Integrand size = 25, antiderivative size = 135 \[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=-\frac {d x \sqrt {1+\frac {b x^n}{a}} \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},1,1+\frac {1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) \sqrt {a+b x^n}}+\frac {c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{a c^2-d^2} \]
c*x*hypergeom([1, 1/n],[1+1/n],-b*c^2*x^n/(a*c^2-d^2))/(a*c^2-d^2)-d*x*App ellF1(1/n,1/2,1,1+1/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)/(a+b*x^n)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(320\) vs. \(2(135)=270\).
Time = 0.55 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.37 \[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=-\frac {2 a d \left (a c^2-d^2\right ) (1+n) x \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},1,1+\frac {1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{\sqrt {a+b x^n} \left (a c^2-d^2+b c^2 x^n\right ) \left (-2 a b c^2 n x^n \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {1}{2},2,2+\frac {1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )+\left (a c^2-d^2\right ) \left (-b n x^n \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {3}{2},1,2+\frac {1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )+2 a (1+n) \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},1,1+\frac {1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )\right )\right )}+\frac {c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{a c^2-d^2} \]
(-2*a*d*(a*c^2 - d^2)*(1 + n)*x*AppellF1[n^(-1), 1/2, 1, 1 + n^(-1), -((b* x^n)/a), -((b*c^2*x^n)/(a*c^2 - d^2))])/(Sqrt[a + b*x^n]*(a*c^2 - d^2 + b* c^2*x^n)*(-2*a*b*c^2*n*x^n*AppellF1[1 + n^(-1), 1/2, 2, 2 + n^(-1), -((b*x ^n)/a), -((b*c^2*x^n)/(a*c^2 - d^2))] + (a*c^2 - d^2)*(-(b*n*x^n*AppellF1[ 1 + n^(-1), 3/2, 1, 2 + n^(-1), -((b*x^n)/a), -((b*c^2*x^n)/(a*c^2 - d^2)) ]) + 2*a*(1 + n)*AppellF1[n^(-1), 1/2, 1, 1 + n^(-1), -((b*x^n)/a), -((b*c ^2*x^n)/(a*c^2 - d^2))]))) + (c*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b*c^2*x^n)/(a*c^2 - d^2))])/(a*c^2 - d^2)
Time = 0.33 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2587, 778, 937, 936}
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.
\(\displaystyle \int \frac {1}{d \sqrt {a+b x^n}+a c+b c x^n} \, dx\) |
\(\Big \downarrow \) 2587 |
\(\displaystyle a c \int \frac {1}{a b c^2 x^n+a \left (a c^2-d^2\right )}dx-a d \int \frac {1}{\sqrt {b x^n+a} \left (a b c^2 x^n+a \left (a c^2-d^2\right )\right )}dx\) |
\(\Big \downarrow \) 778 |
\(\displaystyle \frac {c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{a c^2-d^2}-a d \int \frac {1}{\sqrt {b x^n+a} \left (a b c^2 x^n+a \left (a c^2-d^2\right )\right )}dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{a c^2-d^2}-\frac {a d \sqrt {\frac {b x^n}{a}+1} \int \frac {1}{\sqrt {\frac {b x^n}{a}+1} \left (a b c^2 x^n+a \left (a c^2-d^2\right )\right )}dx}{\sqrt {a+b x^n}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{a c^2-d^2}-\frac {d x \sqrt {\frac {b x^n}{a}+1} \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},1,1+\frac {1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) \sqrt {a+b x^n}}\) |
-((d*x*Sqrt[1 + (b*x^n)/a]*AppellF1[n^(-1), 1/2, 1, 1 + n^(-1), -((b*x^n)/ a), -((b*c^2*x^n)/(a*c^2 - d^2))])/((a*c^2 - d^2)*Sqrt[a + b*x^n])) + (c*x *Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b*c^2*x^n)/(a*c^2 - d^2))])/( a*c^2 - d^2)
3.6.62.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] && !IGtQ[p , 0] && !IntegerQ[1/n] && !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || GtQ[a, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) ], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((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[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q }, x] && NeQ[b*c - a*d, 0] && NeQ[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]
\[\int \frac {1}{a c +b c \,x^{n}+d \sqrt {a +b \,x^{n}}}d x\]
\[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int { \frac {1}{b c x^{n} + a c + \sqrt {b x^{n} + a} d} \,d x } \]
integral((b*c*x^n + a*c - sqrt(b*x^n + a)*d)/(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)
\[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int \frac {1}{a c + b c x^{n} + d \sqrt {a + b x^{n}}}\, dx \]
\[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int { \frac {1}{b c x^{n} + a c + \sqrt {b x^{n} + a} d} \,d x } \]
\[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int { \frac {1}{b c x^{n} + a c + \sqrt {b x^{n} + a} d} \,d x } \]
Timed out. \[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int \frac {1}{a\,c+d\,\sqrt {a+b\,x^n}+b\,c\,x^n} \,d x \]