\(\int \frac {x^m}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx\) [563]

Optimal result

Integrand size = 29, antiderivative size = 167 \[ \int \frac {x^m}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=-\frac {d x^{1+m} \sqrt {1+\frac {b x^n}{a}} \operatorname {AppellF1}\left (\frac {1+m}{n},\frac {1}{2},1,\frac {1+m+n}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) (1+m) \sqrt {a+b x^n}}+\frac {c x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) (1+m)} \]

[Out]

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)

Rubi [A] (verified)

Time = 0.13 (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 = {2187, 371, 525, 524} \[ \int \frac {x^m}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\frac {c x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{(m+1) \left (a c^2-d^2\right )}-\frac {d x^{m+1} \sqrt {\frac {b x^n}{a}+1} \operatorname {AppellF1}\left (\frac {m+1}{n},\frac {1}{2},1,\frac {m+n+1}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{(m+1) \left (a c^2-d^2\right ) \sqrt {a+b x^n}} \]

[In]

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

[Out]

-((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)*Sqrt[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))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 524

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] /; Free
Q[{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])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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])

Rule 2187

Int[(u_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[c, Int[u/(c^2 - a*e
^2 + c*d*x^n), x], x] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = (a c) \int \frac {x^m}{a^2 c^2-a d^2+a b c^2 x^n} \, dx-(a d) \int \frac {x^m}{\sqrt {a+b x^n} \left (a^2 c^2-a d^2+a b c^2 x^n\right )} \, dx \\ & = \frac {c x^{1+m} \, _2F_1\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)}-\frac {\left (a d \sqrt {1+\frac {b x^n}{a}}\right ) \int \frac {x^m}{\sqrt {1+\frac {b x^n}{a}} \left (a^2 c^2-a d^2+a b c^2 x^n\right )} \, dx}{\sqrt {a+b x^n}} \\ & = -\frac {d x^{1+m} \sqrt {1+\frac {b x^n}{a}} F_1\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} \, _2F_1\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)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.93 \[ \int \frac {x^m}{a c+b c x^n+d \sqrt {a+b x^n}} \, dx=\frac {x^{1+m} \left (-d \sqrt {1+\frac {b x^n}{a}} \operatorname {AppellF1}\left (\frac {1+m}{n},\frac {1}{2},1,\frac {1+m+n}{n},-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )+c \sqrt {a+b x^n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b c^2 x^n}{a c^2-d^2}\right )\right )}{\left (a c^2-d^2\right ) (1+m) \sqrt {a+b x^n}} \]

[In]

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

[Out]

(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]*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)*Sqrt[a + b*x^n])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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