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

Optimal. Leaf size=167 \[ \frac{c x^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b c^2 x^n}{a c^2-d^2}\right )}{(m+1) \left (a c^2-d^2\right )}-\frac{d x^{m+1} \sqrt{\frac{b x^n}{a}+1} F_1\left (\frac{m+1}{n};\frac{1}{2},1;\frac{m+n+1}{n};-\frac{b x^n}{a},-\frac{b c^2 x^n}{a c^2-d^2}\right )}{(m+1) \left (a c^2-d^2\right ) \sqrt{a+b x^n}} \]

[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))

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Rubi [A]  time = 0.196985, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2156, 364, 511, 510} \[ \frac{c x^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b c^2 x^n}{a c^2-d^2}\right )}{(m+1) \left (a c^2-d^2\right )}-\frac{d x^{m+1} \sqrt{\frac{b x^n}{a}+1} F_1\left (\frac{m+1}{n};\frac{1}{2},1;\frac{m+n+1}{n};-\frac{b x^n}{a},-\frac{b c^2 x^n}{a c^2-d^2}\right )}{(m+1) \left (a c^2-d^2\right ) \sqrt{a+b x^n}} \]

Antiderivative was successfully verified.

[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 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]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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])

Rubi steps

\begin{align*} \int \frac{x^m}{a c+b c x^n+d \sqrt{a+b x^n}} \, dx &=(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]  time = 0.292834, size = 156, normalized size = 0.93 \[ \frac{x^{m+1} \left (c \sqrt{a+b x^n} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b c^2 x^n}{a c^2-d^2}\right )-d \sqrt{\frac{b x^n}{a}+1} F_1\left (\frac{m+1}{n};\frac{1}{2},1;\frac{m+n+1}{n};-\frac{b x^n}{a},-\frac{b c^2 x^n}{a c^2-d^2}\right )\right )}{(m+1) \left (a c^2-d^2\right ) \sqrt{a+b x^n}} \]

Antiderivative was successfully verified.

[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])

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Maple [F]  time = 0.011, size = 0, normalized size = 0. \begin{align*} \int{{x}^{m} \left ( ac+bc{x}^{n}+d\sqrt{a+b{x}^{n}} \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{b c x^{n} + a c + \sqrt{b x^{n} + a} d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{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} +{\left (2 \, a b c^{2} - b d^{2}\right )} x^{n}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{a c + b c x^{n} + d \sqrt{a + b x^{n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{b c x^{n} + a c + \sqrt{b x^{n} + a} d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)