Integrand size = 22, antiderivative size = 80 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {(b c-a d)^2 x^2}{2 b^3}+\frac {d (2 b c-a d) x^4}{4 b^2}+\frac {d^2 x^6}{6 b}-\frac {a (b c-a d)^2 \log \left (a+b x^2\right )}{2 b^4} \] Output:
1/2*(-a*d+b*c)^2*x^2/b^3+1/4*d*(-a*d+2*b*c)*x^4/b^2+1/6*d^2*x^6/b-1/2*a*(- a*d+b*c)^2*ln(b*x^2+a)/b^4
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {b x^2 \left (6 a^2 d^2-3 a b d \left (4 c+d x^2\right )+2 b^2 \left (3 c^2+3 c d x^2+d^2 x^4\right )\right )-6 a (b c-a d)^2 \log \left (a+b x^2\right )}{12 b^4} \] Input:
Integrate[(x^3*(c + d*x^2)^2)/(a + b*x^2),x]
Output:
(b*x^2*(6*a^2*d^2 - 3*a*b*d*(4*c + d*x^2) + 2*b^2*(3*c^2 + 3*c*d*x^2 + d^2 *x^4)) - 6*a*(b*c - a*d)^2*Log[a + b*x^2])/(12*b^4)
Time = 0.23 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {354, 86, 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.
\(\displaystyle \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^2 \left (d x^2+c\right )^2}{b x^2+a}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {d^2 x^4}{b}+\frac {d (2 b c-a d) x^2}{b^2}+\frac {(b c-a d)^2}{b^3}-\frac {a (a d-b c)^2}{b^3 \left (b x^2+a\right )}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {a (b c-a d)^2 \log \left (a+b x^2\right )}{b^4}+\frac {x^2 (b c-a d)^2}{b^3}+\frac {d x^4 (2 b c-a d)}{2 b^2}+\frac {d^2 x^6}{3 b}\right )\) |
Input:
Int[(x^3*(c + d*x^2)^2)/(a + b*x^2),x]
Output:
(((b*c - a*d)^2*x^2)/b^3 + (d*(2*b*c - a*d)*x^4)/(2*b^2) + (d^2*x^6)/(3*b) - (a*(b*c - a*d)^2*Log[a + b*x^2])/b^4)/2
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 0.46 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.18
method | result | size |
norman | \(\frac {d^{2} x^{6}}{6 b}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{2}}{2 b^{3}}-\frac {d \left (a d -2 b c \right ) x^{4}}{4 b^{2}}-\frac {a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b \,x^{2}+a \right )}{2 b^{4}}\) | \(94\) |
default | \(\frac {\frac {1}{3} b^{2} d^{2} x^{6}-\frac {1}{2} x^{4} a b \,d^{2}+x^{4} b^{2} c d +a^{2} d^{2} x^{2}-2 a b c d \,x^{2}+b^{2} c^{2} x^{2}}{2 b^{3}}-\frac {a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b \,x^{2}+a \right )}{2 b^{4}}\) | \(102\) |
parallelrisch | \(-\frac {-2 b^{3} d^{2} x^{6}+3 x^{4} a \,b^{2} d^{2}-6 x^{4} b^{3} c d -6 x^{2} a^{2} b \,d^{2}+12 x^{2} a \,b^{2} c d -6 x^{2} b^{3} c^{2}+6 \ln \left (b \,x^{2}+a \right ) a^{3} d^{2}-12 \ln \left (b \,x^{2}+a \right ) a^{2} b c d +6 \ln \left (b \,x^{2}+a \right ) a \,b^{2} c^{2}}{12 b^{4}}\) | \(123\) |
risch | \(\frac {d^{2} x^{6}}{6 b}-\frac {x^{4} a \,d^{2}}{4 b^{2}}+\frac {x^{4} c d}{2 b}+\frac {a^{2} d^{2} x^{2}}{2 b^{3}}-\frac {a c d \,x^{2}}{b^{2}}+\frac {c^{2} x^{2}}{2 b}-\frac {a^{3} \ln \left (b \,x^{2}+a \right ) d^{2}}{2 b^{4}}+\frac {a^{2} \ln \left (b \,x^{2}+a \right ) c d}{b^{3}}-\frac {a \ln \left (b \,x^{2}+a \right ) c^{2}}{2 b^{2}}\) | \(124\) |
Input:
int(x^3*(d*x^2+c)^2/(b*x^2+a),x,method=_RETURNVERBOSE)
Output:
1/6*d^2*x^6/b+1/2*(a^2*d^2-2*a*b*c*d+b^2*c^2)/b^3*x^2-1/4*d*(a*d-2*b*c)/b^ 2*x^4-1/2*a/b^4*(a^2*d^2-2*a*b*c*d+b^2*c^2)*ln(b*x^2+a)
Time = 0.07 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.28 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {2 \, b^{3} d^{2} x^{6} + 3 \, {\left (2 \, b^{3} c d - a b^{2} d^{2}\right )} x^{4} + 6 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{2} - 6 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \log \left (b x^{2} + a\right )}{12 \, b^{4}} \] Input:
integrate(x^3*(d*x^2+c)^2/(b*x^2+a),x, algorithm="fricas")
Output:
1/12*(2*b^3*d^2*x^6 + 3*(2*b^3*c*d - a*b^2*d^2)*x^4 + 6*(b^3*c^2 - 2*a*b^2 *c*d + a^2*b*d^2)*x^2 - 6*(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*log(b*x^2 + a))/b^4
Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.04 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=- \frac {a \left (a d - b c\right )^{2} \log {\left (a + b x^{2} \right )}}{2 b^{4}} + x^{4} \left (- \frac {a d^{2}}{4 b^{2}} + \frac {c d}{2 b}\right ) + x^{2} \left (\frac {a^{2} d^{2}}{2 b^{3}} - \frac {a c d}{b^{2}} + \frac {c^{2}}{2 b}\right ) + \frac {d^{2} x^{6}}{6 b} \] Input:
integrate(x**3*(d*x**2+c)**2/(b*x**2+a),x)
Output:
-a*(a*d - b*c)**2*log(a + b*x**2)/(2*b**4) + x**4*(-a*d**2/(4*b**2) + c*d/ (2*b)) + x**2*(a**2*d**2/(2*b**3) - a*c*d/b**2 + c**2/(2*b)) + d**2*x**6/( 6*b)
Time = 0.06 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.26 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {2 \, b^{2} d^{2} x^{6} + 3 \, {\left (2 \, b^{2} c d - a b d^{2}\right )} x^{4} + 6 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2}}{12 \, b^{3}} - \frac {{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} \] Input:
integrate(x^3*(d*x^2+c)^2/(b*x^2+a),x, algorithm="maxima")
Output:
1/12*(2*b^2*d^2*x^6 + 3*(2*b^2*c*d - a*b*d^2)*x^4 + 6*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^2)/b^3 - 1/2*(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*log(b*x^2 + a)/b^4
Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.34 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {2 \, b^{2} d^{2} x^{6} + 6 \, b^{2} c d x^{4} - 3 \, a b d^{2} x^{4} + 6 \, b^{2} c^{2} x^{2} - 12 \, a b c d x^{2} + 6 \, a^{2} d^{2} x^{2}}{12 \, b^{3}} - \frac {{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{4}} \] Input:
integrate(x^3*(d*x^2+c)^2/(b*x^2+a),x, algorithm="giac")
Output:
1/12*(2*b^2*d^2*x^6 + 6*b^2*c*d*x^4 - 3*a*b*d^2*x^4 + 6*b^2*c^2*x^2 - 12*a *b*c*d*x^2 + 6*a^2*d^2*x^2)/b^3 - 1/2*(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)* log(abs(b*x^2 + a))/b^4
Time = 0.48 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.32 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=x^2\,\left (\frac {c^2}{2\,b}+\frac {a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{2\,b}\right )-x^4\,\left (\frac {a\,d^2}{4\,b^2}-\frac {c\,d}{2\,b}\right )-\frac {\ln \left (b\,x^2+a\right )\,\left (a^3\,d^2-2\,a^2\,b\,c\,d+a\,b^2\,c^2\right )}{2\,b^4}+\frac {d^2\,x^6}{6\,b} \] Input:
int((x^3*(c + d*x^2)^2)/(a + b*x^2),x)
Output:
x^2*(c^2/(2*b) + (a*((a*d^2)/b^2 - (2*c*d)/b))/(2*b)) - x^4*((a*d^2)/(4*b^ 2) - (c*d)/(2*b)) - (log(a + b*x^2)*(a^3*d^2 + a*b^2*c^2 - 2*a^2*b*c*d))/( 2*b^4) + (d^2*x^6)/(6*b)
Time = 0.24 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.52 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {-6 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3} d^{2}+12 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b c d -6 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{2} c^{2}+6 a^{2} b \,d^{2} x^{2}-12 a \,b^{2} c d \,x^{2}-3 a \,b^{2} d^{2} x^{4}+6 b^{3} c^{2} x^{2}+6 b^{3} c d \,x^{4}+2 b^{3} d^{2} x^{6}}{12 b^{4}} \] Input:
int(x^3*(d*x^2+c)^2/(b*x^2+a),x)
Output:
( - 6*log(a + b*x**2)*a**3*d**2 + 12*log(a + b*x**2)*a**2*b*c*d - 6*log(a + b*x**2)*a*b**2*c**2 + 6*a**2*b*d**2*x**2 - 12*a*b**2*c*d*x**2 - 3*a*b**2 *d**2*x**4 + 6*b**3*c**2*x**2 + 6*b**3*c*d*x**4 + 2*b**3*d**2*x**6)/(12*b* *4)