Integrand size = 24, antiderivative size = 179 \[ \int (e x)^{3/2} (c+d x)^2 \left (a+b x^2\right )^p \, dx=\frac {2 d^2 (e x)^{5/2} \left (a+b x^2\right )^{1+p}}{b e (9+4 p)}-\frac {2 \left (5 a d^2-b c^2 (9+4 p)\right ) (e x)^{5/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^2}{a}\right )}{5 b e (9+4 p)}+\frac {4 c d (e x)^{7/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},-p,\frac {11}{4},-\frac {b x^2}{a}\right )}{7 e^2} \] Output:
2*d^2*(e*x)^(5/2)*(b*x^2+a)^(p+1)/b/e/(9+4*p)-2/5*(5*a*d^2-b*c^2*(9+4*p))* (e*x)^(5/2)*(b*x^2+a)^p*hypergeom([5/4, -p],[9/4],-b*x^2/a)/b/e/(9+4*p)/(( 1+b*x^2/a)^p)+4/7*c*d*(e*x)^(7/2)*(b*x^2+a)^p*hypergeom([7/4, -p],[11/4],- b*x^2/a)/e^2/((1+b*x^2/a)^p)
Time = 0.84 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.61 \[ \int (e x)^{3/2} (c+d x)^2 \left (a+b x^2\right )^p \, dx=\frac {2}{315} x (e x)^{3/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (63 c^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^2}{a}\right )+5 d x \left (18 c \operatorname {Hypergeometric2F1}\left (\frac {7}{4},-p,\frac {11}{4},-\frac {b x^2}{a}\right )+7 d x \operatorname {Hypergeometric2F1}\left (\frac {9}{4},-p,\frac {13}{4},-\frac {b x^2}{a}\right )\right )\right ) \] Input:
Integrate[(e*x)^(3/2)*(c + d*x)^2*(a + b*x^2)^p,x]
Output:
(2*x*(e*x)^(3/2)*(a + b*x^2)^p*(63*c^2*Hypergeometric2F1[5/4, -p, 9/4, -(( b*x^2)/a)] + 5*d*x*(18*c*Hypergeometric2F1[7/4, -p, 11/4, -((b*x^2)/a)] + 7*d*x*Hypergeometric2F1[9/4, -p, 13/4, -((b*x^2)/a)])))/(315*(1 + (b*x^2)/ a)^p)
Time = 0.50 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {559, 27, 557, 279, 278}
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.
\(\displaystyle \int (e x)^{3/2} (c+d x)^2 \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 559 |
\(\displaystyle \frac {2 \int -\frac {1}{2} (e x)^{3/2} \left (-b (4 p+9) c^2-2 b d (4 p+9) x c+5 a d^2\right ) \left (b x^2+a\right )^pdx}{b (4 p+9)}+\frac {2 d^2 (e x)^{5/2} \left (a+b x^2\right )^{p+1}}{b e (4 p+9)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 d^2 (e x)^{5/2} \left (a+b x^2\right )^{p+1}}{b e (4 p+9)}-\frac {\int (e x)^{3/2} \left (-b (4 p+9) c^2-2 b d (4 p+9) x c+5 a d^2\right ) \left (b x^2+a\right )^pdx}{b (4 p+9)}\) |
\(\Big \downarrow \) 557 |
\(\displaystyle \frac {2 d^2 (e x)^{5/2} \left (a+b x^2\right )^{p+1}}{b e (4 p+9)}-\frac {\left (5 a d^2-b c^2 (4 p+9)\right ) \int (e x)^{3/2} \left (b x^2+a\right )^pdx-\frac {2 b c d (4 p+9) \int (e x)^{5/2} \left (b x^2+a\right )^pdx}{e}}{b (4 p+9)}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {2 d^2 (e x)^{5/2} \left (a+b x^2\right )^{p+1}}{b e (4 p+9)}-\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (5 a d^2-b c^2 (4 p+9)\right ) \int (e x)^{3/2} \left (\frac {b x^2}{a}+1\right )^pdx-\frac {2 b c d (4 p+9) \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int (e x)^{5/2} \left (\frac {b x^2}{a}+1\right )^pdx}{e}}{b (4 p+9)}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {2 d^2 (e x)^{5/2} \left (a+b x^2\right )^{p+1}}{b e (4 p+9)}-\frac {\frac {2 (e x)^{5/2} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (5 a d^2-b c^2 (4 p+9)\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^2}{a}\right )}{5 e}-\frac {4 b c d (4 p+9) (e x)^{7/2} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},-p,\frac {11}{4},-\frac {b x^2}{a}\right )}{7 e^2}}{b (4 p+9)}\) |
Input:
Int[(e*x)^(3/2)*(c + d*x)^2*(a + b*x^2)^p,x]
Output:
(2*d^2*(e*x)^(5/2)*(a + b*x^2)^(1 + p))/(b*e*(9 + 4*p)) - ((2*(5*a*d^2 - b *c^2*(9 + 4*p))*(e*x)^(5/2)*(a + b*x^2)^p*Hypergeometric2F1[5/4, -p, 9/4, -((b*x^2)/a)])/(5*e*(1 + (b*x^2)/a)^p) - (4*b*c*d*(9 + 4*p)*(e*x)^(7/2)*(a + b*x^2)^p*Hypergeometric2F1[7/4, -p, 11/4, -((b*x^2)/a)])/(7*e^2*(1 + (b *x^2)/a)^p))/(b*(9 + 4*p))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym bol] :> Simp[c Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e Int[(e*x)^( m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d^n*(e*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*e^(n - 1)*( m + n + 2*p + 1))), x] + Simp[1/(b*(m + n + 2*p + 1)) Int[(e*x)^m*(a + b* x^2)^p*ExpandToSum[b*(m + n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1 )*x^n - a*d^n*(m + n - 1)*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && IGtQ[n, 1] && !IntegerQ[m] && NeQ[m + n + 2*p + 1, 0]
\[\int \left (e x \right )^{\frac {3}{2}} \left (d x +c \right )^{2} \left (b \,x^{2}+a \right )^{p}d x\]
Input:
int((e*x)^(3/2)*(d*x+c)^2*(b*x^2+a)^p,x)
Output:
int((e*x)^(3/2)*(d*x+c)^2*(b*x^2+a)^p,x)
\[ \int (e x)^{3/2} (c+d x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{2} \left (e x\right )^{\frac {3}{2}} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((e*x)^(3/2)*(d*x+c)^2*(b*x^2+a)^p,x, algorithm="fricas")
Output:
integral((d^2*e*x^3 + 2*c*d*e*x^2 + c^2*e*x)*sqrt(e*x)*(b*x^2 + a)^p, x)
Timed out. \[ \int (e x)^{3/2} (c+d x)^2 \left (a+b x^2\right )^p \, dx=\text {Timed out} \] Input:
integrate((e*x)**(3/2)*(d*x+c)**2*(b*x**2+a)**p,x)
Output:
Timed out
\[ \int (e x)^{3/2} (c+d x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{2} \left (e x\right )^{\frac {3}{2}} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((e*x)^(3/2)*(d*x+c)^2*(b*x^2+a)^p,x, algorithm="maxima")
Output:
integrate((d*x + c)^2*(e*x)^(3/2)*(b*x^2 + a)^p, x)
\[ \int (e x)^{3/2} (c+d x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{2} \left (e x\right )^{\frac {3}{2}} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((e*x)^(3/2)*(d*x+c)^2*(b*x^2+a)^p,x, algorithm="giac")
Output:
integrate((d*x + c)^2*(e*x)^(3/2)*(b*x^2 + a)^p, x)
Timed out. \[ \int (e x)^{3/2} (c+d x)^2 \left (a+b x^2\right )^p \, dx=\int {\left (e\,x\right )}^{3/2}\,{\left (b\,x^2+a\right )}^p\,{\left (c+d\,x\right )}^2 \,d x \] Input:
int((e*x)^(3/2)*(a + b*x^2)^p*(c + d*x)^2,x)
Output:
int((e*x)^(3/2)*(a + b*x^2)^p*(c + d*x)^2, x)
\[ \int (e x)^{3/2} (c+d x)^2 \left (a+b x^2\right )^p \, dx=\text {too large to display} \] Input:
int((e*x)^(3/2)*(d*x+c)^2*(b*x^2+a)^p,x)
Output:
(2*sqrt(e)*e*( - 320*sqrt(x)*(a + b*x**2)**p*a**2*d**2*p**3 - 800*sqrt(x)* (a + b*x**2)**p*a**2*d**2*p**2 - 420*sqrt(x)*(a + b*x**2)**p*a**2*d**2*p + 256*sqrt(x)*(a + b*x**2)**p*a*b*c**2*p**4 + 1216*sqrt(x)*(a + b*x**2)**p* a*b*c**2*p**3 + 1776*sqrt(x)*(a + b*x**2)**p*a*b*c**2*p**2 + 756*sqrt(x)*( a + b*x**2)**p*a*b*c**2*p + 512*sqrt(x)*(a + b*x**2)**p*a*b*c*d*p**4*x + 1 920*sqrt(x)*(a + b*x**2)**p*a*b*c*d*p**3*x + 1888*sqrt(x)*(a + b*x**2)**p* a*b*c*d*p**2*x + 360*sqrt(x)*(a + b*x**2)**p*a*b*c*d*p*x + 256*sqrt(x)*(a + b*x**2)**p*a*b*d**2*p**4*x**2 + 704*sqrt(x)*(a + b*x**2)**p*a*b*d**2*p** 3*x**2 + 496*sqrt(x)*(a + b*x**2)**p*a*b*d**2*p**2*x**2 + 84*sqrt(x)*(a + b*x**2)**p*a*b*d**2*p*x**2 + 256*sqrt(x)*(a + b*x**2)**p*b**2*c**2*p**4*x* *2 + 1280*sqrt(x)*(a + b*x**2)**p*b**2*c**2*p**3*x**2 + 2080*sqrt(x)*(a + b*x**2)**p*b**2*c**2*p**2*x**2 + 1200*sqrt(x)*(a + b*x**2)**p*b**2*c**2*p* x**2 + 189*sqrt(x)*(a + b*x**2)**p*b**2*c**2*x**2 + 512*sqrt(x)*(a + b*x** 2)**p*b**2*c*d*p**4*x**3 + 2304*sqrt(x)*(a + b*x**2)**p*b**2*c*d*p**3*x**3 + 3328*sqrt(x)*(a + b*x**2)**p*b**2*c*d*p**2*x**3 + 1776*sqrt(x)*(a + b*x **2)**p*b**2*c*d*p*x**3 + 270*sqrt(x)*(a + b*x**2)**p*b**2*c*d*x**3 + 256* sqrt(x)*(a + b*x**2)**p*b**2*d**2*p**4*x**4 + 1024*sqrt(x)*(a + b*x**2)**p *b**2*d**2*p**3*x**4 + 1376*sqrt(x)*(a + b*x**2)**p*b**2*d**2*p**2*x**4 + 704*sqrt(x)*(a + b*x**2)**p*b**2*d**2*p*x**4 + 105*sqrt(x)*(a + b*x**2)**p *b**2*d**2*x**4 + 163840*int((sqrt(x)*(a + b*x**2)**p)/(1024*a*p**5*x +...