3.403 \(\int \frac {x^3 (d+e x^2)^q}{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=210 \[ -\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac {2 c \left (e x^2+d\right )}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{2 (q+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac {2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 (q+1) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.33, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1251, 830, 68} \[ -\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac {2 c \left (e x^2+d\right )}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{2 (q+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac {2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 (q+1) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x]

[Out]

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

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 830

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

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.20, size = 183, normalized size = 0.87 \[ -\frac {\left (d+e x^2\right )^{q+1} \left (\left (d \sqrt {b^2-4 a c}+2 a e-b d\right ) \, _2F_1\left (1,q+1;q+2;\frac {2 c \left (e x^2+d\right )}{2 c d+\left (\sqrt {b^2-4 a c}-b\right ) e}\right )+\left (d \sqrt {b^2-4 a c}-2 a e+b d\right ) \, _2F_1\left (1,q+1;q+2;\frac {2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )\right )}{4 (q+1) \sqrt {b^2-4 a c} \left (e (a e-b d)+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x^{2} + d\right )}^{q} x^{3}}{c x^{4} + b x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

integral((e*x^2 + d)^q*x^3/(c*x^4 + b*x^2 + a), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{q} x^{3}}{c x^{4} + b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^q*x^3/(c*x^4 + b*x^2 + a), x)

________________________________________________________________________________________

maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \left (e \,x^{2}+d \right )^{q}}{c \,x^{4}+b \,x^{2}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x^2+d)^q/(c*x^4+b*x^2+a),x)

[Out]

int(x^3*(e*x^2+d)^q/(c*x^4+b*x^2+a),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{q} x^{3}}{c x^{4} + b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)^q*x^3/(c*x^4 + b*x^2 + a), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3\,{\left (e\,x^2+d\right )}^q}{c\,x^4+b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x)

[Out]

int((x^3*(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**3*(e*x**2+d)**q/(c*x**4+b*x**2+a),x)

[Out]

Timed out

________________________________________________________________________________________