Integrand size = 11, antiderivative size = 62 \[ \int f^{\frac {c}{(a+b x)^2}} \, dx=\frac {f^{\frac {c}{(a+b x)^2}} (a+b x)}{b}-\frac {\sqrt {c} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \sqrt {\log (f)}}{b} \] Output:
f^(c/(b*x+a)^2)*(b*x+a)/b-c^(1/2)*Pi^(1/2)*erfi(c^(1/2)*ln(f)^(1/2)/(b*x+a ))*ln(f)^(1/2)/b
Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00 \[ \int f^{\frac {c}{(a+b x)^2}} \, dx=\frac {f^{\frac {c}{(a+b x)^2}} (a+b x)}{b}-\frac {\sqrt {c} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \sqrt {\log (f)}}{b} \] Input:
Integrate[f^(c/(a + b*x)^2),x]
Output:
(f^(c/(a + b*x)^2)*(a + b*x))/b - (Sqrt[c]*Sqrt[Pi]*Erfi[(Sqrt[c]*Sqrt[Log [f]])/(a + b*x)]*Sqrt[Log[f]])/b
Time = 0.38 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2635, 2640, 2633}
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.
\(\displaystyle \int f^{\frac {c}{(a+b x)^2}} \, dx\) |
\(\Big \downarrow \) 2635 |
\(\displaystyle 2 c \log (f) \int \frac {f^{\frac {c}{(a+b x)^2}}}{(a+b x)^2}dx+\frac {(a+b x) f^{\frac {c}{(a+b x)^2}}}{b}\) |
\(\Big \downarrow \) 2640 |
\(\displaystyle \frac {(a+b x) f^{\frac {c}{(a+b x)^2}}}{b}-\frac {2 c \log (f) \int f^{\frac {c}{(a+b x)^2}}d\frac {1}{a+b x}}{b}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {(a+b x) f^{\frac {c}{(a+b x)^2}}}{b}-\frac {\sqrt {\pi } \sqrt {c} \sqrt {\log (f)} \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right )}{b}\) |
Input:
Int[f^(c/(a + b*x)^2),x]
Output:
(f^(c/(a + b*x)^2)*(a + b*x))/b - (Sqrt[c]*Sqrt[Pi]*Erfi[(Sqrt[c]*Sqrt[Log [f]])/(a + b*x)]*Sqrt[Log[f]])/b
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]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[(c + d*x)*(F^(a + b*(c + d*x)^n)/d), x] - Simp[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]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[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)]
Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.05
method | result | size |
risch | \(f^{\frac {c}{\left (b x +a \right )^{2}}} x +\frac {f^{\frac {c}{\left (b x +a \right )^{2}}} a}{b}-\frac {\ln \left (f \right ) c \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-c \ln \left (f \right )}}{b x +a}\right )}{b \sqrt {-c \ln \left (f \right )}}\) | \(65\) |
Input:
int(f^(c/(b*x+a)^2),x,method=_RETURNVERBOSE)
Output:
f^(c/(b*x+a)^2)*x+1/b*f^(c/(b*x+a)^2)*a-1/b*ln(f)*c*Pi^(1/2)/(-c*ln(f))^(1 /2)*erf((-c*ln(f))^(1/2)/(b*x+a))
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.10 \[ \int f^{\frac {c}{(a+b x)^2}} \, dx=\frac {\sqrt {\pi } b \sqrt {-\frac {c \log \left (f\right )}{b^{2}}} \operatorname {erf}\left (\frac {b \sqrt {-\frac {c \log \left (f\right )}{b^{2}}}}{b x + a}\right ) + {\left (b x + a\right )} f^{\frac {c}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{b} \] Input:
integrate(f^(c/(b*x+a)^2),x, algorithm="fricas")
Output:
(sqrt(pi)*b*sqrt(-c*log(f)/b^2)*erf(b*sqrt(-c*log(f)/b^2)/(b*x + a)) + (b* x + a)*f^(c/(b^2*x^2 + 2*a*b*x + a^2)))/b
\[ \int f^{\frac {c}{(a+b x)^2}} \, dx=\int f^{\frac {c}{\left (a + b x\right )^{2}}}\, dx \] Input:
integrate(f**(c/(b*x+a)**2),x)
Output:
Integral(f**(c/(a + b*x)**2), x)
\[ \int f^{\frac {c}{(a+b x)^2}} \, dx=\int { f^{\frac {c}{{\left (b x + a\right )}^{2}}} \,d x } \] Input:
integrate(f^(c/(b*x+a)^2),x, algorithm="maxima")
Output:
2*b*c*integrate(f^(c/(b^2*x^2 + 2*a*b*x + a^2))*x/(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3), x)*log(f) + f^(c/(b^2*x^2 + 2*a*b*x + a^2))*x
\[ \int f^{\frac {c}{(a+b x)^2}} \, dx=\int { f^{\frac {c}{{\left (b x + a\right )}^{2}}} \,d x } \] Input:
integrate(f^(c/(b*x+a)^2),x, algorithm="giac")
Output:
integrate(f^(c/(b*x + a)^2), x)
Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.85 \[ \int f^{\frac {c}{(a+b x)^2}} \, dx=\frac {f^{\frac {c}{{\left (a+b\,x\right )}^2}}\,\left (a+b\,x\right )}{b}-\frac {c\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {\sqrt {c\,\ln \left (f\right )}}{a+b\,x}\right )\,\ln \left (f\right )}{b\,\sqrt {c\,\ln \left (f\right )}} \] Input:
int(f^(c/(a + b*x)^2),x)
Output:
(f^(c/(a + b*x)^2)*(a + b*x))/b - (c*pi^(1/2)*erfi((c*log(f))^(1/2)/(a + b *x))*log(f))/(b*(c*log(f))^(1/2))
\[ \int f^{\frac {c}{(a+b x)^2}} \, dx =\text {Too large to display} \] Input:
int(f^(c/(b*x+a)^2),x)
Output:
(4*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*log(f)**2*b*c**2*x + 2*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*log(f)*a**3*c - 2*f**(c/(a**2 + 2*a*b*x + b**2*x** 2))*log(f)*a**2*b*c*x - 10*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*log(f)*a*b* *2*c*x**2 - 6*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*log(f)*b**3*c*x**3 - 5*f **(c/(a**2 + 2*a*b*x + b**2*x**2))*a**5 - 17*f**(c/(a**2 + 2*a*b*x + b**2* x**2))*a**4*b*x - 18*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*a**3*b**2*x**2 - 2*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*a**2*b**3*x**3 + 7*f**(c/(a**2 + 2*a *b*x + b**2*x**2))*a*b**4*x**4 + 3*f**(c/(a**2 + 2*a*b*x + b**2*x**2))*b** 5*x**5 + 8*int((f**(c/(a**2 + 2*a*b*x + b**2*x**2))*x)/(a**7 + 7*a**6*b*x + 21*a**5*b**2*x**2 + 35*a**4*b**3*x**3 + 35*a**3*b**4*x**4 + 21*a**2*b**5 *x**5 + 7*a*b**6*x**6 + b**7*x**7),x)*log(f)**3*a**4*b**2*c**3 + 32*int((f **(c/(a**2 + 2*a*b*x + b**2*x**2))*x)/(a**7 + 7*a**6*b*x + 21*a**5*b**2*x* *2 + 35*a**4*b**3*x**3 + 35*a**3*b**4*x**4 + 21*a**2*b**5*x**5 + 7*a*b**6* x**6 + b**7*x**7),x)*log(f)**3*a**3*b**3*c**3*x + 48*int((f**(c/(a**2 + 2* a*b*x + b**2*x**2))*x)/(a**7 + 7*a**6*b*x + 21*a**5*b**2*x**2 + 35*a**4*b* *3*x**3 + 35*a**3*b**4*x**4 + 21*a**2*b**5*x**5 + 7*a*b**6*x**6 + b**7*x** 7),x)*log(f)**3*a**2*b**4*c**3*x**2 + 32*int((f**(c/(a**2 + 2*a*b*x + b**2 *x**2))*x)/(a**7 + 7*a**6*b*x + 21*a**5*b**2*x**2 + 35*a**4*b**3*x**3 + 35 *a**3*b**4*x**4 + 21*a**2*b**5*x**5 + 7*a*b**6*x**6 + b**7*x**7),x)*log(f) **3*a*b**5*c**3*x**3 + 8*int((f**(c/(a**2 + 2*a*b*x + b**2*x**2))*x)/(a...