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

Optimal. Leaf size=298 \[ \frac {d x \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {1}{n};\frac {5}{2},\frac {5}{2};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {a+b x^n+c x^{2 n}}}+\frac {e x^{n+1} \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} F_1\left (1+\frac {1}{n};\frac {5}{2},\frac {5}{2};2+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 (n+1) \sqrt {a+b x^n+c x^{2 n}}} \]

[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)

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Rubi [A]  time = 0.35, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1432, 1348, 429, 1385, 510} \[ \frac {d x \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {1}{n};\frac {5}{2},\frac {5}{2};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {a+b x^n+c x^{2 n}}}+\frac {e x^{n+1} \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} F_1\left (1+\frac {1}{n};\frac {5}{2},\frac {5}{2};2+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 (n+1) \sqrt {a+b x^n+c x^{2 n}}} \]

Antiderivative was successfully verified.

[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 429

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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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 1348

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]))^F
racPart[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 1385

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 1432

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

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Mathematica [B]  time = 6.57, size = 6752, normalized size = 22.66 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e x^{n} + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int \frac {e \,x^{n}+d}{\left (b \,x^{n}+c \,x^{2 n}+a \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e x^{n} + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {d+e\,x^n}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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