\(\int \frac {x^2 (a+b x^2)^p}{\sqrt {c+d x}} \, dx\) [89]

Optimal result

Integrand size = 22, antiderivative size = 370 \[ \int \frac {x^2 \left (a+b x^2\right )^p}{\sqrt {c+d x}} \, dx=\frac {2 \sqrt {c+d x} \left (a+b x^2\right )^{1+p}}{b d (5+4 p)}-\frac {2 \left (a d^2-4 b c^2 (1+p)\right ) \sqrt {c+d x} \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{b d^3 (5+4 p)}-\frac {8 c (1+p) (c+d x)^{3/2} \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{3 d^3 (5+4 p)} \] Output:

2*(d*x+c)^(1/2)*(b*x^2+a)^(p+1)/b/d/(5+4*p)-2*(a*d^2-4*b*c^2*(p+1))*(d*x+c 
)^(1/2)*(b*x^2+a)^p*AppellF1(1/2,-p,-p,3/2,(d*x+c)/(c-(-a)^(1/2)*d/b^(1/2) 
),(d*x+c)/(c+(-a)^(1/2)*d/b^(1/2)))/b/d^3/(5+4*p)/((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)-8/3*c*(p+1)*(d*x 
+c)^(3/2)*(b*x^2+a)^p*AppellF1(3/2,-p,-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)))/d^3/(5+4*p)/((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] (warning: unable to verify)

Time = 0.24 (sec) , antiderivative size = 495, normalized size of antiderivative = 1.34 \[ \int \frac {x^2 \left (a+b x^2\right )^p}{\sqrt {c+d x}} \, dx=-\frac {2 (c+d x)^{5/2} \left (a+b x^2\right )^p \left (\frac {a d^2-b c d x+\sqrt {-a b d^2} (c+d x)}{\left (-b c+\sqrt {-a b d^2}\right ) (c+d x)}\right )^{-p} \left (\frac {-a d^2+b c d x+\sqrt {-a b d^2} (c+d x)}{\left (b c+\sqrt {-a b d^2}\right ) (c+d x)}\right )^{-p} \left (-\left (\left (3+16 p+16 p^2\right ) \operatorname {AppellF1}\left (-\frac {5}{2}-2 p,-p,-p,-\frac {3}{2}-2 p,\frac {b c^2+a d^2}{\left (b c-\sqrt {-a b d^2}\right ) (c+d x)},\frac {b c^2+a d^2}{\left (b c+\sqrt {-a b d^2}\right ) (c+d x)}\right )\right )+\frac {c (5+4 p) \left ((2+8 p) \operatorname {AppellF1}\left (-\frac {3}{2}-2 p,-p,-p,-\frac {1}{2}-2 p,\frac {b c^2+a d^2}{\left (b c-\sqrt {-a b d^2}\right ) (c+d x)},\frac {b c^2+a d^2}{\left (b c+\sqrt {-a b d^2}\right ) (c+d x)}\right )-\frac {c (3+4 p) \operatorname {AppellF1}\left (-\frac {1}{2}-2 p,-p,-p,\frac {1}{2}-2 p,\frac {b c^2+a d^2}{\left (b c-\sqrt {-a b d^2}\right ) (c+d x)},\frac {b c^2+a d^2}{\left (b c+\sqrt {-a b d^2}\right ) (c+d x)}\right )}{c+d x}\right )}{c+d x}\right )}{d^3 (1+4 p) (3+4 p) (5+4 p)} \] Input:

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

Output:

(-2*(c + d*x)^(5/2)*(a + b*x^2)^p*(-((3 + 16*p + 16*p^2)*AppellF1[-5/2 - 2 
*p, -p, -p, -3/2 - 2*p, (b*c^2 + a*d^2)/((b*c - Sqrt[-(a*b*d^2)])*(c + d*x 
)), (b*c^2 + a*d^2)/((b*c + Sqrt[-(a*b*d^2)])*(c + d*x))]) + (c*(5 + 4*p)* 
((2 + 8*p)*AppellF1[-3/2 - 2*p, -p, -p, -1/2 - 2*p, (b*c^2 + a*d^2)/((b*c 
- Sqrt[-(a*b*d^2)])*(c + d*x)), (b*c^2 + a*d^2)/((b*c + Sqrt[-(a*b*d^2)])* 
(c + d*x))] - (c*(3 + 4*p)*AppellF1[-1/2 - 2*p, -p, -p, 1/2 - 2*p, (b*c^2 
+ a*d^2)/((b*c - Sqrt[-(a*b*d^2)])*(c + d*x)), (b*c^2 + a*d^2)/((b*c + Sqr 
t[-(a*b*d^2)])*(c + d*x))])/(c + d*x)))/(c + d*x)))/(d^3*(1 + 4*p)*(3 + 4* 
p)*(5 + 4*p)*((a*d^2 - b*c*d*x + Sqrt[-(a*b*d^2)]*(c + d*x))/((-(b*c) + Sq 
rt[-(a*b*d^2)])*(c + d*x)))^p*((-(a*d^2) + b*c*d*x + Sqrt[-(a*b*d^2)]*(c + 
 d*x))/((b*c + Sqrt[-(a*b*d^2)])*(c + d*x)))^p)
 

Rubi [A] (verified)

Time = 1.25 (sec) , antiderivative size = 668, normalized size of antiderivative = 1.81, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {624, 596, 624, 514, 150, 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.

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

\(\Big \downarrow \) 624

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

\(\Big \downarrow \) 596

\(\displaystyle \frac {\frac {2 \sqrt {c+d x} \left (a+b x^2\right )^{p+1}}{b (4 p+5)}-\frac {\int \frac {(a d-b c x) \left (b x^2+a\right )^p}{\sqrt {c+d x}}dx}{b (4 p+5)}}{d}-\frac {c \int \frac {x \left (b x^2+a\right )^p}{\sqrt {c+d x}}dx}{d}\)

\(\Big \downarrow \) 624

\(\displaystyle \frac {\frac {2 \sqrt {c+d x} \left (a+b x^2\right )^{p+1}}{b (4 p+5)}-\frac {\int \frac {(a d-b c x) \left (b x^2+a\right )^p}{\sqrt {c+d x}}dx}{b (4 p+5)}}{d}-\frac {c \left (\frac {\int \sqrt {c+d x} \left (b x^2+a\right )^pdx}{d}-\frac {c \int \frac {\left (b x^2+a\right )^p}{\sqrt {c+d x}}dx}{d}\right )}{d}\)

\(\Big \downarrow \) 514

\(\displaystyle \frac {\frac {2 \sqrt {c+d x} \left (a+b x^2\right )^{p+1}}{b (4 p+5)}-\frac {\int \frac {(a d-b c x) \left (b x^2+a\right )^p}{\sqrt {c+d x}}dx}{b (4 p+5)}}{d}-\frac {c \left (\frac {\left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \int \sqrt {c+d x} \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^pd(c+d x)}{d^2}-\frac {c \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \int \frac {\left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p}{\sqrt {c+d x}}d(c+d x)}{d^2}\right )}{d}\)

\(\Big \downarrow \) 150

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

\(\Big \downarrow \) 719

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

\(\Big \downarrow \) 514

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

-((c*((-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)/Sqrt[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)/Sq 
rt[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)))/d) + ((2*Sqrt[c + d*x]*(a + b*x^2)^(1 + 
 p))/(b*(5 + 4*p)) - ((2*(b*c^2 + a*d^2)*Sqrt[c + d*x]*(a + b*x^2)^p*Appel 
lF1[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)/Sqrt[b]))^p) - (2*b*c*(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))/(b*(5 + 
 4*p)))/d
 

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 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]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 2*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 - 8*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((s 
qrt(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 + 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**3 + 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**2 + 272*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 + 15*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 + 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 ...