\(\int (d+e x)^m \log (c (a+b x^2)^p) \, dx\) [207]

Optimal result

Integrand size = 20, antiderivative size = 205 \[ \int (d+e x)^m \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {\sqrt {b} p (d+e x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{e \left (\sqrt {b} d-\sqrt {-a} e\right ) (1+m) (2+m)}+\frac {\sqrt {b} p (d+e x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{e \left (\sqrt {b} d+\sqrt {-a} e\right ) (1+m) (2+m)}+\frac {(d+e x)^{1+m} \log \left (c \left (a+b x^2\right )^p\right )}{e (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.86 \[ \int (d+e x)^m \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {(d+e x)^{1+m} \left (\frac {\sqrt {b} p (d+e x) \left (\left (\sqrt {b} d+\sqrt {-a} e\right ) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )+\left (\sqrt {b} d-\sqrt {-a} e\right ) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )\right )}{\left (b d^2+a e^2\right ) (2+m)}+\log \left (c \left (a+b x^2\right )^p\right )\right )}{e (1+m)} \] Input:

Integrate[(d + e*x)^m*Log[c*(a + b*x^2)^p],x]
 

Output:

((d + e*x)^(1 + m)*((Sqrt[b]*p*(d + e*x)*((Sqrt[b]*d + Sqrt[-a]*e)*Hyperge 
ometric2F1[1, 2 + m, 3 + m, (Sqrt[b]*(d + e*x))/(Sqrt[b]*d - Sqrt[-a]*e)] 
+ (Sqrt[b]*d - Sqrt[-a]*e)*Hypergeometric2F1[1, 2 + m, 3 + m, (Sqrt[b]*(d 
+ e*x))/(Sqrt[b]*d + Sqrt[-a]*e)]))/((b*d^2 + a*e^2)*(2 + m)) + Log[c*(a + 
 b*x^2)^p]))/(e*(1 + m))
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2913, 615, 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.

\(\displaystyle \int (d+e x)^m \log \left (c \left (a+b x^2\right )^p\right ) \, dx\)

\(\Big \downarrow \) 2913

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

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

Input:

Int[(d + e*x)^m*Log[c*(a + b*x^2)^p],x]
 

Output:

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

Defintions of rubi rules used

rule 615

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
 x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] 
 /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
 

rule 2009

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

rule 2913

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_. 
)*(x_))^(r_.), x_Symbol] :> Simp[(f + g*x)^(r + 1)*((a + b*Log[c*(d + e*x^n 
)^p])/(g*(r + 1))), x] - Simp[b*e*n*(p/(g*(r + 1)))   Int[x^(n - 1)*((f + g 
*x)^(r + 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x 
] && (IGtQ[r, 0] || RationalQ[n]) && NeQ[r, -1]
 

Maple [F]

\[\int \left (e x +d \right )^{m} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )d x\]

Input:

int((e*x+d)^m*ln(c*(b*x^2+a)^p),x)
 

Output:

int((e*x+d)^m*ln(c*(b*x^2+a)^p),x)
 

Fricas [F]

\[ \int (d+e x)^m \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\int { {\left (e x + d\right )}^{m} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \,d x } \] Input:

integrate((e*x+d)^m*log(c*(b*x^2+a)^p),x, algorithm="fricas")
 

Output:

integral((e*x + d)^m*log((b*x^2 + a)^p*c), x)
 

Sympy [F(-1)]

Timed out. \[ \int (d+e x)^m \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\text {Timed out} \] Input:

integrate((e*x+d)**m*ln(c*(b*x**2+a)**p),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (d+e x)^m \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\int { {\left (e x + d\right )}^{m} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \,d x } \] Input:

integrate((e*x+d)^m*log(c*(b*x^2+a)^p),x, algorithm="maxima")
 

Output:

(e*p*x + d*p)*(e*x + d)^m*log(b*x^2 + a)/(e*(m + 1)) + integrate(-(2*b*d*p 
*x - (e*(m + 1)*log(c) - 2*e*p)*b*x^2 - a*e*(m + 1)*log(c))*(e*x + d)^m/(b 
*e*(m + 1)*x^2 + a*e*(m + 1)), x)
 

Giac [F]

\[ \int (d+e x)^m \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\int { {\left (e x + d\right )}^{m} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \,d x } \] Input:

integrate((e*x+d)^m*log(c*(b*x^2+a)^p),x, algorithm="giac")
 

Output:

integrate((e*x + d)^m*log((b*x^2 + a)^p*c), x)
 

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^m \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\int \ln \left (c\,{\left (b\,x^2+a\right )}^p\right )\,{\left (d+e\,x\right )}^m \,d x \] Input:

int(log(c*(a + b*x^2)^p)*(d + e*x)^m,x)
 

Output:

int(log(c*(a + b*x^2)^p)*(d + e*x)^m, x)
 

Reduce [F]

\[ \int (d+e x)^m \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {\left (e x +d \right )^{m} \mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) d \,m^{2}+\left (e x +d \right )^{m} \mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) d m +\left (e x +d \right )^{m} \mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) e \,m^{2} x +\left (e x +d \right )^{m} \mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) e m x -4 \left (e x +d \right )^{m} d m p -2 \left (e x +d \right )^{m} d p -2 \left (e x +d \right )^{m} e m p x +4 \left (\int \frac {\left (e x +d \right )^{m}}{b e m \,x^{3}+b d m \,x^{2}+b e \,x^{3}+a e m x +b d \,x^{2}+a d m +a e x +a d}d x \right ) a d e \,m^{3} p +8 \left (\int \frac {\left (e x +d \right )^{m}}{b e m \,x^{3}+b d m \,x^{2}+b e \,x^{3}+a e m x +b d \,x^{2}+a d m +a e x +a d}d x \right ) a d e \,m^{2} p +4 \left (\int \frac {\left (e x +d \right )^{m}}{b e m \,x^{3}+b d m \,x^{2}+b e \,x^{3}+a e m x +b d \,x^{2}+a d m +a e x +a d}d x \right ) a d e m p +2 \left (\int \frac {\left (e x +d \right )^{m} x}{b e m \,x^{3}+b d m \,x^{2}+b e \,x^{3}+a e m x +b d \,x^{2}+a d m +a e x +a d}d x \right ) a \,e^{2} m^{3} p +4 \left (\int \frac {\left (e x +d \right )^{m} x}{b e m \,x^{3}+b d m \,x^{2}+b e \,x^{3}+a e m x +b d \,x^{2}+a d m +a e x +a d}d x \right ) a \,e^{2} m^{2} p +2 \left (\int \frac {\left (e x +d \right )^{m} x}{b e m \,x^{3}+b d m \,x^{2}+b e \,x^{3}+a e m x +b d \,x^{2}+a d m +a e x +a d}d x \right ) a \,e^{2} m p -2 \left (\int \frac {\left (e x +d \right )^{m} x}{b e m \,x^{3}+b d m \,x^{2}+b e \,x^{3}+a e m x +b d \,x^{2}+a d m +a e x +a d}d x \right ) b \,d^{2} m^{3} p -4 \left (\int \frac {\left (e x +d \right )^{m} x}{b e m \,x^{3}+b d m \,x^{2}+b e \,x^{3}+a e m x +b d \,x^{2}+a d m +a e x +a d}d x \right ) b \,d^{2} m^{2} p -2 \left (\int \frac {\left (e x +d \right )^{m} x}{b e m \,x^{3}+b d m \,x^{2}+b e \,x^{3}+a e m x +b d \,x^{2}+a d m +a e x +a d}d x \right ) b \,d^{2} m p}{e m \left (m^{2}+2 m +1\right )} \] Input:

int((e*x+d)^m*log(c*(b*x^2+a)^p),x)
 

Output:

((d + e*x)**m*log((a + b*x**2)**p*c)*d*m**2 + (d + e*x)**m*log((a + b*x**2 
)**p*c)*d*m + (d + e*x)**m*log((a + b*x**2)**p*c)*e*m**2*x + (d + e*x)**m* 
log((a + b*x**2)**p*c)*e*m*x - 4*(d + e*x)**m*d*m*p - 2*(d + e*x)**m*d*p - 
 2*(d + e*x)**m*e*m*p*x + 4*int((d + e*x)**m/(a*d*m + a*d + a*e*m*x + a*e* 
x + b*d*m*x**2 + b*d*x**2 + b*e*m*x**3 + b*e*x**3),x)*a*d*e*m**3*p + 8*int 
((d + e*x)**m/(a*d*m + a*d + a*e*m*x + a*e*x + b*d*m*x**2 + b*d*x**2 + b*e 
*m*x**3 + b*e*x**3),x)*a*d*e*m**2*p + 4*int((d + e*x)**m/(a*d*m + a*d + a* 
e*m*x + a*e*x + b*d*m*x**2 + b*d*x**2 + b*e*m*x**3 + b*e*x**3),x)*a*d*e*m* 
p + 2*int(((d + e*x)**m*x)/(a*d*m + a*d + a*e*m*x + a*e*x + b*d*m*x**2 + b 
*d*x**2 + b*e*m*x**3 + b*e*x**3),x)*a*e**2*m**3*p + 4*int(((d + e*x)**m*x) 
/(a*d*m + a*d + a*e*m*x + a*e*x + b*d*m*x**2 + b*d*x**2 + b*e*m*x**3 + b*e 
*x**3),x)*a*e**2*m**2*p + 2*int(((d + e*x)**m*x)/(a*d*m + a*d + a*e*m*x + 
a*e*x + b*d*m*x**2 + b*d*x**2 + b*e*m*x**3 + b*e*x**3),x)*a*e**2*m*p - 2*i 
nt(((d + e*x)**m*x)/(a*d*m + a*d + a*e*m*x + a*e*x + b*d*m*x**2 + b*d*x**2 
 + b*e*m*x**3 + b*e*x**3),x)*b*d**2*m**3*p - 4*int(((d + e*x)**m*x)/(a*d*m 
 + a*d + a*e*m*x + a*e*x + b*d*m*x**2 + b*d*x**2 + b*e*m*x**3 + b*e*x**3), 
x)*b*d**2*m**2*p - 2*int(((d + e*x)**m*x)/(a*d*m + a*d + a*e*m*x + a*e*x + 
 b*d*m*x**2 + b*d*x**2 + b*e*m*x**3 + b*e*x**3),x)*b*d**2*m*p)/(e*m*(m**2 
+ 2*m + 1))