c}\right ) + x \left (\frac {3 A d e^{2}}{c} - \frac {B a e^{3}}{c^{2}} + \frac {3 B d^{2} e}{c}\right ) + \left (- \frac {A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} - B c d^{3}}{2 c^{2}} - \frac {\sqrt {- a c^{5}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e\right )}{2 a c^{5}}\right ) \log {\left (x + \frac {A a^{2} e^{3} - 3 A a c d^{2} e + 3 B a^{2} d e^{2} - B a c d^{3} + 2 a c^{2} \left (- \frac {A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} - B c d^{3}}{2 c^{2}} - \frac {\sqrt {- a c^{5}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e\right )}{2 a c^{5}}\right )}{- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e} \right )} + \left (- \frac {A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} - B c d^{3}}{2 c^{2}} + \frac {\sqrt {- a c^{5}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e\right )}{2 a c^{5}}\right ) \log {\left (x + \frac {A a^{2} e^{3} - 3 A a c d^{2} e + 3 B a^{2} d e^{2} - B a c d^{3} + 2 a c^{2} \left (- \frac {A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} - B c d^{3}}{2 c^{2}} + \frac {\sqrt {- a c^{5}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e\right )}{2 a c^{5}}\right )}{- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e} \right )} \] Input:
integrate((B*x+A)*(e*x+d)**3/(c*x**2+a),x)
Output:
B*e**3*x**3/(3*c) + x**2*(A*e**3/(2*c) + 3*B*d*e**2/(2*c)) + x*(3*A*d*e**2 /c - B*a*e**3/c**2 + 3*B*d**2*e/c) + (-(A*a*e**3 - 3*A*c*d**2*e + 3*B*a*d* e**2 - B*c*d**3)/(2*c**2) - sqrt(-a*c**5)*(-3*A*a*c*d*e**2 + A*c**2*d**3 + B*a**2*e**3 - 3*B*a*c*d**2*e)/(2*a*c**5))*log(x + (A*a**2*e**3 - 3*A*a*c* d**2*e + 3*B*a**2*d*e**2 - B*a*c*d**3 + 2*a*c**2*(-(A*a*e**3 - 3*A*c*d**2* e + 3*B*a*d*e**2 - B*c*d**3)/(2*c**2) - sqrt(-a*c**5)*(-3*A*a*c*d*e**2 + A *c**2*d**3 + B*a**2*e**3 - 3*B*a*c*d**2*e)/(2*a*c**5)))/(-3*A*a*c*d*e**2 + A*c**2*d**3 + B*a**2*e**3 - 3*B*a*c*d**2*e)) + (-(A*a*e**3 - 3*A*c*d**2*e + 3*B*a*d*e**2 - B*c*d**3)/(2*c**2) + sqrt(-a*c**5)*(-3*A*a*c*d*e**2 + A* c**2*d**3 + B*a**2*e**3 - 3*B*a*c*d**2*e)/(2*a*c**5))*log(x + (A*a**2*e**3 - 3*A*a*c*d**2*e + 3*B*a**2*d*e**2 - B*a*c*d**3 + 2*a*c**2*(-(A*a*e**3 - 3*A*c*d**2*e + 3*B*a*d*e**2 - B*c*d**3)/(2*c**2) + sqrt(-a*c**5)*(-3*A*a*c *d*e**2 + A*c**2*d**3 + B*a**2*e**3 - 3*B*a*c*d**2*e)/(2*a*c**5)))/(-3*A*a *c*d*e**2 + A*c**2*d**3 + B*a**2*e**3 - 3*B*a*c*d**2*e))
Time = 0.13 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) (d+e x)^3}{a+c x^2} \, dx=\frac {{\left (B c d^{3} + 3 \, A c d^{2} e - 3 \, B a d e^{2} - A a e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {{\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {2 \, B c e^{3} x^{3} + 3 \, {\left (3 \, B c d e^{2} + A c e^{3}\right )} x^{2} + 6 \, {\left (3 \, B c d^{2} e + 3 \, A c d e^{2} - B a e^{3}\right )} x}{6 \, c^{2}} \] Input:
integrate((B*x+A)*(e*x+d)^3/(c*x^2+a),x, algorithm="maxima")
Output:
1/2*(B*c*d^3 + 3*A*c*d^2*e - 3*B*a*d*e^2 - A*a*e^3)*log(c*x^2 + a)/c^2 + ( A*c^2*d^3 - 3*B*a*c*d^2*e - 3*A*a*c*d*e^2 + B*a^2*e^3)*arctan(c*x/sqrt(a*c ))/(sqrt(a*c)*c^2) + 1/6*(2*B*c*e^3*x^3 + 3*(3*B*c*d*e^2 + A*c*e^3)*x^2 + 6*(3*B*c*d^2*e + 3*A*c*d*e^2 - B*a*e^3)*x)/c^2
Time = 0.11 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.02 \[ \int \frac {(A+B x) (d+e x)^3}{a+c x^2} \, dx=\frac {{\left (B c d^{3} + 3 \, A c d^{2} e - 3 \, B a d e^{2} - A a e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {{\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {2 \, B c^{2} e^{3} x^{3} + 9 \, B c^{2} d e^{2} x^{2} + 3 \, A c^{2} e^{3} x^{2} + 18 \, B c^{2} d^{2} e x + 18 \, A c^{2} d e^{2} x - 6 \, B a c e^{3} x}{6 \, c^{3}} \] Input:
integrate((B*x+A)*(e*x+d)^3/(c*x^2+a),x, algorithm="giac")
Output:
1/2*(B*c*d^3 + 3*A*c*d^2*e - 3*B*a*d*e^2 - A*a*e^3)*log(c*x^2 + a)/c^2 + ( A*c^2*d^3 - 3*B*a*c*d^2*e - 3*A*a*c*d*e^2 + B*a^2*e^3)*arctan(c*x/sqrt(a*c ))/(sqrt(a*c)*c^2) + 1/6*(2*B*c^2*e^3*x^3 + 9*B*c^2*d*e^2*x^2 + 3*A*c^2*e^ 3*x^2 + 18*B*c^2*d^2*e*x + 18*A*c^2*d*e^2*x - 6*B*a*c*e^3*x)/c^3
Time = 0.11 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) (d+e x)^3}{a+c x^2} \, dx=x\,\left (\frac {3\,d\,e\,\left (A\,e+B\,d\right )}{c}-\frac {B\,a\,e^3}{c^2}\right )+\frac {x^2\,\left (A\,e^3+3\,B\,d\,e^2\right )}{2\,c}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (B\,a^2\,e^3-3\,B\,a\,c\,d^2\,e-3\,A\,a\,c\,d\,e^2+A\,c^2\,d^3\right )}{\sqrt {a}\,c^{5/2}}+\frac {\ln \left (c\,x^2+a\right )\,\left (-12\,B\,a^2\,c^3\,d\,e^2-4\,A\,a^2\,c^3\,e^3+4\,B\,a\,c^4\,d^3+12\,A\,a\,c^4\,d^2\,e\right )}{8\,a\,c^5}+\frac {B\,e^3\,x^3}{3\,c} \] Input:
int(((A + B*x)*(d + e*x)^3)/(a + c*x^2),x)
Output:
x*((3*d*e*(A*e + B*d))/c - (B*a*e^3)/c^2) + (x^2*(A*e^3 + 3*B*d*e^2))/(2*c ) + (atan((c^(1/2)*x)/a^(1/2))*(A*c^2*d^3 + B*a^2*e^3 - 3*A*a*c*d*e^2 - 3* B*a*c*d^2*e))/(a^(1/2)*c^(5/2)) + (log(a + c*x^2)*(4*B*a*c^4*d^3 - 4*A*a^2 *c^3*e^3 - 12*B*a^2*c^3*d*e^2 + 12*A*a*c^4*d^2*e))/(8*a*c^5) + (B*e^3*x^3) /(3*c)
Time = 0.27 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.43 \[ \int \frac {(A+B x) (d+e x)^3}{a+c x^2} \, dx=\frac {6 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a b \,e^{3}-18 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) a c d \,e^{2}-18 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) b c \,d^{2} e +6 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {a}}\right ) c^{2} d^{3}-3 \,\mathrm {log}\left (c \,x^{2}+a \right ) a^{2} c \,e^{3}-9 \,\mathrm {log}\left (c \,x^{2}+a \right ) a b c d \,e^{2}+9 \,\mathrm {log}\left (c \,x^{2}+a \right ) a \,c^{2} d^{2} e +3 \,\mathrm {log}\left (c \,x^{2}+a \right ) b \,c^{2} d^{3}-6 a b c \,e^{3} x +18 a \,c^{2} d \,e^{2} x +3 a \,c^{2} e^{3} x^{2}+18 b \,c^{2} d^{2} e x +9 b \,c^{2} d \,e^{2} x^{2}+2 b \,c^{2} e^{3} x^{3}}{6 c^{3}} \] Input:
int((B*x+A)*(e*x+d)^3/(c*x^2+a),x)
Output:
(6*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*b*e**3 - 18*sqrt(c)*sqr t(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a*c*d*e**2 - 18*sqrt(c)*sqrt(a)*atan((c *x)/(sqrt(c)*sqrt(a)))*b*c*d**2*e + 6*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)* sqrt(a)))*c**2*d**3 - 3*log(a + c*x**2)*a**2*c*e**3 - 9*log(a + c*x**2)*a* b*c*d*e**2 + 9*log(a + c*x**2)*a*c**2*d**2*e + 3*log(a + c*x**2)*b*c**2*d* *3 - 6*a*b*c*e**3*x + 18*a*c**2*d*e**2*x + 3*a*c**2*e**3*x**2 + 18*b*c**2* d**2*e*x + 9*b*c**2*d*e**2*x**2 + 2*b*c**2*e**3*x**3)/(6*c**3)