\(\int \frac {d+e x^n}{(a+b x^n+c x^{2 n})^{5/2}} \, dx\) [89]

Optimal result

Integrand size = 26, antiderivative size = 298 \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\frac {e x^{1+n} \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {5}{2},\frac {5}{2},2+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 (1+n) \sqrt {a+b x^n+c x^{2 n}}}+\frac {d x \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{n},\frac {5}{2},\frac {5}{2},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {a+b x^n+c x^{2 n}}} \]

[Out]

e*x^(1+n)*AppellF1(1+1/n,5/2,5/2,2+1/n,-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))*(1+2*c
*x^n/(b-(-4*a*c+b^2)^(1/2)))^(1/2)*(1+2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))^(1/2)/a^2/(1+n)/(a+b*x^n+c*x^(2*n))^(1/2
)+d*x*AppellF1(1/n,5/2,5/2,1+1/n,-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))*(1+2*c*x^n/(
b-(-4*a*c+b^2)^(1/2)))^(1/2)*(1+2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))^(1/2)/a^2/(a+b*x^n+c*x^(2*n))^(1/2)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1446, 1362, 440, 1399, 524} \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\frac {d x \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {1}{n},\frac {5}{2},\frac {5}{2},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {a+b x^n+c x^{2 n}}}+\frac {e x^{n+1} \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {5}{2},\frac {5}{2},2+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 (n+1) \sqrt {a+b x^n+c x^{2 n}}} \]

[In]

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

[Out]

(e*x^(1 + n)*Sqrt[1 + (2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[
1 + n^(-1), 5/2, 5/2, 2 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a^
2*(1 + n)*Sqrt[a + b*x^n + c*x^(2*n)]) + (d*x*Sqrt[1 + (2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^n)/(
b + Sqrt[b^2 - 4*a*c])]*AppellF1[n^(-1), 5/2, 5/2, 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/
(b + Sqrt[b^2 - 4*a*c])])/(a^2*Sqrt[a + b*x^n + c*x^(2*n)])

Rule 440

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 1362

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n + c*x^(2*n))
^FracPart[p]/((1 + 2*c*(x^n/(b + Rt[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^n/(b - Rt[b^2 - 4*a*c, 2])))^Fr
acPart[p])), Int[(1 + 2*c*(x^n/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2*c*(x^n/(b - Sqrt[b^2 - 4*a*c])))^p, x], x] /
; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rule 1399

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a +
b*x^n + c*x^(2*n))^FracPart[p]/((1 + 2*c*(x^n/(b + Rt[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^n/(b - Rt[b^
2 - 4*a*c, 2])))^FracPart[p])), Int[(d*x)^m*(1 + 2*c*(x^n/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2*c*(x^n/(b - Sqrt[
b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n]

Rule 1446

Int[((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegran
d[(d + e*x^n)*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4
*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {d}{\left (a+b x^n+c x^{2 n}\right )^{5/2}}+\frac {e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}}\right ) \, dx \\ & = d \int \frac {1}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx+e \int \frac {x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx \\ & = \frac {\left (d \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}}\right ) \int \frac {1}{\left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{5/2} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{5/2}} \, dx}{a^2 \sqrt {a+b x^n+c x^{2 n}}}+\frac {\left (e \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}}\right ) \int \frac {x^n}{\left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{5/2} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{5/2}} \, dx}{a^2 \sqrt {a+b x^n+c x^{2 n}}} \\ & = \frac {e x^{1+n} \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} F_1\left (1+\frac {1}{n};\frac {5}{2},\frac {5}{2};2+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 (1+n) \sqrt {a+b x^n+c x^{2 n}}}+\frac {d x \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} F_1\left (\frac {1}{n};\frac {5}{2},\frac {5}{2};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {a+b x^n+c x^{2 n}}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(701\) vs. \(2(298)=596\).

Time = 5.36 (sec) , antiderivative size = 701, normalized size of antiderivative = 2.35 \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\frac {x \left (-2 c \left (2 a b^2 e+4 a b c d (2-5 n)+8 a^2 c e (-1+2 n)+b^3 d (-2+3 n)\right ) x^n \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \left (a+x^n \left (b+c x^n\right )\right ) \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {1}{2},\frac {1}{2},2+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+(1+n) \left (2 \left (b^3 d (-2+3 n) x^n \left (b+c x^n\right )^2+4 a^3 c \left (b e (-2+3 n)+c d (-2+8 n)+2 c e (-1+3 n) x^n\right )+2 a b \left (b+c x^n\right ) \left (-2 c^2 d (-2+5 n) x^{2 n}+b c x^n \left (d (5-11 n)+e x^n\right )+b^2 \left (d (-1+2 n)+e x^n\right )\right )+a^2 \left (-b^3 e (-2+n)+8 b c^2 e (-2+3 n) x^{2 n}-2 b^2 c \left (d (-5+14 n)-3 e (-1+n) x^n\right )+8 c^3 x^{2 n} \left (d (-1+3 n)+e (-1+2 n) x^n\right )\right )\right )+\left (2 a b^3 e (-2+n)-8 a^2 b c e (-2+3 n)+b^4 d \left (4-8 n+3 n^2\right )+16 a^2 c^2 d \left (1-4 n+3 n^2\right )-4 a b^2 c d \left (5-14 n+6 n^2\right )\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \left (a+x^n \left (b+c x^n\right )\right ) \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},\frac {1}{2},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{3 a^2 \left (b^2-4 a c\right )^2 n^2 (1+n) \left (a+x^n \left (b+c x^n\right )\right )^{3/2}} \]

[In]

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

[Out]

(x*(-2*c*(2*a*b^2*e + 4*a*b*c*d*(2 - 5*n) + 8*a^2*c*e*(-1 + 2*n) + b^3*d*(-2 + 3*n))*x^n*Sqrt[(b - Sqrt[b^2 -
4*a*c] + 2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]*(a
+ x^n*(b + c*x^n))*AppellF1[1 + n^(-1), 1/2, 1/2, 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-
b + Sqrt[b^2 - 4*a*c])] + (1 + n)*(2*(b^3*d*(-2 + 3*n)*x^n*(b + c*x^n)^2 + 4*a^3*c*(b*e*(-2 + 3*n) + c*d*(-2 +
8*n) + 2*c*e*(-1 + 3*n)*x^n) + 2*a*b*(b + c*x^n)*(-2*c^2*d*(-2 + 5*n)*x^(2*n) + b*c*x^n*(d*(5 - 11*n) + e*x^n
) + b^2*(d*(-1 + 2*n) + e*x^n)) + a^2*(-(b^3*e*(-2 + n)) + 8*b*c^2*e*(-2 + 3*n)*x^(2*n) - 2*b^2*c*(d*(-5 + 14*
n) - 3*e*(-1 + n)*x^n) + 8*c^3*x^(2*n)*(d*(-1 + 3*n) + e*(-1 + 2*n)*x^n))) + (2*a*b^3*e*(-2 + n) - 8*a^2*b*c*e
*(-2 + 3*n) + b^4*d*(4 - 8*n + 3*n^2) + 16*a^2*c^2*d*(1 - 4*n + 3*n^2) - 4*a*b^2*c*d*(5 - 14*n + 6*n^2))*Sqrt[
(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b + Sqrt[b^
2 - 4*a*c])]*(a + x^n*(b + c*x^n))*AppellF1[n^(-1), 1/2, 1/2, 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]),
(2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])])))/(3*a^2*(b^2 - 4*a*c)^2*n^2*(1 + n)*(a + x^n*(b + c*x^n))^(3/2))

Maple [F]

\[\int \frac {d +e \,x^{n}}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{\frac {5}{2}}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((d+e*x^n)/(a+b*x^n+c*x^(2*n))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\int \frac {d + e x^{n}}{\left (a + b x^{n} + c x^{2 n}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((d+e*x**n)/(a+b*x**n+c*x**(2*n))**(5/2),x)

[Out]

Integral((d + e*x**n)/(a + b*x**n + c*x**(2*n))**(5/2), x)

Maxima [F]

\[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\int { \frac {e x^{n} + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((d+e*x^n)/(a+b*x^n+c*x^(2*n))^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\int { \frac {e x^{n} + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((d+e*x^n)/(a+b*x^n+c*x^(2*n))^(5/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\int \frac {d+e\,x^n}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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