\(\int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx\) [68]

Optimal result

Integrand size = 25, antiderivative size = 135 \[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=-\frac {d x \sqrt {1+\frac {b x^n}{a}} \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},1,1+\frac {1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) \sqrt {a+b x^n}}+\frac {c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{a c^2-d^2} \] Output:

-d*x*(1+b*x^n/a)^(1/2)*AppellF1(1/n,1/2,1,1+1/n,-b*x^n/a,-b*c^2*x^n/(a*c^2 
-d^2))/(a*c^2-d^2)/(a+b*x^n)^(1/2)+c*x*hypergeom([1, 1/n],[1+1/n],-b*c^2*x 
^n/(a*c^2-d^2))/(a*c^2-d^2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(320\) vs. \(2(135)=270\).

Time = 1.05 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.37 \[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=-\frac {2 a d \left (a c^2-d^2\right ) (1+n) x \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},1,1+\frac {1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{\sqrt {a+b x^n} \left (a c^2-d^2+b c^2 x^n\right ) \left (-2 a b c^2 n x^n \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {1}{2},2,2+\frac {1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )+\left (a c^2-d^2\right ) \left (-b n x^n \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {3}{2},1,2+\frac {1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )+2 a (1+n) \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},1,1+\frac {1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )\right )\right )}+\frac {c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{a c^2-d^2} \] Input:

Integrate[(a*c + b*c*x^n + d*Sqrt[a + b*x^n])^(-1),x]
 

Output:

(-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)
 

Rubi [A] (verified)

Time = 0.55 (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

\(\displaystyle \frac {c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{a c^2-d^2}-\frac {d x \sqrt {\frac {b x^n}{a}+1} \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},1,1+\frac {1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) \sqrt {a+b x^n}}\)

Input:

Int[(a*c + b*c*x^n + d*Sqrt[a + b*x^n])^(-1),x]
 

Output:

-((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)
 

Defintions of rubi rules used

rule 778

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])
 

rule 936

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])
 

rule 937

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])
 

rule 2587

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]
 

Maple [F]

\[\int \frac {1}{a c +b c \,x^{n}+d \sqrt {a +b \,x^{n}}}d x\]

Input:

int(1/(a*c+b*c*x^n+d*(a+b*x^n)^(1/2)),x)
 

Output:

int(1/(a*c+b*c*x^n+d*(a+b*x^n)^(1/2)),x)
 

Fricas [F]

\[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int { \frac {1}{b c x^{n} + a c + \sqrt {b x^{n} + a} d} \,d x } \] Input:

integrate(1/(a*c+b*c*x^n+d*(a+b*x^n)^(1/2)),x, algorithm="fricas")
 

Output:

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)
 

Sympy [F]

\[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int \frac {1}{a c + b c x^{n} + d \sqrt {a + b x^{n}}}\, dx \] Input:

integrate(1/(a*c+b*c*x**n+d*(a+b*x**n)**(1/2)),x)
 

Output:

Integral(1/(a*c + b*c*x**n + d*sqrt(a + b*x**n)), x)
 

Maxima [F]

\[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int { \frac {1}{b c x^{n} + a c + \sqrt {b x^{n} + a} d} \,d x } \] Input:

integrate(1/(a*c+b*c*x^n+d*(a+b*x^n)^(1/2)),x, algorithm="maxima")
 

Output:

integrate(1/(b*c*x^n + a*c + sqrt(b*x^n + a)*d), x)
 

Giac [F]

\[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int { \frac {1}{b c x^{n} + a c + \sqrt {b x^{n} + a} d} \,d x } \] Input:

integrate(1/(a*c+b*c*x^n+d*(a+b*x^n)^(1/2)),x, algorithm="giac")
 

Output:

integrate(1/(b*c*x^n + a*c + sqrt(b*x^n + a)*d), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\int \frac {1}{a\,c+d\,\sqrt {a+b\,x^n}+b\,c\,x^n} \,d x \] Input:

int(1/(a*c + d*(a + b*x^n)^(1/2) + b*c*x^n),x)
 

Output:

int(1/(a*c + d*(a + b*x^n)^(1/2) + b*c*x^n), x)
 

Reduce [F]

\[ \int \frac {1}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\left (\int \frac {x^{n}}{x^{2 n} b^{2} c^{2}+2 x^{n} a b \,c^{2}-x^{n} b \,d^{2}+a^{2} c^{2}-a \,d^{2}}d x \right ) b c -\left (\int \frac {\sqrt {x^{n} b +a}}{x^{2 n} b^{2} c^{2}+2 x^{n} a b \,c^{2}-x^{n} b \,d^{2}+a^{2} c^{2}-a \,d^{2}}d x \right ) d +\left (\int \frac {1}{x^{2 n} b^{2} c^{2}+2 x^{n} a b \,c^{2}-x^{n} b \,d^{2}+a^{2} c^{2}-a \,d^{2}}d x \right ) a c \] Input:

int(1/(a*c+b*c*x^n+d*(a+b*x^n)^(1/2)),x)
 

Output:

int(x**n/(x**(2*n)*b**2*c**2 + 2*x**n*a*b*c**2 - x**n*b*d**2 + a**2*c**2 - 
 a*d**2),x)*b*c - int(sqrt(x**n*b + a)/(x**(2*n)*b**2*c**2 + 2*x**n*a*b*c* 
*2 - x**n*b*d**2 + a**2*c**2 - a*d**2),x)*d + int(1/(x**(2*n)*b**2*c**2 + 
2*x**n*a*b*c**2 - x**n*b*d**2 + a**2*c**2 - a*d**2),x)*a*c