3.349 \(\int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^{10}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 96, 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} \[ \frac {2 F^{a+\frac {b}{(c+d x)^3}}}{3 b^2 d \log ^2(F) (c+d x)^3}-\frac {2 F^{a+\frac {b}{(c+d x)^3}}}{3 b^3 d \log ^3(F)}-\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F) (c+d x)^6} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x)^3)/(c + d*x)^10,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))/(3*b^2*d*(c + d*x)^3*Log[F]^2) - F^(
a + b/(c + d*x)^3)/(3*b*d*(c + d*x)^6*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 \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^{10}} \, dx &=-\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d (c+d x)^6 \log (F)}-\frac {2 \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^7} \, dx}{b \log (F)}\\ &=\frac {2 F^{a+\frac {b}{(c+d x)^3}}}{3 b^2 d (c+d x)^3 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d (c+d x)^6 \log (F)}+\frac {2 \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^4} \, dx}{b^2 \log ^2(F)}\\ &=-\frac {2 F^{a+\frac {b}{(c+d x)^3}}}{3 b^3 d \log ^3(F)}+\frac {2 F^{a+\frac {b}{(c+d x)^3}}}{3 b^2 d (c+d x)^3 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d (c+d x)^6 \log (F)}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 64, normalized size = 0.67 \[ -\frac {F^{a+\frac {b}{(c+d x)^3}} \left (b^2 \log ^2(F)-2 b \log (F) (c+d x)^3+2 (c+d x)^6\right )}{3 b^3 d \log ^3(F) (c+d x)^6} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x)^3)/(c + d*x)^10,x]

[Out]

-1/3*(F^(a + b/(c + d*x)^3)*(2*(c + d*x)^6 - 2*b*(c + d*x)^3*Log[F] + b^2*Log[F]^2))/(b^3*d*(c + d*x)^6*Log[F]
^3)

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fricas [B]  time = 0.41, size = 265, normalized size = 2.76 \[ -\frac {{\left (2 \, d^{6} x^{6} + 12 \, c d^{5} x^{5} + 30 \, c^{2} d^{4} x^{4} + 40 \, c^{3} d^{3} x^{3} + 30 \, c^{4} d^{2} x^{2} + 12 \, c^{5} d x + 2 \, c^{6} + b^{2} \log \relax (F)^{2} - 2 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \relax (F)\right )} F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{3 \, {\left (b^{3} d^{7} x^{6} + 6 \, b^{3} c d^{6} x^{5} + 15 \, b^{3} c^{2} d^{5} x^{4} + 20 \, b^{3} c^{3} d^{4} x^{3} + 15 \, b^{3} c^{4} d^{3} x^{2} + 6 \, b^{3} c^{5} d^{2} x + b^{3} c^{6} d\right )} \log \relax (F)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^10,x, algorithm="fricas")

[Out]

-1/3*(2*d^6*x^6 + 12*c*d^5*x^5 + 30*c^2*d^4*x^4 + 40*c^3*d^3*x^3 + 30*c^4*d^2*x^2 + 12*c^5*d*x + 2*c^6 + b^2*l
og(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))*F^((a*d^3*x^3 + 3*a*c*d^2*x^2 + 3*a*c^2*
d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))/((b^3*d^7*x^6 + 6*b^3*c*d^6*x^5 + 15*b^3*c^2*d^5*x
^4 + 20*b^3*c^3*d^4*x^3 + 15*b^3*c^4*d^3*x^2 + 6*b^3*c^5*d^2*x + b^3*c^6*d)*log(F)^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{3}}}}{{\left (d x + c\right )}^{10}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^10,x, algorithm="giac")

[Out]

integrate(F^(a + b/(d*x + c)^3)/(d*x + c)^10, x)

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maple [B]  time = 0.14, size = 434, normalized size = 4.52 \[ \frac {-\frac {2 d^{8} x^{9} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{3 b^{3} \ln \relax (F )^{3}}-\frac {6 c \,d^{7} x^{8} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{b^{3} \ln \relax (F )^{3}}-\frac {24 c^{2} d^{6} x^{7} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{b^{3} \ln \relax (F )^{3}}+\frac {2 \left (-84 c^{3}+b \ln \relax (F )\right ) d^{5} x^{6} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{3 b^{3} \ln \relax (F )^{3}}+\frac {4 \left (-21 c^{3}+b \ln \relax (F )\right ) c \,d^{4} x^{5} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{b^{3} \ln \relax (F )^{3}}+\frac {2 \left (-42 c^{3}+5 b \ln \relax (F )\right ) c^{2} d^{3} x^{4} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{b^{3} \ln \relax (F )^{3}}-\frac {\left (168 c^{6}-40 b \,c^{3} \ln \relax (F )+b^{2} \ln \relax (F )^{2}\right ) d^{2} x^{3} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{3 b^{3} \ln \relax (F )^{3}}-\frac {\left (24 c^{6}-10 b \,c^{3} \ln \relax (F )+b^{2} \ln \relax (F )^{2}\right ) c d \,x^{2} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{b^{3} \ln \relax (F )^{3}}-\frac {\left (6 c^{6}-4 b \,c^{3} \ln \relax (F )+b^{2} \ln \relax (F )^{2}\right ) c^{2} x \,{\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{b^{3} \ln \relax (F )^{3}}-\frac {\left (2 c^{6}-2 b \,c^{3} \ln \relax (F )+b^{2} \ln \relax (F )^{2}\right ) c^{3} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{3 b^{3} d \ln \relax (F )^{3}}}{\left (d x +c \right )^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+1/(d*x+c)^3*b)/(d*x+c)^10,x)

[Out]

(-2/3*d^8/ln(F)^3/b^3*x^9*exp((a+1/(d*x+c)^3*b)*ln(F))-c^2*(6*c^6-4*b*c^3*ln(F)+b^2*ln(F)^2)/b^3/ln(F)^3*x*exp
((a+1/(d*x+c)^3*b)*ln(F))-1/3*d^2*(168*c^6-40*b*c^3*ln(F)+b^2*ln(F)^2)/ln(F)^3/b^3*x^3*exp((a+1/(d*x+c)^3*b)*l
n(F))+2/3*d^5*(-84*c^3+b*ln(F))/ln(F)^3/b^3*x^6*exp((a+1/(d*x+c)^3*b)*ln(F))-24*d^6*c^2/ln(F)^3/b^3*x^7*exp((a
+1/(d*x+c)^3*b)*ln(F))-6*d^7*c/ln(F)^3/b^3*x^8*exp((a+1/(d*x+c)^3*b)*ln(F))-1/3*(2*c^6-2*b*c^3*ln(F)+b^2*ln(F)
^2)*c^3/b^3/ln(F)^3/d*exp((a+1/(d*x+c)^3*b)*ln(F))-c*d*(24*c^6-10*b*c^3*ln(F)+b^2*ln(F)^2)/ln(F)^3/b^3*x^2*exp
((a+1/(d*x+c)^3*b)*ln(F))+4*c*d^4*(-21*c^3+b*ln(F))/ln(F)^3/b^3*x^5*exp((a+1/(d*x+c)^3*b)*ln(F))+2*c^2*d^3*(-4
2*c^3+5*b*ln(F))/ln(F)^3/b^3*x^4*exp((a+1/(d*x+c)^3*b)*ln(F)))/(d*x+c)^9

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maxima [B]  time = 0.89, size = 300, normalized size = 3.12 \[ -\frac {{\left (2 \, F^{a} d^{6} x^{6} + 12 \, F^{a} c d^{5} x^{5} + 30 \, F^{a} c^{2} d^{4} x^{4} + 2 \, F^{a} c^{6} - 2 \, F^{a} b c^{3} \log \relax (F) + F^{a} b^{2} \log \relax (F)^{2} + 2 \, {\left (20 \, F^{a} c^{3} d^{3} - F^{a} b d^{3} \log \relax (F)\right )} x^{3} + 6 \, {\left (5 \, F^{a} c^{4} d^{2} - F^{a} b c d^{2} \log \relax (F)\right )} x^{2} + 6 \, {\left (2 \, F^{a} c^{5} d - F^{a} b c^{2} d \log \relax (F)\right )} x\right )} F^{\frac {b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{3 \, {\left (b^{3} d^{7} x^{6} \log \relax (F)^{3} + 6 \, b^{3} c d^{6} x^{5} \log \relax (F)^{3} + 15 \, b^{3} c^{2} d^{5} x^{4} \log \relax (F)^{3} + 20 \, b^{3} c^{3} d^{4} x^{3} \log \relax (F)^{3} + 15 \, b^{3} c^{4} d^{3} x^{2} \log \relax (F)^{3} + 6 \, b^{3} c^{5} d^{2} x \log \relax (F)^{3} + b^{3} c^{6} d \log \relax (F)^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^10,x, algorithm="maxima")

[Out]

-1/3*(2*F^a*d^6*x^6 + 12*F^a*c*d^5*x^5 + 30*F^a*c^2*d^4*x^4 + 2*F^a*c^6 - 2*F^a*b*c^3*log(F) + F^a*b^2*log(F)^
2 + 2*(20*F^a*c^3*d^3 - F^a*b*d^3*log(F))*x^3 + 6*(5*F^a*c^4*d^2 - F^a*b*c*d^2*log(F))*x^2 + 6*(2*F^a*c^5*d -
F^a*b*c^2*d*log(F))*x)*F^(b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))/(b^3*d^7*x^6*log(F)^3 + 6*b^3*c*d^6*x^5
*log(F)^3 + 15*b^3*c^2*d^5*x^4*log(F)^3 + 20*b^3*c^3*d^4*x^3*log(F)^3 + 15*b^3*c^4*d^3*x^2*log(F)^3 + 6*b^3*c^
5*d^2*x*log(F)^3 + b^3*c^6*d*log(F)^3)

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mupad [B]  time = 4.08, size = 263, normalized size = 2.74 \[ -\frac {F^a\,F^{\frac {b}{c^3+3\,c^2\,d\,x+3\,c\,d^2\,x^2+d^3\,x^3}}\,\left (\frac {2\,x^6}{3\,b^3\,d\,{\ln \relax (F)}^3}+\frac {b^2\,{\ln \relax (F)}^2-2\,b\,c^3\,\ln \relax (F)+2\,c^6}{3\,b^3\,d^7\,{\ln \relax (F)}^3}+\frac {4\,c\,x^5}{b^3\,d^2\,{\ln \relax (F)}^3}+\frac {10\,c^2\,x^4}{b^3\,d^3\,{\ln \relax (F)}^3}-\frac {2\,x^3\,\left (b\,\ln \relax (F)-20\,c^3\right )}{3\,b^3\,d^4\,{\ln \relax (F)}^3}-\frac {2\,c^2\,x\,\left (b\,\ln \relax (F)-2\,c^3\right )}{b^3\,d^6\,{\ln \relax (F)}^3}-\frac {2\,c\,x^2\,\left (b\,\ln \relax (F)-5\,c^3\right )}{b^3\,d^5\,{\ln \relax (F)}^3}\right )}{x^6+\frac {c^6}{d^6}+\frac {6\,c\,x^5}{d}+\frac {6\,c^5\,x}{d^5}+\frac {15\,c^2\,x^4}{d^2}+\frac {20\,c^3\,x^3}{d^3}+\frac {15\,c^4\,x^2}{d^4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b/(c + d*x)^3)/(c + d*x)^10,x)

[Out]

-(F^a*F^(b/(c^3 + d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x))*((2*x^6)/(3*b^3*d*log(F)^3) + (b^2*log(F)^2 + 2*c^6 - 2*
b*c^3*log(F))/(3*b^3*d^7*log(F)^3) + (4*c*x^5)/(b^3*d^2*log(F)^3) + (10*c^2*x^4)/(b^3*d^3*log(F)^3) - (2*x^3*(
b*log(F) - 20*c^3))/(3*b^3*d^4*log(F)^3) - (2*c^2*x*(b*log(F) - 2*c^3))/(b^3*d^6*log(F)^3) - (2*c*x^2*(b*log(F
) - 5*c^3))/(b^3*d^5*log(F)^3)))/(x^6 + c^6/d^6 + (6*c*x^5)/d + (6*c^5*x)/d^5 + (15*c^2*x^4)/d^2 + (20*c^3*x^3
)/d^3 + (15*c^4*x^2)/d^4)

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sympy [B]  time = 0.49, size = 270, normalized size = 2.81 \[ \frac {F^{a + \frac {b}{\left (c + d x\right )^{3}}} \left (- b^{2} \log {\relax (F )}^{2} + 2 b c^{3} \log {\relax (F )} + 6 b c^{2} d x \log {\relax (F )} + 6 b c d^{2} x^{2} \log {\relax (F )} + 2 b d^{3} x^{3} \log {\relax (F )} - 2 c^{6} - 12 c^{5} d x - 30 c^{4} d^{2} x^{2} - 40 c^{3} d^{3} x^{3} - 30 c^{2} d^{4} x^{4} - 12 c d^{5} x^{5} - 2 d^{6} x^{6}\right )}{3 b^{3} c^{6} d \log {\relax (F )}^{3} + 18 b^{3} c^{5} d^{2} x \log {\relax (F )}^{3} + 45 b^{3} c^{4} d^{3} x^{2} \log {\relax (F )}^{3} + 60 b^{3} c^{3} d^{4} x^{3} \log {\relax (F )}^{3} + 45 b^{3} c^{2} d^{5} x^{4} \log {\relax (F )}^{3} + 18 b^{3} c d^{6} x^{5} \log {\relax (F )}^{3} + 3 b^{3} d^{7} x^{6} \log {\relax (F )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b/(d*x+c)**3)/(d*x+c)**10,x)

[Out]

F**(a + b/(c + d*x)**3)*(-b**2*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*log(F) - 2*c**6 - 12*c**5*d*x - 30*c**4*d**2*x**2 - 40*c**3*d**3*x**3 - 30*c**2*d**4*x**4 - 12*c*d
**5*x**5 - 2*d**6*x**6)/(3*b**3*c**6*d*log(F)**3 + 18*b**3*c**5*d**2*x*log(F)**3 + 45*b**3*c**4*d**3*x**2*log(
F)**3 + 60*b**3*c**3*d**4*x**3*log(F)**3 + 45*b**3*c**2*d**5*x**4*log(F)**3 + 18*b**3*c*d**6*x**5*log(F)**3 +
3*b**3*d**7*x**6*log(F)**3)

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