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

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

[Out]

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

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Rubi [A]  time = 0.05, 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} \[ \frac {F^a (c+d x)^4 \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{4/3} \text {Gamma}\left (-\frac {4}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.03, size = 49, normalized size = 1.00 \[ \frac {F^a (c+d x)^4 \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{4/3} \Gamma \left (-\frac {4}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.52, size = 178, normalized size = 3.63 \[ -\frac {3 \, F^{a} b d \left (-\frac {b \log \relax (F)}{d^{3}}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -\frac {b \log \relax (F)}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) \log \relax (F) - {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4} + 3 \, {\left (b d x + b c\right )} \log \relax (F)\right )} F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*F^a*b*d*(-b*log(F)/d^3)^(1/3)*gamma(2/3, -b*log(F)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))*log(F) -
(d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4 + 3*(b*d*x + b*c)*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)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, {\left (F^{a} d^{3} x^{4} + 4 \, F^{a} c d^{2} x^{3} + 6 \, F^{a} c^{2} d x^{2} + {\left (4 \, F^{a} c^{3} + 3 \, F^{a} b \log \relax (F)\right )} x\right )} F^{\frac {b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}} + \int -\frac {3 \, {\left (F^{a} b c^{4} \log \relax (F) - 3 \, F^{a} b^{2} d x \log \relax (F)^{2}\right )} F^{\frac {b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{4 \, {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(F^a*d^3*x^4 + 4*F^a*c*d^2*x^3 + 6*F^a*c^2*d*x^2 + (4*F^a*c^3 + 3*F^a*b*log(F))*x)*F^(b/(d^3*x^3 + 3*c*d^2
*x^2 + 3*c^2*d*x + c^3)) + integrate(-3/4*(F^a*b*c^4*log(F) - 3*F^a*b^2*d*x*log(F)^2)*F^(b/(d^3*x^3 + 3*c*d^2*
x^2 + 3*c^2*d*x + c^3))/(d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4), x)

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mupad [B]  time = 3.92, size = 128, normalized size = 2.61 \[ \frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^3}}\,{\left (c+d\,x\right )}^4}{4\,d}-\frac {3\,F^a\,\Gamma \left (\frac {2}{3}\right )\,{\left (c+d\,x\right )}^4\,{\left (-\frac {b\,\ln \relax (F)}{{\left (c+d\,x\right )}^3}\right )}^{4/3}}{4\,d}+\frac {3\,F^a\,\Gamma \left (\frac {2}{3},-\frac {b\,\ln \relax (F)}{{\left (c+d\,x\right )}^3}\right )\,{\left (c+d\,x\right )}^4\,{\left (-\frac {b\,\ln \relax (F)}{{\left (c+d\,x\right )}^3}\right )}^{4/3}}{4\,d}+\frac {3\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^3}}\,b\,\ln \relax (F)\,\left (c+d\,x\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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