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

Optimal. Leaf size=49 \[ \frac {F^a \Gamma \left (\frac {4}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d (c+d x)^4 \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{4/3}} \]

[Out]

1/3*F^a*GAMMA(4/3,-b*ln(F)/(d*x+c)^3)/d/(d*x+c)^4/(-b*ln(F)/(d*x+c)^3)^(4/3)

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Rubi [A]  time = 0.04, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2218} \[ \frac {F^a \text {Gamma}\left (\frac {4}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d (c+d x)^4 \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^a*Gamma[4/3, -((b*Log[F])/(c + d*x)^3)])/(3*d*(c + d*x)^4*(-((b*Log[F])/(c + d*x)^3))^(4/3))

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps

\begin {align*} \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^5} \, dx &=\frac {F^a \Gamma \left (\frac {4}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d (c+d x)^4 \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{4/3}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 49, normalized size = 1.00 \[ \frac {F^a \Gamma \left (\frac {4}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d (c+d x)^4 \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^a*Gamma[4/3, -((b*Log[F])/(c + d*x)^3)])/(3*d*(c + d*x)^4*(-((b*Log[F])/(c + d*x)^3))^(4/3))

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fricas [B]  time = 0.44, size = 155, normalized size = 3.16 \[ \frac {{\left (d^{3} x + c d^{2}\right )} F^{a} \left (-\frac {b \log \relax (F)}{d^{3}}\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -\frac {b \log \relax (F)}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) - 3 \, F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}} b \log \relax (F)}{9 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \log \relax (F)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*((d^3*x + c*d^2)*F^a*(-b*log(F)/d^3)^(2/3)*gamma(1/3, -b*log(F)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))
- 3*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*log
(F))/((b^2*d^2*x + b^2*c*d)*log(F)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{3}}}}{{\left (d x + c\right )}^{5}}\,{d x} \]

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]

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

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {F^{a +\frac {b}{\left (d x +c \right )^{3}}}}{\left (d x +c \right )^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+1/(d*x+c)^3*b)/(d*x+c)^5,x)

[Out]

int(F^(a+1/(d*x+c)^3*b)/(d*x+c)^5,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{3}}}}{{\left (d x + c\right )}^{5}}\,{d x} \]

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]

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

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mupad [B]  time = 4.13, size = 114, normalized size = 2.33 \[ \frac {F^a\,\Gamma \left (\frac {1}{3},-\frac {b\,\ln \relax (F)}{{\left (c+d\,x\right )}^3}\right )}{9\,d\,{\left (c+d\,x\right )}^4\,{\left (-\frac {b\,\ln \relax (F)}{{\left (c+d\,x\right )}^3}\right )}^{4/3}}-\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^3}}}{3\,b\,d\,\ln \relax (F)\,\left (c+d\,x\right )}-\frac {2\,\pi \,\sqrt {3}\,F^a}{27\,d\,\Gamma \left (\frac {2}{3}\right )\,{\left (c+d\,x\right )}^4\,{\left (-\frac {b\,\ln \relax (F)}{{\left (c+d\,x\right )}^3}\right )}^{4/3}} \]

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^a*igamma(1/3, -(b*log(F))/(c + d*x)^3))/(9*d*(c + d*x)^4*(-(b*log(F))/(c + d*x)^3)^(4/3)) - (F^a*F^(b/(c +
d*x)^3))/(3*b*d*log(F)*(c + d*x)) - (2*3^(1/2)*F^a*pi)/(27*d*gamma(2/3)*(c + d*x)^4*(-(b*log(F))/(c + d*x)^3)^
(4/3))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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