3.284 \(\int F^{a+b (c+d x)^3} (c+d x)^8 \, dx\)

Optimal. Leaf size=96 \[ -\frac{2 (c+d x)^3 F^{a+b (c+d x)^3}}{3 b^2 d \log ^2(F)}+\frac{2 F^{a+b (c+d x)^3}}{3 b^3 d \log ^3(F)}+\frac{(c+d x)^6 F^{a+b (c+d x)^3}}{3 b d \log (F)} \]

[Out]

(2*F^(a + b*(c + d*x)^3))/(3*b^3*d*Log[F]^3) - (2*F^(a + b*(c + d*x)^3)*(c + d*x)^3)/(3*b^2*d*Log[F]^2) + (F^(
a + b*(c + d*x)^3)*(c + d*x)^6)/(3*b*d*Log[F])

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Rubi [A]  time = 0.208719, antiderivative size = 96, 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, Rules used = {2212, 2209} \[ -\frac{2 (c+d x)^3 F^{a+b (c+d x)^3}}{3 b^2 d \log ^2(F)}+\frac{2 F^{a+b (c+d x)^3}}{3 b^3 d \log ^3(F)}+\frac{(c+d x)^6 F^{a+b (c+d x)^3}}{3 b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^3)*(c + d*x)^8,x]

[Out]

(2*F^(a + b*(c + d*x)^3))/(3*b^3*d*Log[F]^3) - (2*F^(a + b*(c + d*x)^3)*(c + d*x)^3)/(3*b^2*d*Log[F]^2) + (F^(
a + b*(c + d*x)^3)*(c + d*x)^6)/(3*b*d*Log[F])

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])

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]

Rubi steps

\begin{align*} \int F^{a+b (c+d x)^3} (c+d x)^8 \, dx &=\frac{F^{a+b (c+d x)^3} (c+d x)^6}{3 b d \log (F)}-\frac{2 \int F^{a+b (c+d x)^3} (c+d x)^5 \, dx}{b \log (F)}\\ &=-\frac{2 F^{a+b (c+d x)^3} (c+d x)^3}{3 b^2 d \log ^2(F)}+\frac{F^{a+b (c+d x)^3} (c+d x)^6}{3 b d \log (F)}+\frac{2 \int F^{a+b (c+d x)^3} (c+d x)^2 \, dx}{b^2 \log ^2(F)}\\ &=\frac{2 F^{a+b (c+d x)^3}}{3 b^3 d \log ^3(F)}-\frac{2 F^{a+b (c+d x)^3} (c+d x)^3}{3 b^2 d \log ^2(F)}+\frac{F^{a+b (c+d x)^3} (c+d x)^6}{3 b d \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.0421616, size = 56, normalized size = 0.58 \[ \frac{F^{a+b (c+d x)^3} \left (b^2 \log ^2(F) (c+d x)^6-2 b \log (F) (c+d x)^3+2\right )}{3 b^3 d \log ^3(F)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^3)*(c + d*x)^8,x]

[Out]

(F^(a + b*(c + d*x)^3)*(2 - 2*b*(c + d*x)^3*Log[F] + b^2*(c + d*x)^6*Log[F]^2))/(3*b^3*d*Log[F]^3)

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Maple [B]  time = 0.008, size = 200, normalized size = 2.1 \begin{align*}{\frac{ \left ({d}^{6}{x}^{6} \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}+6\,c{d}^{5}{x}^{5} \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}+15\,{c}^{2}{d}^{4}{x}^{4} \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}+20\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{c}^{3}{d}^{3}{x}^{3}+15\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{c}^{4}{d}^{2}{x}^{2}+6\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{c}^{5}dx+ \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{c}^{6}-2\,\ln \left ( F \right ) b{d}^{3}{x}^{3}-6\,\ln \left ( F \right ) bc{d}^{2}{x}^{2}-6\,\ln \left ( F \right ) b{c}^{2}dx-2\,\ln \left ( F \right ) b{c}^{3}+2 \right ){F}^{b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a}}{3\, \left ( \ln \left ( F \right ) \right ) ^{3}{b}^{3}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b*(d*x+c)^3)*(d*x+c)^8,x)

[Out]

1/3*(d^6*x^6*ln(F)^2*b^2+6*c*d^5*x^5*ln(F)^2*b^2+15*c^2*d^4*x^4*ln(F)^2*b^2+20*ln(F)^2*b^2*c^3*d^3*x^3+15*ln(F
)^2*b^2*c^4*d^2*x^2+6*ln(F)^2*b^2*c^5*d*x+ln(F)^2*b^2*c^6-2*ln(F)*b*d^3*x^3-6*ln(F)*b*c*d^2*x^2-6*ln(F)*b*c^2*
d*x-2*ln(F)*b*c^3+2)*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)^3/b^3/d

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Maxima [B]  time = 1.58906, size = 416, normalized size = 4.33 \begin{align*} \frac{{\left (F^{b c^{3} + a} b^{2} d^{6} x^{6} \log \left (F\right )^{2} + 6 \, F^{b c^{3} + a} b^{2} c d^{5} x^{5} \log \left (F\right )^{2} + 15 \, F^{b c^{3} + a} b^{2} c^{2} d^{4} x^{4} \log \left (F\right )^{2} + F^{b c^{3} + a} b^{2} c^{6} \log \left (F\right )^{2} - 2 \, F^{b c^{3} + a} b c^{3} \log \left (F\right ) + 2 \,{\left (10 \, F^{b c^{3} + a} b^{2} c^{3} d^{3} \log \left (F\right )^{2} - F^{b c^{3} + a} b d^{3} \log \left (F\right )\right )} x^{3} + 3 \,{\left (5 \, F^{b c^{3} + a} b^{2} c^{4} d^{2} \log \left (F\right )^{2} - 2 \, F^{b c^{3} + a} b c d^{2} \log \left (F\right )\right )} x^{2} + 6 \,{\left (F^{b c^{3} + a} b^{2} c^{5} d \log \left (F\right )^{2} - F^{b c^{3} + a} b c^{2} d \log \left (F\right )\right )} x + 2 \, 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^{3} d \log \left (F\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^3)*(d*x+c)^8,x, algorithm="maxima")

[Out]

1/3*(F^(b*c^3 + a)*b^2*d^6*x^6*log(F)^2 + 6*F^(b*c^3 + a)*b^2*c*d^5*x^5*log(F)^2 + 15*F^(b*c^3 + a)*b^2*c^2*d^
4*x^4*log(F)^2 + F^(b*c^3 + a)*b^2*c^6*log(F)^2 - 2*F^(b*c^3 + a)*b*c^3*log(F) + 2*(10*F^(b*c^3 + a)*b^2*c^3*d
^3*log(F)^2 - F^(b*c^3 + a)*b*d^3*log(F))*x^3 + 3*(5*F^(b*c^3 + a)*b^2*c^4*d^2*log(F)^2 - 2*F^(b*c^3 + a)*b*c*
d^2*log(F))*x^2 + 6*(F^(b*c^3 + a)*b^2*c^5*d*log(F)^2 - F^(b*c^3 + a)*b*c^2*d*log(F))*x + 2*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^3*d*log(F)^3)

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Fricas [A]  time = 1.53499, size = 371, normalized size = 3.86 \begin{align*} \frac{{\left ({\left (b^{2} d^{6} x^{6} + 6 \, b^{2} c d^{5} x^{5} + 15 \, b^{2} c^{2} d^{4} x^{4} + 20 \, b^{2} c^{3} d^{3} x^{3} + 15 \, b^{2} c^{4} d^{2} x^{2} + 6 \, b^{2} c^{5} d x + b^{2} c^{6}\right )} \log \left (F\right )^{2} - 2 \,{\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 ) + 2\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^{3} d \log \left (F\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^3)*(d*x+c)^8,x, algorithm="fricas")

[Out]

1/3*((b^2*d^6*x^6 + 6*b^2*c*d^5*x^5 + 15*b^2*c^2*d^4*x^4 + 20*b^2*c^3*d^3*x^3 + 15*b^2*c^4*d^2*x^2 + 6*b^2*c^5
*d*x + b^2*c^6)*log(F)^2 - 2*(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*log(F) + 2)*F^(b*d^3*x^3 + 3*b*
c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)/(b^3*d*log(F)^3)

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Sympy [A]  time = 0.275429, size = 306, normalized size = 3.19 \begin{align*} \begin{cases} \frac{F^{a + b \left (c + d x\right )^{3}} \left (b^{2} c^{6} \log{\left (F \right )}^{2} + 6 b^{2} c^{5} d x \log{\left (F \right )}^{2} + 15 b^{2} c^{4} d^{2} x^{2} \log{\left (F \right )}^{2} + 20 b^{2} c^{3} d^{3} x^{3} \log{\left (F \right )}^{2} + 15 b^{2} c^{2} d^{4} x^{4} \log{\left (F \right )}^{2} + 6 b^{2} c d^{5} x^{5} \log{\left (F \right )}^{2} + b^{2} d^{6} x^{6} \log{\left (F \right )}^{2} - 2 b c^{3} \log{\left (F \right )} - 6 b c^{2} d x \log{\left (F \right )} - 6 b c d^{2} x^{2} \log{\left (F \right )} - 2 b d^{3} x^{3} \log{\left (F \right )} + 2\right )}{3 b^{3} d \log{\left (F \right )}^{3}} & \text{for}\: 3 b^{3} d \log{\left (F \right )}^{3} \neq 0 \\c^{8} x + 4 c^{7} d x^{2} + \frac{28 c^{6} d^{2} x^{3}}{3} + 14 c^{5} d^{3} x^{4} + 14 c^{4} d^{4} x^{5} + \frac{28 c^{3} d^{5} x^{6}}{3} + 4 c^{2} d^{6} x^{7} + c d^{7} x^{8} + \frac{d^{8} x^{9}}{9} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b*(d*x+c)**3)*(d*x+c)**8,x)

[Out]

Piecewise((F**(a + b*(c + d*x)**3)*(b**2*c**6*log(F)**2 + 6*b**2*c**5*d*x*log(F)**2 + 15*b**2*c**4*d**2*x**2*l
og(F)**2 + 20*b**2*c**3*d**3*x**3*log(F)**2 + 15*b**2*c**2*d**4*x**4*log(F)**2 + 6*b**2*c*d**5*x**5*log(F)**2
+ b**2*d**6*x**6*log(F)**2 - 2*b*c**3*log(F) - 6*b*c**2*d*x*log(F) - 6*b*c*d**2*x**2*log(F) - 2*b*d**3*x**3*lo
g(F) + 2)/(3*b**3*d*log(F)**3), Ne(3*b**3*d*log(F)**3, 0)), (c**8*x + 4*c**7*d*x**2 + 28*c**6*d**2*x**3/3 + 14
*c**5*d**3*x**4 + 14*c**4*d**4*x**5 + 28*c**3*d**5*x**6/3 + 4*c**2*d**6*x**7 + c*d**7*x**8 + d**8*x**9/9, True
))

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Giac [B]  time = 1.28072, size = 952, normalized size = 9.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^3)*(d*x+c)^8,x, algorithm="giac")

[Out]

1/3*(b^2*d^6*x^6*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*c^3*log(F) + a*log(F))*lo
g(F)^2 + 6*b^2*c*d^5*x^5*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*c^3*log(F) + a*lo
g(F))*log(F)^2 + 15*b^2*c^2*d^4*x^4*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*c^3*lo
g(F) + a*log(F))*log(F)^2 + 20*b^2*c^3*d^3*x^3*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*c^3*log(F) + a*log(F))*log(F)^2 + 15*b^2*c^4*d^2*x^2*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*c^3*log(F) + a*log(F))*log(F)^2 + 6*b^2*c^5*d*x*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*c^3*log(F) + a*log(F))*log(F)^2 + b^2*c^6*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*c^3*log(F) + a*log(F))*log(F)^2 - 2*b*d^3*x^3*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*c^3*log(F) + a*log(F))*log(F) - 6*b*c*d^2*x^2*e^(b*d^3*x^3*log(F) + 3*b*c*d^2*x^2*lo
g(F) + 3*b*c^2*d*x*log(F) + b*c^3*log(F) + a*log(F))*log(F) - 6*b*c^2*d*x*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*c^3*log(F) + a*log(F))*log(F) - 2*b*c^3*e^(b*d^3*x^3*log(F) + 3*b*c*d^2*x^2*lo
g(F) + 3*b*c^2*d*x*log(F) + b*c^3*log(F) + a*log(F))*log(F) + 2*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*c^3*log(F) + a*log(F)))/(b^3*d*log(F)^3)