3.317 \(\int F^{a+\frac{b}{(c+d x)^2}} (c+d x)^5 \, dx\)

Optimal. Leaf size=121 \[ -\frac{b^3 F^a \log ^3(F) \text{Ei}\left (\frac{b \log (F)}{(c+d x)^2}\right )}{12 d}+\frac{b^2 \log ^2(F) (c+d x)^2 F^{a+\frac{b}{(c+d x)^2}}}{12 d}+\frac{(c+d x)^6 F^{a+\frac{b}{(c+d x)^2}}}{6 d}+\frac{b \log (F) (c+d x)^4 F^{a+\frac{b}{(c+d x)^2}}}{12 d} \]

[Out]

(F^(a + b/(c + d*x)^2)*(c + d*x)^6)/(6*d) + (b*F^(a + b/(c + d*x)^2)*(c + d*x)^4*Log[F])/(12*d) + (b^2*F^(a +
b/(c + d*x)^2)*(c + d*x)^2*Log[F]^2)/(12*d) - (b^3*F^a*ExpIntegralEi[(b*Log[F])/(c + d*x)^2]*Log[F]^3)/(12*d)

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Rubi [A]  time = 0.185648, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2214, 2210} \[ -\frac{b^3 F^a \log ^3(F) \text{Ei}\left (\frac{b \log (F)}{(c+d x)^2}\right )}{12 d}+\frac{b^2 \log ^2(F) (c+d x)^2 F^{a+\frac{b}{(c+d x)^2}}}{12 d}+\frac{(c+d x)^6 F^{a+\frac{b}{(c+d x)^2}}}{6 d}+\frac{b \log (F) (c+d x)^4 F^{a+\frac{b}{(c+d x)^2}}}{12 d} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x)^2)*(c + d*x)^5,x]

[Out]

(F^(a + b/(c + d*x)^2)*(c + d*x)^6)/(6*d) + (b*F^(a + b/(c + d*x)^2)*(c + d*x)^4*Log[F])/(12*d) + (b^2*F^(a +
b/(c + d*x)^2)*(c + d*x)^2*Log[F]^2)/(12*d) - (b^3*F^a*ExpIntegralEi[(b*Log[F])/(c + d*x)^2]*Log[F]^3)/(12*d)

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), 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[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 2210

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

Rubi steps

\begin{align*} \int F^{a+\frac{b}{(c+d x)^2}} (c+d x)^5 \, dx &=\frac{F^{a+\frac{b}{(c+d x)^2}} (c+d x)^6}{6 d}+\frac{1}{3} (b \log (F)) \int F^{a+\frac{b}{(c+d x)^2}} (c+d x)^3 \, dx\\ &=\frac{F^{a+\frac{b}{(c+d x)^2}} (c+d x)^6}{6 d}+\frac{b F^{a+\frac{b}{(c+d x)^2}} (c+d x)^4 \log (F)}{12 d}+\frac{1}{6} \left (b^2 \log ^2(F)\right ) \int F^{a+\frac{b}{(c+d x)^2}} (c+d x) \, dx\\ &=\frac{F^{a+\frac{b}{(c+d x)^2}} (c+d x)^6}{6 d}+\frac{b F^{a+\frac{b}{(c+d x)^2}} (c+d x)^4 \log (F)}{12 d}+\frac{b^2 F^{a+\frac{b}{(c+d x)^2}} (c+d x)^2 \log ^2(F)}{12 d}+\frac{1}{6} \left (b^3 \log ^3(F)\right ) \int \frac{F^{a+\frac{b}{(c+d x)^2}}}{c+d x} \, dx\\ &=\frac{F^{a+\frac{b}{(c+d x)^2}} (c+d x)^6}{6 d}+\frac{b F^{a+\frac{b}{(c+d x)^2}} (c+d x)^4 \log (F)}{12 d}+\frac{b^2 F^{a+\frac{b}{(c+d x)^2}} (c+d x)^2 \log ^2(F)}{12 d}-\frac{b^3 F^a \text{Ei}\left (\frac{b \log (F)}{(c+d x)^2}\right ) \log ^3(F)}{12 d}\\ \end{align*}

Mathematica [A]  time = 0.166385, size = 96, normalized size = 0.79 \[ \frac{F^a \left (b \log (F) \left (b \log (F) \left ((c+d x)^2 F^{\frac{b}{(c+d x)^2}}-b \log (F) \text{Ei}\left (\frac{b \log (F)}{(c+d x)^2}\right )\right )+(c+d x)^4 F^{\frac{b}{(c+d x)^2}}\right )+2 (c+d x)^6 F^{\frac{b}{(c+d x)^2}}\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x)^2)*(c + d*x)^5,x]

[Out]

(F^a*(2*F^(b/(c + d*x)^2)*(c + d*x)^6 + b*Log[F]*(F^(b/(c + d*x)^2)*(c + d*x)^4 + b*Log[F]*(F^(b/(c + d*x)^2)*
(c + d*x)^2 - b*ExpIntegralEi[(b*Log[F])/(c + d*x)^2]*Log[F]))))/(12*d)

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Maple [B]  time = 0.05, size = 395, normalized size = 3.3 \begin{align*}{\frac{{d}^{5}{F}^{a}{x}^{6}}{6}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{d}^{4}{F}^{a}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}c{x}^{5}+{\frac{5\,{d}^{3}{F}^{a}{c}^{2}{x}^{4}}{2}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{10\,{d}^{2}{F}^{a}{c}^{3}{x}^{3}}{3}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{5\,d{F}^{a}{c}^{4}{x}^{2}}{2}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{F}^{a}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}{c}^{5}x+{\frac{{F}^{a}{c}^{6}}{6\,d}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{{d}^{3}{F}^{a}b\ln \left ( F \right ){x}^{4}}{12}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{{d}^{2}{F}^{a}b\ln \left ( F \right ) c{x}^{3}}{3}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{d{F}^{a}b\ln \left ( F \right ){c}^{2}{x}^{2}}{2}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{{F}^{a}b\ln \left ( F \right ){c}^{3}x}{3}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{{F}^{a}b\ln \left ( F \right ){c}^{4}}{12\,d}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{d{F}^{a}{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}{x}^{2}}{12}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{{F}^{a}{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}cx}{6}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{{F}^{a}{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}{c}^{2}}{12\,d}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{{F}^{a}{b}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}}{12\,d}{\it Ei} \left ( 1,-{\frac{b\ln \left ( F \right ) }{ \left ( dx+c \right ) ^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b/(d*x+c)^2)*(d*x+c)^5,x)

[Out]

1/6*d^5*F^a*F^(b/(d*x+c)^2)*x^6+d^4*F^a*F^(b/(d*x+c)^2)*c*x^5+5/2*d^3*F^a*F^(b/(d*x+c)^2)*c^2*x^4+10/3*d^2*F^a
*F^(b/(d*x+c)^2)*c^3*x^3+5/2*d*F^a*F^(b/(d*x+c)^2)*c^4*x^2+F^a*F^(b/(d*x+c)^2)*c^5*x+1/6/d*F^a*F^(b/(d*x+c)^2)
*c^6+1/12*d^3*F^a*b*ln(F)*F^(b/(d*x+c)^2)*x^4+1/3*d^2*F^a*b*ln(F)*F^(b/(d*x+c)^2)*c*x^3+1/2*d*F^a*b*ln(F)*F^(b
/(d*x+c)^2)*c^2*x^2+1/3*F^a*b*ln(F)*F^(b/(d*x+c)^2)*c^3*x+1/12/d*F^a*b*ln(F)*F^(b/(d*x+c)^2)*c^4+1/12*d*F^a*b^
2*ln(F)^2*F^(b/(d*x+c)^2)*x^2+1/6*F^a*b^2*ln(F)^2*F^(b/(d*x+c)^2)*c*x+1/12/d*F^a*b^2*ln(F)^2*F^(b/(d*x+c)^2)*c
^2+1/12/d*F^a*b^3*ln(F)^3*Ei(1,-b*ln(F)/(d*x+c)^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{12} \,{\left (2 \, F^{a} d^{5} x^{6} + 12 \, F^{a} c d^{4} x^{5} +{\left (30 \, F^{a} c^{2} d^{3} + F^{a} b d^{3} \log \left (F\right )\right )} x^{4} + 4 \,{\left (10 \, F^{a} c^{3} d^{2} + F^{a} b c d^{2} \log \left (F\right )\right )} x^{3} +{\left (30 \, F^{a} c^{4} d + 6 \, F^{a} b c^{2} d \log \left (F\right ) + F^{a} b^{2} d \log \left (F\right )^{2}\right )} x^{2} + 2 \,{\left (6 \, F^{a} c^{5} + 2 \, F^{a} b c^{3} \log \left (F\right ) + F^{a} b^{2} c \log \left (F\right )^{2}\right )} x\right )} F^{\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}} + \int \frac{{\left (F^{a} b^{3} d^{2} x^{2} \log \left (F\right )^{3} + 2 \, F^{a} b^{3} c d x \log \left (F\right )^{3} - 2 \, F^{a} b c^{6} \log \left (F\right ) - F^{a} b^{2} c^{4} \log \left (F\right )^{2}\right )} F^{\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{6 \,{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^5,x, algorithm="maxima")

[Out]

1/12*(2*F^a*d^5*x^6 + 12*F^a*c*d^4*x^5 + (30*F^a*c^2*d^3 + F^a*b*d^3*log(F))*x^4 + 4*(10*F^a*c^3*d^2 + F^a*b*c
*d^2*log(F))*x^3 + (30*F^a*c^4*d + 6*F^a*b*c^2*d*log(F) + F^a*b^2*d*log(F)^2)*x^2 + 2*(6*F^a*c^5 + 2*F^a*b*c^3
*log(F) + F^a*b^2*c*log(F)^2)*x)*F^(b/(d^2*x^2 + 2*c*d*x + c^2)) + integrate(1/6*(F^a*b^3*d^2*x^2*log(F)^3 + 2
*F^a*b^3*c*d*x*log(F)^3 - 2*F^a*b*c^6*log(F) - F^a*b^2*c^4*log(F)^2)*F^(b/(d^2*x^2 + 2*c*d*x + c^2))/(d^3*x^3
+ 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)

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Fricas [A]  time = 1.66616, size = 486, normalized size = 4.02 \begin{align*} -\frac{F^{a} b^{3}{\rm Ei}\left (\frac{b \log \left (F\right )}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) \log \left (F\right )^{3} -{\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} +{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (F\right )^{2} +{\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4}\right )} \log \left (F\right )\right )} F^{\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^5,x, algorithm="fricas")

[Out]

-1/12*(F^a*b^3*Ei(b*log(F)/(d^2*x^2 + 2*c*d*x + c^2))*log(F)^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*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(F)^2 + (b*d^4
*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4)*log(F))*F^((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(
d^2*x^2 + 2*c*d*x + c^2)))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b/(d*x+c)**2)*(d*x+c)**5,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{5} F^{a + \frac{b}{{\left (d x + c\right )}^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^5,x, algorithm="giac")

[Out]

integrate((d*x + c)^5*F^(a + b/(d*x + c)^2), x)