3.285 \(\int F^{a+b (c+d x)^3} (c+d x)^5 \, dx\)

Optimal. Leaf size=62 \[ \frac {(c+d x)^3 F^{a+b (c+d x)^3}}{3 b d \log (F)}-\frac {F^{a+b (c+d x)^3}}{3 b^2 d \log ^2(F)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2212, 2209} \[ \frac {(c+d x)^3 F^{a+b (c+d x)^3}}{3 b d \log (F)}-\frac {F^{a+b (c+d x)^3}}{3 b^2 d \log ^2(F)} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^3)*(c + d*x)^5,x]

[Out]

-F^(a + b*(c + d*x)^3)/(3*b^2*d*Log[F]^2) + (F^(a + b*(c + d*x)^3)*(c + d*x)^3)/(3*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)^3} (c+d x)^5 \, dx &=\frac {F^{a+b (c+d x)^3} (c+d x)^3}{3 b d \log (F)}-\frac {\int F^{a+b (c+d x)^3} (c+d x)^2 \, dx}{b \log (F)}\\ &=-\frac {F^{a+b (c+d x)^3}}{3 b^2 d \log ^2(F)}+\frac {F^{a+b (c+d x)^3} (c+d x)^3}{3 b d \log (F)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 40, normalized size = 0.65 \[ \frac {F^{a+b (c+d x)^3} \left (b \log (F) (c+d x)^3-1\right )}{3 b^2 d \log ^2(F)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^3)*(c + d*x)^5,x]

[Out]

(F^(a + b*(c + d*x)^3)*(-1 + b*(c + d*x)^3*Log[F]))/(3*b^2*d*Log[F]^2)

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 84, normalized size = 1.35 \[ \frac {{\left ({\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \relax (F) - 1\right )} F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{3 \, b^{2} d \log \relax (F)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^3)*(d*x+c)^5,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.39, size = 891, normalized size = 14.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^3)*(d*x+c)^5,x, algorithm="giac")

[Out]

1/6*(2*(2*(((d*x + c)^3*b*log(abs(F)) - 1)*(pi^2*b^2*sgn(F) - pi^2*b^2 + 2*b^2*log(abs(F))^2)/((pi^2*b^2*sgn(F
) - pi^2*b^2 + 2*b^2*log(abs(F))^2)^2 + 4*(pi*b^2*log(abs(F))*sgn(F) - pi*b^2*log(abs(F)))^2) + (pi*(d*x + c)^
3*b*sgn(F) - pi*(d*x + c)^3*b)*(pi*b^2*log(abs(F))*sgn(F) - pi*b^2*log(abs(F)))/((pi^2*b^2*sgn(F) - pi^2*b^2 +
2*b^2*log(abs(F))^2)^2 + 4*(pi*b^2*log(abs(F))*sgn(F) - pi*b^2*log(abs(F)))^2))*cos(-1/2*pi*b*d^3*x^3*sgn(F)
+ 1/2*pi*b*d^3*x^3 - 3/2*pi*b*c*d^2*x^2*sgn(F) + 3/2*pi*b*c*d^2*x^2 - 3/2*pi*b*c^2*d*x*sgn(F) + 3/2*pi*b*c^2*d
*x - 1/2*pi*b*c^3*sgn(F) + 1/2*pi*b*c^3 - 1/2*pi*a*sgn(F) + 1/2*pi*a) + ((pi*(d*x + c)^3*b*sgn(F) - pi*(d*x +
c)^3*b)*(pi^2*b^2*sgn(F) - pi^2*b^2 + 2*b^2*log(abs(F))^2)/((pi^2*b^2*sgn(F) - pi^2*b^2 + 2*b^2*log(abs(F))^2)
^2 + 4*(pi*b^2*log(abs(F))*sgn(F) - pi*b^2*log(abs(F)))^2) - 4*((d*x + c)^3*b*log(abs(F)) - 1)*(pi*b^2*log(abs
(F))*sgn(F) - pi*b^2*log(abs(F)))/((pi^2*b^2*sgn(F) - pi^2*b^2 + 2*b^2*log(abs(F))^2)^2 + 4*(pi*b^2*log(abs(F)
)*sgn(F) - pi*b^2*log(abs(F)))^2))*sin(-1/2*pi*b*d^3*x^3*sgn(F) + 1/2*pi*b*d^3*x^3 - 3/2*pi*b*c*d^2*x^2*sgn(F)
+ 3/2*pi*b*c*d^2*x^2 - 3/2*pi*b*c^2*d*x*sgn(F) + 3/2*pi*b*c^2*d*x - 1/2*pi*b*c^3*sgn(F) + 1/2*pi*b*c^3 - 1/2*
pi*a*sgn(F) + 1/2*pi*a))*e^((d*x + c)^3*b*log(abs(F)) + a*log(abs(F))) - ((2*(d*x + c)^3*b*i*log(abs(F)) - pi*
(d*x + c)^3*b*sgn(F) + pi*(d*x + c)^3*b - 2*i)*e^(1/2*(pi*(d*x + c)^3*b*(sgn(F) - 1) + pi*a*(sgn(F) - 1))*i)/(
2*pi*b^2*i*log(abs(F))*sgn(F) - 2*pi*b^2*i*log(abs(F)) + pi^2*b^2*sgn(F) - pi^2*b^2 + 2*b^2*log(abs(F))^2) + (
2*(d*x + c)^3*b*i*log(abs(F)) + pi*(d*x + c)^3*b*sgn(F) - pi*(d*x + c)^3*b - 2*i)*e^(-1/2*(pi*(d*x + c)^3*b*(s
gn(F) - 1) + pi*a*(sgn(F) - 1))*i)/(2*pi*b^2*i*log(abs(F))*sgn(F) - 2*pi*b^2*i*log(abs(F)) - pi^2*b^2*sgn(F) +
pi^2*b^2 - 2*b^2*log(abs(F))^2))*e^((d*x + c)^3*b*log(abs(F)) + a*log(abs(F)))/i)/d

________________________________________________________________________________________

maple [A]  time = 0.01, size = 89, normalized size = 1.44 \[ \frac {\left (b \,d^{3} x^{3} \ln \relax (F )+3 b c \,d^{2} x^{2} \ln \relax (F )+3 b \,c^{2} d x \ln \relax (F )+b \,c^{3} \ln \relax (F )-1\right ) F^{b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a}}{3 b^{2} d \ln \relax (F )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+(d*x+c)^3*b)*(d*x+c)^5,x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 2.09, size = 133, normalized size = 2.15 \[ \frac {{\left (F^{b c^{3} + a} b d^{3} x^{3} \log \relax (F) + 3 \, F^{b c^{3} + a} b c d^{2} x^{2} \log \relax (F) + 3 \, F^{b c^{3} + a} b c^{2} d x \log \relax (F) + F^{b c^{3} + a} b c^{3} \log \relax (F) - F^{b c^{3} + a}\right )} e^{\left (b d^{3} x^{3} \log \relax (F) + 3 \, b c d^{2} x^{2} \log \relax (F) + 3 \, b c^{2} d x \log \relax (F)\right )}}{3 \, b^{2} d \log \relax (F)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^3)*(d*x+c)^5,x, algorithm="maxima")

[Out]

1/3*(F^(b*c^3 + a)*b*d^3*x^3*log(F) + 3*F^(b*c^3 + a)*b*c*d^2*x^2*log(F) + 3*F^(b*c^3 + a)*b*c^2*d*x*log(F) +
F^(b*c^3 + a)*b*c^3*log(F) - F^(b*c^3 + a))*e^(b*d^3*x^3*log(F) + 3*b*c*d^2*x^2*log(F) + 3*b*c^2*d*x*log(F))/(
b^2*d*log(F)^2)

________________________________________________________________________________________

mupad [B]  time = 3.63, size = 95, normalized size = 1.53 \[ \frac {F^{b\,d^3\,x^3}\,F^{3\,b\,c^2\,d\,x}\,F^a\,F^{b\,c^3}\,F^{3\,b\,c\,d^2\,x^2}\,\left (b\,\ln \relax (F)\,c^3+3\,b\,\ln \relax (F)\,c^2\,d\,x+3\,b\,\ln \relax (F)\,c\,d^2\,x^2+b\,\ln \relax (F)\,d^3\,x^3-1\right )}{3\,b^2\,d\,{\ln \relax (F)}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b*(c + d*x)^3)*(c + d*x)^5,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.24, size = 144, normalized size = 2.32 \[ \begin {cases} \frac {F^{a + b \left (c + d x\right )^{3}} \left (b c^{3} \log {\relax (F )} + 3 b c^{2} d x \log {\relax (F )} + 3 b c d^{2} x^{2} \log {\relax (F )} + b d^{3} x^{3} \log {\relax (F )} - 1\right )}{3 b^{2} d \log {\relax (F )}^{2}} & \text {for}\: 3 b^{2} d \log {\relax (F )}^{2} \neq 0 \\c^{5} x + \frac {5 c^{4} d x^{2}}{2} + \frac {10 c^{3} d^{2} x^{3}}{3} + \frac {5 c^{2} d^{3} x^{4}}{2} + c d^{4} x^{5} + \frac {d^{5} x^{6}}{6} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b*(d*x+c)**3)*(d*x+c)**5,x)

[Out]

Piecewise((F**(a + b*(c + d*x)**3)*(b*c**3*log(F) + 3*b*c**2*d*x*log(F) + 3*b*c*d**2*x**2*log(F) + b*d**3*x**3
*log(F) - 1)/(3*b**2*d*log(F)**2), Ne(3*b**2*d*log(F)**2, 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))

________________________________________________________________________________________