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*(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])

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Rubi [A]  time = 0.278476, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095 \[ \frac{F^{a+b (c+d x)^2}}{b^3 d \log ^3(F)}-\frac{(c+d x)^2 F^{a+b (c+d x)^2}}{b^2 d \log ^2(F)}+\frac{(c+d x)^4 F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

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])

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Rubi in Sympy [A]  time = 15.989, size = 76, normalized size = 0.84 \[ \frac{F^{a + b \left (c + d x\right )^{2}} \left (c + d x\right )^{4}}{2 b d \log{\left (F \right )}} - \frac{F^{a + b \left (c + d x\right )^{2}} \left (c + d x\right )^{2}}{b^{2} d \log{\left (F \right )}^{2}} + \frac{F^{a + b \left (c + d x\right )^{2}}}{b^{3} d \log{\left (F \right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(F**(a+b*(d*x+c)**2)*(d*x+c)**5,x)

[Out]

F**(a + b*(c + d*x)**2)*(c + d*x)**4/(2*b*d*log(F)) - F**(a + b*(c + d*x)**2)*(c
+ d*x)**2/(b**2*d*log(F)**2) + F**(a + b*(c + d*x)**2)/(b**3*d*log(F)**3)

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Mathematica [A]  time = 0.0494947, 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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Maple [A]  time = 0.01, size = 138, normalized size = 1.5 \[{\frac{ \left ({d}^{4}{x}^{4}{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}+4\,{d}^{3}c{x}^{3}{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}+6\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{c}^{2}{d}^{2}{x}^{2}+4\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{c}^{3}dx+ \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{c}^{4}-2\,\ln \left ( F \right ) b{d}^{2}{x}^{2}-4\,\ln \left ( F \right ) bcdx-2\,\ln \left ( F \right ) b{c}^{2}+2 \right ){F}^{b{d}^{2}{x}^{2}+2\,bcdx+b{c}^{2}+a}}{2\, \left ( \ln \left ( F \right ) \right ) ^{3}{b}^{3}d}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[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

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Maxima [A]  time = 1.36954, size = 2430, normalized size = 26.7 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [A]  time = 0.246327, size = 162, normalized size = 1.78 \[ \frac{{\left ({\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (F\right )^{2} - 2 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right ) + 2\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{2 \, b^{3} d \log \left (F\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((d*x + c)^5*F^((d*x + c)^2*b + a),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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Sympy [A]  time = 0.603353, size = 214, normalized size = 2.35 \[ \begin{cases} \frac{F^{a + b \left (c + d x\right )^{2}} \left (b^{2} c^{4} \log{\left (F \right )}^{2} + 4 b^{2} c^{3} d x \log{\left (F \right )}^{2} + 6 b^{2} c^{2} d^{2} x^{2} \log{\left (F \right )}^{2} + 4 b^{2} c d^{3} x^{3} \log{\left (F \right )}^{2} + b^{2} d^{4} x^{4} \log{\left (F \right )}^{2} - 2 b c^{2} \log{\left (F \right )} - 4 b c d x \log{\left (F \right )} - 2 b d^{2} x^{2} \log{\left (F \right )} + 2\right )}{2 b^{3} d \log{\left (F \right )}^{3}} & \text{for}\: 2 b^{3} d \log{\left (F \right )}^{3} \neq 0 \\c^{5} x + \frac{5 c^{4} d x^{2}}{2} + \frac{10 c^{3} d^{2} x^{3}}{3} + \frac{5 c^{2} d^{3} x^{4}}{2} + c d^{4} x^{5} + \frac{d^{5} x^{6}}{6} & \text{otherwise} \end{cases} \]

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*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(2*b**3*d*log(F)**3, 0)), (c**5*x + 5*c**4*d*x**2/2 + 1
0*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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GIAC/XCAS [A]  time = 0.250627, size = 111, normalized size = 1.22 \[ \frac{{\left (b^{2} d^{4}{\left (x + \frac{c}{d}\right )}^{4}{\rm ln}\left (F\right )^{2} - 2 \, b d^{2}{\left (x + \frac{c}{d}\right )}^{2}{\rm ln}\left (F\right ) + 2\right )} e^{\left (b d^{2} x^{2}{\rm ln}\left (F\right ) + 2 \, b c d x{\rm ln}\left (F\right ) + b c^{2}{\rm ln}\left (F\right ) + a{\rm ln}\left (F\right )\right )}}{2 \, b^{3} d{\rm ln}\left (F\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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