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

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

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2214, 2206, 2211, 2204} \[ -\frac {2 \sqrt {\pi } b^{3/2} F^a \log ^{\frac {3}{2}}(F) \text {Erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{3 d}+\frac {(c+d x)^3 F^{a+\frac {b}{(c+d x)^2}}}{3 d}+\frac {2 b \log (F) (c+d x) F^{a+\frac {b}{(c+d x)^2}}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.08, size = 79, normalized size = 0.77 \[ \frac {F^a \left ((c+d x) F^{\frac {b}{(c+d x)^2}} \left (2 b \log (F)+(c+d x)^2\right )-2 \sqrt {\pi } b^{3/2} \log ^{\frac {3}{2}}(F) \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 130, normalized size = 1.27 \[ \frac {2 \, \sqrt {\pi } F^{a} b d \sqrt {-\frac {b \log \relax (F)}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \relax (F)}{d^{2}}}}{d x + c}\right ) \log \relax (F) + {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3} + 2 \, {\left (b d x + b c\right )} \log \relax (F)\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*sqrt(pi)*F^a*b*d*sqrt(-b*log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)/(d*x + c))*log(F) + (d^3*x^3 + 3*c*d^2*x
^2 + 3*c^2*d*x + c^3 + 2*(b*d*x + b*c)*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 \[ \int {\left (d x + c\right )}^{2} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 169, normalized size = 1.66 \[ \frac {d^{2} x^{3} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{3}+c d \,x^{2} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}-\frac {2 \sqrt {\pi }\, b^{2} F^{a} \erf \left (\frac {\sqrt {-b \ln \relax (F )}}{d x +c}\right ) \ln \relax (F )^{2}}{3 \sqrt {-b \ln \relax (F )}\, d}+\frac {2 b x \,F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )}{3}+c^{2} x \,F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}+\frac {2 b c \,F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} \ln \relax (F )}{3 d}+\frac {c^{3} F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*F^a*d^2*F^(1/(d*x+c)^2*b)*x^3+F^a*d*F^(1/(d*x+c)^2*b)*c*x^2+F^a*F^(1/(d*x+c)^2*b)*c^2*x+1/3*F^a/d*F^(1/(d*
x+c)^2*b)*c^3+2/3*F^a*b*ln(F)*F^(1/(d*x+c)^2*b)*x+2/3*F^a/d*b*ln(F)*F^(1/(d*x+c)^2*b)*c-2/3*F^a/d*b^2*ln(F)^2*
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 \[ \frac {1}{3} \, {\left (F^{a} d^{2} x^{3} + 3 \, F^{a} c d x^{2} + {\left (3 \, F^{a} c^{2} + 2 \, F^{a} b \log \relax (F)\right )} x\right )} F^{\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}} + \int \frac {2 \, {\left (2 \, F^{a} b^{2} d x \log \relax (F)^{2} - F^{a} b c^{3} \log \relax (F)\right )} F^{\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{3 \, {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(F^a*d^2*x^3 + 3*F^a*c*d*x^2 + (3*F^a*c^2 + 2*F^a*b*log(F))*x)*F^(b/(d^2*x^2 + 2*c*d*x + c^2)) + integrate
(2/3*(2*F^a*b^2*d*x*log(F)^2 - F^a*b*c^3*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 = 4.00, size = 97, normalized size = 0.95 \[ \frac {\left (\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}}{3}+\frac {2\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b\,\ln \relax (F)}{3\,{\left (c+d\,x\right )}^2}\right )\,{\left (c+d\,x\right )}^3}{d}-\frac {2\,F^a\,b^2\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \relax (F)}{\sqrt {b\,\ln \relax (F)}\,\left (c+d\,x\right )}\right )\,{\ln \relax (F)}^2}{3\,d\,\sqrt {b\,\ln \relax (F)}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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