\(\int (f x)^m \log (c (d+\frac {e}{\sqrt {x}})^p) \, dx\) [62]

Optimal result

Integrand size = 20, antiderivative size = 80 \[ \int (f x)^m \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right ) \, dx=\frac {e p \sqrt {x} (f x)^m \operatorname {Hypergeometric2F1}\left (1,-1-2 m,-2 m,-\frac {e}{d \sqrt {x}}\right )}{d \left (1+3 m+2 m^2\right )}+\frac {(f x)^{1+m} \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right )}{f (1+m)} \] Output:

e*p*x^(1/2)*(f*x)^m*hypergeom([1, -1-2*m],[-2*m],-e/d/x^(1/2))/d/(2*m^2+3* 
m+1)+(f*x)^(1+m)*ln(c*(d+e/x^(1/2))^p)/f/(1+m)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96 \[ \int (f x)^m \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right ) \, dx=\frac {\sqrt {x} (f x)^m \left (e p \operatorname {Hypergeometric2F1}\left (1,-1-2 m,-2 m,-\frac {e}{d \sqrt {x}}\right )+d (1+2 m) \sqrt {x} \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right )\right )}{d (1+m) (1+2 m)} \] Input:

Integrate[(f*x)^m*Log[c*(d + e/Sqrt[x])^p],x]
 

Output:

(Sqrt[x]*(f*x)^m*(e*p*Hypergeometric2F1[1, -1 - 2*m, -2*m, -(e/(d*Sqrt[x]) 
)] + d*(1 + 2*m)*Sqrt[x]*Log[c*(d + e/Sqrt[x])^p]))/(d*(1 + m)*(1 + 2*m))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2905, 30, 795, 864, 74}

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.

\(\displaystyle \int (f x)^m \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right ) \, dx\)

\(\Big \downarrow \) 2905

\(\displaystyle \frac {e p \int \frac {(f x)^{m+1}}{\left (d+\frac {e}{\sqrt {x}}\right ) x^{3/2}}dx}{2 f (m+1)}+\frac {(f x)^{m+1} \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right )}{f (m+1)}\)

\(\Big \downarrow \) 30

\(\displaystyle \frac {e p x^{-m} (f x)^m \int \frac {x^{m-\frac {1}{2}}}{d+\frac {e}{\sqrt {x}}}dx}{2 (m+1)}+\frac {(f x)^{m+1} \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right )}{f (m+1)}\)

\(\Big \downarrow \) 795

\(\displaystyle \frac {e p x^{-m} (f x)^m \int \frac {x^m}{\sqrt {x} d+e}dx}{2 (m+1)}+\frac {(f x)^{m+1} \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right )}{f (m+1)}\)

\(\Big \downarrow \) 864

\(\displaystyle \frac {e p x^{-m} (f x)^m \int \frac {x^{\frac {1}{2} (2 m+1)}}{\sqrt {x} d+e}d\sqrt {x}}{m+1}+\frac {(f x)^{m+1} \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right )}{f (m+1)}\)

\(\Big \downarrow \) 74

\(\displaystyle \frac {(f x)^{m+1} \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right )}{f (m+1)}+\frac {p x (f x)^m \operatorname {Hypergeometric2F1}\left (1,2 (m+1),2 m+3,-\frac {d \sqrt {x}}{e}\right )}{2 (m+1)^2}\)

Input:

Int[(f*x)^m*Log[c*(d + e/Sqrt[x])^p],x]
 

Output:

(p*x*(f*x)^m*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, -((d*Sqrt[x])/e)])/( 
2*(1 + m)^2) + ((f*x)^(1 + m)*Log[c*(d + e/Sqrt[x])^p])/(f*(1 + m))
 

Defintions of rubi rules used

rule 30

Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I 
ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) 
Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & 
&  !IntegerQ[p]
 

rule 74

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x 
)^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] 
/; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] 
 &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
 

rule 795

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* 
(b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
 

rule 864

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denomi 
nator[n]}, Simp[k   Subst[Int[x^(k*(m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x 
^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]
 

rule 2905

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^ 
(m_.), x_Symbol] :> Simp[(f*x)^(m + 1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m 
+ 1))), x] - Simp[b*e*n*(p/(f*(m + 1)))   Int[x^(n - 1)*((f*x)^(m + 1)/(d + 
 e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]
 

Maple [F]

\[\int \left (f x \right )^{m} \ln \left (c \left (d +\frac {e}{\sqrt {x}}\right )^{p}\right )d x\]

Input:

int((f*x)^m*ln(c*(d+e/x^(1/2))^p),x)
 

Output:

int((f*x)^m*ln(c*(d+e/x^(1/2))^p),x)
 

Fricas [F]

\[ \int (f x)^m \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{p}\right ) \,d x } \] Input:

integrate((f*x)^m*log(c*(d+e/x^(1/2))^p),x, algorithm="fricas")
 

Output:

integral((f*x)^m*log(c*((d*x + e*sqrt(x))/x)^p), x)
 

Sympy [F]

\[ \int (f x)^m \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right ) \, dx=\int \left (f x\right )^{m} \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{p} \right )}\, dx \] Input:

integrate((f*x)**m*ln(c*(d+e/x**(1/2))**p),x)
 

Output:

Integral((f*x)**m*log(c*(d + e/sqrt(x))**p), x)
 

Maxima [F]

\[ \int (f x)^m \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{p}\right ) \,d x } \] Input:

integrate((f*x)^m*log(c*(d+e/x^(1/2))^p),x, algorithm="maxima")
 

Output:

d^2*f^m*p*integrate(1/2*x*x^m/(d*e*(m + 1)*sqrt(x) + e^2*(m + 1)), x) + 1/ 
2*(2*(2*m^2 + 5*m + 3)*e*f^m*x*x^m*log((d*sqrt(x) + e)^p) - 2*(2*m^2 + 5*m 
 + 3)*e*f^m*x*x^m*log(x^(1/2*p)) - 2*(m*p + p)*d*f^m*x^(3/2)*x^m + (2*(2*m 
^2 + 5*m + 3)*e*f^m*log(c) + (2*m*p + 3*p)*e*f^m)*x*x^m)/((2*m^3 + 7*m^2 + 
 8*m + 3)*e)
 

Giac [F]

\[ \int (f x)^m \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right ) \, dx=\int { \left (f x\right )^{m} \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{p}\right ) \,d x } \] Input:

integrate((f*x)^m*log(c*(d+e/x^(1/2))^p),x, algorithm="giac")
 

Output:

integrate((f*x)^m*log(c*(d + e/sqrt(x))^p), x)
 

Mupad [F(-1)]

Timed out. \[ \int (f x)^m \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right ) \, dx=\int \ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^p\right )\,{\left (f\,x\right )}^m \,d x \] Input:

int(log(c*(d + e/x^(1/2))^p)*(f*x)^m,x)
 

Output:

int(log(c*(d + e/x^(1/2))^p)*(f*x)^m, x)
 

Reduce [F]

\[ \int (f x)^m \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^p\right ) \, dx=\frac {f^{m} \left (2 x^{m +\frac {1}{2}} d e m p +4 x^{m} \mathrm {log}\left (\frac {\left (\sqrt {x}\, d +e \right )^{p} c}{x^{\frac {p}{2}}}\right ) d^{2} m^{2} x +2 x^{m} \mathrm {log}\left (\frac {\left (\sqrt {x}\, d +e \right )^{p} c}{x^{\frac {p}{2}}}\right ) d^{2} m x -4 x^{m} \mathrm {log}\left (\frac {\left (\sqrt {x}\, d +e \right )^{p} c}{x^{\frac {p}{2}}}\right ) e^{2} m^{2}-2 x^{m} \mathrm {log}\left (\frac {\left (\sqrt {x}\, d +e \right )^{p} c}{x^{\frac {p}{2}}}\right ) e^{2} m -2 x^{m} e^{2} m p -x^{m} e^{2} p +4 \left (\int \frac {x^{m} \mathrm {log}\left (\frac {\left (\sqrt {x}\, d +e \right )^{p} c}{x^{\frac {p}{2}}}\right )}{x}d x \right ) e^{2} m^{3}+2 \left (\int \frac {x^{m} \mathrm {log}\left (\frac {\left (\sqrt {x}\, d +e \right )^{p} c}{x^{\frac {p}{2}}}\right )}{x}d x \right ) e^{2} m^{2}\right )}{2 d^{2} m \left (2 m^{2}+3 m +1\right )} \] Input:

int((f*x)^m*log(c*(d+e/x^(1/2))^p),x)
 

Output:

(f**m*(2*x**((2*m + 1)/2)*d*e*m*p + 4*x**m*log(((sqrt(x)*d + e)**p*c)/x**( 
p/2))*d**2*m**2*x + 2*x**m*log(((sqrt(x)*d + e)**p*c)/x**(p/2))*d**2*m*x - 
 4*x**m*log(((sqrt(x)*d + e)**p*c)/x**(p/2))*e**2*m**2 - 2*x**m*log(((sqrt 
(x)*d + e)**p*c)/x**(p/2))*e**2*m - 2*x**m*e**2*m*p - x**m*e**2*p + 4*int( 
(x**m*log(((sqrt(x)*d + e)**p*c)/x**(p/2)))/x,x)*e**2*m**3 + 2*int((x**m*l 
og(((sqrt(x)*d + e)**p*c)/x**(p/2)))/x,x)*e**2*m**2))/(2*d**2*m*(2*m**2 + 
3*m + 1))