3.407 \(\int \frac {x^6 (d+e x^2)^q}{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=339 \[ \frac {x \left (-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{c^2 \left (b-\sqrt {b^2-4 a c}\right )}+\frac {x \left (\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{c^2 \left (\sqrt {b^2-4 a c}+b\right )}-\frac {b x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \, _2F_1\left (\frac {1}{2},-q;\frac {3}{2};-\frac {e x^2}{d}\right )}{c^2}+\frac {x^3 \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \, _2F_1\left (\frac {3}{2},-q;\frac {5}{2};-\frac {e x^2}{d}\right )}{3 c} \]

[Out]

-b*x*(e*x^2+d)^q*hypergeom([1/2, -q],[3/2],-e*x^2/d)/c^2/((1+e*x^2/d)^q)+1/3*x^3*(e*x^2+d)^q*hypergeom([3/2, -
q],[5/2],-e*x^2/d)/c/((1+e*x^2/d)^q)+x*(e*x^2+d)^q*AppellF1(1/2,1,-q,3/2,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-e*x^
2/d)*(b^2-a*c-b*(-3*a*c+b^2)/(-4*a*c+b^2)^(1/2))/c^2/((1+e*x^2/d)^q)/(b-(-4*a*c+b^2)^(1/2))+x*(e*x^2+d)^q*Appe
llF1(1/2,1,-q,3/2,-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)),-e*x^2/d)*(b^2-a*c+b*(-3*a*c+b^2)/(-4*a*c+b^2)^(1/2))/c^2/((
1+e*x^2/d)^q)/(b+(-4*a*c+b^2)^(1/2))

________________________________________________________________________________________

Rubi [A]  time = 0.63, antiderivative size = 339, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {1303, 246, 245, 365, 364, 1692, 430, 429} \[ \frac {x \left (-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{c^2 \left (b-\sqrt {b^2-4 a c}\right )}+\frac {x \left (\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{c^2 \left (\sqrt {b^2-4 a c}+b\right )}-\frac {b x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \, _2F_1\left (\frac {1}{2},-q;\frac {3}{2};-\frac {e x^2}{d}\right )}{c^2}+\frac {x^3 \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \, _2F_1\left (\frac {3}{2},-q;\frac {5}{2};-\frac {e x^2}{d}\right )}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^6*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x]

[Out]

((b^2 - a*c - (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a*c])*x*(d + e*x^2)^q*AppellF1[1/2, 1, -q, 3/2, (-2*c*x^2)/(b - S
qrt[b^2 - 4*a*c]), -((e*x^2)/d)])/(c^2*(b - Sqrt[b^2 - 4*a*c])*(1 + (e*x^2)/d)^q) + ((b^2 - a*c + (b*(b^2 - 3*
a*c))/Sqrt[b^2 - 4*a*c])*x*(d + e*x^2)^q*AppellF1[1/2, 1, -q, 3/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), -((e*x^
2)/d)])/(c^2*(b + Sqrt[b^2 - 4*a*c])*(1 + (e*x^2)/d)^q) - (b*x*(d + e*x^2)^q*Hypergeometric2F1[1/2, -q, 3/2, -
((e*x^2)/d)])/(c^2*(1 + (e*x^2)/d)^q) + (x^3*(d + e*x^2)^q*Hypergeometric2F1[3/2, -q, 5/2, -((e*x^2)/d)])/(3*c
*(1 + (e*x^2)/d)^q)

Rule 245

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 364

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

Rule 365

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

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 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 1303

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x^2)^q, (f*x)^m/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[b^2
- 4*a*c, 0] &&  !IntegerQ[q] && IntegerQ[m]

Rule 1692

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {x^6 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx &=\int \left (-\frac {b \left (d+e x^2\right )^q}{c^2}+\frac {x^2 \left (d+e x^2\right )^q}{c}+\frac {\left (a b+\left (b^2-a c\right ) x^2\right ) \left (d+e x^2\right )^q}{c^2 \left (a+b x^2+c x^4\right )}\right ) \, dx\\ &=\frac {\int \frac {\left (a b+\left (b^2-a c\right ) x^2\right ) \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx}{c^2}-\frac {b \int \left (d+e x^2\right )^q \, dx}{c^2}+\frac {\int x^2 \left (d+e x^2\right )^q \, dx}{c}\\ &=\frac {\int \left (\frac {\left (b^2-a c+\frac {b \left (-b^2+3 a c\right )}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^q}{b-\sqrt {b^2-4 a c}+2 c x^2}+\frac {\left (b^2-a c-\frac {b \left (-b^2+3 a c\right )}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^q}{b+\sqrt {b^2-4 a c}+2 c x^2}\right ) \, dx}{c^2}-\frac {\left (b \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q}\right ) \int \left (1+\frac {e x^2}{d}\right )^q \, dx}{c^2}+\frac {\left (\left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q}\right ) \int x^2 \left (1+\frac {e x^2}{d}\right )^q \, dx}{c}\\ &=-\frac {b x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \, _2F_1\left (\frac {1}{2},-q;\frac {3}{2};-\frac {e x^2}{d}\right )}{c^2}+\frac {x^3 \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \, _2F_1\left (\frac {3}{2},-q;\frac {5}{2};-\frac {e x^2}{d}\right )}{3 c}+\frac {\left (b^2-a c-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right ) \int \frac {\left (d+e x^2\right )^q}{b-\sqrt {b^2-4 a c}+2 c x^2} \, dx}{c^2}+\frac {\left (b^2-a c+\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right ) \int \frac {\left (d+e x^2\right )^q}{b+\sqrt {b^2-4 a c}+2 c x^2} \, dx}{c^2}\\ &=-\frac {b x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \, _2F_1\left (\frac {1}{2},-q;\frac {3}{2};-\frac {e x^2}{d}\right )}{c^2}+\frac {x^3 \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \, _2F_1\left (\frac {3}{2},-q;\frac {5}{2};-\frac {e x^2}{d}\right )}{3 c}+\frac {\left (\left (b^2-a c-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q}\right ) \int \frac {\left (1+\frac {e x^2}{d}\right )^q}{b-\sqrt {b^2-4 a c}+2 c x^2} \, dx}{c^2}+\frac {\left (\left (b^2-a c+\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q}\right ) \int \frac {\left (1+\frac {e x^2}{d}\right )^q}{b+\sqrt {b^2-4 a c}+2 c x^2} \, dx}{c^2}\\ &=\frac {\left (b^2-a c-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{c^2 \left (b-\sqrt {b^2-4 a c}\right )}+\frac {\left (b^2-a c+\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{c^2 \left (b+\sqrt {b^2-4 a c}\right )}-\frac {b x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \, _2F_1\left (\frac {1}{2},-q;\frac {3}{2};-\frac {e x^2}{d}\right )}{c^2}+\frac {x^3 \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \, _2F_1\left (\frac {3}{2},-q;\frac {5}{2};-\frac {e x^2}{d}\right )}{3 c}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 0.49, size = 0, normalized size = 0.00 \[ \int \frac {x^6 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(x^6*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x]

[Out]

Integrate[(x^6*(d + e*x^2)^q)/(a + b*x^2 + c*x^4), x]

________________________________________________________________________________________

fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x^{2} + d\right )}^{q} x^{6}}{c x^{4} + b x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

integral((e*x^2 + d)^q*x^6/(c*x^4 + b*x^2 + a), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{q} x^{6}}{c x^{4} + b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^q*x^6/(c*x^4 + b*x^2 + a), x)

________________________________________________________________________________________

maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {x^{6} \left (e \,x^{2}+d \right )^{q}}{c \,x^{4}+b \,x^{2}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(e*x^2+d)^q/(c*x^4+b*x^2+a),x)

[Out]

int(x^6*(e*x^2+d)^q/(c*x^4+b*x^2+a),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{q} x^{6}}{c x^{4} + b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)^q*x^6/(c*x^4 + b*x^2 + a), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^6\,{\left (e\,x^2+d\right )}^q}{c\,x^4+b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x)

[Out]

int((x^6*(d + e*x^2)^q)/(a + b*x^2 + c*x^4), x)

________________________________________________________________________________________

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(x**6*(e*x**2+d)**q/(c*x**4+b*x**2+a),x)

[Out]

Timed out

________________________________________________________________________________________