\(\int F^{a+b (c+d x)^3} (c+d x)^5 \, dx\) [285]

Optimal result

Integrand size = 21, antiderivative size = 62 \[ \int F^{a+b (c+d x)^3} (c+d x)^5 \, dx=-\frac {F^{a+b (c+d x)^3}}{3 b^2 d \log ^2(F)}+\frac {F^{a+b (c+d x)^3} (c+d x)^3}{3 b d \log (F)} \]

[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] (verified)

Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2243, 2240} \[ \int F^{a+b (c+d x)^3} (c+d x)^5 \, dx=\frac {(c+d x)^3 F^{a+b (c+d x)^3}}{3 b d \log (F)}-\frac {F^{a+b (c+d x)^3}}{3 b^2 d \log ^2(F)} \]

[In]

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

[Out]

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

Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.65 \[ \int F^{a+b (c+d x)^3} (c+d x)^5 \, dx=\frac {F^{a+b (c+d x)^3} \left (-1+b (c+d x)^3 \log (F)\right )}{3 b^2 d \log ^2(F)} \]

[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)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.44

method	result	size
gosper	\(\frac {\left (\ln \left (F \right ) b \,d^{3} x^{3}+3 \ln \left (F \right ) b c \,d^{2} x^{2}+3 \ln \left (F \right ) b \,c^{2} d x +\ln \left (F \right ) b \,c^{3}-1\right ) F^{b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a}}{3 b^{2} d \ln \left (F \right )^{2}}\)	\(89\)
risch	\(\frac {\left (\ln \left (F \right ) b \,d^{3} x^{3}+3 \ln \left (F \right ) b c \,d^{2} x^{2}+3 \ln \left (F \right ) b \,c^{2} d x +\ln \left (F \right ) b \,c^{3}-1\right ) F^{b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a}}{3 b^{2} d \ln \left (F \right )^{2}}\)	\(89\)
parallelrisch	\(\frac {d^{3} F^{a +b \left (d x +c \right )^{3}} x^{3} b \ln \left (F \right )+3 c \,d^{2} F^{a +b \left (d x +c \right )^{3}} x^{2} b \ln \left (F \right )+3 c^{2} F^{a +b \left (d x +c \right )^{3}} x b \ln \left (F \right ) d +\ln \left (F \right ) F^{a +b \left (d x +c \right )^{3}} b \,c^{3}-F^{a +b \left (d x +c \right )^{3}}}{3 b^{2} d \ln \left (F \right )^{2}}\)	\(120\)
norman	\(\frac {c^{2} x \,{\mathrm e}^{\left (a +b \left (d x +c \right )^{3}\right ) \ln \left (F \right )}}{\ln \left (F \right ) b}+\frac {d c \,x^{2} {\mathrm e}^{\left (a +b \left (d x +c \right )^{3}\right ) \ln \left (F \right )}}{\ln \left (F \right ) b}+\frac {\left (\ln \left (F \right ) b \,c^{3}-1\right ) {\mathrm e}^{\left (a +b \left (d x +c \right )^{3}\right ) \ln \left (F \right )}}{3 b^{2} d \ln \left (F \right )^{2}}+\frac {d^{2} x^{3} {\mathrm e}^{\left (a +b \left (d x +c \right )^{3}\right ) \ln \left (F \right )}}{3 \ln \left (F \right ) b}\)	\(123\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.35 \[ \int F^{a+b (c+d x)^3} (c+d x)^5 \, dx=\frac {{\left ({\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right ) - 1\right )} F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{3 \, b^{2} d \log \left (F\right )^{2}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (49) = 98\).

Time = 0.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.31 \[ \int F^{a+b (c+d x)^3} (c+d x)^5 \, dx=\begin {cases} \frac {F^{a + b \left (c + d x\right )^{3}} \left (b c^{3} \log {\left (F \right )} + 3 b c^{2} d x \log {\left (F \right )} + 3 b c d^{2} x^{2} \log {\left (F \right )} + b d^{3} x^{3} \log {\left (F \right )} - 1\right )}{3 b^{2} d \log {\left (F \right )}^{2}} & \text {for}\: b^{2} d \log {\left (F \right )}^{2} \neq 0 \\c^{5} x + \frac {5 c^{4} d x^{2}}{2} + \frac {10 c^{3} d^{2} x^{3}}{3} + \frac {5 c^{2} d^{3} x^{4}}{2} + c d^{4} x^{5} + \frac {d^{5} x^{6}}{6} & \text {otherwise} \end {cases} \]

[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(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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (58) = 116\).

Time = 0.35 (sec) , antiderivative size = 133, normalized size of antiderivative = 2.15 \[ \int F^{a+b (c+d x)^3} (c+d x)^5 \, dx=\frac {{\left (F^{b c^{3} + a} b d^{3} x^{3} \log \left (F\right ) + 3 \, F^{b c^{3} + a} b c d^{2} x^{2} \log \left (F\right ) + 3 \, F^{b c^{3} + a} b c^{2} d x \log \left (F\right ) + F^{b c^{3} + a} b c^{3} \log \left (F\right ) - F^{b c^{3} + a}\right )} e^{\left (b d^{3} x^{3} \log \left (F\right ) + 3 \, b c d^{2} x^{2} \log \left (F\right ) + 3 \, b c^{2} d x \log \left (F\right )\right )}}{3 \, b^{2} d \log \left (F\right )^{2}} \]

[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)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 1014, normalized size of antiderivative = 16.35 \[ \int F^{a+b (c+d x)^3} (c+d x)^5 \, dx=\text {Too large to display} \]

[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))) - I*((pi*(d*x + c)^3*b*sgn(F) - pi*(d*x
+ c)^3*b - 2*I*(d*x + c)^3*b*log(abs(F)) + 2*I)*e^(1/2*I*pi*b*d^3*x^3*sgn(F) - 1/2*I*pi*b*d^3*x^3 + 3/2*I*pi*
b*c*d^2*x^2*sgn(F) - 3/2*I*pi*b*c*d^2*x^2 + 3/2*I*pi*b*c^2*d*x*sgn(F) - 3/2*I*pi*b*c^2*d*x + 1/2*I*pi*b*c^3*sg
n(F) - 1/2*I*pi*b*c^3 + 1/2*I*pi*a*sgn(F) - 1/2*I*pi*a)/(pi^2*b^2*sgn(F) + 2*I*pi*b^2*log(abs(F))*sgn(F) - pi^
2*b^2 - 2*I*pi*b^2*log(abs(F)) + 2*b^2*log(abs(F))^2) + (pi*(d*x + c)^3*b*sgn(F) - pi*(d*x + c)^3*b + 2*I*(d*x
+ c)^3*b*log(abs(F)) - 2*I)*e^(-1/2*I*pi*b*d^3*x^3*sgn(F) + 1/2*I*pi*b*d^3*x^3 - 3/2*I*pi*b*c*d^2*x^2*sgn(F)
+ 3/2*I*pi*b*c*d^2*x^2 - 3/2*I*pi*b*c^2*d*x*sgn(F) + 3/2*I*pi*b*c^2*d*x - 1/2*I*pi*b*c^3*sgn(F) + 1/2*I*pi*b*c
^3 - 1/2*I*pi*a*sgn(F) + 1/2*I*pi*a)/(pi^2*b^2*sgn(F) - 2*I*pi*b^2*log(abs(F))*sgn(F) - pi^2*b^2 + 2*I*pi*b^2*
log(abs(F)) + 2*b^2*log(abs(F))^2))*e^((d*x + c)^3*b*log(abs(F)) + a*log(abs(F))))/d

Mupad [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.53 \[ \int F^{a+b (c+d x)^3} (c+d x)^5 \, dx=\frac {F^{b\,d^3\,x^3}\,F^{3\,b\,c^2\,d\,x}\,F^a\,F^{b\,c^3}\,F^{3\,b\,c\,d^2\,x^2}\,\left (b\,\ln \left (F\right )\,c^3+3\,b\,\ln \left (F\right )\,c^2\,d\,x+3\,b\,\ln \left (F\right )\,c\,d^2\,x^2+b\,\ln \left (F\right )\,d^3\,x^3-1\right )}{3\,b^2\,d\,{\ln \left (F\right )}^2} \]

[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)