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

Optimal. Leaf size=135 \[ \frac {c x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b c^2 x^n}{a c^2-d^2}\right )}{a c^2-d^2}-\frac {d x \sqrt {\frac {b x^n}{a}+1} F_1\left (\frac {1}{n};\frac {1}{2},1;1+\frac {1}{n};-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) \sqrt {a+b x^n}} \]

[Out]

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*AppellF1(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)

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2156, 245, 430, 429} \[ \frac {c x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b c^2 x^n}{a c^2-d^2}\right )}{a c^2-d^2}-\frac {d x \sqrt {\frac {b x^n}{a}+1} F_1\left (\frac {1}{n};\frac {1}{2},1;1+\frac {1}{n};-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) \sqrt {a+b x^n}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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)

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-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 429

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 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[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 2156

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

________________________________________________________________________________________

Mathematica [B]  time = 0.60, size = 320, normalized size = 2.37 \[ \frac {c x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b c^2 x^n}{a c^2-d^2}\right )}{a c^2-d^2}-\frac {2 a d (n+1) x \left (a c^2-d^2\right ) F_1\left (\frac {1}{n};\frac {1}{2},1;1+\frac {1}{n};-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )}{\sqrt {a+b x^n} \left (a c^2+b c^2 x^n-d^2\right ) \left (\left (a c^2-d^2\right ) \left (2 a (n+1) F_1\left (\frac {1}{n};\frac {1}{2},1;1+\frac {1}{n};-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )-b n x^n F_1\left (1+\frac {1}{n};\frac {3}{2},1;2+\frac {1}{n};-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )\right )-2 a b c^2 n x^n F_1\left (1+\frac {1}{n};\frac {1}{2},2;2+\frac {1}{n};-\frac {b x^n}{a},-\frac {b c^2 x^n}{a c^2-d^2}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-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)

________________________________________________________________________________________

fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b c x^{n} + a c - \sqrt {b x^{n} + a} d}{b^{2} c^{2} x^{2 \, n} + a^{2} c^{2} - a d^{2} + {\left (2 \, a b c^{2} - b d^{2}\right )} x^{n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{b c x^{n} + a c + \sqrt {b x^{n} + a} d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \frac {1}{b c \,x^{n}+a c +\sqrt {b \,x^{n}+a}\, d}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{b c x^{n} + a c + \sqrt {b x^{n} + a} d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{a\,c+d\,\sqrt {a+b\,x^n}+b\,c\,x^n} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a c + b c x^{n} + d \sqrt {a + b x^{n}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________