Integrand size = 14, antiderivative size = 49 \[ \int \left (\frac {a+b+c x^2}{d}\right )^m \, dx=\frac {d x \left (\frac {a+b}{d}+\frac {c x^2}{d}\right )^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {3}{2}+m,\frac {3}{2},-\frac {c x^2}{a+b}\right )}{a+b} \] Output:
d*x*((a+b)/d+c*x^2/d)^(1+m)*hypergeom([1, 3/2+m],[3/2],-c*x^2/(a+b))/(a+b)
Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08 \[ \int \left (\frac {a+b+c x^2}{d}\right )^m \, dx=x \left (\frac {a+b+c x^2}{d}\right )^m \left (1+\frac {c x^2}{a+b}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},-\frac {c x^2}{a+b}\right ) \] Input:
Integrate[((a + b + c*x^2)/d)^m,x]
Output:
(x*((a + b + c*x^2)/d)^m*Hypergeometric2F1[1/2, -m, 3/2, -((c*x^2)/(a + b) )])/(1 + (c*x^2)/(a + b))^m
Time = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2072, 238, 237}
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.
\(\displaystyle \int \left (\frac {a+b+c x^2}{d}\right )^m \, dx\) |
\(\Big \downarrow \) 2072 |
\(\displaystyle \int \left (\frac {a+b}{d}+\frac {c x^2}{d}\right )^mdx\) |
\(\Big \downarrow \) 238 |
\(\displaystyle \left (\frac {c x^2}{a+b}+1\right )^{-m} \left (\frac {a+b}{d}+\frac {c x^2}{d}\right )^m \int \left (\frac {c x^2}{a+b}+1\right )^mdx\) |
\(\Big \downarrow \) 237 |
\(\displaystyle x \left (\frac {c x^2}{a+b}+1\right )^{-m} \left (\frac {a+b}{d}+\frac {c x^2}{d}\right )^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},-\frac {c x^2}{a+b}\right )\) |
Input:
Int[((a + b + c*x^2)/d)^m,x]
Output:
(x*((a + b)/d + (c*x^2)/d)^m*Hypergeometric2F1[1/2, -m, 3/2, -((c*x^2)/(a + b))])/(1 + (c*x^2)/(a + b))^m
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] && !IntegerQ[2*p ] && GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) ^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(1 + b*(x^2/a))^p, x], x] / ; FreeQ[{a, b, p}, x] && !IntegerQ[2*p] && !GtQ[a, 0]
Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && B inomialQ[u, x] && !BinomialMatchQ[u, x]
\[\int \left (\frac {c \,x^{2}+a +b}{d}\right )^{m}d x\]
Input:
int(((c*x^2+a+b)/d)^m,x)
Output:
int(((c*x^2+a+b)/d)^m,x)
\[ \int \left (\frac {a+b+c x^2}{d}\right )^m \, dx=\int { \left (\frac {c x^{2} + a + b}{d}\right )^{m} \,d x } \] Input:
integrate(((c*x^2+a+b)/d)^m,x, algorithm="fricas")
Output:
integral(((c*x^2 + a + b)/d)^m, x)
\[ \int \left (\frac {a+b+c x^2}{d}\right )^m \, dx=\int \left (\frac {a + b + c x^{2}}{d}\right )^{m}\, dx \] Input:
integrate(((c*x**2+a+b)/d)**m,x)
Output:
Integral(((a + b + c*x**2)/d)**m, x)
\[ \int \left (\frac {a+b+c x^2}{d}\right )^m \, dx=\int { \left (\frac {c x^{2} + a + b}{d}\right )^{m} \,d x } \] Input:
integrate(((c*x^2+a+b)/d)^m,x, algorithm="maxima")
Output:
integrate(((c*x^2 + a + b)/d)^m, x)
\[ \int \left (\frac {a+b+c x^2}{d}\right )^m \, dx=\int { \left (\frac {c x^{2} + a + b}{d}\right )^{m} \,d x } \] Input:
integrate(((c*x^2+a+b)/d)^m,x, algorithm="giac")
Output:
integrate(((c*x^2 + a + b)/d)^m, x)
Time = 23.47 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.10 \[ \int \left (\frac {a+b+c x^2}{d}\right )^m \, dx=\frac {x\,{\left (\frac {a+b}{d}+\frac {c\,x^2}{d}\right )}^m\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},-m;\ \frac {3}{2};\ -\frac {c\,x^2}{a+b}\right )}{{\left (\frac {c\,x^2}{a+b}+1\right )}^m} \] Input:
int(((a + b + c*x^2)/d)^m,x)
Output:
(x*((a + b)/d + (c*x^2)/d)^m*hypergeom([1/2, -m], 3/2, -(c*x^2)/(a + b)))/ ((c*x^2)/(a + b) + 1)^m
\[ \int \left (\frac {a+b+c x^2}{d}\right )^m \, dx=\frac {\left (c \,x^{2}+a +b \right )^{m} x +4 \left (\int \frac {\left (c \,x^{2}+a +b \right )^{m}}{2 c m \,x^{2}+c \,x^{2}+2 a m +2 b m +a +b}d x \right ) a \,m^{2}+2 \left (\int \frac {\left (c \,x^{2}+a +b \right )^{m}}{2 c m \,x^{2}+c \,x^{2}+2 a m +2 b m +a +b}d x \right ) a m +4 \left (\int \frac {\left (c \,x^{2}+a +b \right )^{m}}{2 c m \,x^{2}+c \,x^{2}+2 a m +2 b m +a +b}d x \right ) b \,m^{2}+2 \left (\int \frac {\left (c \,x^{2}+a +b \right )^{m}}{2 c m \,x^{2}+c \,x^{2}+2 a m +2 b m +a +b}d x \right ) b m}{d^{m} \left (2 m +1\right )} \] Input:
int(((c*x^2+a+b)/d)^m,x)
Output:
((a + b + c*x**2)**m*x + 4*int((a + b + c*x**2)**m/(2*a*m + a + 2*b*m + b + 2*c*m*x**2 + c*x**2),x)*a*m**2 + 2*int((a + b + c*x**2)**m/(2*a*m + a + 2*b*m + b + 2*c*m*x**2 + c*x**2),x)*a*m + 4*int((a + b + c*x**2)**m/(2*a*m + a + 2*b*m + b + 2*c*m*x**2 + c*x**2),x)*b*m**2 + 2*int((a + b + c*x**2) **m/(2*a*m + a + 2*b*m + b + 2*c*m*x**2 + c*x**2),x)*b*m)/(d**m*(2*m + 1))