3.337 \(\int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx\)

Optimal. Leaf size=183 \[ -\frac {105 \sqrt {\pi } F^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{32 b^{9/2} d \log ^{\frac {9}{2}}(F)}+\frac {105 F^{a+\frac {b}{(c+d x)^2}}}{16 b^4 d \log ^4(F) (c+d x)}-\frac {35 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d \log ^3(F) (c+d x)^3}+\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d \log ^2(F) (c+d x)^5}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^7} \]

[Out]

105/16*F^(a+b/(d*x+c)^2)/b^4/d/(d*x+c)/ln(F)^4-35/8*F^(a+b/(d*x+c)^2)/b^3/d/(d*x+c)^3/ln(F)^3+7/4*F^(a+b/(d*x+
c)^2)/b^2/d/(d*x+c)^5/ln(F)^2-1/2*F^(a+b/(d*x+c)^2)/b/d/(d*x+c)^7/ln(F)-105/32*F^a*erfi(b^(1/2)*ln(F)^(1/2)/(d
*x+c))*Pi^(1/2)/b^(9/2)/d/ln(F)^(9/2)

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Rubi [A]  time = 0.26, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2212, 2211, 2204} \[ -\frac {105 \sqrt {\pi } F^a \text {Erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{32 b^{9/2} d \log ^{\frac {9}{2}}(F)}+\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d \log ^2(F) (c+d x)^5}-\frac {35 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d \log ^3(F) (c+d x)^3}+\frac {105 F^{a+\frac {b}{(c+d x)^2}}}{16 b^4 d \log ^4(F) (c+d x)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^7} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x)^2)/(c + d*x)^10,x]

[Out]

(-105*F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)])/(32*b^(9/2)*d*Log[F]^(9/2)) + (105*F^(a + b/(c + d*
x)^2))/(16*b^4*d*(c + d*x)*Log[F]^4) - (35*F^(a + b/(c + d*x)^2))/(8*b^3*d*(c + d*x)^3*Log[F]^3) + (7*F^(a + b
/(c + d*x)^2))/(4*b^2*d*(c + d*x)^5*Log[F]^2) - F^(a + b/(c + d*x)^2)/(2*b*d*(c + d*x)^7*Log[F])

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 2211

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 2212

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*} \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx &=-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)}-\frac {7 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^8} \, dx}{2 b \log (F)}\\ &=\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^5 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)}+\frac {35 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx}{4 b^2 \log ^2(F)}\\ &=-\frac {35 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x)^3 \log ^3(F)}+\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^5 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)}-\frac {105 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4} \, dx}{8 b^3 \log ^3(F)}\\ &=\frac {105 F^{a+\frac {b}{(c+d x)^2}}}{16 b^4 d (c+d x) \log ^4(F)}-\frac {35 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x)^3 \log ^3(F)}+\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^5 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)}+\frac {105 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx}{16 b^4 \log ^4(F)}\\ &=\frac {105 F^{a+\frac {b}{(c+d x)^2}}}{16 b^4 d (c+d x) \log ^4(F)}-\frac {35 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x)^3 \log ^3(F)}+\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^5 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)}-\frac {105 \operatorname {Subst}\left (\int F^{a+b x^2} \, dx,x,\frac {1}{c+d x}\right )}{16 b^4 d \log ^4(F)}\\ &=-\frac {105 F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{32 b^{9/2} d \log ^{\frac {9}{2}}(F)}+\frac {105 F^{a+\frac {b}{(c+d x)^2}}}{16 b^4 d (c+d x) \log ^4(F)}-\frac {35 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x)^3 \log ^3(F)}+\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^5 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 127, normalized size = 0.69 \[ \frac {F^a \left (\frac {2 \sqrt {b} \sqrt {\log (F)} F^{\frac {b}{(c+d x)^2}} \left (-8 b^3 \log ^3(F)+28 b^2 \log ^2(F) (c+d x)^2-70 b \log (F) (c+d x)^4+105 (c+d x)^6\right )}{(c+d x)^7}-105 \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )\right )}{32 b^{9/2} d \log ^{\frac {9}{2}}(F)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x)^2)/(c + d*x)^10,x]

[Out]

(F^a*(-105*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)] + (2*Sqrt[b]*F^(b/(c + d*x)^2)*Sqrt[Log[F]]*(105*(c
+ d*x)^6 - 70*b*(c + d*x)^4*Log[F] + 28*b^2*(c + d*x)^2*Log[F]^2 - 8*b^3*Log[F]^3))/(c + d*x)^7))/(32*b^(9/2)
*d*Log[F]^(9/2))

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fricas [B]  time = 0.45, size = 439, normalized size = 2.40 \[ \frac {105 \, \sqrt {\pi } {\left (d^{8} x^{7} + 7 \, c d^{7} x^{6} + 21 \, c^{2} d^{6} x^{5} + 35 \, c^{3} d^{5} x^{4} + 35 \, c^{4} d^{4} x^{3} + 21 \, c^{5} d^{3} x^{2} + 7 \, c^{6} d^{2} x + c^{7} d\right )} F^{a} \sqrt {-\frac {b \log \relax (F)}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \relax (F)}{d^{2}}}}{d x + c}\right ) - 2 \, {\left (8 \, b^{4} \log \relax (F)^{4} - 28 \, {\left (b^{3} d^{2} x^{2} + 2 \, b^{3} c d x + b^{3} c^{2}\right )} \log \relax (F)^{3} + 70 \, {\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \relax (F)^{2} - 105 \, {\left (b d^{6} x^{6} + 6 \, b c d^{5} x^{5} + 15 \, b c^{2} d^{4} x^{4} + 20 \, b c^{3} d^{3} x^{3} + 15 \, b c^{4} d^{2} x^{2} + 6 \, b c^{5} d x + b c^{6}\right )} \log \relax (F)\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{32 \, {\left (b^{5} d^{8} x^{7} + 7 \, b^{5} c d^{7} x^{6} + 21 \, b^{5} c^{2} d^{6} x^{5} + 35 \, b^{5} c^{3} d^{5} x^{4} + 35 \, b^{5} c^{4} d^{4} x^{3} + 21 \, b^{5} c^{5} d^{3} x^{2} + 7 \, b^{5} c^{6} d^{2} x + b^{5} c^{7} d\right )} \log \relax (F)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^10,x, algorithm="fricas")

[Out]

1/32*(105*sqrt(pi)*(d^8*x^7 + 7*c*d^7*x^6 + 21*c^2*d^6*x^5 + 35*c^3*d^5*x^4 + 35*c^4*d^4*x^3 + 21*c^5*d^3*x^2
+ 7*c^6*d^2*x + c^7*d)*F^a*sqrt(-b*log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)/(d*x + c)) - 2*(8*b^4*log(F)^4 - 28*(
b^3*d^2*x^2 + 2*b^3*c*d*x + b^3*c^2)*log(F)^3 + 70*(b^2*d^4*x^4 + 4*b^2*c*d^3*x^3 + 6*b^2*c^2*d^2*x^2 + 4*b^2*
c^3*d*x + b^2*c^4)*log(F)^2 - 105*(b*d^6*x^6 + 6*b*c*d^5*x^5 + 15*b*c^2*d^4*x^4 + 20*b*c^3*d^3*x^3 + 15*b*c^4*
d^2*x^2 + 6*b*c^5*d*x + b*c^6)*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^
5*d^8*x^7 + 7*b^5*c*d^7*x^6 + 21*b^5*c^2*d^6*x^5 + 35*b^5*c^3*d^5*x^4 + 35*b^5*c^4*d^4*x^3 + 21*b^5*c^5*d^3*x^
2 + 7*b^5*c^6*d^2*x + b^5*c^7*d)*log(F)^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{10}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^10,x, algorithm="giac")

[Out]

integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^10, x)

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maple [A]  time = 0.20, size = 175, normalized size = 0.96 \[ -\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{2 \left (d x +c \right )^{7} b d \ln \relax (F )}+\frac {7 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{4 \left (d x +c \right )^{5} b^{2} d \ln \relax (F )^{2}}-\frac {35 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{8 \left (d x +c \right )^{3} b^{3} d \ln \relax (F )^{3}}+\frac {105 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{16 \left (d x +c \right ) b^{4} d \ln \relax (F )^{4}}-\frac {105 \sqrt {\pi }\, F^{a} \erf \left (\frac {\sqrt {-b \ln \relax (F )}}{d x +c}\right )}{32 \sqrt {-b \ln \relax (F )}\, b^{4} d \ln \relax (F )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+1/(d*x+c)^2*b)/(d*x+c)^10,x)

[Out]

-1/2*F^a/d*F^(1/(d*x+c)^2*b)/(d*x+c)^7/b/ln(F)+7/4*F^a/d/b^2/ln(F)^2*F^(1/(d*x+c)^2*b)/(d*x+c)^5-35/8*F^a/d/b^
3/ln(F)^3*F^(1/(d*x+c)^2*b)/(d*x+c)^3+105/16*F^a/d/b^4/ln(F)^4*F^(1/(d*x+c)^2*b)/(d*x+c)-105/32*F^a/d/b^4/ln(F
)^4*Pi^(1/2)/(-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 \[ \int \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{10}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^10,x, algorithm="maxima")

[Out]

integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^10, x)

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mupad [B]  time = 4.62, size = 160, normalized size = 0.87 \[ -\frac {\frac {F^a\,\left (105\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \relax (F)}{\sqrt {b\,\ln \relax (F)}\,\left (c+d\,x\right )}\right )-\frac {210\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,\sqrt {b\,\ln \relax (F)}}{c+d\,x}\right )}{32\,\sqrt {b\,\ln \relax (F)}}-\frac {7\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^2\,{\ln \relax (F)}^2}{4\,{\left (c+d\,x\right )}^5}+\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^3\,{\ln \relax (F)}^3}{2\,{\left (c+d\,x\right )}^7}+\frac {35\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b\,\ln \relax (F)}{8\,{\left (c+d\,x\right )}^3}}{b^4\,d\,{\ln \relax (F)}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b/(c + d*x)^2)/(c + d*x)^10,x)

[Out]

-((F^a*(105*pi^(1/2)*erfi((b*log(F))/((b*log(F))^(1/2)*(c + d*x))) - (210*F^(b/(c + d*x)^2)*(b*log(F))^(1/2))/
(c + d*x)))/(32*(b*log(F))^(1/2)) - (7*F^a*F^(b/(c + d*x)^2)*b^2*log(F)^2)/(4*(c + d*x)^5) + (F^a*F^(b/(c + d*
x)^2)*b^3*log(F)^3)/(2*(c + d*x)^7) + (35*F^a*F^(b/(c + d*x)^2)*b*log(F))/(8*(c + d*x)^3))/(b^4*d*log(F)^4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b/(d*x+c)**2)/(d*x+c)**10,x)

[Out]

Timed out

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