3.227 \(\int f^{\frac {c}{(a+b x)^2}} x \, dx\)

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

[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 = 0.12, 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 = {2226, 2206, 2211, 2204, 2214, 2210} \[ \frac {\sqrt {\pi } a \sqrt {c} \sqrt {\log (f)} \text {Erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right )}{b^2}-\frac {c \log (f) \text {Ei}\left (\frac {c \log (f)}{(a+b x)^2}\right )}{2 b^2}+\frac {(a+b x)^2 f^{\frac {c}{(a+b x)^2}}}{2 b^2}-\frac {a (a+b x) f^{\frac {c}{(a+b x)^2}}}{b^2} \]

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 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 2210

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 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]))

Rule 2226

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)) \operatorname {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 = 0.05, size = 89, normalized size = 0.80 \[ \frac {\left (b^2 x^2-a^2\right ) f^{\frac {c}{(a+b x)^2}}+2 \sqrt {\pi } a \sqrt {c} \sqrt {\log (f)} \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right )-c \log (f) \text {Ei}\left (\frac {c \log (f)}{(a+b x)^2}\right )}{2 b^2} \]

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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fricas [A]  time = 0.42, size = 107, normalized size = 0.96 \[ -\frac {2 \, \sqrt {\pi } a b \sqrt {-\frac {c \log \relax (f)}{b^{2}}} \operatorname {erf}\left (\frac {b \sqrt {-\frac {c \log \relax (f)}{b^{2}}}}{b x + a}\right ) + c {\rm Ei}\left (\frac {c \log \relax (f)}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right ) \log \relax (f) - {\left (b^{2} x^{2} - a^{2}\right )} f^{\frac {c}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{2 \, b^{2}} \]

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

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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maple [A]  time = 0.07, size = 93, normalized size = 0.84 \[ \frac {x^{2} f^{\frac {c}{\left (b x +a \right )^{2}}}}{2}+\frac {\sqrt {\pi }\, a c \erf \left (\frac {\sqrt {-c \ln \relax (f )}}{b x +a}\right ) \ln \relax (f )}{\sqrt {-c \ln \relax (f )}\, b^{2}}-\frac {a^{2} f^{\frac {c}{\left (b x +a \right )^{2}}}}{2 b^{2}}+\frac {c \Ei \left (1, -\frac {c \ln \relax (f )}{\left (b x +a \right )^{2}}\right ) \ln \relax (f )}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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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