3.348 \(\int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^7} \, dx\)

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

[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)/b/d/(d*x+c)^3/ln(F)

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Rubi [A]  time = 0.09, 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} \[ \frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b^2 d \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^3}}}{3 b d \log (F) (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x)^3)/(c + d*x)^7,x]

[Out]

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

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Mathematica [A]  time = 0.03, size = 47, normalized size = 0.76 \[ \frac {F^{a+\frac {b}{(c+d x)^3}} \left ((c+d x)^3-b \log (F)\right )}{3 b^2 d \log ^2(F) (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x)^3)/(c + d*x)^7,x]

[Out]

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

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fricas [B]  time = 0.42, size = 148, normalized size = 2.39 \[ \frac {{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3} - b \log \relax (F)\right )} F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{3 \, {\left (b^{2} d^{4} x^{3} + 3 \, b^{2} c d^{3} x^{2} + 3 \, b^{2} c^{2} d^{2} x + b^{2} c^{3} d\right )} \log \relax (F)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^7,x, algorithm="fricas")

[Out]

1/3*(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3 - b*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^2*d^4*x^3 + 3*b^2*c*d^3*x^2 + 3*b^2*c^2*d^2*x + b^2*c^3*d)*
log(F)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{3}}}}{{\left (d x + c\right )}^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^7,x, algorithm="giac")

[Out]

integrate(F^(a + b/(d*x + c)^3)/(d*x + c)^7, x)

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maple [B]  time = 0.09, size = 261, normalized size = 4.21 \[ \frac {\frac {d^{5} x^{6} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{3 b^{2} \ln \relax (F )^{2}}+\frac {2 c \,d^{4} x^{5} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{b^{2} \ln \relax (F )^{2}}+\frac {5 c^{2} d^{3} x^{4} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{b^{2} \ln \relax (F )^{2}}-\frac {\left (-20 c^{3}+b \ln \relax (F )\right ) d^{2} x^{3} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{3 b^{2} \ln \relax (F )^{2}}-\frac {\left (-5 c^{3}+b \ln \relax (F )\right ) c d \,x^{2} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{b^{2} \ln \relax (F )^{2}}-\frac {\left (-2 c^{3}+b \ln \relax (F )\right ) c^{2} x \,{\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{b^{2} \ln \relax (F )^{2}}-\frac {\left (-c^{3}+b \ln \relax (F )\right ) c^{3} {\mathrm e}^{\left (a +\frac {b}{\left (d x +c \right )^{3}}\right ) \ln \relax (F )}}{3 b^{2} d \ln \relax (F )^{2}}}{\left (d x +c \right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+1/(d*x+c)^3*b)/(d*x+c)^7,x)

[Out]

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

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maxima [B]  time = 0.65, size = 144, normalized size = 2.32 \[ \frac {{\left (F^{a} d^{3} x^{3} + 3 \, F^{a} c d^{2} x^{2} + 3 \, F^{a} c^{2} d x + F^{a} c^{3} - F^{a} b \log \relax (F)\right )} F^{\frac {b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{3 \, {\left (b^{2} d^{4} x^{3} \log \relax (F)^{2} + 3 \, b^{2} c d^{3} x^{2} \log \relax (F)^{2} + 3 \, b^{2} c^{2} d^{2} x \log \relax (F)^{2} + b^{2} c^{3} d \log \relax (F)^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^7,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.89, size = 136, normalized size = 2.19 \[ \frac {F^a\,F^{\frac {b}{c^3+3\,c^2\,d\,x+3\,c\,d^2\,x^2+d^3\,x^3}}\,\left (\frac {x^3}{3\,b^2\,d\,{\ln \relax (F)}^2}-\frac {b\,\ln \relax (F)-c^3}{3\,b^2\,d^4\,{\ln \relax (F)}^2}+\frac {c\,x^2}{b^2\,d^2\,{\ln \relax (F)}^2}+\frac {c^2\,x}{b^2\,d^3\,{\ln \relax (F)}^2}\right )}{x^3+\frac {c^3}{d^3}+\frac {3\,c\,x^2}{d}+\frac {3\,c^2\,x}{d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b/(c + d*x)^3)/(c + d*x)^7,x)

[Out]

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

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sympy [B]  time = 0.36, size = 114, normalized size = 1.84 \[ \frac {F^{a + \frac {b}{\left (c + d x\right )^{3}}} \left (- b \log {\relax (F )} + c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}\right )}{3 b^{2} c^{3} d \log {\relax (F )}^{2} + 9 b^{2} c^{2} d^{2} x \log {\relax (F )}^{2} + 9 b^{2} c d^{3} x^{2} \log {\relax (F )}^{2} + 3 b^{2} d^{4} x^{3} \log {\relax (F )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b/(d*x+c)**3)/(d*x+c)**7,x)

[Out]

F**(a + b/(c + d*x)**3)*(-b*log(F) + c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3)/(3*b**2*c**3*d*log(F)**2 +
9*b**2*c**2*d**2*x*log(F)**2 + 9*b**2*c*d**3*x**2*log(F)**2 + 3*b**2*d**4*x**3*log(F)**2)

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