3.318 \(\int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx\)

Optimal. Leaf size=87 \[ -\frac {b^2 F^a \log ^2(F) \text {Ei}\left (\frac {b \log (F)}{(c+d x)^2}\right )}{4 d}+\frac {(c+d x)^4 F^{a+\frac {b}{(c+d x)^2}}}{4 d}+\frac {b \log (F) (c+d x)^2 F^{a+\frac {b}{(c+d x)^2}}}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 87, 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 = {2214, 2210} \[ -\frac {b^2 F^a \log ^2(F) \text {Ei}\left (\frac {b \log (F)}{(c+d x)^2}\right )}{4 d}+\frac {(c+d x)^4 F^{a+\frac {b}{(c+d x)^2}}}{4 d}+\frac {b \log (F) (c+d x)^2 F^{a+\frac {b}{(c+d x)^2}}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^(a + b/(c + d*x)^2)*(c + d*x)^4)/(4*d) + (b*F^(a + b/(c + d*x)^2)*(c + d*x)^2*Log[F])/(4*d) - (b^2*F^a*ExpI
ntegralEi[(b*Log[F])/(c + d*x)^2]*Log[F]^2)/(4*d)

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]

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]))

Rubi steps

\begin {align*} \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)^4}{4 d}+\frac {1}{2} (b \log (F)) \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)^4}{4 d}+\frac {b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \log (F)}{4 d}+\frac {1}{2} \left (b^2 \log ^2(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)^4}{4 d}+\frac {b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \log (F)}{4 d}-\frac {b^2 F^a \text {Ei}\left (\frac {b \log (F)}{(c+d x)^2}\right ) \log ^2(F)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 71, normalized size = 0.82 \[ \frac {F^a \left (b \log (F) \left ((c+d x)^2 F^{\frac {b}{(c+d x)^2}}-b \log (F) \text {Ei}\left (\frac {b \log (F)}{(c+d x)^2}\right )\right )+(c+d x)^4 F^{\frac {b}{(c+d x)^2}}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^a*(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])))/(4*d)

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fricas [A]  time = 0.46, size = 145, normalized size = 1.67 \[ -\frac {F^{a} b^{2} {\rm Ei}\left (\frac {b \log \relax (F)}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) \log \relax (F)^{2} - {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4} + {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \relax (F)\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^3,x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^3,x, algorithm="giac")

[Out]

integrate((d*x + c)^3*F^(a + b/(d*x + c)^2), x)

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maple [B]  time = 0.07, size = 208, normalized size = 2.39 \[ \frac {d^{3} x^{4} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{4}+c \,d^{2} x^{3} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}+\frac {b d \,x^{2} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )}{4}+\frac {3 c^{2} d \,x^{2} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{2}+\frac {b c x \,F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )}{2}+c^{3} x \,F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}+\frac {b^{2} F^{a} \Ei \left (1, -\frac {b \ln \relax (F )}{\left (d x +c \right )^{2}}\right ) \ln \relax (F )^{2}}{4 d}+\frac {b \,c^{2} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )}{4 d}+\frac {c^{4} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+1/(d*x+c)^2*b)*(d*x+c)^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)*(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 + F^a*b*d*log(F))*x^2 + 2*(2*F^a*c^3 + F^a*b*c*log(F))*x)*F^
(b/(d^2*x^2 + 2*c*d*x + c^2)) + integrate(1/2*(F^a*b^2*d^2*x^2*log(F)^2 + 2*F^a*b^2*c*d*x*log(F)^2 - F^a*b*c^4
*log(F))*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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mupad [B]  time = 3.70, size = 76, normalized size = 0.87 \[ \frac {F^a\,b^2\,{\ln \relax (F)}^2\,\left (\frac {\mathrm {expint}\left (-\frac {b\,\ln \relax (F)}{{\left (c+d\,x\right )}^2}\right )}{2}+F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,\left (\frac {{\left (c+d\,x\right )}^2}{2\,b\,\ln \relax (F)}+\frac {{\left (c+d\,x\right )}^4}{2\,b^2\,{\ln \relax (F)}^2}\right )\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b/(c + d*x)^2)*(c + d*x)^3,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int F^{a + \frac {b}{\left (c + d x\right )^{2}}} \left (c + d x\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b/(d*x+c)**2)*(d*x+c)**3,x)

[Out]

Integral(F**(a + b/(c + d*x)**2)*(c + d*x)**3, x)

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