Integrand size = 21, antiderivative size = 91 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\frac {F^{a+b (c+d x)^2}}{b^3 d \log ^3(F)}-\frac {F^{a+b (c+d x)^2} (c+d x)^2}{b^2 d \log ^2(F)}+\frac {F^{a+b (c+d x)^2} (c+d x)^4}{2 b d \log (F)} \] Output:
F^(a+b*(d*x+c)^2)/b^3/d/ln(F)^3-F^(a+b*(d*x+c)^2)*(d*x+c)^2/b^2/d/ln(F)^2+ 1/2*F^(a+b*(d*x+c)^2)*(d*x+c)^4/b/d/ln(F)
Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\frac {F^{a+b (c+d x)^2} \left (2-2 b (c+d x)^2 \log (F)+b^2 (c+d x)^4 \log ^2(F)\right )}{2 b^3 d \log ^3(F)} \] Input:
Integrate[F^(a + b*(c + d*x)^2)*(c + d*x)^5,x]
Output:
(F^(a + b*(c + d*x)^2)*(2 - 2*b*(c + d*x)^2*Log[F] + b^2*(c + d*x)^4*Log[F ]^2))/(2*b^3*d*Log[F]^3)
Time = 0.65 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2641, 2641, 2638}
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.
\(\displaystyle \int (c+d x)^5 F^{a+b (c+d x)^2} \, dx\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle \frac {(c+d x)^4 F^{a+b (c+d x)^2}}{2 b d \log (F)}-\frac {2 \int F^{b (c+d x)^2+a} (c+d x)^3dx}{b \log (F)}\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle \frac {(c+d x)^4 F^{a+b (c+d x)^2}}{2 b d \log (F)}-\frac {2 \left (\frac {(c+d x)^2 F^{a+b (c+d x)^2}}{2 b d \log (F)}-\frac {\int F^{b (c+d x)^2+a} (c+d x)dx}{b \log (F)}\right )}{b \log (F)}\) |
\(\Big \downarrow \) 2638 |
\(\displaystyle \frac {(c+d x)^4 F^{a+b (c+d x)^2}}{2 b d \log (F)}-\frac {2 \left (\frac {(c+d x)^2 F^{a+b (c+d x)^2}}{2 b d \log (F)}-\frac {F^{a+b (c+d x)^2}}{2 b^2 d \log ^2(F)}\right )}{b \log (F)}\) |
Input:
Int[F^(a + b*(c + d*x)^2)*(c + d*x)^5,x]
Output:
(F^(a + b*(c + d*x)^2)*(c + d*x)^4)/(2*b*d*Log[F]) - (2*(-1/2*F^(a + b*(c + d*x)^2)/(b^2*d*Log[F]^2) + (F^(a + b*(c + d*x)^2)*(c + d*x)^2)/(2*b*d*Lo g[F])))/(b*Log[F])
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_ .), x_Symbol] :> Simp[(e + f*x)^n*(F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n *Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] && EqQ [d*e - c*f, 0]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[(c + d*x)^(m - n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*L og[F])), x] - Simp[(m - n + 1)/(b*n*Log[F]) Int[(c + d*x)^(m - n)*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/ n)] && LtQ[0, (m + 1)/n, 5] && IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n , 0])
Time = 0.14 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.41
method | result | size |
orering | \(\frac {\left (d^{4} x^{4} b^{2} \ln \left (F \right )^{2}+4 d^{3} c \,x^{3} b^{2} \ln \left (F \right )^{2}+6 \ln \left (F \right )^{2} b^{2} c^{2} d^{2} x^{2}+4 \ln \left (F \right )^{2} b^{2} c^{3} d x +\ln \left (F \right )^{2} b^{2} c^{4}-2 \ln \left (F \right ) b \,d^{2} x^{2}-4 \ln \left (F \right ) b c d x -2 \ln \left (F \right ) b \,c^{2}+2\right ) F^{a +b \left (d x +c \right )^{2}}}{2 d \,b^{3} \ln \left (F \right )^{3}}\) | \(128\) |
gosper | \(\frac {\left (d^{4} x^{4} b^{2} \ln \left (F \right )^{2}+4 d^{3} c \,x^{3} b^{2} \ln \left (F \right )^{2}+6 \ln \left (F \right )^{2} b^{2} c^{2} d^{2} x^{2}+4 \ln \left (F \right )^{2} b^{2} c^{3} d x +\ln \left (F \right )^{2} b^{2} c^{4}-2 \ln \left (F \right ) b \,d^{2} x^{2}-4 \ln \left (F \right ) b c d x -2 \ln \left (F \right ) b \,c^{2}+2\right ) F^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}+a}}{2 \ln \left (F \right )^{3} b^{3} d}\) | \(138\) |
risch | \(\frac {\left (d^{4} x^{4} b^{2} \ln \left (F \right )^{2}+4 d^{3} c \,x^{3} b^{2} \ln \left (F \right )^{2}+6 \ln \left (F \right )^{2} b^{2} c^{2} d^{2} x^{2}+4 \ln \left (F \right )^{2} b^{2} c^{3} d x +\ln \left (F \right )^{2} b^{2} c^{4}-2 \ln \left (F \right ) b \,d^{2} x^{2}-4 \ln \left (F \right ) b c d x -2 \ln \left (F \right ) b \,c^{2}+2\right ) F^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}+a}}{2 \ln \left (F \right )^{3} b^{3} d}\) | \(138\) |
norman | \(\frac {d \left (3 \ln \left (F \right ) b \,c^{2}-1\right ) x^{2} {\mathrm e}^{\left (a +b \left (d x +c \right )^{2}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{2} b^{2}}+\frac {\left (\ln \left (F \right )^{2} b^{2} c^{4}-2 \ln \left (F \right ) b \,c^{2}+2\right ) {\mathrm e}^{\left (a +b \left (d x +c \right )^{2}\right ) \ln \left (F \right )}}{2 \ln \left (F \right )^{3} b^{3} d}+\frac {d^{3} x^{4} {\mathrm e}^{\left (a +b \left (d x +c \right )^{2}\right ) \ln \left (F \right )}}{2 \ln \left (F \right ) b}+\frac {2 c \left (\ln \left (F \right ) b \,c^{2}-1\right ) x \,{\mathrm e}^{\left (a +b \left (d x +c \right )^{2}\right ) \ln \left (F \right )}}{\ln \left (F \right )^{2} b^{2}}+\frac {2 d^{2} c \,x^{3} {\mathrm e}^{\left (a +b \left (d x +c \right )^{2}\right ) \ln \left (F \right )}}{\ln \left (F \right ) b}\) | \(183\) |
parallelrisch | \(\frac {d^{4} F^{a +b \left (d x +c \right )^{2}} x^{4} b^{2} \ln \left (F \right )^{2}+4 d^{3} c \,F^{a +b \left (d x +c \right )^{2}} x^{3} b^{2} \ln \left (F \right )^{2}+6 \ln \left (F \right )^{2} x^{2} F^{a +b \left (d x +c \right )^{2}} b^{2} c^{2} d^{2}+4 \ln \left (F \right )^{2} x \,F^{a +b \left (d x +c \right )^{2}} b^{2} c^{3} d +\ln \left (F \right )^{2} F^{a +b \left (d x +c \right )^{2}} b^{2} c^{4}-2 d^{2} F^{a +b \left (d x +c \right )^{2}} x^{2} b \ln \left (F \right )-4 c \,F^{a +b \left (d x +c \right )^{2}} x b \ln \left (F \right ) d -2 \ln \left (F \right ) F^{a +b \left (d x +c \right )^{2}} b \,c^{2}+2 F^{a +b \left (d x +c \right )^{2}}}{2 \ln \left (F \right )^{3} b^{3} d}\) | \(233\) |
Input:
int(F^(a+b*(d*x+c)^2)*(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
1/2/d*(d^4*x^4*b^2*ln(F)^2+4*d^3*c*x^3*b^2*ln(F)^2+6*ln(F)^2*b^2*c^2*d^2*x ^2+4*ln(F)^2*b^2*c^3*d*x+ln(F)^2*b^2*c^4-2*ln(F)*b*d^2*x^2-4*ln(F)*b*c*d*x -2*ln(F)*b*c^2+2)/b^3/ln(F)^3*F^(a+b*(d*x+c)^2)
Time = 0.09 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.32 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\frac {{\left ({\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (F\right )^{2} - 2 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right ) + 2\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{2 \, b^{3} d \log \left (F\right )^{3}} \] Input:
integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^5,x, algorithm="fricas")
Output:
1/2*((b^2*d^4*x^4 + 4*b^2*c*d^3*x^3 + 6*b^2*c^2*d^2*x^2 + 4*b^2*c^3*d*x + b^2*c^4)*log(F)^2 - 2*(b*d^2*x^2 + 2*b*c*d*x + b*c^2)*log(F) + 2)*F^(b*d^2 *x^2 + 2*b*c*d*x + b*c^2 + a)/(b^3*d*log(F)^3)
Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (76) = 152\).
Time = 0.11 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.33 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\begin {cases} \frac {F^{a + b \left (c + d x\right )^{2}} \left (b^{2} c^{4} \log {\left (F \right )}^{2} + 4 b^{2} c^{3} d x \log {\left (F \right )}^{2} + 6 b^{2} c^{2} d^{2} x^{2} \log {\left (F \right )}^{2} + 4 b^{2} c d^{3} x^{3} \log {\left (F \right )}^{2} + b^{2} d^{4} x^{4} \log {\left (F \right )}^{2} - 2 b c^{2} \log {\left (F \right )} - 4 b c d x \log {\left (F \right )} - 2 b d^{2} x^{2} \log {\left (F \right )} + 2\right )}{2 b^{3} d \log {\left (F \right )}^{3}} & \text {for}\: b^{3} d \log {\left (F \right )}^{3} \neq 0 \\c^{5} x + \frac {5 c^{4} d x^{2}}{2} + \frac {10 c^{3} d^{2} x^{3}}{3} + \frac {5 c^{2} d^{3} x^{4}}{2} + c d^{4} x^{5} + \frac {d^{5} x^{6}}{6} & \text {otherwise} \end {cases} \] Input:
integrate(F**(a+b*(d*x+c)**2)*(d*x+c)**5,x)
Output:
Piecewise((F**(a + b*(c + d*x)**2)*(b**2*c**4*log(F)**2 + 4*b**2*c**3*d*x* log(F)**2 + 6*b**2*c**2*d**2*x**2*log(F)**2 + 4*b**2*c*d**3*x**3*log(F)**2 + b**2*d**4*x**4*log(F)**2 - 2*b*c**2*log(F) - 4*b*c*d*x*log(F) - 2*b*d** 2*x**2*log(F) + 2)/(2*b**3*d*log(F)**3), Ne(b**3*d*log(F)**3, 0)), (c**5*x + 5*c**4*d*x**2/2 + 10*c**3*d**2*x**3/3 + 5*c**2*d**3*x**4/2 + c*d**4*x** 5 + d**5*x**6/6, True))
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.57 (sec) , antiderivative size = 1438, normalized size of antiderivative = 15.80 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\text {Too large to display} \] Input:
integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^5,x, algorithm="maxima")
Output:
-5/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b*c*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F) /(b*d^2))) - 1)*log(F)^2/((b*log(F))^(3/2)*d^2*sqrt(-(b*d^2*x + b*c*d)^2*l og(F)/(b*d^2))) - F^((b*d^2*x + b*c*d)^2/(b*d^2))*b*log(F)/((b*log(F))^(3/ 2)*d))*F^a*c^4/sqrt(b*log(F)) + 5*(sqrt(pi)*(b*d^2*x + b*c*d)*b^2*c^2*(erf (sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^3/((b*log(F))^(5/2 )*d^3*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 2*F^((b*d^2*x + b*c*d)^ 2/(b*d^2))*b^2*c*log(F)^2/((b*log(F))^(5/2)*d^2) - (b*d^2*x + b*c*d)^3*gam ma(3/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^3/((b*log(F))^(5/2)*d^ 5*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)))*F^a*c^3*d/sqrt(b*log(F)) - 5*(sqrt(pi)*(b*d^2*x + b*c*d)*b^3*c^3*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log( F)/(b*d^2))) - 1)*log(F)^4/((b*log(F))^(7/2)*d^4*sqrt(-(b*d^2*x + b*c*d)^2 *log(F)/(b*d^2))) - 3*F^((b*d^2*x + b*c*d)^2/(b*d^2))*b^3*c^2*log(F)^3/((b *log(F))^(7/2)*d^3) - 3*(b*d^2*x + b*c*d)^3*b*c*gamma(3/2, -(b*d^2*x + b*c *d)^2*log(F)/(b*d^2))*log(F)^4/((b*log(F))^(7/2)*d^6*(-(b*d^2*x + b*c*d)^2 *log(F)/(b*d^2))^(3/2)) + b^2*gamma(2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2) )*log(F)^2/((b*log(F))^(7/2)*d^3))*F^a*c^2*d^2/sqrt(b*log(F)) + 5/2*(sqrt( pi)*(b*d^2*x + b*c*d)*b^4*c^4*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2 ))) - 1)*log(F)^5/((b*log(F))^(9/2)*d^5*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/( b*d^2))) - 4*F^((b*d^2*x + b*c*d)^2/(b*d^2))*b^4*c^3*log(F)^4/((b*log(F))^ (9/2)*d^4) - 6*(b*d^2*x + b*c*d)^3*b^2*c^2*gamma(3/2, -(b*d^2*x + b*c*d...
Time = 0.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\frac {{\left (b^{2} d^{4} {\left (x + \frac {c}{d}\right )}^{4} \log \left (F\right )^{2} - 2 \, b d^{2} {\left (x + \frac {c}{d}\right )}^{2} \log \left (F\right ) + 2\right )} e^{\left (b d^{2} x^{2} \log \left (F\right ) + 2 \, b c d x \log \left (F\right ) + b c^{2} \log \left (F\right ) + a \log \left (F\right )\right )}}{2 \, b^{3} d \log \left (F\right )^{3}} \] Input:
integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^5,x, algorithm="giac")
Output:
1/2*(b^2*d^4*(x + c/d)^4*log(F)^2 - 2*b*d^2*(x + c/d)^2*log(F) + 2)*e^(b*d ^2*x^2*log(F) + 2*b*c*d*x*log(F) + b*c^2*log(F) + a*log(F))/(b^3*d*log(F)^ 3)
Time = 0.18 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.56 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,\left (b^2\,c^4\,{\ln \left (F\right )}^2+4\,b^2\,c^3\,d\,x\,{\ln \left (F\right )}^2+6\,b^2\,c^2\,d^2\,x^2\,{\ln \left (F\right )}^2+4\,b^2\,c\,d^3\,x^3\,{\ln \left (F\right )}^2+b^2\,d^4\,x^4\,{\ln \left (F\right )}^2-2\,b\,c^2\,\ln \left (F\right )-4\,b\,c\,d\,x\,\ln \left (F\right )-2\,b\,d^2\,x^2\,\ln \left (F\right )+2\right )}{2\,b^3\,d\,{\ln \left (F\right )}^3} \] Input:
int(F^(a + b*(c + d*x)^2)*(c + d*x)^5,x)
Output:
(F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*(b^2*c^4*log(F)^2 - 2*b*c^2*log (F) - 2*b*d^2*x^2*log(F) + b^2*d^4*x^4*log(F)^2 + 4*b^2*c*d^3*x^3*log(F)^2 + 6*b^2*c^2*d^2*x^2*log(F)^2 - 4*b*c*d*x*log(F) + 4*b^2*c^3*d*x*log(F)^2 + 2))/(2*b^3*d*log(F)^3)
Time = 0.16 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.51 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\frac {f^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}+a} \left (\mathrm {log}\left (f \right )^{2} b^{2} c^{4}+4 \mathrm {log}\left (f \right )^{2} b^{2} c^{3} d x +6 \mathrm {log}\left (f \right )^{2} b^{2} c^{2} d^{2} x^{2}+4 \mathrm {log}\left (f \right )^{2} b^{2} c \,d^{3} x^{3}+\mathrm {log}\left (f \right )^{2} b^{2} d^{4} x^{4}-2 \,\mathrm {log}\left (f \right ) b \,c^{2}-4 \,\mathrm {log}\left (f \right ) b c d x -2 \,\mathrm {log}\left (f \right ) b \,d^{2} x^{2}+2\right )}{2 \mathrm {log}\left (f \right )^{3} b^{3} d} \] Input:
int(F^(a+b*(d*x+c)^2)*(d*x+c)^5,x)
Output:
(f**(a + b*c**2 + 2*b*c*d*x + b*d**2*x**2)*(log(f)**2*b**2*c**4 + 4*log(f) **2*b**2*c**3*d*x + 6*log(f)**2*b**2*c**2*d**2*x**2 + 4*log(f)**2*b**2*c*d **3*x**3 + log(f)**2*b**2*d**4*x**4 - 2*log(f)*b*c**2 - 4*log(f)*b*c*d*x - 2*log(f)*b*d**2*x**2 + 2))/(2*log(f)**3*b**3*d)