\(\int \frac {(a+b (c+d x)^2)^p}{c+d x} \, dx\) [51]

Optimal result

Integrand size = 21, antiderivative size = 52 \[ \int \frac {\left (a+b (c+d x)^2\right )^p}{c+d x} \, dx=-\frac {\left (a+b (c+d x)^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b (c+d x)^2}{a}\right )}{2 a d (1+p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b (c+d x)^2\right )^p}{c+d x} \, dx=-\frac {\left (a+b (c+d x)^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b (c+d x)^2}{a}\right )}{2 a d (1+p)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {895, 243, 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.

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

\(\Big \downarrow \) 895

\(\displaystyle \frac {\int \frac {\left (b (c+d x)^2+a\right )^p}{c+d x}d(c+d x)}{d}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\int \frac {\left (b (c+d x)^2+a\right )^p}{(c+d x)^2}d(c+d x)^2}{2 d}\)

\(\Big \downarrow \) 75

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

Input:

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

Output:

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

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 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 895

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Simp[u^m/(Coeff 
icient[v, x, 1]*v^m)   Subst[Int[x^m*(a + b*x^n)^p, x], x, v], x] /; FreeQ[ 
{a, b, m, n, p}, x] && LinearPairQ[u, v, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

((a + b*c**2 + 2*b*c*d*x + b*d**2*x**2)**p*a + (a + b*c**2 + 2*b*c*d*x + b 
*d**2*x**2)**p*b*c**2 - 2*int(((a + b*c**2 + 2*b*c*d*x + b*d**2*x**2)**p*x 
**2)/(a*c + a*d*x + b*c**3 + 3*b*c**2*d*x + 3*b*c*d**2*x**2 + b*d**3*x**3) 
,x)*a*b*d**3*p - 4*int(((a + b*c**2 + 2*b*c*d*x + b*d**2*x**2)**p*x)/(a*c 
+ a*d*x + b*c**3 + 3*b*c**2*d*x + 3*b*c*d**2*x**2 + b*d**3*x**3),x)*a*b*c* 
d**2*p)/(2*b*c**2*d*p)