Integrand size = 18, antiderivative size = 125 \[ \int x^4 (c+d x) \left (a+b x^2\right )^p \, dx=\frac {a^2 d \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}-\frac {a d \left (a+b x^2\right )^{2+p}}{b^3 (2+p)}+\frac {d \left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)}+\frac {1}{5} c x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right ) \] Output:
1/2*a^2*d*(b*x^2+a)^(p+1)/b^3/(p+1)-a*d*(b*x^2+a)^(2+p)/b^3/(2+p)+1/2*d*(b *x^2+a)^(3+p)/b^3/(3+p)+1/5*c*x^5*(b*x^2+a)^p*hypergeom([5/2, -p],[7/2],-b *x^2/a)/((1+b*x^2/a)^p)
Time = 0.07 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.90 \[ \int x^4 (c+d x) \left (a+b x^2\right )^p \, dx=\frac {1}{10} \left (a+b x^2\right )^p \left (\frac {5 d \left (a+b x^2\right ) \left (2 a^2-2 a b (1+p) x^2+b^2 \left (2+3 p+p^2\right ) x^4\right )}{b^3 (1+p) (2+p) (3+p)}+2 c x^5 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\right ) \] Input:
Integrate[x^4*(c + d*x)*(a + b*x^2)^p,x]
Output:
((a + b*x^2)^p*((5*d*(a + b*x^2)*(2*a^2 - 2*a*b*(1 + p)*x^2 + b^2*(2 + 3*p + p^2)*x^4))/(b^3*(1 + p)*(2 + p)*(3 + p)) + (2*c*x^5*Hypergeometric2F1[5 /2, -p, 7/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/10
Time = 0.43 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.98, 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.
\(\displaystyle \int x^4 (c+d x) \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 542 |
\(\displaystyle c \int x^4 \left (b x^2+a\right )^pdx+d \int x^5 \left (b x^2+a\right )^pdx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle c \int x^4 \left (b x^2+a\right )^pdx+\frac {1}{2} d \int x^4 \left (b x^2+a\right )^pdx^2\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{2} d \int \left (\frac {a^2 \left (b x^2+a\right )^p}{b^2}-\frac {2 a \left (b x^2+a\right )^{p+1}}{b^2}+\frac {\left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2+c \int x^4 \left (b x^2+a\right )^pdx\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {1}{2} d \int \left (\frac {a^2 \left (b x^2+a\right )^p}{b^2}-\frac {2 a \left (b x^2+a\right )^{p+1}}{b^2}+\frac {\left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2+c \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int x^4 \left (\frac {b x^2}{a}+1\right )^pdx\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {1}{2} d \int \left (\frac {a^2 \left (b x^2+a\right )^p}{b^2}-\frac {2 a \left (b x^2+a\right )^{p+1}}{b^2}+\frac {\left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2+\frac {1}{5} c x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} d \left (\frac {a^2 \left (a+b x^2\right )^{p+1}}{b^3 (p+1)}-\frac {2 a \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac {\left (a+b x^2\right )^{p+3}}{b^3 (p+3)}\right )+\frac {1}{5} c x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\) |
Input:
Int[x^4*(c + d*x)*(a + b*x^2)^p,x]
Output:
(d*((a^2*(a + b*x^2)^(1 + p))/(b^3*(1 + p)) - (2*a*(a + b*x^2)^(2 + p))/(b ^3*(2 + p)) + (a + b*x^2)^(3 + p)/(b^3*(3 + p))))/2 + (c*x^5*(a + b*x^2)^p *Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(5*(1 + (b*x^2)/a)^p)
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])
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]
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])
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])
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]
\[\int x^{4} \left (d x +c \right ) \left (b \,x^{2}+a \right )^{p}d x\]
Input:
int(x^4*(d*x+c)*(b*x^2+a)^p,x)
Output:
int(x^4*(d*x+c)*(b*x^2+a)^p,x)
\[ \int x^4 (c+d x) \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )} {\left (b x^{2} + a\right )}^{p} x^{4} \,d x } \] Input:
integrate(x^4*(d*x+c)*(b*x^2+a)^p,x, algorithm="fricas")
Output:
integral((d*x^5 + c*x^4)*(b*x^2 + a)^p, x)
Leaf count of result is larger than twice the leaf count of optimal. 921 vs. \(2 (104) = 208\).
Time = 8.38 (sec) , antiderivative size = 950, normalized size of antiderivative = 7.60 \[ \int x^4 (c+d x) \left (a+b x^2\right )^p \, dx =\text {Too large to display} \] Input:
integrate(x**4*(d*x+c)*(b*x**2+a)**p,x)
Output:
a**p*c*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + d*Piece wise((a**p*x**6/6, Eq(b, 0)), (2*a**2*log(x - sqrt(-a/b))/(4*a**2*b**3 + 8 *a*b**4*x**2 + 4*b**5*x**4) + 2*a**2*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8* a*b**4*x**2 + 4*b**5*x**4) + 3*a**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5* x**4) + 4*a*b*x**2*log(x - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b* *5*x**4) + 4*a*b*x**2*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4 *b**5*x**4) + 4*a*b*x**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b **2*x**4*log(x - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x** 4), Eq(p, -3)), (-2*a**2*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2* a**2*log(x + sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a**2/(2*a*b**3 + 2*b **4*x**2) - 2*a*b*x**2*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a* b*x**2*log(x + sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) + b**2*x**4/(2*a*b**3 + 2*b**4*x**2), Eq(p, -2)), (a**2*log(x - sqrt(-a/b))/(2*b**3) + a**2*log( x + sqrt(-a/b))/(2*b**3) - a*x**2/(2*b**2) + x**4/(4*b), Eq(p, -1)), (2*a* *3*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) - 2* a**2*b*p*x**2*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12 *b**3) + a*b**2*p**2*x**4*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22 *b**3*p + 12*b**3) + a*b**2*p*x**4*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3* p**2 + 22*b**3*p + 12*b**3) + b**3*p**2*x**6*(a + b*x**2)**p/(2*b**3*p*...
\[ \int x^4 (c+d x) \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )} {\left (b x^{2} + a\right )}^{p} x^{4} \,d x } \] Input:
integrate(x^4*(d*x+c)*(b*x^2+a)^p,x, algorithm="maxima")
Output:
integrate((d*x + c)*(b*x^2 + a)^p*x^4, x)
\[ \int x^4 (c+d x) \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )} {\left (b x^{2} + a\right )}^{p} x^{4} \,d x } \] Input:
integrate(x^4*(d*x+c)*(b*x^2+a)^p,x, algorithm="giac")
Output:
integrate((d*x + c)*(b*x^2 + a)^p*x^4, x)
Timed out. \[ \int x^4 (c+d x) \left (a+b x^2\right )^p \, dx=\int x^4\,{\left (b\,x^2+a\right )}^p\,\left (c+d\,x\right ) \,d x \] Input:
int(x^4*(a + b*x^2)^p*(c + d*x),x)
Output:
int(x^4*(a + b*x^2)^p*(c + d*x), x)
\[ \int x^4 (c+d x) \left (a+b x^2\right )^p \, dx =\text {Too large to display} \] Input:
int(x^4*(d*x+c)*(b*x^2+a)^p,x)
Output:
(16*(a + b*x**2)**p*a**3*d*p**3 + 72*(a + b*x**2)**p*a**3*d*p**2 + 92*(a + b*x**2)**p*a**3*d*p + 30*(a + b*x**2)**p*a**3*d - 12*(a + b*x**2)**p*a**2 *b*c*p**4*x - 72*(a + b*x**2)**p*a**2*b*c*p**3*x - 132*(a + b*x**2)**p*a** 2*b*c*p**2*x - 72*(a + b*x**2)**p*a**2*b*c*p*x - 16*(a + b*x**2)**p*a**2*b *d*p**4*x**2 - 72*(a + b*x**2)**p*a**2*b*d*p**3*x**2 - 92*(a + b*x**2)**p* a**2*b*d*p**2*x**2 - 30*(a + b*x**2)**p*a**2*b*d*p*x**2 + 8*(a + b*x**2)** p*a*b**2*c*p**5*x**3 + 52*(a + b*x**2)**p*a*b**2*c*p**4*x**3 + 112*(a + b* x**2)**p*a*b**2*c*p**3*x**3 + 92*(a + b*x**2)**p*a*b**2*c*p**2*x**3 + 24*( a + b*x**2)**p*a*b**2*c*p*x**3 + 8*(a + b*x**2)**p*a*b**2*d*p**5*x**4 + 44 *(a + b*x**2)**p*a*b**2*d*p**4*x**4 + 82*(a + b*x**2)**p*a*b**2*d*p**3*x** 4 + 61*(a + b*x**2)**p*a*b**2*d*p**2*x**4 + 15*(a + b*x**2)**p*a*b**2*d*p* x**4 + 8*(a + b*x**2)**p*b**3*c*p**5*x**5 + 64*(a + b*x**2)**p*b**3*c*p**4 *x**5 + 190*(a + b*x**2)**p*b**3*c*p**3*x**5 + 260*(a + b*x**2)**p*b**3*c* p**2*x**5 + 162*(a + b*x**2)**p*b**3*c*p*x**5 + 36*(a + b*x**2)**p*b**3*c* x**5 + 8*(a + b*x**2)**p*b**3*d*p**5*x**6 + 60*(a + b*x**2)**p*b**3*d*p**4 *x**6 + 170*(a + b*x**2)**p*b**3*d*p**3*x**6 + 225*(a + b*x**2)**p*b**3*d* p**2*x**6 + 137*(a + b*x**2)**p*b**3*d*p*x**6 + 30*(a + b*x**2)**p*b**3*d* x**6 + 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*b*c*p**7 + 1 008*int((a + b*x**2)**p/(8*a*p**3 + 36*a*p**2 + 46*a*p + 15*a + 8*b*p**...