Integrand size = 20, antiderivative size = 198 \[ \int x \sqrt {c+d x} \left (a+b x^2\right )^p \, dx=\frac {\sqrt {c+d x} \left (a+b x^2\right )^{1+p}}{2 b (1+p)}-\frac {\sqrt {c+d x} \left (a+b x^2\right )^{1+p} \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-1-p} \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-1-p} \operatorname {AppellF1}\left (\frac {1}{2},-1-p,-1-p,\frac {3}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{2 b (1+p)} \] Output:
1/2*(d*x+c)^(1/2)*(b*x^2+a)^(p+1)/b/(p+1)-1/2*(d*x+c)^(1/2)*(b*x^2+a)^(p+1 )*(1-(d*x+c)/(c-(-a)^(1/2)*d/b^(1/2)))^(-1-p)*(1-(d*x+c)/(c+(-a)^(1/2)*d/b ^(1/2)))^(-1-p)*AppellF1(1/2,-1-p,-1-p,3/2,(d*x+c)/(c-(-a)^(1/2)*d/b^(1/2) ),(d*x+c)/(c+(-a)^(1/2)*d/b^(1/2)))/b/(p+1)
Time = 0.12 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.25 \[ \int x \sqrt {c+d x} \left (a+b x^2\right )^p \, dx=\frac {2 \left (\frac {d \left (\sqrt {-\frac {a b}{d^2}} d-b x\right )}{b c+\sqrt {-\frac {a b}{d^2}} d^2}\right )^{-p} (c+d x)^{3/2} \left (\frac {a-\sqrt {-\frac {a b}{d^2}} d x}{a+c \sqrt {-\frac {a b}{d^2}}}\right )^{-p} \left (a+b x^2\right )^p \left (-5 c \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {c+d x}{c+\frac {a}{\sqrt {-\frac {a b}{d^2}}}},\frac {b (c+d x)}{b c+\sqrt {-\frac {a b}{d^2}} d^2}\right )+3 (c+d x) \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},\frac {c+d x}{c+\frac {a}{\sqrt {-\frac {a b}{d^2}}}},\frac {b (c+d x)}{b c+\sqrt {-\frac {a b}{d^2}} d^2}\right )\right )}{15 d^2} \] Input:
Integrate[x*Sqrt[c + d*x]*(a + b*x^2)^p,x]
Output:
(2*(c + d*x)^(3/2)*(a + b*x^2)^p*(-5*c*AppellF1[3/2, -p, -p, 5/2, (c + d*x )/(c + a/Sqrt[-((a*b)/d^2)]), (b*(c + d*x))/(b*c + Sqrt[-((a*b)/d^2)]*d^2) ] + 3*(c + d*x)*AppellF1[5/2, -p, -p, 7/2, (c + d*x)/(c + a/Sqrt[-((a*b)/d ^2)]), (b*(c + d*x))/(b*c + Sqrt[-((a*b)/d^2)]*d^2)]))/(15*d^2*((d*(Sqrt[- ((a*b)/d^2)]*d - b*x))/(b*c + Sqrt[-((a*b)/d^2)]*d^2))^p*((a - Sqrt[-((a*b )/d^2)]*d*x)/(a + c*Sqrt[-((a*b)/d^2)]))^p)
Time = 0.68 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.80, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {596, 719, 514, 150}
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.
\(\displaystyle \int x \sqrt {c+d x} \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 596 |
\(\displaystyle \frac {2 \sqrt {c+d x} \left (a+b x^2\right )^{p+1}}{b (4 p+5)}-\frac {\int \frac {(a d-b c x) \left (b x^2+a\right )^p}{\sqrt {c+d x}}dx}{b (4 p+5)}\) |
\(\Big \downarrow \) 719 |
\(\displaystyle \frac {2 \sqrt {c+d x} \left (a+b x^2\right )^{p+1}}{b (4 p+5)}-\frac {\frac {\left (a d^2+b c^2\right ) \int \frac {\left (b x^2+a\right )^p}{\sqrt {c+d x}}dx}{d}-\frac {b c \int \sqrt {c+d x} \left (b x^2+a\right )^pdx}{d}}{b (4 p+5)}\) |
\(\Big \downarrow \) 514 |
\(\displaystyle \frac {2 \sqrt {c+d x} \left (a+b x^2\right )^{p+1}}{b (4 p+5)}-\frac {\frac {\left (a d^2+b c^2\right ) \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \int \frac {\left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p}{\sqrt {c+d x}}d(c+d x)}{d^2}-\frac {b c \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \int \sqrt {c+d x} \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^pd(c+d x)}{d^2}}{b (4 p+5)}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {2 \sqrt {c+d x} \left (a+b x^2\right )^{p+1}}{b (4 p+5)}-\frac {\frac {2 \sqrt {c+d x} \left (a d^2+b c^2\right ) \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{d^2}-\frac {2 b c (c+d x)^{3/2} \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{3 d^2}}{b (4 p+5)}\) |
Input:
Int[x*Sqrt[c + d*x]*(a + b*x^2)^p,x]
Output:
(2*Sqrt[c + d*x]*(a + b*x^2)^(1 + p))/(b*(5 + 4*p)) - ((2*(b*c^2 + a*d^2)* Sqrt[c + d*x]*(a + b*x^2)^p*AppellF1[1/2, -p, -p, 3/2, (c + d*x)/(c - (Sqr t[-a]*d)/Sqrt[b]), (c + d*x)/(c + (Sqrt[-a]*d)/Sqrt[b])])/(d^2*(1 - (c + d *x)/(c - (Sqrt[-a]*d)/Sqrt[b]))^p*(1 - (c + d*x)/(c + (Sqrt[-a]*d)/Sqrt[b] ))^p) - (2*b*c*(c + d*x)^(3/2)*(a + b*x^2)^p*AppellF1[3/2, -p, -p, 5/2, (c + d*x)/(c - (Sqrt[-a]*d)/Sqrt[b]), (c + d*x)/(c + (Sqrt[-a]*d)/Sqrt[b])]) /(3*d^2*(1 - (c + d*x)/(c - (Sqrt[-a]*d)/Sqrt[b]))^p*(1 - (c + d*x)/(c + ( Sqrt[-a]*d)/Sqrt[b]))^p))/(b*(5 + 4*p))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {q = Rt[-a/b, 2]}, Simp[(a + b*x^2)^p/(d*(1 - (c + d*x)/(c - d*q))^p*(1 - ( c + d*x)/(c + d*q))^p) Subst[Int[x^n*Simp[1 - x/(c + d*q), x]^p*Simp[1 - x/(c - d*q), x]^p, x], x, c + d*x], x]] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c^2 + a*d^2, 0]
Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c + d*x)^n*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 2))), x] - Simp[n/(b*(n + 2*p + 2)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^p*(a*d - b*c*x), x], x] /; FreeQ[{a, b, c, d, p}, x] && GtQ[n, 0] && NeQ[n + 2*p + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
\[\int x \sqrt {d x +c}\, \left (b \,x^{2}+a \right )^{p}d x\]
Input:
int(x*(d*x+c)^(1/2)*(b*x^2+a)^p,x)
Output:
int(x*(d*x+c)^(1/2)*(b*x^2+a)^p,x)
\[ \int x \sqrt {c+d x} \left (a+b x^2\right )^p \, dx=\int { \sqrt {d x + c} {\left (b x^{2} + a\right )}^{p} x \,d x } \] Input:
integrate(x*(d*x+c)^(1/2)*(b*x^2+a)^p,x, algorithm="fricas")
Output:
integral(sqrt(d*x + c)*(b*x^2 + a)^p*x, x)
\[ \int x \sqrt {c+d x} \left (a+b x^2\right )^p \, dx=\int x \left (a + b x^{2}\right )^{p} \sqrt {c + d x}\, dx \] Input:
integrate(x*(d*x+c)**(1/2)*(b*x**2+a)**p,x)
Output:
Integral(x*(a + b*x**2)**p*sqrt(c + d*x), x)
\[ \int x \sqrt {c+d x} \left (a+b x^2\right )^p \, dx=\int { \sqrt {d x + c} {\left (b x^{2} + a\right )}^{p} x \,d x } \] Input:
integrate(x*(d*x+c)^(1/2)*(b*x^2+a)^p,x, algorithm="maxima")
Output:
integrate(sqrt(d*x + c)*(b*x^2 + a)^p*x, x)
\[ \int x \sqrt {c+d x} \left (a+b x^2\right )^p \, dx=\int { \sqrt {d x + c} {\left (b x^{2} + a\right )}^{p} x \,d x } \] Input:
integrate(x*(d*x+c)^(1/2)*(b*x^2+a)^p,x, algorithm="giac")
Output:
integrate(sqrt(d*x + c)*(b*x^2 + a)^p*x, x)
Timed out. \[ \int x \sqrt {c+d x} \left (a+b x^2\right )^p \, dx=\int x\,{\left (b\,x^2+a\right )}^p\,\sqrt {c+d\,x} \,d x \] Input:
int(x*(a + b*x^2)^p*(c + d*x)^(1/2),x)
Output:
int(x*(a + b*x^2)^p*(c + d*x)^(1/2), x)
\[ \int x \sqrt {c+d x} \left (a+b x^2\right )^p \, dx=\text {too large to display} \] Input:
int(x*(d*x+c)^(1/2)*(b*x^2+a)^p,x)
Output:
(2*(4*sqrt(c + d*x)*(a + b*x**2)**p*a*d*p + 4*sqrt(c + d*x)*(a + b*x**2)** p*a*d + sqrt(c + d*x)*(a + b*x**2)**p*b*c*x + 4*sqrt(c + d*x)*(a + b*x**2) **p*b*d*p*x**2 + 3*sqrt(c + d*x)*(a + b*x**2)**p*b*d*x**2 - 64*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(16*a*c*p**2 + 32*a*c*p + 15*a*c + 16*a*d*p* *2*x + 32*a*d*p*x + 15*a*d*x + 16*b*c*p**2*x**2 + 32*b*c*p*x**2 + 15*b*c*x **2 + 16*b*d*p**2*x**3 + 32*b*d*p*x**3 + 15*b*d*x**3),x)*a*b*d**2*p**3 - 1 60*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(16*a*c*p**2 + 32*a*c*p + 15*a *c + 16*a*d*p**2*x + 32*a*d*p*x + 15*a*d*x + 16*b*c*p**2*x**2 + 32*b*c*p*x **2 + 15*b*c*x**2 + 16*b*d*p**2*x**3 + 32*b*d*p*x**3 + 15*b*d*x**3),x)*a*b *d**2*p**2 - 124*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(16*a*c*p**2 + 3 2*a*c*p + 15*a*c + 16*a*d*p**2*x + 32*a*d*p*x + 15*a*d*x + 16*b*c*p**2*x** 2 + 32*b*c*p*x**2 + 15*b*c*x**2 + 16*b*d*p**2*x**3 + 32*b*d*p*x**3 + 15*b* d*x**3),x)*a*b*d**2*p - 30*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(16*a* c*p**2 + 32*a*c*p + 15*a*c + 16*a*d*p**2*x + 32*a*d*p*x + 15*a*d*x + 16*b* c*p**2*x**2 + 32*b*c*p*x**2 + 15*b*c*x**2 + 16*b*d*p**2*x**3 + 32*b*d*p*x* *3 + 15*b*d*x**3),x)*a*b*d**2 - 32*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2 )/(16*a*c*p**2 + 32*a*c*p + 15*a*c + 16*a*d*p**2*x + 32*a*d*p*x + 15*a*d*x + 16*b*c*p**2*x**2 + 32*b*c*p*x**2 + 15*b*c*x**2 + 16*b*d*p**2*x**3 + 32* b*d*p*x**3 + 15*b*d*x**3),x)*b**2*c**2*p**3 - 80*int((sqrt(c + d*x)*(a + b *x**2)**p*x**2)/(16*a*c*p**2 + 32*a*c*p + 15*a*c + 16*a*d*p**2*x + 32*a...