\(\int \frac {x^3 (a+b x^2)^p}{(c+d x)^2} \, dx\) [40]

Optimal result

Integrand size = 20, antiderivative size = 223 \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\frac {\left (a+b x^2\right )^{1+p}}{2 b d^2 (1+p)}-\frac {3 c^2 \left (a+b x^2\right )^{1+p}}{2 b d^2 p \left (c^2-d^2 x^2\right )}-\frac {2 d x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},-p,2,\frac {7}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{5 c^3}+\frac {c^2 \left (3 a d^2+b c^2 (3+2 p)\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{2 d^2 \left (b c^2+a d^2\right )^2 p (1+p)} \] Output:

1/2*(b*x^2+a)^(p+1)/b/d^2/(p+1)-3/2*c^2*(b*x^2+a)^(p+1)/b/d^2/p/(-d^2*x^2+ 
c^2)-2/5*d*x^5*(b*x^2+a)^p*AppellF1(5/2,2,-p,7/2,d^2*x^2/c^2,-b*x^2/a)/c^3 
/((1+b*x^2/a)^p)+1/2*c^2*(3*a*d^2+b*c^2*(3+2*p))*(b*x^2+a)^(p+1)*hypergeom 
([2, p+1],[2+p],d^2*(b*x^2+a)/(a*d^2+b*c^2))/d^2/(a*d^2+b*c^2)^2/p/(p+1)
 

Mathematica [A] (warning: unable to verify)

Time = 0.68 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.54 \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\frac {\left (a+b x^2\right )^p \left (-\frac {2 c^3 \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{(-1+2 p) (c+d x)}+\frac {3 c^2 \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{p}+d \left (\frac {d \left (a+b x^2-a \left (1+\frac {b x^2}{a}\right )^{-p}\right )}{b+b p}-4 c x \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )\right )\right )}{2 d^4} \] Input:

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

Output:

((a + b*x^2)^p*((-2*c^3*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (c - Sqrt[-(a/b 
)]*d)/(c + d*x), (c + Sqrt[-(a/b)]*d)/(c + d*x)])/((-1 + 2*p)*((d*(-Sqrt[- 
(a/b)] + x))/(c + d*x))^p*((d*(Sqrt[-(a/b)] + x))/(c + d*x))^p*(c + d*x)) 
+ (3*c^2*AppellF1[-2*p, -p, -p, 1 - 2*p, (c - Sqrt[-(a/b)]*d)/(c + d*x), ( 
c + Sqrt[-(a/b)]*d)/(c + d*x)])/(p*((d*(-Sqrt[-(a/b)] + x))/(c + d*x))^p*( 
(d*(Sqrt[-(a/b)] + x))/(c + d*x))^p) + d*((d*(a + b*x^2 - a/(1 + (b*x^2)/a 
)^p))/(b + b*p) - (4*c*x*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 
 + (b*x^2)/a)^p)))/(2*d^4)
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.39, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {603, 25, 2185, 27, 719, 238, 237, 504, 334, 333, 353, 78}

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.

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

\(\Big \downarrow \) 603

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2185

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 719

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

\(\Big \downarrow \) 238

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

\(\Big \downarrow \) 237

\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \int \frac {\left (b x^2+a\right )^p}{c+d x}dx}{d}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 504

\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \left (c \int \frac {\left (b x^2+a\right )^p}{c^2-d^2 x^2}dx-d \int \frac {x \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\right )}{d}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \left (c \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \frac {\left (\frac {b x^2}{a}+1\right )^p}{c^2-d^2 x^2}dx-d \int \frac {x \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\right )}{d}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 333

\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-d \int \frac {x \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\right )}{d}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-\frac {1}{2} d \int \frac {\left (b x^2+a\right )^p}{c^2-d^2 x^2}dx^2\right )}{d}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {\frac {c \left (\frac {c \left (3 a d^2+b c^2 (2 p+3)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-\frac {d \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{2 (p+1) \left (a d^2+b c^2\right )}\right )}{d}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a d^2+b c^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )}{d^2}+\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^2 (p+1)}}{a d^2+b c^2}+\frac {c^3 \left (a+b x^2\right )^{p+1}}{d^2 (c+d x) \left (a d^2+b c^2\right )}\)

Input:

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

Output:

(c^3*(a + b*x^2)^(1 + p))/(d^2*(b*c^2 + a*d^2)*(c + d*x)) + (((b*c^2 + a*d 
^2)*(a + b*x^2)^(1 + p))/(2*b*d^2*(1 + p)) + (c*(-(((2*a*d^2 + b*c^2*(3 + 
2*p))*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(d*(1 
 + (b*x^2)/a)^p)) + (c*(3*a*d^2 + b*c^2*(3 + 2*p))*((x*(a + b*x^2)^p*Appel 
lF1[1/2, -p, 1, 3/2, -((b*x^2)/a), (d^2*x^2)/c^2])/(c*(1 + (b*x^2)/a)^p) - 
 (d*(a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (d^2*(a + b*x^2 
))/(b*c^2 + a*d^2)])/(2*(b*c^2 + a*d^2)*(1 + p))))/d))/d^2)/(b*c^2 + a*d^2 
)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[ 
(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && 
NeQ[b*c - a*d, 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 353

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

rule 504

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c   I 
nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d   Int[x*((a + b*x^2)^p/(c 
^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, p}, x]
 

rule 603

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> With[{Qx = PolynomialQuotient[x^m, c + d*x, x], R = PolynomialRemainde 
r[x^m, c + d*x, x]}, Simp[d*R*(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*ExpandToSum[(n + 1)*(b*c^2 + a*d^2)*Qx + b*c*R*(n + 
1) - b*d*R*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IGt 
Q[m, 1] && 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]
 

rule 2185

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x 
)^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p 
)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d 
, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && 
True) &&  !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 
 1/2, 0]))
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{-1,[0,3,1,0]%%%} / %%%{1,[0,0,0,3]%%%} Error: Bad Argument 
 Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(4*(a + b*x**2)**p*a*c*p**2 + 8*(a + b*x**2)**p*a*c*p + 3*(a + b*x**2)**p* 
a*c - 8*(a + b*x**2)**p*a*d*p**2*x - 4*(a + b*x**2)**p*a*d*p*x + 3*(a + b* 
x**2)**p*a*d*x - 4*(a + b*x**2)**p*b*c*p**3*x**2 - 8*(a + b*x**2)**p*b*c*p 
**2*x**2 - 3*(a + b*x**2)**p*b*c*p*x**2 + 4*(a + b*x**2)**p*b*d*p**3*x**3 
+ 4*(a + b*x**2)**p*b*d*p**2*x**3 + (a + b*x**2)**p*b*d*p*x**3 + 48*int((a 
 + b*x**2)**p/(4*a*c**2*p**2 + 4*a*c**2*p + a*c**2 + 8*a*c*d*p**2*x + 8*a* 
c*d*p*x + 2*a*c*d*x + 4*a*d**2*p**2*x**2 + 4*a*d**2*p*x**2 + a*d**2*x**2 + 
 4*b*c**2*p**2*x**2 + 4*b*c**2*p*x**2 + b*c**2*x**2 + 8*b*c*d*p**2*x**3 + 
8*b*c*d*p*x**3 + 2*b*c*d*x**3 + 4*b*d**2*p**2*x**4 + 4*b*d**2*p*x**4 + b*d 
**2*x**4),x)*a**2*c**2*d*p**4 + 96*int((a + b*x**2)**p/(4*a*c**2*p**2 + 4* 
a*c**2*p + a*c**2 + 8*a*c*d*p**2*x + 8*a*c*d*p*x + 2*a*c*d*x + 4*a*d**2*p* 
*2*x**2 + 4*a*d**2*p*x**2 + a*d**2*x**2 + 4*b*c**2*p**2*x**2 + 4*b*c**2*p* 
x**2 + b*c**2*x**2 + 8*b*c*d*p**2*x**3 + 8*b*c*d*p*x**3 + 2*b*c*d*x**3 + 4 
*b*d**2*p**2*x**4 + 4*b*d**2*p*x**4 + b*d**2*x**4),x)*a**2*c**2*d*p**3 + 6 
0*int((a + b*x**2)**p/(4*a*c**2*p**2 + 4*a*c**2*p + a*c**2 + 8*a*c*d*p**2* 
x + 8*a*c*d*p*x + 2*a*c*d*x + 4*a*d**2*p**2*x**2 + 4*a*d**2*p*x**2 + a*d** 
2*x**2 + 4*b*c**2*p**2*x**2 + 4*b*c**2*p*x**2 + b*c**2*x**2 + 8*b*c*d*p**2 
*x**3 + 8*b*c*d*p*x**3 + 2*b*c*d*x**3 + 4*b*d**2*p**2*x**4 + 4*b*d**2*p*x* 
*4 + b*d**2*x**4),x)*a**2*c**2*d*p**2 + 12*int((a + b*x**2)**p/(4*a*c**2*p 
**2 + 4*a*c**2*p + a*c**2 + 8*a*c*d*p**2*x + 8*a*c*d*p*x + 2*a*c*d*x + ...