\(\int \frac {x (a+b x^2)^p}{(c+d x)^{3/2}} \, dx\) [96]

Optimal result

Integrand size = 20, antiderivative size = 198 \[ \int \frac {x \left (a+b x^2\right )^p}{(c+d x)^{3/2}} \, dx=\frac {\left (a+b x^2\right )^{1+p}}{2 b (1+p) (c+d x)^{3/2}}-\frac {\left (a+b x^2\right )^{1+p} \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-1-p} \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-1-p} \operatorname {AppellF1}\left (-\frac {3}{2},-1-p,-1-p,-\frac {1}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{2 b (1+p) (c+d x)^{3/2}} \] Output:

1/2*(b*x^2+a)^(p+1)/b/(p+1)/(d*x+c)^(3/2)-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(-3/2,-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)^(3/2)
 

Mathematica [A] (warning: unable to verify)

Time = 0.12 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.11 \[ \int \frac {x \left (a+b x^2\right )^p}{(c+d x)^{3/2}} \, dx=\frac {2 \left (\frac {d \left (\sqrt {-\frac {a}{b}}-x\right )}{c+\sqrt {-\frac {a}{b}} d}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{-c+\sqrt {-\frac {a}{b}} d}\right )^{-p} \left (a+b x^2\right )^p \left (c \operatorname {AppellF1}\left (-\frac {1}{2},-p,-p,\frac {1}{2},\frac {c+d x}{c-\sqrt {-\frac {a}{b}} d},\frac {c+d x}{c+\sqrt {-\frac {a}{b}} d}\right )+(c+d x) \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},\frac {c+d x}{c-\sqrt {-\frac {a}{b}} d},\frac {c+d x}{c+\sqrt {-\frac {a}{b}} d}\right )\right )}{d^2 \sqrt {c+d x}} \] Input:

Integrate[(x*(a + b*x^2)^p)/(c + d*x)^(3/2),x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.74 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {594, 27, 719, 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.

\(\displaystyle \int \frac {x \left (a+b x^2\right )^p}{(c+d x)^{3/2}} \, dx\)

\(\Big \downarrow \) 594

\(\displaystyle \frac {2 c \left (a+b x^2\right )^{p+1}}{\sqrt {c+d x} \left (a d^2+b c^2\right )}-\frac {2 \int -\frac {(a d-b c (4 p+3) x) \left (b x^2+a\right )^p}{2 \sqrt {c+d x}}dx}{a d^2+b c^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(a d-b c (4 p+3) x) \left (b x^2+a\right )^p}{\sqrt {c+d x}}dx}{a d^2+b c^2}+\frac {2 c \left (a+b x^2\right )^{p+1}}{\sqrt {c+d x} \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {\frac {\left (a d^2+b c^2 (4 p+3)\right ) \int \frac {\left (b x^2+a\right )^p}{\sqrt {c+d x}}dx}{d}-\frac {b c (4 p+3) \int \sqrt {c+d x} \left (b x^2+a\right )^pdx}{d}}{a d^2+b c^2}+\frac {2 c \left (a+b x^2\right )^{p+1}}{\sqrt {c+d x} \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 514

\(\displaystyle \frac {\frac {\left (a+b x^2\right )^p \left (a d^2+b c^2 (4 p+3)\right ) \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \int \frac {\left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p}{\sqrt {c+d x}}d(c+d x)}{d^2}-\frac {b c (4 p+3) \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \int \sqrt {c+d x} \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^pd(c+d x)}{d^2}}{a d^2+b c^2}+\frac {2 c \left (a+b x^2\right )^{p+1}}{\sqrt {c+d x} \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {\frac {2 \sqrt {c+d x} \left (a+b x^2\right )^p \left (a d^2+b c^2 (4 p+3)\right ) \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{d^2}-\frac {2 b c (4 p+3) (c+d x)^{3/2} \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{3 d^2}}{a d^2+b c^2}+\frac {2 c \left (a+b x^2\right )^{p+1}}{\sqrt {c+d x} \left (a d^2+b c^2\right )}\)

Input:

Int[(x*(a + b*x^2)^p)/(c + d*x)^(3/2),x]
 

Output:

(2*c*(a + b*x^2)^(1 + p))/((b*c^2 + a*d^2)*Sqrt[c + d*x]) + ((2*(a*d^2 + b 
*c^2*(3 + 4*p))*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 + (Sqr 
t[-a]*d)/Sqrt[b]))^p) - (2*b*c*(3 + 4*p)*(c + d*x)^(3/2)*(a + b*x^2)^p*App 
ellF1[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))/(b*c^2 + a*d^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 594

Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))) 
, x] + Simp[1/((n + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^(n + 1)*(a + b*x^2) 
^p*(a*d*(n + 1) + b*c*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] 
 && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 719

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

Maple [F]

\[\int \frac {x \left (b \,x^{2}+a \right )^{p}}{\left (d x +c \right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(x*(b*x^2+a)^p/(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(x*(b*x**2+a)**p/(d*x+c)**(3/2),x)
 

Output:

Integral(x*(a + b*x**2)**p/(c + d*x)**(3/2), x)
 

Maxima [F]

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

integrate(x*(b*x^2+a)^p/(d*x+c)^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^p*x/(d*x + c)^(3/2), x)
 

Giac [F]

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

integrate(x*(b*x^2+a)^p/(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^p*x/(d*x + c)^(3/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x \left (a+b x^2\right )^p}{(c+d x)^{3/2}} \, dx=\text {too large to display} \] Input:

int(x*(b*x^2+a)^p/(d*x+c)^(3/2),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**2*p + a*c**2 + 8* 
a*c*d*p*x + 2*a*c*d*x + 4*a*d**2*p*x**2 + a*d**2*x**2 + 4*b*c**2*p*x**2 + 
b*c**2*x**2 + 8*b*c*d*p*x**3 + 2*b*c*d*x**3 + 4*b*d**2*p*x**4 + b*d**2*x** 
4),x)*a*b*c*d**2*p**2 + int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(4*a*c**2 
*p + a*c**2 + 8*a*c*d*p*x + 2*a*c*d*x + 4*a*d**2*p*x**2 + a*d**2*x**2 + 4* 
b*c**2*p*x**2 + b*c**2*x**2 + 8*b*c*d*p*x**3 + 2*b*c*d*x**3 + 4*b*d**2*p*x 
**4 + b*d**2*x**4),x)*a*b*c*d**2 - 16*int((sqrt(c + d*x)*(a + b*x**2)**p*x 
**2)/(4*a*c**2*p + a*c**2 + 8*a*c*d*p*x + 2*a*c*d*x + 4*a*d**2*p*x**2 + a* 
d**2*x**2 + 4*b*c**2*p*x**2 + b*c**2*x**2 + 8*b*c*d*p*x**3 + 2*b*c*d*x**3 
+ 4*b*d**2*p*x**4 + b*d**2*x**4),x)*a*b*d**3*p**2*x + int((sqrt(c + d*x)*( 
a + b*x**2)**p*x**2)/(4*a*c**2*p + a*c**2 + 8*a*c*d*p*x + 2*a*c*d*x + 4*a* 
d**2*p*x**2 + a*d**2*x**2 + 4*b*c**2*p*x**2 + b*c**2*x**2 + 8*b*c*d*p*x**3 
 + 2*b*c*d*x**3 + 4*b*d**2*p*x**4 + b*d**2*x**4),x)*a*b*d**3*x - 16*int((s 
qrt(c + d*x)*(a + b*x**2)**p*x**2)/(4*a*c**2*p + a*c**2 + 8*a*c*d*p*x + 2* 
a*c*d*x + 4*a*d**2*p*x**2 + a*d**2*x**2 + 4*b*c**2*p*x**2 + b*c**2*x**2 + 
8*b*c*d*p*x**3 + 2*b*c*d*x**3 + 4*b*d**2*p*x**4 + b*d**2*x**4),x)*b**2*c** 
3*p**2 - 12*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(4*a*c**2*p + a*c**2 
+ 8*a*c*d*p*x + 2*a*c*d*x + 4*a*d**2*p*x**2 + a*d**2*x**2 + 4*b*c**2*p*x** 
2 + b*c**2*x**2 + 8*b*c*d*p*x**3 + 2*b*c*d*x**3 + 4*b*d**2*p*x**4 + b*d...