\(\int (c+d x)^{3/2} (a+b x^2)^p \, dx\) [85]

Optimal result

Integrand size = 19, antiderivative size = 150 \[ \int (c+d x)^{3/2} \left (a+b x^2\right )^p \, dx=\frac {2 (c+d x)^{5/2} \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{5 d} \] Output:

2/5*(d*x+c)^(5/2)*(b*x^2+a)^p*AppellF1(5/2,-p,-p,7/2,(d*x+c)/(c-(-a)^(1/2) 
*d/b^(1/2)),(d*x+c)/(c+(-a)^(1/2)*d/b^(1/2)))/d/((1-(d*x+c)/(c-(-a)^(1/2)* 
d/b^(1/2)))^p)/((1-(d*x+c)/(c+(-a)^(1/2)*d/b^(1/2)))^p)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.15 \[ \int (c+d x)^{3/2} \left (a+b x^2\right )^p \, dx=\frac {2 \left (\frac {d \left (\sqrt {-\frac {a b}{d^2}} d-b x\right )}{b c+\sqrt {-\frac {a b}{d^2}} d^2}\right )^{-p} (c+d x)^{5/2} \left (\frac {a-\sqrt {-\frac {a b}{d^2}} d x}{a+c \sqrt {-\frac {a b}{d^2}}}\right )^{-p} \left (a+b x^2\right )^p \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},\frac {c+d x}{c+\frac {a}{\sqrt {-\frac {a b}{d^2}}}},\frac {b (c+d x)}{b c+\sqrt {-\frac {a b}{d^2}} d^2}\right )}{5 d} \] Input:

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

Output:

(2*(c + d*x)^(5/2)*(a + b*x^2)^p*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* 
d*((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] (verified)

Time = 0.40 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {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.

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

\(\Big \downarrow \) 514

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

(2*(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)/Sqrt[b])])/(5*d*(1 - ( 
c + d*x)/(c - (Sqrt[-a]*d)/Sqrt[b]))^p*(1 - (c + d*x)/(c + (Sqrt[-a]*d)/Sq 
rt[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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((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)
 

Giac [F]

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

integrate((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)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(16*sqrt(c + d*x)*(a + b*x**2)**p*a*d*p + 18*sqrt(c + d*x)*(a + b*x**2)**p 
*a*d + 8*sqrt(c + d*x)*(a + b*x**2)**p*b*c*p*x + 12*sqrt(c + d*x)*(a + b*x 
**2)**p*b*c*x + 8*sqrt(c + d*x)*(a + b*x**2)**p*b*d*p*x**2 + 6*sqrt(c + d* 
x)*(a + b*x**2)**p*b*d*x**2 - 256*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2) 
/(16*a*c*p**2 + 32*a*c*p + 15*a*c + 16*a*d*p**2*x + 32*a*d*p*x + 15*a*d*x 
+ 16*b*c*p**2*x**2 + 32*b*c*p*x**2 + 15*b*c*x**2 + 16*b*d*p**2*x**3 + 32*b 
*d*p*x**3 + 15*b*d*x**3),x)*a*b*d**2*p**4 - 1024*int((sqrt(c + d*x)*(a + b 
*x**2)**p*x**2)/(16*a*c*p**2 + 32*a*c*p + 15*a*c + 16*a*d*p**2*x + 32*a*d* 
p*x + 15*a*d*x + 16*b*c*p**2*x**2 + 32*b*c*p*x**2 + 15*b*c*x**2 + 16*b*d*p 
**2*x**3 + 32*b*d*p*x**3 + 15*b*d*x**3),x)*a*b*d**2*p**3 - 1408*int((sqrt( 
c + d*x)*(a + b*x**2)**p*x**2)/(16*a*c*p**2 + 32*a*c*p + 15*a*c + 16*a*d*p 
**2*x + 32*a*d*p*x + 15*a*d*x + 16*b*c*p**2*x**2 + 32*b*c*p*x**2 + 15*b*c* 
x**2 + 16*b*d*p**2*x**3 + 32*b*d*p*x**3 + 15*b*d*x**3),x)*a*b*d**2*p**2 - 
768*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(16*a*c*p**2 + 32*a*c*p + 15* 
a*c + 16*a*d*p**2*x + 32*a*d*p*x + 15*a*d*x + 16*b*c*p**2*x**2 + 32*b*c*p* 
x**2 + 15*b*c*x**2 + 16*b*d*p**2*x**3 + 32*b*d*p*x**3 + 15*b*d*x**3),x)*a* 
b*d**2*p - 135*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(16*a*c*p**2 + 32* 
a*c*p + 15*a*c + 16*a*d*p**2*x + 32*a*d*p*x + 15*a*d*x + 16*b*c*p**2*x**2 
+ 32*b*c*p*x**2 + 15*b*c*x**2 + 16*b*d*p**2*x**3 + 32*b*d*p*x**3 + 15*b*d* 
x**3),x)*a*b*d**2 + 48*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(16*a*c...