3.58 \(\int \frac {1}{(d+e x^n) \sqrt {a+c x^{2 n}}} \, dx\)

Optimal. Leaf size=171 \[ \frac {x \sqrt {\frac {c x^{2 n}}{a}+1} F_1\left (\frac {1}{2 n};\frac {1}{2},1;\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d \sqrt {a+c x^{2 n}}}-\frac {e x^{n+1} \sqrt {\frac {c x^{2 n}}{a}+1} F_1\left (\frac {n+1}{2 n};\frac {1}{2},1;\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d^2 (n+1) \sqrt {a+c x^{2 n}}} \]

[Out]

x*AppellF1(1/2/n,1,1/2,1+1/2/n,e^2*x^(2*n)/d^2,-c*x^(2*n)/a)*(1+c*x^(2*n)/a)^(1/2)/d/(a+c*x^(2*n))^(1/2)-e*x^(
1+n)*AppellF1(1/2*(1+n)/n,1,1/2,3/2+1/2/n,e^2*x^(2*n)/d^2,-c*x^(2*n)/a)*(1+c*x^(2*n)/a)^(1/2)/d^2/(1+n)/(a+c*x
^(2*n))^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1438, 430, 429, 511, 510} \[ \frac {x \sqrt {\frac {c x^{2 n}}{a}+1} F_1\left (\frac {1}{2 n};\frac {1}{2},1;\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d \sqrt {a+c x^{2 n}}}-\frac {e x^{n+1} \sqrt {\frac {c x^{2 n}}{a}+1} F_1\left (\frac {n+1}{2 n};\frac {1}{2},1;\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d^2 (n+1) \sqrt {a+c x^{2 n}}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x^n)*Sqrt[a + c*x^(2*n)]),x]

[Out]

(x*Sqrt[1 + (c*x^(2*n))/a]*AppellF1[1/(2*n), 1/2, 1, (2 + n^(-1))/2, -((c*x^(2*n))/a), (e^2*x^(2*n))/d^2])/(d*
Sqrt[a + c*x^(2*n)]) - (e*x^(1 + n)*Sqrt[1 + (c*x^(2*n))/a]*AppellF1[(1 + n)/(2*n), 1/2, 1, (3 + n^(-1))/2, -(
(c*x^(2*n))/a), (e^2*x^(2*n))/d^2])/(d^2*(1 + n)*Sqrt[a + c*x^(2*n)])

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 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])

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 1438

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^(2
*n))^p, (d/(d^2 - e^2*x^(2*n)) - (e*x^n)/(d^2 - e^2*x^(2*n)))^(-q), x], x] /; FreeQ[{a, c, d, e, n, p}, x] &&
EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[q, 0]

Rubi steps

\begin {align*} \int \frac {1}{\left (d+e x^n\right ) \sqrt {a+c x^{2 n}}} \, dx &=\int \left (\frac {d}{\sqrt {a+c x^{2 n}} \left (d^2-e^2 x^{2 n}\right )}+\frac {e x^n}{\sqrt {a+c x^{2 n}} \left (-d^2+e^2 x^{2 n}\right )}\right ) \, dx\\ &=d \int \frac {1}{\sqrt {a+c x^{2 n}} \left (d^2-e^2 x^{2 n}\right )} \, dx+e \int \frac {x^n}{\sqrt {a+c x^{2 n}} \left (-d^2+e^2 x^{2 n}\right )} \, dx\\ &=\frac {\left (d \sqrt {1+\frac {c x^{2 n}}{a}}\right ) \int \frac {1}{\sqrt {1+\frac {c x^{2 n}}{a}} \left (d^2-e^2 x^{2 n}\right )} \, dx}{\sqrt {a+c x^{2 n}}}+\frac {\left (e \sqrt {1+\frac {c x^{2 n}}{a}}\right ) \int \frac {x^n}{\sqrt {1+\frac {c x^{2 n}}{a}} \left (-d^2+e^2 x^{2 n}\right )} \, dx}{\sqrt {a+c x^{2 n}}}\\ &=\frac {x \sqrt {1+\frac {c x^{2 n}}{a}} F_1\left (\frac {1}{2 n};\frac {1}{2},1;\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d \sqrt {a+c x^{2 n}}}-\frac {e x^{1+n} \sqrt {1+\frac {c x^{2 n}}{a}} F_1\left (\frac {1+n}{2 n};\frac {1}{2},1;\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d^2 (1+n) \sqrt {a+c x^{2 n}}}\\ \end {align*}

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Mathematica [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d+e x^n\right ) \sqrt {a+c x^{2 n}}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[1/((d + e*x^n)*Sqrt[a + c*x^(2*n)]),x]

[Out]

Integrate[1/((d + e*x^n)*Sqrt[a + c*x^(2*n)]), x]

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fricas [F]  time = 1.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2 \, n} + a}}{a e x^{n} + a d + {\left (c e x^{n} + c d\right )} x^{2 \, n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^(2*n) + a)/(a*e*x^n + a*d + (c*e*x^n + c*d)*x^(2*n)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{2 \, n} + a} {\left (e x^{n} + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(c*x^(2*n) + a)*(e*x^n + d)), x)

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \,x^{n}+d \right ) \sqrt {c \,x^{2 n}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{2 \, n} + a} {\left (e x^{n} + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(c*x^(2*n) + a)*(e*x^n + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {a+c\,x^{2\,n}}\,\left (d+e\,x^n\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + c x^{2 n}} \left (d + e x^{n}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(a + c*x**(2*n))*(d + e*x**n)), x)

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