3.329 \(\int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^6 \, dx\)

Optimal. Leaf size=170 \[ -\frac {8 \sqrt {\pi } b^{7/2} F^a \log ^{\frac {7}{2}}(F) \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{105 d}+\frac {8 b^3 \log ^3(F) (c+d x) F^{a+\frac {b}{(c+d x)^2}}}{105 d}+\frac {4 b^2 \log ^2(F) (c+d x)^3 F^{a+\frac {b}{(c+d x)^2}}}{105 d}+\frac {(c+d x)^7 F^{a+\frac {b}{(c+d x)^2}}}{7 d}+\frac {2 b \log (F) (c+d x)^5 F^{a+\frac {b}{(c+d x)^2}}}{35 d} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2214, 2206, 2211, 2204} \[ -\frac {8 \sqrt {\pi } b^{7/2} F^a \log ^{\frac {7}{2}}(F) \text {Erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{105 d}+\frac {8 b^3 \log ^3(F) (c+d x) F^{a+\frac {b}{(c+d x)^2}}}{105 d}+\frac {4 b^2 \log ^2(F) (c+d x)^3 F^{a+\frac {b}{(c+d x)^2}}}{105 d}+\frac {(c+d x)^7 F^{a+\frac {b}{(c+d x)^2}}}{7 d}+\frac {2 b \log (F) (c+d x)^5 F^{a+\frac {b}{(c+d x)^2}}}{35 d} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x)^2)*(c + d*x)^6,x]

[Out]

(F^(a + b/(c + d*x)^2)*(c + d*x)^7)/(7*d) + (2*b*F^(a + b/(c + d*x)^2)*(c + d*x)^5*Log[F])/(35*d) + (4*b^2*F^(
a + b/(c + d*x)^2)*(c + d*x)^3*Log[F]^2)/(105*d) + (8*b^3*F^(a + b/(c + d*x)^2)*(c + d*x)*Log[F]^3)/(105*d) -
(8*b^(7/2)*F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)]*Log[F]^(7/2))/(105*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 2206

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[((c + d*x)*F^(a + b*(c + d*x)^n))/d, x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

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

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Mathematica [A]  time = 0.13, size = 113, normalized size = 0.66 \[ \frac {F^a \left ((c+d x) F^{\frac {b}{(c+d x)^2}} \left (8 b^3 \log ^3(F)+4 b^2 \log ^2(F) (c+d x)^2+6 b \log (F) (c+d x)^4+15 (c+d x)^6\right )-8 \sqrt {\pi } b^{7/2} \log ^{\frac {7}{2}}(F) \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x)^2)*(c + d*x)^6,x]

[Out]

(F^a*(-8*b^(7/2)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)]*Log[F]^(7/2) + F^(b/(c + d*x)^2)*(c + d*x)*(1
5*(c + d*x)^6 + 6*b*(c + d*x)^4*Log[F] + 4*b^2*(c + d*x)^2*Log[F]^2 + 8*b^3*Log[F]^3)))/(105*d)

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fricas [A]  time = 0.41, size = 293, normalized size = 1.72 \[ \frac {8 \, \sqrt {\pi } F^{a} b^{3} d \sqrt {-\frac {b \log \relax (F)}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \relax (F)}{d^{2}}}}{d x + c}\right ) \log \relax (F)^{3} + {\left (15 \, d^{7} x^{7} + 105 \, c d^{6} x^{6} + 315 \, c^{2} d^{5} x^{5} + 525 \, c^{3} d^{4} x^{4} + 525 \, c^{4} d^{3} x^{3} + 315 \, c^{5} d^{2} x^{2} + 105 \, c^{6} d x + 15 \, c^{7} + 8 \, {\left (b^{3} d x + b^{3} c\right )} \log \relax (F)^{3} + 4 \, {\left (b^{2} d^{3} x^{3} + 3 \, b^{2} c d^{2} x^{2} + 3 \, b^{2} c^{2} d x + b^{2} c^{3}\right )} \log \relax (F)^{2} + 6 \, {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + 10 \, b c^{3} d^{2} x^{2} + 5 \, b c^{4} d x + b c^{5}\right )} \log \relax (F)\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^6,x, algorithm="fricas")

[Out]

1/105*(8*sqrt(pi)*F^a*b^3*d*sqrt(-b*log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)/(d*x + c))*log(F)^3 + (15*d^7*x^7 +
105*c*d^6*x^6 + 315*c^2*d^5*x^5 + 525*c^3*d^4*x^4 + 525*c^4*d^3*x^3 + 315*c^5*d^2*x^2 + 105*c^6*d*x + 15*c^7 +
8*(b^3*d*x + b^3*c)*log(F)^3 + 4*(b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*log(F)^2 + 6*(b*d^
5*x^5 + 5*b*c*d^4*x^4 + 10*b*c^2*d^3*x^3 + 10*b*c^3*d^2*x^2 + 5*b*c^4*d*x + b*c^5)*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 \[ \int {\left (d x + c\right )}^{6} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^6,x, algorithm="giac")

[Out]

integrate((d*x + c)^6*F^(a + b/(d*x + c)^2), x)

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maple [B]  time = 0.10, size = 543, normalized size = 3.19 \[ \frac {d^{6} x^{7} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{7}+c \,d^{5} x^{6} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}+\frac {2 b \,d^{4} x^{5} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )}{35}+3 c^{2} d^{4} x^{5} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}+\frac {2 b c \,d^{3} x^{4} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )}{7}+5 c^{3} d^{3} x^{4} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}+\frac {4 b^{2} d^{2} x^{3} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )^{2}}{105}+\frac {4 b \,c^{2} d^{2} x^{3} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )}{7}+5 c^{4} d^{2} x^{3} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}+\frac {4 b^{2} c d \,x^{2} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )^{2}}{35}+\frac {4 b \,c^{3} d \,x^{2} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )}{7}+3 c^{5} d \,x^{2} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}-\frac {8 \sqrt {\pi }\, b^{4} F^{a} \erf \left (\frac {\sqrt {-b \ln \relax (F )}}{d x +c}\right ) \ln \relax (F )^{4}}{105 \sqrt {-b \ln \relax (F )}\, d}+\frac {8 b^{3} x \,F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )^{3}}{105}+\frac {4 b^{2} c^{2} x \,F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )^{2}}{35}+\frac {2 b \,c^{4} x \,F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )}{7}+c^{6} x \,F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}+\frac {8 b^{3} c \,F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )^{3}}{105 d}+\frac {4 b^{2} c^{3} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )^{2}}{105 d}+\frac {2 b \,c^{5} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )}{35 d}+\frac {c^{7} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{7 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+1/(d*x+c)^2*b)*(d*x+c)^6,x)

[Out]

1/7*F^a*d^6*F^(1/(d*x+c)^2*b)*x^7+F^a*d^5*F^(1/(d*x+c)^2*b)*c*x^6+3*F^a*d^4*F^(1/(d*x+c)^2*b)*c^2*x^5+5*F^a*d^
3*F^(1/(d*x+c)^2*b)*c^3*x^4+5*F^a*d^2*F^(1/(d*x+c)^2*b)*c^4*x^3+3*F^a*d*F^(1/(d*x+c)^2*b)*c^5*x^2+F^a*F^(1/(d*
x+c)^2*b)*c^6*x+1/7*F^a/d*F^(1/(d*x+c)^2*b)*c^7+2/35*F^a*d^4*b*ln(F)*F^(1/(d*x+c)^2*b)*x^5+2/7*F^a*d^3*b*ln(F)
*F^(1/(d*x+c)^2*b)*c*x^4+4/7*F^a*d^2*b*ln(F)*F^(1/(d*x+c)^2*b)*c^2*x^3+4/7*F^a*d*b*ln(F)*F^(1/(d*x+c)^2*b)*c^3
*x^2+2/7*F^a*b*ln(F)*F^(1/(d*x+c)^2*b)*c^4*x+2/35*F^a/d*b*ln(F)*F^(1/(d*x+c)^2*b)*c^5+4/105*F^a*d^2*b^2*ln(F)^
2*F^(1/(d*x+c)^2*b)*x^3+4/35*F^a*d*b^2*ln(F)^2*F^(1/(d*x+c)^2*b)*c*x^2+4/35*F^a*b^2*ln(F)^2*F^(1/(d*x+c)^2*b)*
c^2*x+4/105*F^a/d*b^2*ln(F)^2*F^(1/(d*x+c)^2*b)*c^3+8/105*F^a*b^3*ln(F)^3*F^(1/(d*x+c)^2*b)*x+8/105*F^a/d*b^3*
ln(F)^3*F^(1/(d*x+c)^2*b)*c-8/105*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 \[ \frac {1}{105} \, {\left (15 \, F^{a} d^{6} x^{7} + 105 \, F^{a} c d^{5} x^{6} + 3 \, {\left (105 \, F^{a} c^{2} d^{4} + 2 \, F^{a} b d^{4} \log \relax (F)\right )} x^{5} + 15 \, {\left (35 \, F^{a} c^{3} d^{3} + 2 \, F^{a} b c d^{3} \log \relax (F)\right )} x^{4} + {\left (525 \, F^{a} c^{4} d^{2} + 60 \, F^{a} b c^{2} d^{2} \log \relax (F) + 4 \, F^{a} b^{2} d^{2} \log \relax (F)^{2}\right )} x^{3} + 3 \, {\left (105 \, F^{a} c^{5} d + 20 \, F^{a} b c^{3} d \log \relax (F) + 4 \, F^{a} b^{2} c d \log \relax (F)^{2}\right )} x^{2} + {\left (105 \, F^{a} c^{6} + 30 \, F^{a} b c^{4} \log \relax (F) + 12 \, F^{a} b^{2} c^{2} \log \relax (F)^{2} + 8 \, F^{a} b^{3} \log \relax (F)^{3}\right )} x\right )} F^{\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}} + \int \frac {2 \, {\left (8 \, F^{a} b^{4} d x \log \relax (F)^{4} - 15 \, F^{a} b c^{7} \log \relax (F) - 6 \, F^{a} b^{2} c^{5} \log \relax (F)^{2} - 4 \, F^{a} b^{3} c^{3} \log \relax (F)^{3}\right )} F^{\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{105 \, {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^6,x, algorithm="maxima")

[Out]

1/105*(15*F^a*d^6*x^7 + 105*F^a*c*d^5*x^6 + 3*(105*F^a*c^2*d^4 + 2*F^a*b*d^4*log(F))*x^5 + 15*(35*F^a*c^3*d^3
+ 2*F^a*b*c*d^3*log(F))*x^4 + (525*F^a*c^4*d^2 + 60*F^a*b*c^2*d^2*log(F) + 4*F^a*b^2*d^2*log(F)^2)*x^3 + 3*(10
5*F^a*c^5*d + 20*F^a*b*c^3*d*log(F) + 4*F^a*b^2*c*d*log(F)^2)*x^2 + (105*F^a*c^6 + 30*F^a*b*c^4*log(F) + 12*F^
a*b^2*c^2*log(F)^2 + 8*F^a*b^3*log(F)^3)*x)*F^(b/(d^2*x^2 + 2*c*d*x + c^2)) + integrate(2/105*(8*F^a*b^4*d*x*l
og(F)^4 - 15*F^a*b*c^7*log(F) - 6*F^a*b^2*c^5*log(F)^2 - 4*F^a*b^3*c^3*log(F)^3)*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 = 4.41, size = 199, normalized size = 1.17 \[ \frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,{\left (c+d\,x\right )}^7}{7\,d}-\frac {8\,F^a\,\sqrt {\pi }\,{\left (c+d\,x\right )}^7\,{\left (-\frac {b\,\ln \relax (F)}{{\left (c+d\,x\right )}^2}\right )}^{7/2}}{105\,d}+\frac {4\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^2\,{\ln \relax (F)}^2\,{\left (c+d\,x\right )}^3}{105\,d}+\frac {2\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b\,\ln \relax (F)\,{\left (c+d\,x\right )}^5}{35\,d}+\frac {8\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^3\,{\ln \relax (F)}^3\,\left (c+d\,x\right )}{105\,d}+\frac {8\,F^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-\frac {b\,\ln \relax (F)}{{\left (c+d\,x\right )}^2}}\right )\,{\left (c+d\,x\right )}^7\,{\left (-\frac {b\,\ln \relax (F)}{{\left (c+d\,x\right )}^2}\right )}^{7/2}}{105\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b/(c + d*x)^2)*(c + d*x)^6,x)

[Out]

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

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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)**6,x)

[Out]

Timed out

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