\(\int \frac {x (a+b x^2)^p}{\sqrt {c+d x}} \, dx\) [1876]

Optimal result

Integrand size = 20, antiderivative size = 198 \[ \int \frac {x \left (a+b x^2\right )^p}{\sqrt {c+d x}} \, dx=\frac {\left (a+b x^2\right )^{1+p}}{2 b (1+p) \sqrt {c+d x}}-\frac {\left (a+b x^2\right )^{1+p} \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-1-p} \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-1-p} \operatorname {AppellF1}\left (-\frac {1}{2},-1-p,-1-p,\frac {1}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{2 b (1+p) \sqrt {c+d x}} \] Output:

1/2*(b*x^2+a)^(p+1)/b/(p+1)/(d*x+c)^(1/2)-1/2*(b*x^2+a)^(p+1)*(1-(d*x+c)/( 
c-(-a)^(1/2)*d/b^(1/2)))^(-1-p)*(1-(d*x+c)/(c+(-a)^(1/2)*d/b^(1/2)))^(-1-p 
)*AppellF1(-1/2,-1-p,-1-p,1/2,(d*x+c)/(c-(-a)^(1/2)*d/b^(1/2)),(d*x+c)/(c+ 
(-a)^(1/2)*d/b^(1/2)))/b/(p+1)/(d*x+c)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 0.51 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.13 \[ \int \frac {x \left (a+b x^2\right )^p}{\sqrt {c+d x}} \, dx=\frac {2 \left (\frac {d \left (\sqrt {-\frac {a}{b}}-x\right )}{c+\sqrt {-\frac {a}{b}} d}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{-c+\sqrt {-\frac {a}{b}} d}\right )^{-p} \sqrt {c+d x} \left (a+b x^2\right )^p \left (-3 c \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},\frac {c+d x}{c-\sqrt {-\frac {a}{b}} d},\frac {c+d x}{c+\sqrt {-\frac {a}{b}} d}\right )+(c+d x) \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {c+d x}{c-\sqrt {-\frac {a}{b}} d},\frac {c+d x}{c+\sqrt {-\frac {a}{b}} d}\right )\right )}{3 d^2} \] Input:

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

Output:

(2*Sqrt[c + d*x]*(a + b*x^2)^p*(-3*c*AppellF1[1/2, -p, -p, 3/2, (c + d*x)/ 
(c - Sqrt[-(a/b)]*d), (c + d*x)/(c + Sqrt[-(a/b)]*d)] + (c + d*x)*AppellF1 
[3/2, -p, -p, 5/2, (c + d*x)/(c - Sqrt[-(a/b)]*d), (c + d*x)/(c + Sqrt[-(a 
/b)]*d)]))/(3*d^2*((d*(Sqrt[-(a/b)] - x))/(c + Sqrt[-(a/b)]*d))^p*((d*(Sqr 
t[-(a/b)] + x))/(-c + Sqrt[-(a/b)]*d))^p)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.52, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {624, 514, 150}

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.

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

\(\Big \downarrow \) 624

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

\(\Big \downarrow \) 514

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

\(\Big \downarrow \) 150

\(\displaystyle \frac {2 (c+d x)^{3/2} \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{3 d^2}-\frac {2 c \sqrt {c+d x} \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{d^2}\)

Input:

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

Output:

(-2*c*Sqrt[c + d*x]*(a + b*x^2)^p*AppellF1[1/2, -p, -p, 3/2, (c + d*x)/(c 
- (Sqrt[-a]*d)/Sqrt[b]), (c + d*x)/(c + (Sqrt[-a]*d)/Sqrt[b])])/(d^2*(1 - 
(c + d*x)/(c - (Sqrt[-a]*d)/Sqrt[b]))^p*(1 - (c + d*x)/(c + (Sqrt[-a]*d)/S 
qrt[b]))^p) + (2*(c + d*x)^(3/2)*(a + b*x^2)^p*AppellF1[3/2, -p, -p, 5/2, 
(c + d*x)/(c - (Sqrt[-a]*d)/Sqrt[b]), (c + d*x)/(c + (Sqrt[-a]*d)/Sqrt[b]) 
])/(3*d^2*(1 - (c + d*x)/(c - (Sqrt[-a]*d)/Sqrt[b]))^p*(1 - (c + d*x)/(c + 
 (Sqrt[-a]*d)/Sqrt[b]))^p)
 

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 514

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Rt[-a/b, 2]}, Simp[(a + b*x^2)^p/(d*(1 - (c + d*x)/(c - d*q))^p*(1 - ( 
c + d*x)/(c + d*q))^p)   Subst[Int[x^n*Simp[1 - x/(c + d*q), x]^p*Simp[1 - 
x/(c - d*q), x]^p, x], x, c + d*x], x]] /; FreeQ[{a, b, c, d, n, p}, x] && 
NeQ[b*c^2 + a*d^2, 0]
 

rule 624

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[1/d   Int[x^(m - 1)*(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] - Si 
mp[c/d   Int[x^(m - 1)*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, 
 d, n, p}, x] && IGtQ[m, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*sqrt(c + d*x)*(a + b*x**2)**p*a*d + 2*sqrt(c + d*x)*(a + b*x**2)**p*b*c 
*x - 16*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(4*a*c*p + 3*a*c + 4*a*d* 
p*x + 3*a*d*x + 4*b*c*p*x**2 + 3*b*c*x**2 + 4*b*d*p*x**3 + 3*b*d*x**3),x)* 
a*b*d**2*p**2 - 16*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(4*a*c*p + 3*a 
*c + 4*a*d*p*x + 3*a*d*x + 4*b*c*p*x**2 + 3*b*c*x**2 + 4*b*d*p*x**3 + 3*b* 
d*x**3),x)*a*b*d**2*p - 3*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(4*a*c* 
p + 3*a*c + 4*a*d*p*x + 3*a*d*x + 4*b*c*p*x**2 + 3*b*c*x**2 + 4*b*d*p*x**3 
 + 3*b*d*x**3),x)*a*b*d**2 - 16*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/( 
4*a*c*p + 3*a*c + 4*a*d*p*x + 3*a*d*x + 4*b*c*p*x**2 + 3*b*c*x**2 + 4*b*d* 
p*x**3 + 3*b*d*x**3),x)*b**2*c**2*p**2 - 20*int((sqrt(c + d*x)*(a + b*x**2 
)**p*x**2)/(4*a*c*p + 3*a*c + 4*a*d*p*x + 3*a*d*x + 4*b*c*p*x**2 + 3*b*c*x 
**2 + 4*b*d*p*x**3 + 3*b*d*x**3),x)*b**2*c**2*p - 6*int((sqrt(c + d*x)*(a 
+ b*x**2)**p*x**2)/(4*a*c*p + 3*a*c + 4*a*d*p*x + 3*a*d*x + 4*b*c*p*x**2 + 
 3*b*c*x**2 + 4*b*d*p*x**3 + 3*b*d*x**3),x)*b**2*c**2 - 4*int((sqrt(c + d* 
x)*(a + b*x**2)**p)/(4*a*c*p + 3*a*c + 4*a*d*p*x + 3*a*d*x + 4*b*c*p*x**2 
+ 3*b*c*x**2 + 4*b*d*p*x**3 + 3*b*d*x**3),x)*a**2*d**2*p - 3*int((sqrt(c + 
 d*x)*(a + b*x**2)**p)/(4*a*c*p + 3*a*c + 4*a*d*p*x + 3*a*d*x + 4*b*c*p*x* 
*2 + 3*b*c*x**2 + 4*b*d*p*x**3 + 3*b*d*x**3),x)*a**2*d**2 - 8*int((sqrt(c 
+ d*x)*(a + b*x**2)**p)/(4*a*c*p + 3*a*c + 4*a*d*p*x + 3*a*d*x + 4*b*c*p*x 
**2 + 3*b*c*x**2 + 4*b*d*p*x**3 + 3*b*d*x**3),x)*a*b*c**2*p - 6*int((sq...