\(\int F^{a+b (c+d x)^2} (c+d x)^5 \, dx\) [258]

Optimal result

Integrand size = 21, antiderivative size = 91 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\frac {F^{a+b (c+d x)^2}}{b^3 d \log ^3(F)}-\frac {F^{a+b (c+d x)^2} (c+d x)^2}{b^2 d \log ^2(F)}+\frac {F^{a+b (c+d x)^2} (c+d x)^4}{2 b d \log (F)} \]

[Out]

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)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2243, 2240} \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\frac {F^{a+b (c+d x)^2}}{b^3 d \log ^3(F)}-\frac {(c+d x)^2 F^{a+b (c+d x)^2}}{b^2 d \log ^2(F)}+\frac {(c+d x)^4 F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

[In]

Int[F^(a + b*(c + d*x)^2)*(c + d*x)^5,x]

[Out]

F^(a + b*(c + d*x)^2)/(b^3*d*Log[F]^3) - (F^(a + b*(c + d*x)^2)*(c + d*x)^2)/(b^2*d*Log[F]^2) + (F^(a + b*(c +
d*x)^2)*(c + d*x)^4)/(2*b*d*Log[F])

Rule 2240

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]

Rule 2243

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*Log[F])), x] - Dist[(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])

Rubi steps \begin{align*} \text {integral}& = \frac {F^{a+b (c+d x)^2} (c+d x)^4}{2 b d \log (F)}-\frac {2 \int F^{a+b (c+d x)^2} (c+d x)^3 \, dx}{b \log (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)}+\frac {2 \int F^{a+b (c+d x)^2} (c+d x) \, dx}{b^2 \log ^2(F)} \\ & = \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)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\frac {F^{a+b (c+d x)^2} \left (2-2 b (c+d x)^2 \log (F)+b^2 (c+d x)^4 \log ^2(F)\right )}{2 b^3 d \log ^3(F)} \]

[In]

Integrate[F^(a + b*(c + d*x)^2)*(c + d*x)^5,x]

[Out]

(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)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.52

method	result	size
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	\(\frac {d^{4} F^{a +b \left (d x +c \right )^{2}} x^{4} \ln \left (F \right )^{2} b^{2}+4 d^{3} c \,F^{a +b \left (d x +c \right )^{2}} x^{3} \ln \left (F \right )^{2} b^{2}+6 \ln \left (F \right )^{2} x^{2} F^{a +b \left (d x +c \right )^{2}} b^{2} c^{2} d^{2}+4 \ln \left (F \right )^{2} x \,F^{a +b \left (d x +c \right )^{2}} b^{2} c^{3} d +\ln \left (F \right )^{2} F^{a +b \left (d x +c \right )^{2}} b^{2} c^{4}-2 d^{2} F^{a +b \left (d x +c \right )^{2}} x^{2} \ln \left (F \right ) b -4 c \,F^{a +b \left (d x +c \right )^{2}} x \ln \left (F \right ) b d -2 \ln \left (F \right ) F^{a +b \left (d x +c \right )^{2}} b \,c^{2}+2 F^{a +b \left (d x +c \right )^{2}}}{2 \ln \left (F \right )^{3} b^{3} d}\)	\(233\)

[In]

int(F^(a+b*(d*x+c)^2)*(d*x+c)^5,x,method=_RETURNVERBOSE)

[Out]

1/2*(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)*F^(b*d^2*x^2+2*b*c*d*x+b*c^2+a)/ln(F)^3/b^3/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.32 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\frac {{\left ({\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (F\right )^{2} - 2 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right ) + 2\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{2 \, b^{3} d \log \left (F\right )^{3}} \]

[In]

integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^5,x, algorithm="fricas")

[Out]

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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (76) = 152\).

Time = 0.12 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.33 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\begin {cases} \frac {F^{a + b \left (c + d x\right )^{2}} \left (b^{2} c^{4} \log {\left (F \right )}^{2} + 4 b^{2} c^{3} d x \log {\left (F \right )}^{2} + 6 b^{2} c^{2} d^{2} x^{2} \log {\left (F \right )}^{2} + 4 b^{2} c d^{3} x^{3} \log {\left (F \right )}^{2} + b^{2} d^{4} x^{4} \log {\left (F \right )}^{2} - 2 b c^{2} \log {\left (F \right )} - 4 b c d x \log {\left (F \right )} - 2 b d^{2} x^{2} \log {\left (F \right )} + 2\right )}{2 b^{3} d \log {\left (F \right )}^{3}} & \text {for}\: b^{3} d \log {\left (F \right )}^{3} \neq 0 \\c^{5} x + \frac {5 c^{4} d x^{2}}{2} + \frac {10 c^{3} d^{2} x^{3}}{3} + \frac {5 c^{2} d^{3} x^{4}}{2} + c d^{4} x^{5} + \frac {d^{5} x^{6}}{6} & \text {otherwise} \end {cases} \]

[In]

integrate(F**(a+b*(d*x+c)**2)*(d*x+c)**5,x)

[Out]

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*lo
g(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))

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.71 (sec) , antiderivative size = 1438, normalized size of antiderivative = 15.80 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\text {Too large to display} \]

[In]

integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^5,x, algorithm="maxima")

[Out]

-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*log(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*gamma(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)^2*log(F)/(b*d^2))*log(F)^5/((b*log(F))^(9/2)*d^7*(-(b*d^2*x +
b*c*d)^2*log(F)/(b*d^2))^(3/2)) + 4*b^3*c*gamma(2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^3/((b*log(F))^
(9/2)*d^4) - (b*d^2*x + b*c*d)^5*gamma(5/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^5/((b*log(F))^(9/2)*d^
9*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(5/2)))*F^a*c*d^3/sqrt(b*log(F)) - 1/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b^5
*c^5*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^6/((b*log(F))^(11/2)*d^6*sqrt(-(b*d^2*x + b*c
*d)^2*log(F)/(b*d^2))) - 5*F^((b*d^2*x + b*c*d)^2/(b*d^2))*b^5*c^4*log(F)^5/((b*log(F))^(11/2)*d^5) - 10*(b*d^
2*x + b*c*d)^3*b^3*c^3*gamma(3/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^6/((b*log(F))^(11/2)*d^8*(-(b*d^
2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)) + 10*b^4*c^2*gamma(2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^4/((b*
log(F))^(11/2)*d^5) - b^3*gamma(3, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^3/((b*log(F))^(11/2)*d^5) - 5*(
b*d^2*x + b*c*d)^5*b*c*gamma(5/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^6/((b*log(F))^(11/2)*d^10*(-(b*d
^2*x + b*c*d)^2*log(F)/(b*d^2))^(5/2)))*F^a*d^4/sqrt(b*log(F)) + 1/2*sqrt(pi)*F^(b*c^2 + a)*c^5*erf(sqrt(-b*lo
g(F))*d*x - b*c*log(F)/sqrt(-b*log(F)))/(sqrt(-b*log(F))*F^(b*c^2)*d)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\frac {{\left (b^{2} d^{4} {\left (x + \frac {c}{d}\right )}^{4} \log \left (F\right )^{2} - 2 \, b d^{2} {\left (x + \frac {c}{d}\right )}^{2} \log \left (F\right ) + 2\right )} e^{\left (b d^{2} x^{2} \log \left (F\right ) + 2 \, b c d x \log \left (F\right ) + b c^{2} \log \left (F\right ) + a \log \left (F\right )\right )}}{2 \, b^{3} d \log \left (F\right )^{3}} \]

[In]

integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^5,x, algorithm="giac")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.56 \[ \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx=\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,\left (b^2\,c^4\,{\ln \left (F\right )}^2+4\,b^2\,c^3\,d\,x\,{\ln \left (F\right )}^2+6\,b^2\,c^2\,d^2\,x^2\,{\ln \left (F\right )}^2+4\,b^2\,c\,d^3\,x^3\,{\ln \left (F\right )}^2+b^2\,d^4\,x^4\,{\ln \left (F\right )}^2-2\,b\,c^2\,\ln \left (F\right )-4\,b\,c\,d\,x\,\ln \left (F\right )-2\,b\,d^2\,x^2\,\ln \left (F\right )+2\right )}{2\,b^3\,d\,{\ln \left (F\right )}^3} \]

[In]

int(F^(a + b*(c + d*x)^2)*(c + d*x)^5,x)

[Out]

(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)