Integrand size = 24, antiderivative size = 88 \[ \int \sqrt {c+d x} \left (c^2-d^2 x^2\right )^p \, dx=-\frac {2^{\frac {1}{2}+p} \sqrt {c+d x} \left (1+\frac {d x}{c}\right )^{-\frac {3}{2}-p} \left (c^2-d^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{c d (1+p)} \] Output:
-2^(1/2+p)*(d*x+c)^(1/2)*(1+d*x/c)^(-3/2-p)*(-d^2*x^2+c^2)^(p+1)*hypergeom ([p+1, -1/2-p],[2+p],1/2*(-d*x+c)/c)/c/d/(p+1)
Time = 0.54 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.01 \[ \int \sqrt {c+d x} \left (c^2-d^2 x^2\right )^p \, dx=\frac {2^{\frac {1}{2}+p} (-c+d x) \sqrt {c+d x} \left (1+\frac {d x}{c}\right )^{-\frac {1}{2}-p} \left (c^2-d^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{d (1+p)} \] Input:
Integrate[Sqrt[c + d*x]*(c^2 - d^2*x^2)^p,x]
Output:
(2^(1/2 + p)*(-c + d*x)*Sqrt[c + d*x]*(1 + (d*x)/c)^(-1/2 - p)*(c^2 - d^2* x^2)^p*Hypergeometric2F1[-1/2 - p, 1 + p, 2 + p, (c - d*x)/(2*c)])/(d*(1 + p))
Time = 0.36 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {474, 473, 79}
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.
\(\displaystyle \int \sqrt {c+d x} \left (c^2-d^2 x^2\right )^p \, dx\) |
\(\Big \downarrow \) 474 |
\(\displaystyle \frac {\sqrt {c+d x} \int \sqrt {\frac {d x}{c}+1} \left (c^2-d^2 x^2\right )^pdx}{\sqrt {\frac {d x}{c}+1}}\) |
\(\Big \downarrow \) 473 |
\(\displaystyle \sqrt {c+d x} \left (\frac {d x}{c}+1\right )^{-p-\frac {3}{2}} \left (c^2-c d x\right )^{-p-1} \left (c^2-d^2 x^2\right )^{p+1} \int \left (\frac {d x}{c}+1\right )^{p+\frac {1}{2}} \left (c^2-c d x\right )^pdx\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {2^{p+\frac {1}{2}} \sqrt {c+d x} \left (\frac {d x}{c}+1\right )^{-p-\frac {3}{2}} \left (c^2-d^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p-\frac {1}{2},p+1,p+2,\frac {c-d x}{2 c}\right )}{c d (p+1)}\) |
Input:
Int[Sqrt[c + d*x]*(c^2 - d^2*x^2)^p,x]
Output:
-((2^(1/2 + p)*Sqrt[c + d*x]*(1 + (d*x)/c)^(-3/2 - p)*(c^2 - d^2*x^2)^(1 + p)*Hypergeometric2F1[-1/2 - p, 1 + p, 2 + p, (c - d*x)/(2*c)])/(c*d*(1 + p)))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 1))) Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) && !Gt Q[a, 0] && !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(1 + d *(x/c))^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && !(IntegerQ[n] || GtQ[c, 0])
\[\int \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p}d x\]
Input:
int((d*x+c)^(1/2)*(-d^2*x^2+c^2)^p,x)
Output:
int((d*x+c)^(1/2)*(-d^2*x^2+c^2)^p,x)
\[ \int \sqrt {c+d x} \left (c^2-d^2 x^2\right )^p \, dx=\int { \sqrt {d x + c} {\left (-d^{2} x^{2} + c^{2}\right )}^{p} \,d x } \] Input:
integrate((d*x+c)^(1/2)*(-d^2*x^2+c^2)^p,x, algorithm="fricas")
Output:
integral(sqrt(d*x + c)*(-d^2*x^2 + c^2)^p, x)
\[ \int \sqrt {c+d x} \left (c^2-d^2 x^2\right )^p \, dx=\int \left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{p} \sqrt {c + d x}\, dx \] Input:
integrate((d*x+c)**(1/2)*(-d**2*x**2+c**2)**p,x)
Output:
Integral((-(-c + d*x)*(c + d*x))**p*sqrt(c + d*x), x)
\[ \int \sqrt {c+d x} \left (c^2-d^2 x^2\right )^p \, dx=\int { \sqrt {d x + c} {\left (-d^{2} x^{2} + c^{2}\right )}^{p} \,d x } \] Input:
integrate((d*x+c)^(1/2)*(-d^2*x^2+c^2)^p,x, algorithm="maxima")
Output:
integrate(sqrt(d*x + c)*(-d^2*x^2 + c^2)^p, x)
\[ \int \sqrt {c+d x} \left (c^2-d^2 x^2\right )^p \, dx=\int { \sqrt {d x + c} {\left (-d^{2} x^{2} + c^{2}\right )}^{p} \,d x } \] Input:
integrate((d*x+c)^(1/2)*(-d^2*x^2+c^2)^p,x, algorithm="giac")
Output:
integrate(sqrt(d*x + c)*(-d^2*x^2 + c^2)^p, x)
Timed out. \[ \int \sqrt {c+d x} \left (c^2-d^2 x^2\right )^p \, dx=\int {\left (c^2-d^2\,x^2\right )}^p\,\sqrt {c+d\,x} \,d x \] Input:
int((c^2 - d^2*x^2)^p*(c + d*x)^(1/2),x)
Output:
int((c^2 - d^2*x^2)^p*(c + d*x)^(1/2), x)
\[ \int \sqrt {c+d x} \left (c^2-d^2 x^2\right )^p \, dx=\frac {8 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} c p +2 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} c +2 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} d x +64 \left (\int \frac {\sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-4 d^{2} p \,x^{2}-3 d^{2} x^{2}+4 c^{2} p +3 c^{2}}d x \right ) c \,d^{2} p^{3}+80 \left (\int \frac {\sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-4 d^{2} p \,x^{2}-3 d^{2} x^{2}+4 c^{2} p +3 c^{2}}d x \right ) c \,d^{2} p^{2}+24 \left (\int \frac {\sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-4 d^{2} p \,x^{2}-3 d^{2} x^{2}+4 c^{2} p +3 c^{2}}d x \right ) c \,d^{2} p}{d \left (4 p +3\right )} \] Input:
int((d*x+c)^(1/2)*(-d^2*x^2+c^2)^p,x)
Output:
(2*(4*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*c*p + sqrt(c + d*x)*(c**2 - d**2 *x**2)**p*c + sqrt(c + d*x)*(c**2 - d**2*x**2)**p*d*x + 32*int((sqrt(c + d *x)*(c**2 - d**2*x**2)**p*x)/(4*c**2*p + 3*c**2 - 4*d**2*p*x**2 - 3*d**2*x **2),x)*c*d**2*p**3 + 40*int((sqrt(c + d*x)*(c**2 - d**2*x**2)**p*x)/(4*c* *2*p + 3*c**2 - 4*d**2*p*x**2 - 3*d**2*x**2),x)*c*d**2*p**2 + 12*int((sqrt (c + d*x)*(c**2 - d**2*x**2)**p*x)/(4*c**2*p + 3*c**2 - 4*d**2*p*x**2 - 3* d**2*x**2),x)*c*d**2*p))/(d*(4*p + 3))