\(\int \frac {(e x)^m (c+d x)}{\sqrt {a x^n+b x^{1+n}}} \, dx\) [373]

Optimal result

Integrand size = 28, antiderivative size = 145 \[ \int \frac {(e x)^m (c+d x)}{\sqrt {a x^n+b x^{1+n}}} \, dx=\frac {2 d x^{1-n} (e x)^m \sqrt {a x^n+b x^{1+n}}}{b (3+2 m-n)}+\frac {2 \left (b c-\frac {a d (2+2 m-n)}{3+2 m-n}\right ) x^{-n} \left (-\frac {b x}{a}\right )^{\frac {1}{2} (-2 m+n)} (e x)^m \sqrt {a x^n+b x^{1+n}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2 m+n),\frac {3}{2},1+\frac {b x}{a}\right )}{b^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.73 \[ \int \frac {(e x)^m (c+d x)}{\sqrt {a x^n+b x^{1+n}}} \, dx=\frac {2 (e x)^m (a+b x) \left (b d x+(b c (3+2 m-n)+a d (-2-2 m+n)) \left (-\frac {b x}{a}\right )^{\frac {1}{2} (-2 m+n)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2 m+n),\frac {3}{2},1+\frac {b x}{a}\right )\right )}{b^2 (3+2 m-n) \sqrt {x^n (a+b x)}} \] Input:

Integrate[((e*x)^m*(c + d*x))/Sqrt[a*x^n + b*x^(1 + n)],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.17, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1948, 90, 77, 75}

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.

\(\displaystyle \int \frac {(c+d x) (e x)^m}{\sqrt {a x^n+b x^{n+1}}} \, dx\)

\(\Big \downarrow \) 1948

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

\(\Big \downarrow \) 90

\(\displaystyle \frac {\sqrt {a+b x} (e x)^m x^{\frac {1}{2} (n-2 m)} \left (\left (c-\frac {a d (2 m-n+2)}{b (2 m-n+3)}\right ) \int \frac {x^{m-\frac {n}{2}}}{\sqrt {a+b x}}dx+\frac {2 d \sqrt {a+b x} x^{m-\frac {n}{2}+1}}{b (2 m-n+3)}\right )}{\sqrt {a x^n+b x^{n+1}}}\)

\(\Big \downarrow \) 77

\(\displaystyle \frac {\sqrt {a+b x} (e x)^m x^{\frac {1}{2} (n-2 m)} \left (x^{m-\frac {n}{2}} \left (-\frac {b x}{a}\right )^{\frac {1}{2} (n-2 m)} \left (c-\frac {a d (2 m-n+2)}{b (2 m-n+3)}\right ) \int \frac {\left (-\frac {b x}{a}\right )^{m-\frac {n}{2}}}{\sqrt {a+b x}}dx+\frac {2 d \sqrt {a+b x} x^{m-\frac {n}{2}+1}}{b (2 m-n+3)}\right )}{\sqrt {a x^n+b x^{n+1}}}\)

\(\Big \downarrow \) 75

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

Input:

Int[((e*x)^m*(c + d*x))/Sqrt[a*x^n + b*x^(1 + n)],x]
 

Output:

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

Defintions of rubi rules used

rule 75

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

rule 77

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

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 1948

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Simp[e^IntPart[m]*(e*x)^FracPart[m]*( 
(a*x^j + b*x^(j + n))^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x 
^n)^FracPart[p]))   Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; 
FreeQ[{a, b, c, d, e, j, m, n, p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] 
 && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])
 

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (d x +c \right )}{\sqrt {a \,x^{n}+b \,x^{1+n}}}d x\]

Input:

int((e*x)^m*(d*x+c)/(a*x^n+b*x^(1+n))^(1/2),x)
 

Output:

int((e*x)^m*(d*x+c)/(a*x^n+b*x^(1+n))^(1/2),x)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {(e x)^m (c+d x)}{\sqrt {a x^n+b x^{1+n}}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((e*x)^m*(d*x+c)/(a*x^n+b*x^(1+n))^(1/2),x, algorithm="fricas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (has polynomial part)
 

Sympy [F]

\[ \int \frac {(e x)^m (c+d x)}{\sqrt {a x^n+b x^{1+n}}} \, dx=\int \frac {\left (e x\right )^{m} \left (c + d x\right )}{\sqrt {a x^{n} + b x^{n + 1}}}\, dx \] Input:

integrate((e*x)**m*(d*x+c)/(a*x**n+b*x**(1+n))**(1/2),x)
 

Output:

Integral((e*x)**m*(c + d*x)/sqrt(a*x**n + b*x**(n + 1)), x)
 

Maxima [F]

\[ \int \frac {(e x)^m (c+d x)}{\sqrt {a x^n+b x^{1+n}}} \, dx=\int { \frac {{\left (d x + c\right )} \left (e x\right )^{m}}{\sqrt {b x^{n + 1} + a x^{n}}} \,d x } \] Input:

integrate((e*x)^m*(d*x+c)/(a*x^n+b*x^(1+n))^(1/2),x, algorithm="maxima")
 

Output:

integrate((d*x + c)*(e*x)^m/sqrt(b*x^(n + 1) + a*x^n), x)
 

Giac [F]

\[ \int \frac {(e x)^m (c+d x)}{\sqrt {a x^n+b x^{1+n}}} \, dx=\int { \frac {{\left (d x + c\right )} \left (e x\right )^{m}}{\sqrt {b x^{n + 1} + a x^{n}}} \,d x } \] Input:

integrate((e*x)^m*(d*x+c)/(a*x^n+b*x^(1+n))^(1/2),x, algorithm="giac")
 

Output:

integrate((d*x + c)*(e*x)^m/sqrt(b*x^(n + 1) + a*x^n), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m (c+d x)}{\sqrt {a x^n+b x^{1+n}}} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (c+d\,x\right )}{\sqrt {a\,x^n+b\,x^{n+1}}} \,d x \] Input:

int(((e*x)^m*(c + d*x))/(a*x^n + b*x^(n + 1))^(1/2),x)
 

Output:

int(((e*x)^m*(c + d*x))/(a*x^n + b*x^(n + 1))^(1/2), x)
 

Reduce [F]

\[ \int \frac {(e x)^m (c+d x)}{\sqrt {a x^n+b x^{1+n}}} \, dx=e^{m} \left (\left (\int \frac {x^{m}}{x^{\frac {n}{2}} \sqrt {b x +a}}d x \right ) c +\left (\int \frac {x^{m} x}{x^{\frac {n}{2}} \sqrt {b x +a}}d x \right ) d \right ) \] Input:

int((e*x)^m*(d*x+c)/(a*x^n+b*x^(1+n))^(1/2),x)
 

Output:

e**m*(int(x**m/(x**(n/2)*sqrt(a + b*x)),x)*c + int((x**m*x)/(x**(n/2)*sqrt 
(a + b*x)),x)*d)