Integrand size = 20, antiderivative size = 148 \[ \int x^{3/2} (c+d x)^n \left (a+b x^2\right ) \, dx=-\frac {14 b c x^{5/2} (c+d x)^{1+n}}{d^2 (7+2 n) (9+2 n)}+\frac {2 b x^{7/2} (c+d x)^{1+n}}{d (9+2 n)}+\frac {2 \left (35 b c^2+a d^2 (7+2 n) (9+2 n)\right ) x^{5/2} (c+d x)^n \left (\frac {c+d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-n,\frac {7}{2},-\frac {d x}{c}\right )}{5 d^2 (7+2 n) (9+2 n)} \] Output:
-14*b*c*x^(5/2)*(d*x+c)^(1+n)/d^2/(7+2*n)/(9+2*n)+2*b*x^(7/2)*(d*x+c)^(1+n )/d/(9+2*n)+2/5*(35*b*c^2+a*d^2*(7+2*n)*(9+2*n))*x^(5/2)*(d*x+c)^n*hyperge om([5/2, -n],[7/2],-d*x/c)/d^2/(7+2*n)/(9+2*n)/(((d*x+c)/c)^n)
Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.74 \[ \int x^{3/2} (c+d x)^n \left (a+b x^2\right ) \, dx=\frac {2 x^{5/2} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \left (b c^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-2-n,\frac {7}{2},-\frac {d x}{c}\right )-2 b c^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-1-n,\frac {7}{2},-\frac {d x}{c}\right )+\left (b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-n,\frac {7}{2},-\frac {d x}{c}\right )\right )}{5 d^2} \] Input:
Integrate[x^(3/2)*(c + d*x)^n*(a + b*x^2),x]
Output:
(2*x^(5/2)*(c + d*x)^n*(b*c^2*Hypergeometric2F1[5/2, -2 - n, 7/2, -((d*x)/ c)] - 2*b*c^2*Hypergeometric2F1[5/2, -1 - n, 7/2, -((d*x)/c)] + (b*c^2 + a *d^2)*Hypergeometric2F1[5/2, -n, 7/2, -((d*x)/c)]))/(5*d^2*(1 + (d*x)/c)^n )
Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {521, 27, 90, 76, 74}
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.
\(\displaystyle \int x^{3/2} \left (a+b x^2\right ) (c+d x)^n \, dx\) |
\(\Big \downarrow \) 521 |
\(\displaystyle \frac {2 \int \frac {1}{2} x^{3/2} (a d (2 n+9)-7 b c x) (c+d x)^ndx}{d (2 n+9)}+\frac {2 b x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int x^{3/2} (a d (2 n+9)-7 b c x) (c+d x)^ndx}{d (2 n+9)}+\frac {2 b x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {\frac {\left (a d^2 (2 n+7) (2 n+9)+35 b c^2\right ) \int x^{3/2} (c+d x)^ndx}{d (2 n+7)}-\frac {14 b c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{d (2 n+9)}+\frac {2 b x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\) |
\(\Big \downarrow \) 76 |
\(\displaystyle \frac {\frac {(c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (a d^2 (2 n+7) (2 n+9)+35 b c^2\right ) \int x^{3/2} \left (\frac {d x}{c}+1\right )^ndx}{d (2 n+7)}-\frac {14 b c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{d (2 n+9)}+\frac {2 b x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\) |
\(\Big \downarrow \) 74 |
\(\displaystyle \frac {\frac {2 x^{5/2} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (a d^2 (2 n+7) (2 n+9)+35 b c^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-n,\frac {7}{2},-\frac {d x}{c}\right )}{5 d (2 n+7)}-\frac {14 b c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{d (2 n+9)}+\frac {2 b x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\) |
Input:
Int[x^(3/2)*(c + d*x)^n*(a + b*x^2),x]
Output:
(2*b*x^(7/2)*(c + d*x)^(1 + n))/(d*(9 + 2*n)) + ((-14*b*c*x^(5/2)*(c + d*x )^(1 + n))/(d*(7 + 2*n)) + (2*(35*b*c^2 + a*d^2*(7 + 2*n)*(9 + 2*n))*x^(5/ 2)*(c + d*x)^n*Hypergeometric2F1[5/2, -n, 7/2, -((d*x)/c)])/(5*d*(7 + 2*n) *(1 + (d*x)/c)^n))/(d*(9 + 2*n))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x )^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] && !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^IntPart [n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d* (x/c))^n, x], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && !Integer Q[n] && !GtQ[c, 0] && !GtQ[-d/(b*c), 0] && ((RationalQ[m] && !(EqQ[n, -2 ^(-1)] && EqQ[c^2 - d^2, 0])) || !RationalQ[n])
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]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[b^p*(e*x)^(m + 2*p)*((c + d*x)^(n + 1)/(d*e^(2*p)*(m + n + 2*p + 1))), x] + Simp[1/(d*e^(2*p)*(m + n + 2*p + 1)) Int[(e*x)^m*(c + d*x)^n*ExpandToSum[d*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x^2)^p - b^p*(e*x)^ (2*p)) - b^p*(e*c)*(m + 2*p)*(e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && NeQ[m + n + 2*p + 1, 0] && !IntegerQ[m] && !I ntegerQ[n]
\[\int x^{\frac {3}{2}} \left (d x +c \right )^{n} \left (b \,x^{2}+a \right )d x\]
Input:
int(x^(3/2)*(d*x+c)^n*(b*x^2+a),x)
Output:
int(x^(3/2)*(d*x+c)^n*(b*x^2+a),x)
\[ \int x^{3/2} (c+d x)^n \left (a+b x^2\right ) \, dx=\int { {\left (b x^{2} + a\right )} {\left (d x + c\right )}^{n} x^{\frac {3}{2}} \,d x } \] Input:
integrate(x^(3/2)*(d*x+c)^n*(b*x^2+a),x, algorithm="fricas")
Output:
integral((b*x^3 + a*x)*(d*x + c)^n*sqrt(x), x)
Timed out. \[ \int x^{3/2} (c+d x)^n \left (a+b x^2\right ) \, dx=\text {Timed out} \] Input:
integrate(x**(3/2)*(d*x+c)**n*(b*x**2+a),x)
Output:
Timed out
\[ \int x^{3/2} (c+d x)^n \left (a+b x^2\right ) \, dx=\int { {\left (b x^{2} + a\right )} {\left (d x + c\right )}^{n} x^{\frac {3}{2}} \,d x } \] Input:
integrate(x^(3/2)*(d*x+c)^n*(b*x^2+a),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)*(d*x + c)^n*x^(3/2), x)
\[ \int x^{3/2} (c+d x)^n \left (a+b x^2\right ) \, dx=\int { {\left (b x^{2} + a\right )} {\left (d x + c\right )}^{n} x^{\frac {3}{2}} \,d x } \] Input:
integrate(x^(3/2)*(d*x+c)^n*(b*x^2+a),x, algorithm="giac")
Output:
integrate((b*x^2 + a)*(d*x + c)^n*x^(3/2), x)
Timed out. \[ \int x^{3/2} (c+d x)^n \left (a+b x^2\right ) \, dx=\int x^{3/2}\,\left (b\,x^2+a\right )\,{\left (c+d\,x\right )}^n \,d x \] Input:
int(x^(3/2)*(a + b*x^2)*(c + d*x)^n,x)
Output:
int(x^(3/2)*(a + b*x^2)*(c + d*x)^n, x)
\[ \int x^{3/2} (c+d x)^n \left (a+b x^2\right ) \, dx=\text {too large to display} \] Input:
int(x^(3/2)*(d*x+c)^n*(b*x^2+a),x)
Output:
(2*( - 24*sqrt(x)*(c + d*x)**n*a*c**2*d**2*n**3 - 192*sqrt(x)*(c + d*x)**n *a*c**2*d**2*n**2 - 378*sqrt(x)*(c + d*x)**n*a*c**2*d**2*n + 16*sqrt(x)*(c + d*x)**n*a*c*d**3*n**4*x + 136*sqrt(x)*(c + d*x)**n*a*c*d**3*n**3*x + 31 6*sqrt(x)*(c + d*x)**n*a*c*d**3*n**2*x + 126*sqrt(x)*(c + d*x)**n*a*c*d**3 *n*x + 16*sqrt(x)*(c + d*x)**n*a*d**4*n**4*x**2 + 160*sqrt(x)*(c + d*x)**n *a*d**4*n**3*x**2 + 520*sqrt(x)*(c + d*x)**n*a*d**4*n**2*x**2 + 600*sqrt(x )*(c + d*x)**n*a*d**4*n*x**2 + 189*sqrt(x)*(c + d*x)**n*a*d**4*x**2 - 210* sqrt(x)*(c + d*x)**n*b*c**4*n + 140*sqrt(x)*(c + d*x)**n*b*c**3*d*n**2*x + 70*sqrt(x)*(c + d*x)**n*b*c**3*d*n*x - 56*sqrt(x)*(c + d*x)**n*b*c**2*d** 2*n**3*x**2 - 112*sqrt(x)*(c + d*x)**n*b*c**2*d**2*n**2*x**2 - 42*sqrt(x)* (c + d*x)**n*b*c**2*d**2*n*x**2 + 16*sqrt(x)*(c + d*x)**n*b*c*d**3*n**4*x* *3 + 72*sqrt(x)*(c + d*x)**n*b*c*d**3*n**3*x**3 + 92*sqrt(x)*(c + d*x)**n* b*c*d**3*n**2*x**3 + 30*sqrt(x)*(c + d*x)**n*b*c*d**3*n*x**3 + 16*sqrt(x)* (c + d*x)**n*b*d**4*n**4*x**4 + 128*sqrt(x)*(c + d*x)**n*b*d**4*n**3*x**4 + 344*sqrt(x)*(c + d*x)**n*b*d**4*n**2*x**4 + 352*sqrt(x)*(c + d*x)**n*b*d **4*n*x**4 + 105*sqrt(x)*(c + d*x)**n*b*d**4*x**4 + 384*int((sqrt(x)*(c + d*x)**n)/(32*c*n**5*x + 400*c*n**4*x + 1840*c*n**3*x + 3800*c*n**2*x + 337 8*c*n*x + 945*c*x + 32*d*n**5*x**2 + 400*d*n**4*x**2 + 1840*d*n**3*x**2 + 3800*d*n**2*x**2 + 3378*d*n*x**2 + 945*d*x**2),x)*a*c**3*d**2*n**8 + 7872* int((sqrt(x)*(c + d*x)**n)/(32*c*n**5*x + 400*c*n**4*x + 1840*c*n**3*x ...