3.3.27 \(\int f^{\frac {c}{(a+b x)^2}} x \, dx\) [227]

Optimal. Leaf size=111 \[ -\frac {a f^{\frac {c}{(a+b x)^2}} (a+b x)}{b^2}+\frac {f^{\frac {c}{(a+b x)^2}} (a+b x)^2}{2 b^2}+\frac {a \sqrt {c} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \sqrt {\log (f)}}{b^2}-\frac {c \text {Ei}\left (\frac {c \log (f)}{(a+b x)^2}\right ) \log (f)}{2 b^2} \]

[Out]

-a*f^(c/(b*x+a)^2)*(b*x+a)/b^2+1/2*f^(c/(b*x+a)^2)*(b*x+a)^2/b^2-1/2*c*Ei(c*ln(f)/(b*x+a)^2)*ln(f)/b^2+a*erfi(
c^(1/2)*ln(f)^(1/2)/(b*x+a))*c^(1/2)*Pi^(1/2)*ln(f)^(1/2)/b^2

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Rubi [A]
time = 0.08, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2258, 2237, 2242, 2235, 2245, 2241} \begin {gather*} \frac {\sqrt {\pi } a \sqrt {c} \sqrt {\log (f)} \text {Erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right )}{b^2}-\frac {c \log (f) \text {Ei}\left (\frac {c \log (f)}{(a+b x)^2}\right )}{2 b^2}+\frac {(a+b x)^2 f^{\frac {c}{(a+b x)^2}}}{2 b^2}-\frac {a (a+b x) f^{\frac {c}{(a+b x)^2}}}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[f^(c/(a + b*x)^2)*x,x]

[Out]

-((a*f^(c/(a + b*x)^2)*(a + b*x))/b^2) + (f^(c/(a + b*x)^2)*(a + b*x)^2)/(2*b^2) + (a*Sqrt[c]*Sqrt[Pi]*Erfi[(S
qrt[c]*Sqrt[Log[f]])/(a + b*x)]*Sqrt[Log[f]])/b^2 - (c*ExpIntegralEi[(c*Log[f])/(a + b*x)^2]*Log[f])/(2*b^2)

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 2241

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[F^a*(ExpIntegralEi[
b*(c + d*x)^n*Log[F]]/(f*n)), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 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]))

Rule 2258

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rubi steps

\begin {align*} \int f^{\frac {c}{(a+b x)^2}} x \, dx &=\int \left (-\frac {a f^{\frac {c}{(a+b x)^2}}}{b}+\frac {f^{\frac {c}{(a+b x)^2}} (a+b x)}{b}\right ) \, dx\\ &=\frac {\int f^{\frac {c}{(a+b x)^2}} (a+b x) \, dx}{b}-\frac {a \int f^{\frac {c}{(a+b x)^2}} \, dx}{b}\\ &=-\frac {a f^{\frac {c}{(a+b x)^2}} (a+b x)}{b^2}+\frac {f^{\frac {c}{(a+b x)^2}} (a+b x)^2}{2 b^2}+\frac {(c \log (f)) \int \frac {f^{\frac {c}{(a+b x)^2}}}{a+b x} \, dx}{b}-\frac {(2 a c \log (f)) \int \frac {f^{\frac {c}{(a+b x)^2}}}{(a+b x)^2} \, dx}{b}\\ &=-\frac {a f^{\frac {c}{(a+b x)^2}} (a+b x)}{b^2}+\frac {f^{\frac {c}{(a+b x)^2}} (a+b x)^2}{2 b^2}-\frac {c \text {Ei}\left (\frac {c \log (f)}{(a+b x)^2}\right ) \log (f)}{2 b^2}+\frac {(2 a c \log (f)) \text {Subst}\left (\int f^{c x^2} \, dx,x,\frac {1}{a+b x}\right )}{b^2}\\ &=-\frac {a f^{\frac {c}{(a+b x)^2}} (a+b x)}{b^2}+\frac {f^{\frac {c}{(a+b x)^2}} (a+b x)^2}{2 b^2}+\frac {a \sqrt {c} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \sqrt {\log (f)}}{b^2}-\frac {c \text {Ei}\left (\frac {c \log (f)}{(a+b x)^2}\right ) \log (f)}{2 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 89, normalized size = 0.80 \begin {gather*} \frac {f^{\frac {c}{(a+b x)^2}} \left (-a^2+b^2 x^2\right )+2 a \sqrt {c} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \sqrt {\log (f)}-c \text {Ei}\left (\frac {c \log (f)}{(a+b x)^2}\right ) \log (f)}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[f^(c/(a + b*x)^2)*x,x]

[Out]

(f^(c/(a + b*x)^2)*(-a^2 + b^2*x^2) + 2*a*Sqrt[c]*Sqrt[Pi]*Erfi[(Sqrt[c]*Sqrt[Log[f]])/(a + b*x)]*Sqrt[Log[f]]
- c*ExpIntegralEi[(c*Log[f])/(a + b*x)^2]*Log[f])/(2*b^2)

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Maple [A]
time = 0.05, size = 93, normalized size = 0.84

method	result	size
risch	\(\frac {f^{\frac {c}{\left (b x +a \right )^{2}}} x^{2}}{2}-\frac {f^{\frac {c}{\left (b x +a \right )^{2}}} a^{2}}{2 b^{2}}+\frac {\ln \left (f \right ) c \expIntegral \left (1, -\frac {c \ln \left (f \right )}{\left (b x +a \right )^{2}}\right )}{2 b^{2}}+\frac {a \ln \left (f \right ) c \sqrt {\pi }\, \erf \left (\frac {\sqrt {-c \ln \left (f \right )}}{b x +a}\right )}{b^{2} \sqrt {-c \ln \left (f \right )}}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*f^(c/(b*x+a)^2)*x^2-1/2/b^2*f^(c/(b*x+a)^2)*a^2+1/2/b^2*ln(f)*c*Ei(1,-c*ln(f)/(b*x+a)^2)+1/b^2*a*ln(f)*c*P
i^(1/2)/(-c*ln(f))^(1/2)*erf((-c*ln(f))^(1/2)/(b*x+a))

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

[Out]

b*c*integrate(f^(c/(b^2*x^2 + 2*a*b*x + a^2))*x^2/(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3), x)*log(f) + 1/2*f
^(c/(b^2*x^2 + 2*a*b*x + a^2))*x^2

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Fricas [A]
time = 0.40, size = 107, normalized size = 0.96 \begin {gather*} -\frac {2 \, \sqrt {\pi } a b \sqrt {-\frac {c \log \left (f\right )}{b^{2}}} \operatorname {erf}\left (\frac {b \sqrt {-\frac {c \log \left (f\right )}{b^{2}}}}{b x + a}\right ) + c {\rm Ei}\left (\frac {c \log \left (f\right )}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right ) \log \left (f\right ) - {\left (b^{2} x^{2} - a^{2}\right )} f^{\frac {c}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*sqrt(pi)*a*b*sqrt(-c*log(f)/b^2)*erf(b*sqrt(-c*log(f)/b^2)/(b*x + a)) + c*Ei(c*log(f)/(b^2*x^2 + 2*a*b
*x + a^2))*log(f) - (b^2*x^2 - a^2)*f^(c/(b^2*x^2 + 2*a*b*x + a^2)))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(c/(a + b*x)**2)*x, 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^(c/(b*x+a)^2)*x,x, algorithm="giac")

[Out]

integrate(f^(c/(b*x + a)^2)*x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int f^{\frac {c}{{\left (a+b\,x\right )}^2}}\,x \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(c/(a + b*x)^2)*x,x)

[Out]

int(f^(c/(a + b*x)^2)*x, x)

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