\(\int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^8} \, dx\) [336]

Optimal result

Integrand size = 21, antiderivative size = 149 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^8} \, dx=\frac {15 F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{16 b^{7/2} d \log ^{\frac {7}{2}}(F)}-\frac {15 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x) \log ^3(F)}+\frac {5 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^3 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)} \]

[Out]

-15/8*F^(a+b/(d*x+c)^2)/b^3/d/(d*x+c)/ln(F)^3+5/4*F^(a+b/(d*x+c)^2)/b^2/d/(d*x+c)^3/ln(F)^2-1/2*F^(a+b/(d*x+c)
^2)/b/d/(d*x+c)^5/ln(F)+15/16*F^a*erfi(b^(1/2)*ln(F)^(1/2)/(d*x+c))*Pi^(1/2)/b^(7/2)/d/ln(F)^(7/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2243, 2242, 2235} \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^8} \, dx=\frac {15 \sqrt {\pi } F^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{16 b^{7/2} d \log ^{\frac {7}{2}}(F)}-\frac {15 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d \log ^3(F) (c+d x)}+\frac {5 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d \log ^2(F) (c+d x)^3}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^5} \]

[In]

Int[F^(a + b/(c + d*x)^2)/(c + d*x)^8,x]

[Out]

(15*F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)])/(16*b^(7/2)*d*Log[F]^(7/2)) - (15*F^(a + b/(c + d*x)^
2))/(8*b^3*d*(c + d*x)*Log[F]^3) + (5*F^(a + b/(c + d*x)^2))/(4*b^2*d*(c + d*x)^3*Log[F]^2) - F^(a + b/(c + d*
x)^2)/(2*b*d*(c + d*x)^5*Log[F])

Rule 2235

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 2242

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[1/(d*(m + 1))
, Subst[Int[F^(a + b*x^2), x], x, (c + d*x)^(m + 1)], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && EqQ[n, 2*(m + 1
)]

Rule 2243

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

Rubi steps \begin{align*} \text {integral}& = -\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}-\frac {5 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx}{2 b \log (F)} \\ & = \frac {5 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^3 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}+\frac {15 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4} \, dx}{4 b^2 \log ^2(F)} \\ & = -\frac {15 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x) \log ^3(F)}+\frac {5 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^3 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}-\frac {15 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx}{8 b^3 \log ^3(F)} \\ & = -\frac {15 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x) \log ^3(F)}+\frac {5 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^3 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}+\frac {15 \text {Subst}\left (\int F^{a+b x^2} \, dx,x,\frac {1}{c+d x}\right )}{8 b^3 d \log ^3(F)} \\ & = \frac {15 F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{16 b^{7/2} d \log ^{\frac {7}{2}}(F)}-\frac {15 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x) \log ^3(F)}+\frac {5 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^3 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.74 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^8} \, dx=\frac {F^a \left (15 \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )-\frac {2 \sqrt {b} F^{\frac {b}{(c+d x)^2}} \sqrt {\log (F)} \left (15 (c+d x)^4-10 b (c+d x)^2 \log (F)+4 b^2 \log ^2(F)\right )}{(c+d x)^5}\right )}{16 b^{7/2} d \log ^{\frac {7}{2}}(F)} \]

[In]

Integrate[F^(a + b/(c + d*x)^2)/(c + d*x)^8,x]

[Out]

(F^a*(15*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)] - (2*Sqrt[b]*F^(b/(c + d*x)^2)*Sqrt[Log[F]]*(15*(c +
d*x)^4 - 10*b*(c + d*x)^2*Log[F] + 4*b^2*Log[F]^2))/(c + d*x)^5))/(16*b^(7/2)*d*Log[F]^(7/2))

Maple [A] (verified)

Time = 2.39 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.95

method	result	size
risch	\(-\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{2 d \left (d x +c \right )^{5} b \ln \left (F \right )}+\frac {5 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{4 d \,b^{2} \ln \left (F \right )^{2} \left (d x +c \right )^{3}}-\frac {15 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{8 d \,b^{3} \ln \left (F \right )^{3} \left (d x +c \right )}+\frac {15 F^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (F \right )}}{d x +c}\right )}{16 d \,b^{3} \ln \left (F \right )^{3} \sqrt {-b \ln \left (F \right )}}\)	\(142\)

[In]

int(F^(a+b/(d*x+c)^2)/(d*x+c)^8,x,method=_RETURNVERBOSE)

[Out]

-1/2*F^a/d*F^(b/(d*x+c)^2)/(d*x+c)^5/b/ln(F)+5/4*F^a/d/b^2/ln(F)^2*F^(b/(d*x+c)^2)/(d*x+c)^3-15/8*F^a/d/b^3/ln
(F)^3*F^(b/(d*x+c)^2)/(d*x+c)+15/16*F^a/d/b^3/ln(F)^3*Pi^(1/2)/(-b*ln(F))^(1/2)*erf((-b*ln(F))^(1/2)/(d*x+c))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (131) = 262\).

Time = 0.28 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.05 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^8} \, dx=-\frac {15 \, \sqrt {\pi } {\left (d^{6} x^{5} + 5 \, c d^{5} x^{4} + 10 \, c^{2} d^{4} x^{3} + 10 \, c^{3} d^{3} x^{2} + 5 \, c^{4} d^{2} x + c^{5} d\right )} F^{a} \sqrt {-\frac {b \log \left (F\right )}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}}}{d x + c}\right ) + 2 \, {\left (4 \, b^{3} \log \left (F\right )^{3} - 10 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (F\right )^{2} + 15 \, {\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4}\right )} \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{16 \, {\left (b^{4} d^{6} x^{5} + 5 \, b^{4} c d^{5} x^{4} + 10 \, b^{4} c^{2} d^{4} x^{3} + 10 \, b^{4} c^{3} d^{3} x^{2} + 5 \, b^{4} c^{4} d^{2} x + b^{4} c^{5} d\right )} \log \left (F\right )^{4}} \]

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^8,x, algorithm="fricas")

[Out]

-1/16*(15*sqrt(pi)*(d^6*x^5 + 5*c*d^5*x^4 + 10*c^2*d^4*x^3 + 10*c^3*d^3*x^2 + 5*c^4*d^2*x + c^5*d)*F^a*sqrt(-b
*log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)/(d*x + c)) + 2*(4*b^3*log(F)^3 - 10*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^
2)*log(F)^2 + 15*(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)))/((b^4*d^6*x^5 + 5*b^4*c*d^5*x^4 + 10*b^4*c^2*d^4*x^3 + 10*b^
4*c^3*d^3*x^2 + 5*b^4*c^4*d^2*x + b^4*c^5*d)*log(F)^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^8} \, dx=\text {Timed out} \]

[In]

integrate(F**(a+b/(d*x+c)**2)/(d*x+c)**8,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^8} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{8}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^8,x, algorithm="maxima")

[Out]

integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^8, x)

Giac [F]

\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^8} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{8}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^8,x, algorithm="giac")

[Out]

integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^8, x)

Mupad [B] (verification not implemented)

Time = 0.99 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.95 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^8} \, dx=\frac {5\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}}{4\,b^2\,d\,{\ln \left (F\right )}^2\,{\left (c+d\,x\right )}^3}-\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}}{2\,b\,d\,\ln \left (F\right )\,{\left (c+d\,x\right )}^5}-\frac {15\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}}{8\,b^3\,d\,{\ln \left (F\right )}^3\,\left (c+d\,x\right )}+\frac {15\,F^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (F\right )}{\sqrt {b\,\ln \left (F\right )}\,\left (c+d\,x\right )}\right )}{16\,b^3\,d\,{\ln \left (F\right )}^3\,\sqrt {b\,\ln \left (F\right )}} \]

[In]

int(F^(a + b/(c + d*x)^2)/(c + d*x)^8,x)

[Out]

(5*F^a*F^(b/(c + d*x)^2))/(4*b^2*d*log(F)^2*(c + d*x)^3) - (F^a*F^(b/(c + d*x)^2))/(2*b*d*log(F)*(c + d*x)^5)
- (15*F^a*F^(b/(c + d*x)^2))/(8*b^3*d*log(F)^3*(c + d*x)) + (15*F^a*pi^(1/2)*erfi((b*log(F))/((b*log(F))^(1/2)
*(c + d*x))))/(16*b^3*d*log(F)^3*(b*log(F))^(1/2))