3.7.92 \(\int \frac {x^{-1+2 n}}{(a+b x^n)^{5/2} \sqrt {c+d x^n}} \, dx\)

Optimal. Leaf size=95 \[ \frac {2 a \sqrt {c+d x^n}}{3 b n (b c-a d) \left (a+b x^n\right )^{3/2}}-\frac {2 (3 b c-a d) \sqrt {c+d x^n}}{3 b n (b c-a d)^2 \sqrt {a+b x^n}} \]

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Rubi [A]  time = 0.07, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {446, 78, 37} \begin {gather*} \frac {2 a \sqrt {c+d x^n}}{3 b n (b c-a d) \left (a+b x^n\right )^{3/2}}-\frac {2 (3 b c-a d) \sqrt {c+d x^n}}{3 b n (b c-a d)^2 \sqrt {a+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*n)/((a + b*x^n)^(5/2)*Sqrt[c + d*x^n]),x]

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 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 )^{5/2} \sqrt {c+d x^n}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx,x,x^n\right )}{n}\\ &=\frac {2 a \sqrt {c+d x^n}}{3 b (b c-a d) n \left (a+b x^n\right )^{3/2}}+\frac {(3 b c-a d) \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx,x,x^n\right )}{3 b (b c-a d) n}\\ &=\frac {2 a \sqrt {c+d x^n}}{3 b (b c-a d) n \left (a+b x^n\right )^{3/2}}-\frac {2 (3 b c-a d) \sqrt {c+d x^n}}{3 b (b c-a d)^2 n \sqrt {a+b x^n}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 57, normalized size = 0.60 \begin {gather*} \frac {2 \sqrt {c+d x^n} \left (-2 a c+a d x^n-3 b c x^n\right )}{3 n (b c-a d)^2 \left (a+b x^n\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 2*n)/((a + b*x^n)^(5/2)*Sqrt[c + d*x^n]),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.86, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{-1+2 n}}{\left (a+b x^n\right )^{5/2} \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)^(5/2)*Sqrt[c + d*x^n]),x]

[Out]

Defer[IntegrateAlgebraic][x^(-1 + 2*n)/((a + b*x^n)^(5/2)*Sqrt[c + d*x^n]), x]

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fricas [A]  time = 0.74, size = 135, normalized size = 1.42 \begin {gather*} -\frac {2 \, {\left (2 \, a c + {\left (3 \, b c - a d\right )} x^{n}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{3 \, {\left ({\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} n x^{2 \, n} + 2 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} n x^{n} + {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} 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)^(5/2)/(c+d*x^n)^(1/2),x, algorithm="fricas")

[Out]

-2/3*(2*a*c + (3*b*c - a*d)*x^n)*sqrt(b*x^n + a)*sqrt(d*x^n + c)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*n*x^(2
*n) + 2*(a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*n*x^n + (a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*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 {5}{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)^(5/2)/(c+d*x^n)^(1/2),x, algorithm="giac")

[Out]

integrate(x^(2*n - 1)/((b*x^n + a)^(5/2)*sqrt(d*x^n + c)), x)

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maple [F]  time = 0.92, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2 n -1}}{\left (b \,x^{n}+a \right )^{\frac {5}{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)^(5/2)/(d*x^n+c)^(1/2),x)

[Out]

int(x^(2*n-1)/(b*x^n+a)^(5/2)/(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 {5}{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)^(5/2)/(c+d*x^n)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^(2*n - 1)/((b*x^n + a)^(5/2)*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 )}^{5/2}\,\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)^(5/2)*(c + d*x^n)^(1/2)),x)

[Out]

int(x^(2*n - 1)/((a + b*x^n)^(5/2)*(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)**(5/2)/(c+d*x**n)**(1/2),x)

[Out]

Timed out

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