\(\int x (c+d x)^{3/2} (a+b x^2)^p \, dx\) [84]

Optimal result

Integrand size = 20, antiderivative size = 198 \[ \int x (c+d x)^{3/2} \left (a+b x^2\right )^p \, dx=\frac {(c+d x)^{3/2} \left (a+b x^2\right )^{1+p}}{2 b (1+p)}-\frac {(c+d x)^{3/2} \left (a+b x^2\right )^{1+p} \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-1-p} \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-1-p} \operatorname {AppellF1}\left (\frac {3}{2},-1-p,-1-p,\frac {5}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{2 b (1+p)} \] Output:

1/2*(d*x+c)^(3/2)*(b*x^2+a)^(p+1)/b/(p+1)-1/2*(d*x+c)^(3/2)*(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,5/2,(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)
 

Mathematica [A] (warning: unable to verify)

Time = 0.16 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.25 \[ \int x (c+d x)^{3/2} \left (a+b x^2\right )^p \, dx=\frac {2 \left (\frac {d \left (\sqrt {-\frac {a b}{d^2}} d-b x\right )}{b c+\sqrt {-\frac {a b}{d^2}} d^2}\right )^{-p} (c+d x)^{5/2} \left (\frac {a-\sqrt {-\frac {a b}{d^2}} d x}{a+c \sqrt {-\frac {a b}{d^2}}}\right )^{-p} \left (a+b x^2\right )^p \left (-7 c \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},\frac {c+d x}{c+\frac {a}{\sqrt {-\frac {a b}{d^2}}}},\frac {b (c+d x)}{b c+\sqrt {-\frac {a b}{d^2}} d^2}\right )+5 (c+d x) \operatorname {AppellF1}\left (\frac {7}{2},-p,-p,\frac {9}{2},\frac {c+d x}{c+\frac {a}{\sqrt {-\frac {a b}{d^2}}}},\frac {b (c+d x)}{b c+\sqrt {-\frac {a b}{d^2}} d^2}\right )\right )}{35 d^2} \] Input:

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

Output:

(2*(c + d*x)^(5/2)*(a + b*x^2)^p*(-7*c*AppellF1[5/2, -p, -p, 7/2, (c + d*x 
)/(c + a/Sqrt[-((a*b)/d^2)]), (b*(c + d*x))/(b*c + Sqrt[-((a*b)/d^2)]*d^2) 
] + 5*(c + d*x)*AppellF1[7/2, -p, -p, 9/2, (c + d*x)/(c + a/Sqrt[-((a*b)/d 
^2)]), (b*(c + d*x))/(b*c + Sqrt[-((a*b)/d^2)]*d^2)]))/(35*d^2*((d*(Sqrt[- 
((a*b)/d^2)]*d - b*x))/(b*c + Sqrt[-((a*b)/d^2)]*d^2))^p*((a - Sqrt[-((a*b 
)/d^2)]*d*x)/(a + c*Sqrt[-((a*b)/d^2)]))^p)
 

Rubi [A] (warning: unable to verify)

Time = 0.69 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.81, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {596, 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.

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

\(\Big \downarrow \) 596

\(\displaystyle \frac {2 (c+d x)^{3/2} \left (a+b x^2\right )^{p+1}}{b (4 p+7)}-\frac {3 \int (a d-b c x) \sqrt {c+d x} \left (b x^2+a\right )^pdx}{b (4 p+7)}\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {2 (c+d x)^{3/2} \left (a+b x^2\right )^{p+1}}{b (4 p+7)}-\frac {3 \left (\frac {\left (a d^2+b c^2\right ) \int \sqrt {c+d x} \left (b x^2+a\right )^pdx}{d}-\frac {b c \int (c+d x)^{3/2} \left (b x^2+a\right )^pdx}{d}\right )}{b (4 p+7)}\)

\(\Big \downarrow \) 514

\(\displaystyle \frac {2 (c+d x)^{3/2} \left (a+b x^2\right )^{p+1}}{b (4 p+7)}-\frac {3 \left (\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \int \sqrt {c+d x} \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^pd(c+d x)}{d^2}-\frac {b c \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \int (c+d x)^{3/2} \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^pd(c+d x)}{d^2}\right )}{b (4 p+7)}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {2 (c+d x)^{3/2} \left (a+b x^2\right )^{p+1}}{b (4 p+7)}-\frac {3 \left (\frac {2 (c+d x)^{3/2} \left (a d^2+b c^2\right ) \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{3 d^2}-\frac {2 b c (c+d x)^{5/2} \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{5 d^2}\right )}{b (4 p+7)}\)

Input:

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

Output:

(2*(c + d*x)^(3/2)*(a + b*x^2)^(1 + p))/(b*(7 + 4*p)) - (3*((2*(b*c^2 + a* 
d^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) - (2*b*c*(c + d*x)^(5/2)*(a + b*x^2)^p*AppellF1[5/2, -p, -p 
, 7/2, (c + d*x)/(c - (Sqrt[-a]*d)/Sqrt[b]), (c + d*x)/(c + (Sqrt[-a]*d)/S 
qrt[b])])/(5*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*(7 + 4*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 596

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

( - 24*sqrt(c + d*x)*(a + b*x**2)**p*a**2*d**3*p - 30*sqrt(c + d*x)*(a + b 
*x**2)**p*a**2*d**3 + 32*sqrt(c + d*x)*(a + b*x**2)**p*a*b*c**2*d*p**2 + 8 
8*sqrt(c + d*x)*(a + b*x**2)**p*a*b*c**2*d*p + 54*sqrt(c + d*x)*(a + b*x** 
2)**p*a*b*c**2*d + 32*sqrt(c + d*x)*(a + b*x**2)**p*a*b*c*d**2*p**2*x + 40 
*sqrt(c + d*x)*(a + b*x**2)**p*a*b*c*d**2*p*x + 6*sqrt(c + d*x)*(a + b*x** 
2)**p*b**2*c**3*x + 32*sqrt(c + d*x)*(a + b*x**2)**p*b**2*c**2*d*p**2*x**2 
 + 88*sqrt(c + d*x)*(a + b*x**2)**p*b**2*c**2*d*p*x**2 + 48*sqrt(c + d*x)* 
(a + b*x**2)**p*b**2*c**2*d*x**2 + 32*sqrt(c + d*x)*(a + b*x**2)**p*b**2*c 
*d**2*p**2*x**3 + 64*sqrt(c + d*x)*(a + b*x**2)**p*b**2*c*d**2*p*x**3 + 30 
*sqrt(c + d*x)*(a + b*x**2)**p*b**2*c*d**2*x**3 + 3072*int((sqrt(c + d*x)* 
(a + b*x**2)**p*x**2)/(64*a*c*p**3 + 240*a*c*p**2 + 284*a*c*p + 105*a*c + 
64*a*d*p**3*x + 240*a*d*p**2*x + 284*a*d*p*x + 105*a*d*x + 64*b*c*p**3*x** 
2 + 240*b*c*p**2*x**2 + 284*b*c*p*x**2 + 105*b*c*x**2 + 64*b*d*p**3*x**3 + 
 240*b*d*p**2*x**3 + 284*b*d*p*x**3 + 105*b*d*x**3),x)*a**2*b*d**4*p**5 + 
16128*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(64*a*c*p**3 + 240*a*c*p**2 
 + 284*a*c*p + 105*a*c + 64*a*d*p**3*x + 240*a*d*p**2*x + 284*a*d*p*x + 10 
5*a*d*x + 64*b*c*p**3*x**2 + 240*b*c*p**2*x**2 + 284*b*c*p*x**2 + 105*b*c* 
x**2 + 64*b*d*p**3*x**3 + 240*b*d*p**2*x**3 + 284*b*d*p*x**3 + 105*b*d*x** 
3),x)*a**2*b*d**4*p**4 + 31872*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(6 
4*a*c*p**3 + 240*a*c*p**2 + 284*a*c*p + 105*a*c + 64*a*d*p**3*x + 240*a...