3.11.69 \(\int \frac {x^{-1+2 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx\) [1069]

Optimal. Leaf size=89 \[ \frac {\sqrt {a+b x^n} \sqrt {c+d x^n}}{b d n}-\frac {(b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{b^{3/2} d^{3/2} n} \]

[Out]

-(a*d+b*c)*arctanh(d^(1/2)*(a+b*x^n)^(1/2)/b^(1/2)/(c+d*x^n)^(1/2))/b^(3/2)/d^(3/2)/n+(a+b*x^n)^(1/2)*(c+d*x^n
)^(1/2)/b/d/n

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Rubi [A]
time = 0.06, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {457, 81, 65, 223, 212} \begin {gather*} \frac {\sqrt {a+b x^n} \sqrt {c+d x^n}}{b d n}-\frac {(a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{b^{3/2} d^{3/2} n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^n]*Sqrt[c + d*x^n])/(b*d*n) - ((b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^n])/(Sqrt[b]*Sqrt[c + d
*x^n])])/(b^(3/2)*d^(3/2)*n)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^{-1+2 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx &=\frac {\text {Subst}\left (\int \frac {x}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^n\right )}{n}\\ &=\frac {\sqrt {a+b x^n} \sqrt {c+d x^n}}{b d n}-\frac {(b c+a d) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^n\right )}{2 b d n}\\ &=\frac {\sqrt {a+b x^n} \sqrt {c+d x^n}}{b d n}-\frac {(b c+a d) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x^n}\right )}{b^2 d n}\\ &=\frac {\sqrt {a+b x^n} \sqrt {c+d x^n}}{b d n}-\frac {(b c+a d) \text {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x^n}}{\sqrt {c+d x^n}}\right )}{b^2 d n}\\ &=\frac {\sqrt {a+b x^n} \sqrt {c+d x^n}}{b d n}-\frac {(b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{b^{3/2} d^{3/2} n}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 123, normalized size = 1.38 \begin {gather*} \frac {b \sqrt {d} \sqrt {a+b x^n} \left (c+d x^n\right )-\sqrt {b c-a d} (b c+a d) \sqrt {\frac {b \left (c+d x^n\right )}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b c-a d}}\right )}{b^2 d^{3/2} n \sqrt {c+d x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d]*Sqrt[a + b*x^n]*(c + d*x^n) - Sqrt[b*c - a*d]*(b*c + a*d)*Sqrt[(b*(c + d*x^n))/(b*c - a*d)]*ArcSinh
[(Sqrt[d]*Sqrt[a + b*x^n])/Sqrt[b*c - a*d]])/(b^2*d^(3/2)*n*Sqrt[c + d*x^n])

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \frac {x^{-1+2 n}}{\sqrt {a +b \,x^{n}}\, \sqrt {c +d \,x^{n}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.03, size = 281, normalized size = 3.16 \begin {gather*} \left [\frac {4 \, \sqrt {b x^{n} + a} \sqrt {d x^{n} + c} b d + {\left (b c + a d\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2 \, n} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, \sqrt {b d} b d x^{n} + {\left (b c + a d\right )} \sqrt {b d}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{n}\right )}{4 \, b^{2} d^{2} n}, \frac {2 \, \sqrt {b x^{n} + a} \sqrt {d x^{n} + c} b d + {\left (b c + a d\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, \sqrt {-b d} b d x^{n} + {\left (b c + a d\right )} \sqrt {-b d}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{2 \, {\left (b^{2} d^{2} x^{2 \, n} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{n}\right )}}\right )}{2 \, b^{2} d^{2} n}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*sqrt(b*x^n + a)*sqrt(d*x^n + c)*b*d + (b*c + a*d)*sqrt(b*d)*log(8*b^2*d^2*x^(2*n) + b^2*c^2 + 6*a*b*c*
d + a^2*d^2 - 4*(2*sqrt(b*d)*b*d*x^n + (b*c + a*d)*sqrt(b*d))*sqrt(b*x^n + a)*sqrt(d*x^n + c) + 8*(b^2*c*d + a
*b*d^2)*x^n))/(b^2*d^2*n), 1/2*(2*sqrt(b*x^n + a)*sqrt(d*x^n + c)*b*d + (b*c + a*d)*sqrt(-b*d)*arctan(1/2*(2*s
qrt(-b*d)*b*d*x^n + (b*c + a*d)*sqrt(-b*d))*sqrt(b*x^n + a)*sqrt(d*x^n + c)/(b^2*d^2*x^(2*n) + a*b*c*d + (b^2*
c*d + a*b*d^2)*x^n)))/(b^2*d^2*n)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2 n - 1}}{\sqrt {a + b x^{n}} \sqrt {c + d x^{n}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{2\,n-1}}{\sqrt {a+b\,x^n}\,\sqrt {c+d\,x^n}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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