3.275 \(\int \frac {F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx\)

Optimal. Leaf size=102 \[ \frac {2 \sqrt {\pi } b^{3/2} F^a \log ^{\frac {3}{2}}(F) \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{3 d}-\frac {F^{a+b (c+d x)^2}}{3 d (c+d x)^3}-\frac {2 b \log (F) F^{a+b (c+d x)^2}}{3 d (c+d x)} \]

[Out]

-1/3*F^(a+b*(d*x+c)^2)/d/(d*x+c)^3-2/3*b*F^(a+b*(d*x+c)^2)*ln(F)/d/(d*x+c)+2/3*b^(3/2)*F^a*erfi((d*x+c)*b^(1/2
)*ln(F)^(1/2))*ln(F)^(3/2)*Pi^(1/2)/d

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Rubi [A]  time = 0.15, antiderivative size = 102, 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, 2204} \[ \frac {2 \sqrt {\pi } b^{3/2} F^a \log ^{\frac {3}{2}}(F) \text {Erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{3 d}-\frac {F^{a+b (c+d x)^2}}{3 d (c+d x)^3}-\frac {2 b \log (F) F^{a+b (c+d x)^2}}{3 d (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^2)/(c + d*x)^4,x]

[Out]

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

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

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 \frac {F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx &=-\frac {F^{a+b (c+d x)^2}}{3 d (c+d x)^3}+\frac {1}{3} (2 b \log (F)) \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^2} \, dx\\ &=-\frac {F^{a+b (c+d x)^2}}{3 d (c+d x)^3}-\frac {2 b F^{a+b (c+d x)^2} \log (F)}{3 d (c+d x)}+\frac {1}{3} \left (4 b^2 \log ^2(F)\right ) \int F^{a+b (c+d x)^2} \, dx\\ &=-\frac {F^{a+b (c+d x)^2}}{3 d (c+d x)^3}-\frac {2 b F^{a+b (c+d x)^2} \log (F)}{3 d (c+d x)}+\frac {2 b^{3/2} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \log ^{\frac {3}{2}}(F)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 81, normalized size = 0.79 \[ \frac {F^a \left (2 \sqrt {\pi } b^{3/2} \log ^{\frac {3}{2}}(F) \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )-\frac {F^{b (c+d x)^2} \left (2 b \log (F) (c+d x)^2+1\right )}{(c+d x)^3}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^2)/(c + d*x)^4,x]

[Out]

(F^a*(2*b^(3/2)*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]]*Log[F]^(3/2) - (F^(b*(c + d*x)^2)*(1 + 2*b*(c +
d*x)^2*Log[F]))/(c + d*x)^3))/(3*d)

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fricas [A]  time = 0.43, size = 163, normalized size = 1.60 \[ -\frac {2 \, \sqrt {\pi } {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \sqrt {-b d^{2} \log \relax (F)} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \relax (F)} {\left (d x + c\right )}}{d}\right ) \log \relax (F) + {\left (2 \, {\left (b d^{3} x^{2} + 2 \, b c d^{2} x + b c^{2} d\right )} \log \relax (F) + d\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{3 \, {\left (d^{5} x^{3} + 3 \, c d^{4} x^{2} + 3 \, c^{2} d^{3} x + c^{3} d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^4,x, algorithm="fricas")

[Out]

-1/3*(2*sqrt(pi)*(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*sqrt(-b*d^2*log(F))*F^a*erf(sqrt(-b*d^2*log
(F))*(d*x + c)/d)*log(F) + (2*(b*d^3*x^2 + 2*b*c*d^2*x + b*c^2*d)*log(F) + d)*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2
+ a))/(d^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^4,x, algorithm="giac")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^4, x)

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maple [A]  time = 0.08, size = 96, normalized size = 0.94 \[ \frac {2 \sqrt {\pi }\, b^{2} F^{a} \erf \left (\sqrt {-b \ln \relax (F )}\, \left (d x +c \right )\right ) \ln \relax (F )^{2}}{3 \sqrt {-b \ln \relax (F )}\, d}-\frac {2 b \,F^{a} F^{\left (d x +c \right )^{2} b} \ln \relax (F )}{3 \left (d x +c \right ) d}-\frac {F^{a} F^{\left (d x +c \right )^{2} b}}{3 \left (d x +c \right )^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+(d*x+c)^2*b)/(d*x+c)^4,x)

[Out]

-1/3/d/(d*x+c)^3*F^((d*x+c)^2*b)*F^a-2/3/d*b*ln(F)/(d*x+c)*F^((d*x+c)^2*b)*F^a+2/3/d*b^2*ln(F)^2*Pi^(1/2)*F^a/
(-b*ln(F))^(1/2)*erf((-b*ln(F))^(1/2)*(d*x+c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^4,x, algorithm="maxima")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^4, x)

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mupad [B]  time = 5.03, size = 201, normalized size = 1.97 \[ \frac {2\,F^a\,b^2\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \relax (F)\,d^2+b\,c\,\ln \relax (F)\,d}{\sqrt {b\,d^2\,\ln \relax (F)}}\right )\,{\ln \relax (F)}^2}{3\,\sqrt {b\,d^2\,\ln \relax (F)}}-\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,\left (\frac {1}{3\,d}+\frac {2\,b\,c^2\,\ln \relax (F)}{3\,d}\right )+\frac {4\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,b\,c\,x\,\ln \relax (F)}{3}+\frac {2\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,b\,d\,x^2\,\ln \relax (F)}{3}}{c^3+3\,c^2\,d\,x+3\,c\,d^2\,x^2+d^3\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b*(c + d*x)^2)/(c + d*x)^4,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b*(d*x+c)**2)/(d*x+c)**4,x)

[Out]

Integral(F**(a + b*(c + d*x)**2)/(c + d*x)**4, x)

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