Integrand size = 29, antiderivative size = 167 \[ \int \frac {x^m}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=-\frac {d x^{1+m} \sqrt {1+\frac {b x^n}{a}} \operatorname {AppellF1}\left (\frac {1+m}{n},\frac {1}{2},1,\frac {1+m+n}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) (1+m) \sqrt {a+b x^n}}+\frac {c x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) (1+m)} \]
c*x^(1+m)*hypergeom([1, (1+m)/n],[(1+m+n)/n],-b*c^2*x^n/(a*c^2-d^2))/(a*c^ 2-d^2)/(1+m)-d*x^(1+m)*AppellF1((1+m)/n,1/2,1,(1+m+n)/n,-b*x^n/a,-b*c^2*x^ n/(a*c^2-d^2))*(1+b*x^n/a)^(1/2)/(a*c^2-d^2)/(1+m)/(a+b*x^n)^(1/2)
Time = 0.41 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.93 \[ \int \frac {x^m}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\frac {x^{1+m} \left (-d \sqrt {1+\frac {b x^n}{a}} \operatorname {AppellF1}\left (\frac {1+m}{n},\frac {1}{2},1,\frac {1+m+n}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )+c \sqrt {a+b x^n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )\right )}{\left (a c^2-d^2\right ) (1+m) \sqrt {a+b x^n}} \]
(x^(1 + m)*(-(d*Sqrt[1 + (b*x^n)/a]*AppellF1[(1 + m)/n, 1/2, 1, (1 + m + n )/n, -((b*x^n)/a), -((b*c^2*x^n)/(a*c^2 - d^2))]) + c*Sqrt[a + b*x^n]*Hype rgeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((b*c^2*x^n)/(a*c^2 - d^2))])) /((a*c^2 - d^2)*(1 + m)*Sqrt[a + b*x^n])
Time = 0.42 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2587, 888, 1013, 1012}
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.
\(\displaystyle \int \frac {x^m}{d \sqrt {a+b x^n}+a c+b c x^n} \, dx\) |
\(\Big \downarrow \) 2587 |
\(\displaystyle a c \int \frac {x^m}{a b c^2 x^n+a \left (a c^2-d^2\right )}dx-a d \int \frac {x^m}{\sqrt {b x^n+a} \left (a b c^2 x^n+a \left (a c^2-d^2\right )\right )}dx\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {c x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{(m+1) \left (a c^2-d^2\right )}-a d \int \frac {x^m}{\sqrt {b x^n+a} \left (a b c^2 x^n+a \left (a c^2-d^2\right )\right )}dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {c x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{(m+1) \left (a c^2-d^2\right )}-\frac {a d \sqrt {\frac {b x^n}{a}+1} \int \frac {x^m}{\sqrt {\frac {b x^n}{a}+1} \left (a b c^2 x^n+a \left (a c^2-d^2\right )\right )}dx}{\sqrt {a+b x^n}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {c x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{(m+1) \left (a c^2-d^2\right )}-\frac {d x^{m+1} \sqrt {\frac {b x^n}{a}+1} \operatorname {AppellF1}\left (\frac {m+1}{n},\frac {1}{2},1,\frac {m+n+1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{(m+1) \left (a c^2-d^2\right ) \sqrt {a+b x^n}}\) |
-((d*x^(1 + m)*Sqrt[1 + (b*x^n)/a]*AppellF1[(1 + m)/n, 1/2, 1, (1 + m + n) /n, -((b*x^n)/a), -((b*c^2*x^n)/(a*c^2 - d^2))])/((a*c^2 - d^2)*(1 + m)*Sq rt[a + b*x^n])) + (c*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n) /n, -((b*c^2*x^n)/(a*c^2 - d^2))])/((a*c^2 - d^2)*(1 + m))
3.6.63.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ n/a))^FracPart[p]) Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & & NeQ[m, n - 1] && !(IntegerQ[p] || GtQ[a, 0])
Int[(u_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_ Symbol] :> Simp[c Int[u/(c^2 - a*e^2 + c*d*x^n), x], x] - Simp[a*e Int[ u/((c^2 - a*e^2 + c*d*x^n)*Sqrt[a + b*x^n]), x], x] /; FreeQ[{a, b, c, d, e , n}, x] && EqQ[b*c - a*d, 0]
\[\int \frac {x^{m}}{a c +b c \,x^{n}+d \sqrt {a +b \,x^{n}}}d x\]
\[ \int \frac {x^m}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int { \frac {x^{m}}{b c x^{n} + a c + \sqrt {b x^{n} + a} d} \,d x } \]
integral((b*c*x^m*x^n + a*c*x^m - sqrt(b*x^n + a)*d*x^m)/(b^2*c^2*x^(2*n) + a^2*c^2 - a*d^2 + (2*a*b*c^2 - b*d^2)*x^n), x)
\[ \int \frac {x^m}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int \frac {x^{m}}{a c + b c x^{n} + d \sqrt {a + b x^{n}}}\, dx \]
\[ \int \frac {x^m}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int { \frac {x^{m}}{b c x^{n} + a c + \sqrt {b x^{n} + a} d} \,d x } \]
\[ \int \frac {x^m}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int { \frac {x^{m}}{b c x^{n} + a c + \sqrt {b x^{n} + a} d} \,d x } \]
Timed out. \[ \int \frac {x^m}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int \frac {x^m}{a\,c+d\,\sqrt {a+b\,x^n}+b\,c\,x^n} \,d x \]