\(\int \frac {x^3 (a+b x^2)^p}{c+d x} \, dx\) [30]

Optimal result

Integrand size = 20, antiderivative size = 163 \[ \int \frac {x^3 \left (a+b x^2\right )^p}{c+d x} \, dx=-\frac {c \left (a+b x^2\right )^{1+p}}{2 b d^2 (1+p)}-\frac {d x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},-p,1,\frac {7}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{5 c^2}+\frac {c^3 \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{2 d^2 \left (b c^2+a d^2\right ) (1+p)} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.31 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.60 \[ \int \frac {x^3 \left (a+b x^2\right )^p}{c+d x} \, dx=\frac {\left (a+b x^2\right )^p \left (-\frac {3 c^3 \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{p}+\frac {d \left (1+\frac {b x^2}{a}\right )^{-p} \left (6 b c^2 (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+d \left (-3 c \left (b x^2 \left (1+\frac {b x^2}{a}\right )^p+a \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )\right )+2 b d (1+p) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )\right )\right )}{b (1+p)}\right )}{6 d^4} \] Input:

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

Output:

((a + b*x^2)^p*((-3*c^3*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*(6*b*c^2*(1 + p)* 
x*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)] + d*(-3*c*(b*x^2*(1 + (b*x 
^2)/a)^p + a*(-1 + (1 + (b*x^2)/a)^p)) + 2*b*d*(1 + p)*x^3*Hypergeometric2 
F1[3/2, -p, 5/2, -((b*x^2)/a)])))/(b*(1 + p)*(1 + (b*x^2)/a)^p)))/(6*d^4)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {621, 354, 90, 78, 395, 394}

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.

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

\(\Big \downarrow \) 621

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 78

\(\displaystyle \frac {1}{2} c \left (\frac {c^2 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{d^2 (p+1) \left (a d^2+b c^2\right )}-\frac {\left (a+b x^2\right )^{p+1}}{b d^2 (p+1)}\right )-d \int \frac {x^4 \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\)

\(\Big \downarrow \) 395

\(\displaystyle \frac {1}{2} c \left (\frac {c^2 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{d^2 (p+1) \left (a d^2+b c^2\right )}-\frac {\left (a+b x^2\right )^{p+1}}{b d^2 (p+1)}\right )-d \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \frac {x^4 \left (\frac {b x^2}{a}+1\right )^p}{c^2-d^2 x^2}dx\)

\(\Big \downarrow \) 394

\(\displaystyle \frac {1}{2} c \left (\frac {c^2 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{d^2 (p+1) \left (a d^2+b c^2\right )}-\frac {\left (a+b x^2\right )^{p+1}}{b d^2 (p+1)}\right )-\frac {d x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},-p,1,\frac {7}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{5 c^2}\)

Input:

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

Output:

-1/5*(d*x^5*(a + b*x^2)^p*AppellF1[5/2, -p, 1, 7/2, -((b*x^2)/a), (d^2*x^2 
)/c^2])/(c^2*(1 + (b*x^2)/a)^p) + (c*(-((a + b*x^2)^(1 + p)/(b*d^2*(1 + p) 
)) + (c^2*(a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (d^2*(a + 
 b*x^2))/(b*c^2 + a*d^2)])/(d^2*(b*c^2 + a*d^2)*(1 + p))))/2
 

Defintions of rubi rules used

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 90

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

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(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] && IntegerQ 
[(m - 1)/2]
 

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

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

rule 621

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 2*(a + b*x**2)**p*a**2*d**3*p - 2*(a + b*x**2)**p*a**2*d**3 + 2*(a + b 
*x**2)**p*a*b*c**2*d*p + 3*(a + b*x**2)**p*a*b*c**2*d + 4*(a + b*x**2)**p* 
a*b*c*d**2*p**2*x + 4*(a + b*x**2)**p*a*b*c*d**2*p*x + 4*(a + b*x**2)**p*b 
**2*c**3*p**2*x + 10*(a + b*x**2)**p*b**2*c**3*p*x + 6*(a + b*x**2)**p*b** 
2*c**3*x - 4*(a + b*x**2)**p*b**2*c**2*d*p**2*x**2 - 8*(a + b*x**2)**p*b** 
2*c**2*d*p*x**2 - 3*(a + b*x**2)**p*b**2*c**2*d*x**2 + 4*(a + b*x**2)**p*b 
**2*c*d**2*p**2*x**3 + 6*(a + b*x**2)**p*b**2*c*d**2*p*x**3 + 2*(a + b*x** 
2)**p*b**2*c*d**2*x**3 - 16*int((a + b*x**2)**p/(4*a*c*p**2 + 8*a*c*p + 3* 
a*c + 4*a*d*p**2*x + 8*a*d*p*x + 3*a*d*x + 4*b*c*p**2*x**2 + 8*b*c*p*x**2 
+ 3*b*c*x**2 + 4*b*d*p**2*x**3 + 8*b*d*p*x**3 + 3*b*d*x**3),x)*a**2*b*c**2 
*d**2*p**4 - 48*int((a + b*x**2)**p/(4*a*c*p**2 + 8*a*c*p + 3*a*c + 4*a*d* 
p**2*x + 8*a*d*p*x + 3*a*d*x + 4*b*c*p**2*x**2 + 8*b*c*p*x**2 + 3*b*c*x**2 
 + 4*b*d*p**2*x**3 + 8*b*d*p*x**3 + 3*b*d*x**3),x)*a**2*b*c**2*d**2*p**3 - 
 44*int((a + b*x**2)**p/(4*a*c*p**2 + 8*a*c*p + 3*a*c + 4*a*d*p**2*x + 8*a 
*d*p*x + 3*a*d*x + 4*b*c*p**2*x**2 + 8*b*c*p*x**2 + 3*b*c*x**2 + 4*b*d*p** 
2*x**3 + 8*b*d*p*x**3 + 3*b*d*x**3),x)*a**2*b*c**2*d**2*p**2 - 12*int((a + 
 b*x**2)**p/(4*a*c*p**2 + 8*a*c*p + 3*a*c + 4*a*d*p**2*x + 8*a*d*p*x + 3*a 
*d*x + 4*b*c*p**2*x**2 + 8*b*c*p*x**2 + 3*b*c*x**2 + 4*b*d*p**2*x**3 + 8*b 
*d*p*x**3 + 3*b*d*x**3),x)*a**2*b*c**2*d**2*p - 16*int((a + b*x**2)**p/(4* 
a*c*p**2 + 8*a*c*p + 3*a*c + 4*a*d*p**2*x + 8*a*d*p*x + 3*a*d*x + 4*b*c...