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

Optimal. Leaf size=91 \[ \frac {F^{a+b (c+d x)^2}}{b^3 d \log ^3(F)}-\frac {(c+d x)^2 F^{a+b (c+d x)^2}}{b^2 d \log ^2(F)}+\frac {(c+d x)^4 F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

[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)

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Rubi [A]  time = 0.18, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2212, 2209} \[ -\frac {(c+d x)^2 F^{a+b (c+d x)^2}}{b^2 d \log ^2(F)}+\frac {F^{a+b (c+d x)^2}}{b^3 d \log ^3(F)}+\frac {(c+d x)^4 F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

Antiderivative was successfully verified.

[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 2209

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 2212

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*} \int F^{a+b (c+d x)^2} (c+d x)^5 \, dx &=\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*}

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Mathematica [A]  time = 0.03, size = 56, normalized size = 0.62 \[ \frac {F^{a+b (c+d x)^2} \left (b^2 \log ^2(F) (c+d x)^4-2 b \log (F) (c+d x)^2+2\right )}{2 b^3 d \log ^3(F)} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.43, size = 120, normalized size = 1.32 \[ \frac {{\left ({\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \relax (F)^{2} - 2 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \relax (F) + 2\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{2 \, b^{3} d \log \relax (F)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.30, size = 82, normalized size = 0.90 \[ \frac {{\left (b^{2} d^{4} {\left (x + \frac {c}{d}\right )}^{4} \log \relax (F)^{2} - 2 \, b d^{2} {\left (x + \frac {c}{d}\right )}^{2} \log \relax (F) + 2\right )} e^{\left (b d^{2} x^{2} \log \relax (F) + 2 \, b c d x \log \relax (F) + b c^{2} \log \relax (F) + a \log \relax (F)\right )}}{2 \, b^{3} d \log \relax (F)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 138, normalized size = 1.52 \[ \frac {\left (b^{2} d^{4} x^{4} \ln \relax (F )^{2}+4 b^{2} c \,d^{3} x^{3} \ln \relax (F )^{2}+6 b^{2} c^{2} d^{2} x^{2} \ln \relax (F )^{2}+4 b^{2} c^{3} d x \ln \relax (F )^{2}+b^{2} c^{4} \ln \relax (F )^{2}-2 b \,d^{2} x^{2} \ln \relax (F )-4 b c d x \ln \relax (F )-2 b \,c^{2} \ln \relax (F )+2\right ) F^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}+a}}{2 b^{3} d \ln \relax (F )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*d^4*x^4*ln(F)^2+4*b^2*c*d^3*x^3*ln(F)^2+6*b^2*c^2*d^2*x^2*ln(F)^2+4*b^2*c^3*d*x*ln(F)^2+b^2*c^4*ln(F)
^2-2*b*d^2*x^2*ln(F)-4*b*c*d*x*ln(F)-2*b*c^2*ln(F)+2)*F^(b*d^2*x^2+2*b*c*d*x+b*c^2+a)/ln(F)^3/b^3/d

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maxima [C]  time = 6.92, size = 1438, normalized size = 15.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.66, size = 142, normalized size = 1.56 \[ \frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,\left (b^2\,c^4\,{\ln \relax (F)}^2+4\,b^2\,c^3\,d\,x\,{\ln \relax (F)}^2+6\,b^2\,c^2\,d^2\,x^2\,{\ln \relax (F)}^2+4\,b^2\,c\,d^3\,x^3\,{\ln \relax (F)}^2+b^2\,d^4\,x^4\,{\ln \relax (F)}^2-2\,b\,c^2\,\ln \relax (F)-4\,b\,c\,d\,x\,\ln \relax (F)-2\,b\,d^2\,x^2\,\ln \relax (F)+2\right )}{2\,b^3\,d\,{\ln \relax (F)}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.26, size = 214, normalized size = 2.35 \[ \begin {cases} \frac {F^{a + b \left (c + d x\right )^{2}} \left (b^{2} c^{4} \log {\relax (F )}^{2} + 4 b^{2} c^{3} d x \log {\relax (F )}^{2} + 6 b^{2} c^{2} d^{2} x^{2} \log {\relax (F )}^{2} + 4 b^{2} c d^{3} x^{3} \log {\relax (F )}^{2} + b^{2} d^{4} x^{4} \log {\relax (F )}^{2} - 2 b c^{2} \log {\relax (F )} - 4 b c d x \log {\relax (F )} - 2 b d^{2} x^{2} \log {\relax (F )} + 2\right )}{2 b^{3} d \log {\relax (F )}^{3}} & \text {for}\: 2 b^{3} d \log {\relax (F )}^{3} \neq 0 \\c^{5} x + \frac {5 c^{4} d x^{2}}{2} + \frac {10 c^{3} d^{2} x^{3}}{3} + \frac {5 c^{2} d^{3} x^{4}}{2} + c d^{4} x^{5} + \frac {d^{5} x^{6}}{6} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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(2*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))

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