Integrand size = 17, antiderivative size = 94 \[ \int \frac {\left (a+b x^2\right )^2}{c+d x} \, dx=-\frac {b c \left (b c^2+2 a d^2\right ) x}{d^4}+\frac {b \left (b c^2+2 a d^2\right ) x^2}{2 d^3}-\frac {b^2 c x^3}{3 d^2}+\frac {b^2 x^4}{4 d}+\frac {\left (b c^2+a d^2\right )^2 \log (c+d x)}{d^5} \] Output:
-b*c*(2*a*d^2+b*c^2)*x/d^4+1/2*b*(2*a*d^2+b*c^2)*x^2/d^3-1/3*b^2*c*x^3/d^2 +1/4*b^2*x^4/d+(a*d^2+b*c^2)^2*ln(d*x+c)/d^5
Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+b x^2\right )^2}{c+d x} \, dx=\frac {b d x \left (12 a d^2 (-2 c+d x)+b \left (-12 c^3+6 c^2 d x-4 c d^2 x^2+3 d^3 x^3\right )\right )+12 \left (b c^2+a d^2\right )^2 \log (c+d x)}{12 d^5} \] Input:
Integrate[(a + b*x^2)^2/(c + d*x),x]
Output:
(b*d*x*(12*a*d^2*(-2*c + d*x) + b*(-12*c^3 + 6*c^2*d*x - 4*c*d^2*x^2 + 3*d ^3*x^3)) + 12*(b*c^2 + a*d^2)^2*Log[c + d*x])/(12*d^5)
Time = 0.42 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {476, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2}{c+d x} \, dx\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \int \left (\frac {\left (a d^2+b c^2\right )^2}{d^4 (c+d x)}-\frac {b c \left (2 a d^2+b c^2\right )}{d^4}+\frac {b x \left (2 a d^2+b c^2\right )}{d^3}-\frac {b^2 c x^2}{d^2}+\frac {b^2 x^3}{d}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a d^2+b c^2\right )^2 \log (c+d x)}{d^5}-\frac {b c x \left (2 a d^2+b c^2\right )}{d^4}+\frac {b x^2 \left (2 a d^2+b c^2\right )}{2 d^3}-\frac {b^2 c x^3}{3 d^2}+\frac {b^2 x^4}{4 d}\) |
Input:
Int[(a + b*x^2)^2/(c + d*x),x]
Output:
-((b*c*(b*c^2 + 2*a*d^2)*x)/d^4) + (b*(b*c^2 + 2*a*d^2)*x^2)/(2*d^3) - (b^ 2*c*x^3)/(3*d^2) + (b^2*x^4)/(4*d) + ((b*c^2 + a*d^2)^2*Log[c + d*x])/d^5
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.02
method | result | size |
default | \(-\frac {b \left (-\frac {b \,d^{3} x^{4}}{4}+\frac {x^{3} b c \,d^{2}}{3}-\frac {\left (2 a \,d^{2}+b \,c^{2}\right ) x^{2} d}{2}+c \left (2 a \,d^{2}+b \,c^{2}\right ) x \right )}{d^{4}}+\frac {\left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right ) \ln \left (d x +c \right )}{d^{5}}\) | \(96\) |
norman | \(\frac {b^{2} x^{4}}{4 d}+\frac {b \left (2 a \,d^{2}+b \,c^{2}\right ) x^{2}}{2 d^{3}}-\frac {b^{2} c \,x^{3}}{3 d^{2}}-\frac {b c \left (2 a \,d^{2}+b \,c^{2}\right ) x}{d^{4}}+\frac {\left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right ) \ln \left (d x +c \right )}{d^{5}}\) | \(101\) |
risch | \(\frac {b^{2} x^{4}}{4 d}-\frac {b^{2} c \,x^{3}}{3 d^{2}}+\frac {b a \,x^{2}}{d}+\frac {b^{2} c^{2} x^{2}}{2 d^{3}}-\frac {2 b a c x}{d^{2}}-\frac {b^{2} c^{3} x}{d^{4}}+\frac {\ln \left (d x +c \right ) a^{2}}{d}+\frac {2 \ln \left (d x +c \right ) b \,c^{2} a}{d^{3}}+\frac {\ln \left (d x +c \right ) b^{2} c^{4}}{d^{5}}\) | \(114\) |
parallelrisch | \(\frac {3 b^{2} d^{4} x^{4}-4 b^{2} c \,d^{3} x^{3}+12 a b \,d^{4} x^{2}+6 d^{2} c^{2} x^{2} b^{2}+12 \ln \left (d x +c \right ) a^{2} d^{4}+24 \ln \left (d x +c \right ) a b \,c^{2} d^{2}+12 \ln \left (d x +c \right ) b^{2} c^{4}-24 a b c \,d^{3} x -12 b^{2} c^{3} d x}{12 d^{5}}\) | \(117\) |
Input:
int((b*x^2+a)^2/(d*x+c),x,method=_RETURNVERBOSE)
Output:
-b/d^4*(-1/4*b*d^3*x^4+1/3*x^3*b*c*d^2-1/2*(2*a*d^2+b*c^2)*x^2*d+c*(2*a*d^ 2+b*c^2)*x)+1/d^5*(a^2*d^4+2*a*b*c^2*d^2+b^2*c^4)*ln(d*x+c)
Time = 0.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right )^2}{c+d x} \, dx=\frac {3 \, b^{2} d^{4} x^{4} - 4 \, b^{2} c d^{3} x^{3} + 6 \, {\left (b^{2} c^{2} d^{2} + 2 \, a b d^{4}\right )} x^{2} - 12 \, {\left (b^{2} c^{3} d + 2 \, a b c d^{3}\right )} x + 12 \, {\left (b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}\right )} \log \left (d x + c\right )}{12 \, d^{5}} \] Input:
integrate((b*x^2+a)^2/(d*x+c),x, algorithm="fricas")
Output:
1/12*(3*b^2*d^4*x^4 - 4*b^2*c*d^3*x^3 + 6*(b^2*c^2*d^2 + 2*a*b*d^4)*x^2 - 12*(b^2*c^3*d + 2*a*b*c*d^3)*x + 12*(b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4)*lo g(d*x + c))/d^5
Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right )^2}{c+d x} \, dx=- \frac {b^{2} c x^{3}}{3 d^{2}} + \frac {b^{2} x^{4}}{4 d} + x^{2} \left (\frac {a b}{d} + \frac {b^{2} c^{2}}{2 d^{3}}\right ) + x \left (- \frac {2 a b c}{d^{2}} - \frac {b^{2} c^{3}}{d^{4}}\right ) + \frac {\left (a d^{2} + b c^{2}\right )^{2} \log {\left (c + d x \right )}}{d^{5}} \] Input:
integrate((b*x**2+a)**2/(d*x+c),x)
Output:
-b**2*c*x**3/(3*d**2) + b**2*x**4/(4*d) + x**2*(a*b/d + b**2*c**2/(2*d**3) ) + x*(-2*a*b*c/d**2 - b**2*c**3/d**4) + (a*d**2 + b*c**2)**2*log(c + d*x) /d**5
Time = 0.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right )^2}{c+d x} \, dx=\frac {3 \, b^{2} d^{3} x^{4} - 4 \, b^{2} c d^{2} x^{3} + 6 \, {\left (b^{2} c^{2} d + 2 \, a b d^{3}\right )} x^{2} - 12 \, {\left (b^{2} c^{3} + 2 \, a b c d^{2}\right )} x}{12 \, d^{4}} + \frac {{\left (b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}\right )} \log \left (d x + c\right )}{d^{5}} \] Input:
integrate((b*x^2+a)^2/(d*x+c),x, algorithm="maxima")
Output:
1/12*(3*b^2*d^3*x^4 - 4*b^2*c*d^2*x^3 + 6*(b^2*c^2*d + 2*a*b*d^3)*x^2 - 12 *(b^2*c^3 + 2*a*b*c*d^2)*x)/d^4 + (b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4)*log( d*x + c)/d^5
Time = 0.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+b x^2\right )^2}{c+d x} \, dx=\frac {3 \, b^{2} d^{3} x^{4} - 4 \, b^{2} c d^{2} x^{3} + 6 \, b^{2} c^{2} d x^{2} + 12 \, a b d^{3} x^{2} - 12 \, b^{2} c^{3} x - 24 \, a b c d^{2} x}{12 \, d^{4}} + \frac {{\left (b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}\right )} \log \left ({\left | d x + c \right |}\right )}{d^{5}} \] Input:
integrate((b*x^2+a)^2/(d*x+c),x, algorithm="giac")
Output:
1/12*(3*b^2*d^3*x^4 - 4*b^2*c*d^2*x^3 + 6*b^2*c^2*d*x^2 + 12*a*b*d^3*x^2 - 12*b^2*c^3*x - 24*a*b*c*d^2*x)/d^4 + (b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4)* log(abs(d*x + c))/d^5
Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+b x^2\right )^2}{c+d x} \, dx=x^2\,\left (\frac {b^2\,c^2}{2\,d^3}+\frac {a\,b}{d}\right )+\frac {\ln \left (c+d\,x\right )\,\left (a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4\right )}{d^5}+\frac {b^2\,x^4}{4\,d}-\frac {b^2\,c\,x^3}{3\,d^2}-\frac {c\,x\,\left (\frac {b^2\,c^2}{d^3}+\frac {2\,a\,b}{d}\right )}{d} \] Input:
int((a + b*x^2)^2/(c + d*x),x)
Output:
x^2*((b^2*c^2)/(2*d^3) + (a*b)/d) + (log(c + d*x)*(a^2*d^4 + b^2*c^4 + 2*a *b*c^2*d^2))/d^5 + (b^2*x^4)/(4*d) - (b^2*c*x^3)/(3*d^2) - (c*x*((b^2*c^2) /d^3 + (2*a*b)/d))/d
Time = 0.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a+b x^2\right )^2}{c+d x} \, dx=\frac {12 \,\mathrm {log}\left (d x +c \right ) a^{2} d^{4}+24 \,\mathrm {log}\left (d x +c \right ) a b \,c^{2} d^{2}+12 \,\mathrm {log}\left (d x +c \right ) b^{2} c^{4}-24 a b c \,d^{3} x +12 a b \,d^{4} x^{2}-12 b^{2} c^{3} d x +6 b^{2} c^{2} d^{2} x^{2}-4 b^{2} c \,d^{3} x^{3}+3 b^{2} d^{4} x^{4}}{12 d^{5}} \] Input:
int((b*x^2+a)^2/(d*x+c),x)
Output:
(12*log(c + d*x)*a**2*d**4 + 24*log(c + d*x)*a*b*c**2*d**2 + 12*log(c + d* x)*b**2*c**4 - 24*a*b*c*d**3*x + 12*a*b*d**4*x**2 - 12*b**2*c**3*d*x + 6*b **2*c**2*d**2*x**2 - 4*b**2*c*d**3*x**3 + 3*b**2*d**4*x**4)/(12*d**5)