3.6.88 \(\int \frac {c+d x^{n/2}+e x^n+f x^{3 n/2}}{(a+b x^n)^2} \, dx\) [588]

Optimal. Leaf size=162 \[ \frac {x \left (b c-a e+(b d-a f) x^{n/2}\right )}{a b n \left (a+b x^n\right )}-\frac {(b d (2-n)-a f (2+n)) x^{\frac {2+n}{2}} \, _2F_1\left (1,\frac {1}{2} \left (1+\frac {2}{n}\right );\frac {1}{2} \left (3+\frac {2}{n}\right );-\frac {b x^n}{a}\right )}{a^2 b n (2+n)}+\frac {(a e-b c (1-n)) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^2 b n} \]

[Out]

x*(b*c-a*e+(-a*f+b*d)*x^(1/2*n))/a/b/n/(a+b*x^n)-(b*d*(2-n)-a*f*(2+n))*x^(1+1/2*n)*hypergeom([1, 1/2+1/n],[3/2
+1/n],-b*x^n/a)/a^2/b/n/(2+n)+(a*e-b*c*(1-n))*x*hypergeom([1, 1/n],[1+1/n],-b*x^n/a)/a^2/b/n

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Rubi [A]
time = 0.09, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1906, 1432, 251, 371} \begin {gather*} \frac {x (a e-b c (1-n)) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^2 b n}-\frac {x^{\frac {n+2}{2}} (b d (2-n)-a f (n+2)) \, _2F_1\left (1,\frac {1}{2} \left (1+\frac {2}{n}\right );\frac {1}{2} \left (3+\frac {2}{n}\right );-\frac {b x^n}{a}\right )}{a^2 b n (n+2)}+\frac {x \left (x^{n/2} (b d-a f)-a e+b c\right )}{a b n \left (a+b x^n\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^(n/2) + e*x^n + f*x^((3*n)/2))/(a + b*x^n)^2,x]

[Out]

(x*(b*c - a*e + (b*d - a*f)*x^(n/2)))/(a*b*n*(a + b*x^n)) - ((b*d*(2 - n) - a*f*(2 + n))*x^((2 + n)/2)*Hyperge
ometric2F1[1, (1 + 2/n)/2, (3 + 2/n)/2, -((b*x^n)/a)])/(a^2*b*n*(2 + n)) + ((a*e - b*c*(1 - n))*x*Hypergeometr
ic2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a^2*b*n)

Rule 251

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1432

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Dist[d, Int[1/(a + c*x^(2*n)), x], x] + D
ist[e, Int[x^n/(a + c*x^(2*n)), x], x] /; FreeQ[{a, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &
& (PosQ[a*c] ||  !IntegerQ[n])

Rule 1906

Int[(P3_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{A = Coeff[P3, x^(n/2), 0], B = Coeff[P3, x^(n/2),
1], C = Coeff[P3, x^(n/2), 2], D = Coeff[P3, x^(n/2), 3]}, Simp[-(x*(b*A - a*C + (b*B - a*D)*x^(n/2))*(a + b*x
^n)^(p + 1))/(a*b*n*(p + 1)), x] - Dist[1/(2*a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*Simp[2*a*C - 2*b*A*(n*(p
+ 1) + 1) + (a*D*(n + 2) - b*B*(n*(2*p + 3) + 2))*x^(n/2), x], x], x]] /; FreeQ[{a, b, n}, x] && PolyQ[P3, x^(
n/2), 3] && ILtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {c+d x^{n/2}+e x^n+f x^{3 n/2}}{\left (a+b x^n\right )^2} \, dx &=\frac {x \left (b c-a e+(b d-a f) x^{n/2}\right )}{a b n \left (a+b x^n\right )}+\frac {\int \frac {2 (a e-b c (1-n))-(b d (2-n)-a f (2+n)) x^{n/2}}{a+b x^n} \, dx}{2 a b n}\\ &=\frac {x \left (b c-a e+(b d-a f) x^{n/2}\right )}{a b n \left (a+b x^n\right )}+\frac {(a e-b c (1-n)) \int \frac {1}{a+b x^n} \, dx}{a b n}-\frac {(b d (2-n)-a f (2+n)) \int \frac {x^{n/2}}{a+b x^n} \, dx}{2 a b n}\\ &=\frac {x \left (b c-a e+(b d-a f) x^{n/2}\right )}{a b n \left (a+b x^n\right )}-\frac {(b d (2-n)-a f (2+n)) x^{\frac {2+n}{2}} \, _2F_1\left (1,\frac {1}{2} \left (1+\frac {2}{n}\right );\frac {1}{2} \left (3+\frac {2}{n}\right );-\frac {b x^n}{a}\right )}{a^2 b n (2+n)}+\frac {(a e-b c (1-n)) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^2 b n}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 151, normalized size = 0.93 \begin {gather*} \frac {x \left ((b d (-2+n)+a f (2+n)) x^{n/2} \left (a+b x^n\right ) \, _2F_1\left (1,\frac {1}{2}+\frac {1}{n};\frac {3}{2}+\frac {1}{n};-\frac {b x^n}{a}\right )+(2+n) \left (a \left (b \left (c+d x^{n/2}\right )-a \left (e+f x^{n/2}\right )\right )+(a e+b c (-1+n)) \left (a+b x^n\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )\right )\right )}{a^2 b n (2+n) \left (a+b x^n\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^(n/2) + e*x^n + f*x^((3*n)/2))/(a + b*x^n)^2,x]

[Out]

(x*((b*d*(-2 + n) + a*f*(2 + n))*x^(n/2)*(a + b*x^n)*Hypergeometric2F1[1, 1/2 + n^(-1), 3/2 + n^(-1), -((b*x^n
)/a)] + (2 + n)*(a*(b*(c + d*x^(n/2)) - a*(e + f*x^(n/2))) + (a*e + b*c*(-1 + n))*(a + b*x^n)*Hypergeometric2F
1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])))/(a^2*b*n*(2 + n)*(a + b*x^n))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {c +d \,x^{\frac {n}{2}}+e \,x^{n}+f \,x^{\frac {3 n}{2}}}{\left (a +b \,x^{n}\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [F]
time = 0.38, size = 46, normalized size = 0.28 \begin {gather*} {\rm integral}\left (\frac {f x^{\frac {3}{2} \, n} + d x^{\frac {1}{2} \, n} + e x^{n} + c}{b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((f*x^(3/2*n) + d*x^(1/2*n) + e*x^n + c)/(b^2*x^(2*n) + 2*a*b*x^n + a^2), x)

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Sympy [F(-1)] Timed out
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((c+d*x**(1/2*n)+e*x**n+f*x**(3/2*n))/(a+b*x**n)**2,x)

[Out]

Timed out

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {c+e\,x^n+d\,x^{n/2}+f\,x^{\frac {3\,n}{2}}}{{\left (a+b\,x^n\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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