3.4.31 \(\int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \, dx\) [331]

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

[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 = 0.08, 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 = {2245, 2237, 2242, 2235} \begin {gather*} -\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} \end {gather*}

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 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 2237

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 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 2245

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 ) \text {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 = 0.06, size = 79, normalized size = 0.77 \begin {gather*} \frac {F^a \left (-2 b^{3/2} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right ) \log ^{\frac {3}{2}}(F)+F^{\frac {b}{(c+d x)^2}} (c+d x) \left ((c+d x)^2+2 b \log (F)\right )\right )}{3 d} \end {gather*}

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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Maple [A]
time = 0.08, size = 169, normalized size = 1.66

method	result	size
risch	\(\frac {F^{a} d^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} x^{3}}{3}+F^{a} d \,F^{\frac {b}{\left (d x +c \right )^{2}}} c \,x^{2}+F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2} x +\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{3}}{3 d}+\frac {2 F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} x}{3}+\frac {2 F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c}{3 d}-\frac {2 F^{a} b^{2} \ln \left (F \right )^{2} \sqrt {\pi }\, \erf \left (\frac {\sqrt {-b \ln \left (F \right )}}{d x +c}\right )}{3 d \sqrt {-b \ln \left (F \right )}}\)	\(169\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*F^a*d^2*F^(b/(d*x+c)^2)*x^3+F^a*d*F^(b/(d*x+c)^2)*c*x^2+F^a*F^(b/(d*x+c)^2)*c^2*x+1/3*F^a/d*F^(b/(d*x+c)^2
)*c^3+2/3*F^a*b*ln(F)*F^(b/(d*x+c)^2)*x+2/3*F^a/d*b*ln(F)*F^(b/(d*x+c)^2)*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 0.40, size = 130, normalized size = 1.27 \begin {gather*} \frac {2 \, \sqrt {\pi } F^{a} b d \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 ) \log \left (F\right ) + {\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 \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}}}}{3 \, d} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [B]
time = 4.00, size = 97, normalized size = 0.95 \begin {gather*} \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 \left (F\right )}{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 \left (F\right )}{\sqrt {b\,\ln \left (F\right )}\,\left (c+d\,x\right )}\right )\,{\ln \left (F\right )}^2}{3\,d\,\sqrt {b\,\ln \left (F\right )}} \end {gather*}

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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