\(\int \frac {(d x)^m (a+b x)^p}{\sqrt {c x^2}} \, dx\) [466]

Optimal result

Integrand size = 22, antiderivative size = 57 \[ \int \frac {(d x)^m (a+b x)^p}{\sqrt {c x^2}} \, dx=\frac {(d x)^{1+m} (a+b x)^p \left (1+\frac {b x}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (m,-p,1+m,-\frac {b x}{a}\right )}{d m \sqrt {c x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93 \[ \int \frac {(d x)^m (a+b x)^p}{\sqrt {c x^2}} \, dx=\frac {x (d x)^m (a+b x)^p \left (1+\frac {b x}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (m,-p,1+m,-\frac {b x}{a}\right )}{m \sqrt {c x^2}} \] Input:

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

Output:

(x*(d*x)^m*(a + b*x)^p*Hypergeometric2F1[m, -p, 1 + m, -((b*x)/a)])/(m*Sqr 
t[c*x^2]*(1 + (b*x)/a)^p)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {30, 76, 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.

\(\displaystyle \int \frac {(d x)^m (a+b x)^p}{\sqrt {c x^2}} \, dx\)

\(\Big \downarrow \) 30

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

\(\Big \downarrow \) 76

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

\(\Big \downarrow \) 74

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

Input:

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

Output:

(x*(d*x)^m*(a + b*x)^p*Hypergeometric2F1[m, -p, 1 + m, -((b*x)/a)])/(m*Sqr 
t[c*x^2]*(1 + (b*x)/a)^p)
 

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 76

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sqrt(c*x^2)*(b*x + a)^p*(d*x)^m/(c*x^2), x)
 

Sympy [F]

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

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

Output:

Integral((d*x)**m*(a + b*x)**p/sqrt(c*x**2), x)
 

Maxima [F]

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

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

Output:

integrate((b*x + a)^p*(d*x)^m/sqrt(c*x^2), x)
 

Giac [F]

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

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

Output:

integrate((b*x + a)^p*(d*x)^m/sqrt(c*x^2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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