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

Optimal result

Integrand size = 18, antiderivative size = 100 \[ \int x^3 (c+d x) \left (a+b x^2\right )^p \, dx=-\frac {a c \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {c \left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)}+\frac {1}{5} d x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int x^3 (c+d x) \left (a+b x^2\right )^p \, dx=\frac {1}{10} \left (a+b x^2\right )^p \left (-\frac {5 c \left (a+b x^2\right ) \left (a-b (1+p) x^2\right )}{b^2 (1+p) (2+p)}+2 d x^5 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\right ) \] Input:

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

Output:

((a + b*x^2)^p*((-5*c*(a + b*x^2)*(a - b*(1 + p)*x^2))/(b^2*(1 + p)*(2 + p 
)) + (2*d*x^5*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(1 + (b*x^2)/ 
a)^p))/10
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {542, 243, 53, 279, 278, 2009}

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.

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

\(\Big \downarrow \) 542

\(\displaystyle c \int x^3 \left (b x^2+a\right )^pdx+d \int x^4 \left (b x^2+a\right )^pdx\)

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 53

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 53

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 278

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

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 542

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d*x^4 + c*x^3)*(b*x^2 + a)^p, x)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (82) = 164\).

Time = 6.80 (sec) , antiderivative size = 364, normalized size of antiderivative = 3.64 \[ \int x^3 (c+d x) \left (a+b x^2\right )^p \, dx=\frac {a^{p} d x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + c \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} - \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

a**p*d*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + c*Piece 
wise((a**p*x**4/4, Eq(b, 0)), (a*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x* 
*2) + a*log(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + a/(2*a*b**2 + 2*b** 
3*x**2) + b*x**2*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + b*x**2*log 
(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2), Eq(p, -2)), (-a*log(x - sqrt(-a 
/b))/(2*b**2) - a*log(x + sqrt(-a/b))/(2*b**2) + x**2/(2*b), Eq(p, -1)), ( 
-a**2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + a*b*p*x**2*(a + 
b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*p*x**4*(a + b*x**2)**p 
/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*x**4*(a + b*x**2)**p/(2*b**2*p** 
2 + 6*b**2*p + 4*b**2), True))
 

Maxima [F]

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

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

Output:

d*integrate((b*x^2 + a)^p*x^4, x) + 1/2*(b^2*(p + 1)*x^4 + a*b*p*x^2 - a^2 
)*(b*x^2 + a)^p*c/((p^2 + 3*p + 2)*b^2)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

( - 8*(a + b*x**2)**p*a**2*c*p**3 - 36*(a + b*x**2)**p*a**2*c*p**2 - 46*(a 
 + b*x**2)**p*a**2*c*p - 15*(a + b*x**2)**p*a**2*c - 12*(a + b*x**2)**p*a* 
*2*d*p**3*x - 36*(a + b*x**2)**p*a**2*d*p**2*x - 24*(a + b*x**2)**p*a**2*d 
*p*x + 8*(a + b*x**2)**p*a*b*c*p**4*x**2 + 36*(a + b*x**2)**p*a*b*c*p**3*x 
**2 + 46*(a + b*x**2)**p*a*b*c*p**2*x**2 + 15*(a + b*x**2)**p*a*b*c*p*x**2 
 + 8*(a + b*x**2)**p*a*b*d*p**4*x**3 + 28*(a + b*x**2)**p*a*b*d*p**3*x**3 
+ 28*(a + b*x**2)**p*a*b*d*p**2*x**3 + 8*(a + b*x**2)**p*a*b*d*p*x**3 + 8* 
(a + b*x**2)**p*b**2*c*p**4*x**4 + 44*(a + b*x**2)**p*b**2*c*p**3*x**4 + 8 
2*(a + b*x**2)**p*b**2*c*p**2*x**4 + 61*(a + b*x**2)**p*b**2*c*p*x**4 + 15 
*(a + b*x**2)**p*b**2*c*x**4 + 8*(a + b*x**2)**p*b**2*d*p**4*x**5 + 40*(a 
+ b*x**2)**p*b**2*d*p**3*x**5 + 70*(a + b*x**2)**p*b**2*d*p**2*x**5 + 50*( 
a + b*x**2)**p*b**2*d*p*x**5 + 12*(a + b*x**2)**p*b**2*d*x**5 + 96*int((a 
+ b*x**2)**p/(8*a*p**3 + 36*a*p**2 + 46*a*p + 15*a + 8*b*p**3*x**2 + 36*b* 
p**2*x**2 + 46*b*p*x**2 + 15*b*x**2),x)*a**3*d*p**6 + 720*int((a + b*x**2) 
**p/(8*a*p**3 + 36*a*p**2 + 46*a*p + 15*a + 8*b*p**3*x**2 + 36*b*p**2*x**2 
 + 46*b*p*x**2 + 15*b*x**2),x)*a**3*d*p**5 + 2040*int((a + b*x**2)**p/(8*a 
*p**3 + 36*a*p**2 + 46*a*p + 15*a + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b* 
p*x**2 + 15*b*x**2),x)*a**3*d*p**4 + 2700*int((a + b*x**2)**p/(8*a*p**3 + 
36*a*p**2 + 46*a*p + 15*a + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b*p*x**2 + 
 15*b*x**2),x)*a**3*d*p**3 + 1644*int((a + b*x**2)**p/(8*a*p**3 + 36*a*...