\(\int f^{\frac {c}{(a+b x)^2}} x \, dx\) [160]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 13, antiderivative size = 111 \[ \int f^{\frac {c}{(a+b x)^2}} x \, dx=-\frac {a f^{\frac {c}{(a+b x)^2}} (a+b x)}{b^2}+\frac {f^{\frac {c}{(a+b x)^2}} (a+b x)^2}{2 b^2}+\frac {a \sqrt {c} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \sqrt {\log (f)}}{b^2}-\frac {c \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{(a+b x)^2}\right ) \log (f)}{2 b^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.80 \[ \int f^{\frac {c}{(a+b x)^2}} x \, dx=\frac {f^{\frac {c}{(a+b x)^2}} \left (-a^2+b^2 x^2\right )+2 a \sqrt {c} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \sqrt {\log (f)}-c \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{(a+b x)^2}\right ) \log (f)}{2 b^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2656, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2656

\(\displaystyle \int \left (\frac {(a+b x) f^{\frac {c}{(a+b x)^2}}}{b}-\frac {a f^{\frac {c}{(a+b x)^2}}}{b}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {\pi } a \sqrt {c} \sqrt {\log (f)} \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right )}{b^2}-\frac {c \log (f) \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{(a+b x)^2}\right )}{2 b^2}+\frac {(a+b x)^2 f^{\frac {c}{(a+b x)^2}}}{2 b^2}-\frac {a (a+b x) f^{\frac {c}{(a+b x)^2}}}{b^2}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2656
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(Px_), x_Symbol] :> Int[ 
ExpandLinearProduct[F^(a + b*(c + d*x)^n), Px, c, d, x], x] /; FreeQ[{F, a, 
 b, c, d, n}, x] && PolynomialQ[Px, x]
 
Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.84

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96 \[ \int f^{\frac {c}{(a+b x)^2}} x \, dx=-\frac {2 \, \sqrt {\pi } a b \sqrt {-\frac {c \log \left (f\right )}{b^{2}}} \operatorname {erf}\left (\frac {b \sqrt {-\frac {c \log \left (f\right )}{b^{2}}}}{b x + a}\right ) + c {\rm Ei}\left (\frac {c \log \left (f\right )}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right ) \log \left (f\right ) - {\left (b^{2} x^{2} - a^{2}\right )} f^{\frac {c}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{2 \, b^{2}} \] Input:

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

Output:

-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
 

Sympy [F]

\[ \int f^{\frac {c}{(a+b x)^2}} x \, dx=\int f^{\frac {c}{\left (a + b x\right )^{2}}} x\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int f^{\frac {c}{(a+b x)^2}} x \, dx=\int { f^{\frac {c}{{\left (b x + a\right )}^{2}}} x \,d x } \] Input:

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

Output:

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
 

Giac [F]

\[ \int f^{\frac {c}{(a+b x)^2}} x \, dx=\int { f^{\frac {c}{{\left (b x + a\right )}^{2}}} x \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int f^{\frac {c}{(a+b x)^2}} x \, dx=\int f^{\frac {c}{{\left (a+b\,x\right )}^2}}\,x \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int f^{\frac {c}{(a+b x)^2}} x \, dx=\text {too large to display} \] Input:

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

Output:

(8*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*log(f)**3*a*b*c**3*x + 4*f**(c/(a** 
2 + 2*a*b*x + b**2*x**2))*log(f)**2*a**4*c**2 - 12*f**(c/(a**2 + 2*a*b*x + 
 b**2*x**2))*log(f)**2*a**3*b*c**2*x - 20*f**(c/(a**2 + 2*a*b*x + b**2*x** 
2))*log(f)**2*a**2*b**2*c**2*x**2 - 12*f**(c/(a**2 + 2*a*b*x + b**2*x**2)) 
*log(f)**2*a*b**3*c**2*x**3 - 14*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*log(f 
)*a**6*c - 30*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*log(f)*a**5*b*c*x - 16*f 
**(c/(a**2 + 2*a*b*x + b**2*x**2))*log(f)*a**4*b**2*c*x**2 + 8*f**(c/(a**2 
 + 2*a*b*x + b**2*x**2))*log(f)*a**3*b**3*c*x**3 + 17*f**(c/(a**2 + 2*a*b* 
x + b**2*x**2))*log(f)*a**2*b**4*c*x**4 + 12*f**(c/(a**2 + 2*a*b*x + b**2* 
x**2))*log(f)*a*b**5*c*x**5 + 3*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*log(f) 
*b**6*c*x**6 + 13*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*a**8 + 52*f**(c/(a** 
2 + 2*a*b*x + b**2*x**2))*a**7*b*x + 78*f**(c/(a**2 + 2*a*b*x + b**2*x**2) 
)*a**6*b**2*x**2 + 52*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*a**5*b**3*x**3 + 
 13*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*a**4*b**4*x**4 + 6*int((f**(c/(a** 
2 + 2*a*b*x + b**2*x**2))*x**4)/(a**5 + 5*a**4*b*x + 10*a**3*b**2*x**2 + 1 
0*a**2*b**3*x**3 + 5*a*b**4*x**4 + b**5*x**5),x)*log(f)**2*a**4*b**5*c**2 
+ 24*int((f**(c/(a**2 + 2*a*b*x + b**2*x**2))*x**4)/(a**5 + 5*a**4*b*x + 1 
0*a**3*b**2*x**2 + 10*a**2*b**3*x**3 + 5*a*b**4*x**4 + b**5*x**5),x)*log(f 
)**2*a**3*b**6*c**2*x + 36*int((f**(c/(a**2 + 2*a*b*x + b**2*x**2))*x**4)/ 
(a**5 + 5*a**4*b*x + 10*a**3*b**2*x**2 + 10*a**2*b**3*x**3 + 5*a*b**4*x...