3.7.91 \(\int \frac {x^{-1+2 n}}{(a+b x^n)^{3/2} \sqrt {c+d x^n}} \, dx\)

Optimal. Leaf size=91 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{b^{3/2} \sqrt {d} n}+\frac {2 a \sqrt {c+d x^n}}{b n (b c-a d) \sqrt {a+b x^n}} \]

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Rubi [A]  time = 0.09, antiderivative size = 91, 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 = {446, 78, 63, 217, 206} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{b^{3/2} \sqrt {d} n}+\frac {2 a \sqrt {c+d x^n}}{b n (b c-a d) \sqrt {a+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

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 78

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

Rule 206

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

Rule 217

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 446

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

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Mathematica [A]  time = 0.90, size = 122, normalized size = 1.34 \begin {gather*} \frac {2 \left (\frac {a b \left (c+d x^n\right )}{(b c-a d) \sqrt {a+b x^n}}+\frac {\sqrt {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 )}{\sqrt {d}}\right )}{b^2 n \sqrt {c+d x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.32, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{-1+2 n}}{\left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

Defer[IntegrateAlgebraic][x^(-1 + 2*n)/((a + b*x^n)^(3/2)*Sqrt[c + d*x^n]), x]

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fricas [B]  time = 0.60, size = 408, normalized size = 4.48 \begin {gather*} \left [\frac {4 \, \sqrt {b x^{n} + a} \sqrt {d x^{n} + c} a b d + {\left ({\left (b^{2} c - a b d\right )} \sqrt {b d} x^{n} + {\left (a b c - a^{2} d\right )} \sqrt {b d}\right )} \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 )}{2 \, {\left ({\left (b^{4} c d - a b^{3} d^{2}\right )} n x^{n} + {\left (a b^{3} c d - a^{2} b^{2} d^{2}\right )} n\right )}}, \frac {2 \, \sqrt {b x^{n} + a} \sqrt {d x^{n} + c} a b d - {\left ({\left (b^{2} c - a b d\right )} \sqrt {-b d} x^{n} + {\left (a b c - a^{2} d\right )} \sqrt {-b d}\right )} \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 )}{{\left (b^{4} c d - a b^{3} d^{2}\right )} n x^{n} + {\left (a b^{3} c d - a^{2} b^{2} d^{2}\right )} n}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)/(a+b*x^n)^(3/2)/(c+d*x^n)^(1/2),x, algorithm="fricas")

[Out]

[1/2*(4*sqrt(b*x^n + a)*sqrt(d*x^n + c)*a*b*d + ((b^2*c - a*b*d)*sqrt(b*d)*x^n + (a*b*c - a^2*d)*sqrt(b*d))*lo
g(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^4*c*d - a*b^3*d^2)*n*x^n + (a*b^3*c*d - a^2*b^2*d^2)
*n), (2*sqrt(b*x^n + a)*sqrt(d*x^n + c)*a*b*d - ((b^2*c - a*b*d)*sqrt(-b*d)*x^n + (a*b*c - a^2*d)*sqrt(-b*d))*
arctan(1/2*(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)/(b^2*d^2*x^(2*n) +
a*b*c*d + (b^2*c*d + a*b*d^2)*x^n)))/((b^4*c*d - a*b^3*d^2)*n*x^n + (a*b^3*c*d - a^2*b^2*d^2)*n)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2 \, n - 1}}{{\left (b x^{n} + a\right )}^{\frac {3}{2}} \sqrt {d x^{n} + c}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)/(a+b*x^n)^(3/2)/(c+d*x^n)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.95, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2 n -1}}{\left (b \,x^{n}+a \right )^{\frac {3}{2}} \sqrt {d \,x^{n}+c}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2 \, n - 1}}{{\left (b x^{n} + a\right )}^{\frac {3}{2}} \sqrt {d x^{n} + c}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)/(a+b*x^n)^(3/2)/(c+d*x^n)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^(2*n - 1)/((b*x^n + a)^(3/2)*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}}{{\left (a+b\,x^n\right )}^{3/2}\,\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)^(3/2)*(c + d*x^n)^(1/2)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+2*n)/(a+b*x**n)**(3/2)/(c+d*x**n)**(1/2),x)

[Out]

Timed out

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