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

Optimal. Leaf size=322 \[ -\frac{c \left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{2 a^2 (q+1) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}-\frac{c \left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 a^2 (q+1) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}+\frac{b \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{e x^2}{d}+1\right )}{2 a^2 d (q+1)}+\frac{e \left (d+e x^2\right )^{q+1} \, _2F_1\left (2,q+1;q+2;\frac{e x^2}{d}+1\right )}{2 a d^2 (q+1)} \]

[Out]

-(c*(b + (b^2 - 2*a*c)/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*a^2*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(1 + q)) - (c*(b - (b^2
- 2*a*c)/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*a^2*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(1 + q)) + (b*(d + e*x^2)^(1 + q)*Hype
rgeometric2F1[1, 1 + q, 2 + q, 1 + (e*x^2)/d])/(2*a^2*d*(1 + q)) + (e*(d + e*x^2)^(1 + q)*Hypergeometric2F1[2,
1 + q, 2 + q, 1 + (e*x^2)/d])/(2*a*d^2*(1 + q))

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Rubi [A]  time = 0.66344, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {1251, 960, 65, 830, 68} \[ -\frac{c \left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{2 a^2 (q+1) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}-\frac{c \left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 a^2 (q+1) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}+\frac{b \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{e x^2}{d}+1\right )}{2 a^2 d (q+1)}+\frac{e \left (d+e x^2\right )^{q+1} \, _2F_1\left (2,q+1;q+2;\frac{e x^2}{d}+1\right )}{2 a d^2 (q+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*(b + (b^2 - 2*a*c)/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*a^2*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(1 + q)) - (c*(b - (b^2
- 2*a*c)/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*a^2*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(1 + q)) + (b*(d + e*x^2)^(1 + q)*Hype
rgeometric2F1[1, 1 + q, 2 + q, 1 + (e*x^2)/d])/(2*a^2*d*(1 + q)) + (e*(d + e*x^2)^(1 + q)*Hypergeometric2F1[2,
1 + q, 2 + q, 1 + (e*x^2)/d])/(2*a*d^2*(1 + q))

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]

Rule 960

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] &&
ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

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 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]

Rubi steps

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

Mathematica [A]  time = 0.444902, size = 259, normalized size = 0.8 \[ \frac{\left (d+e x^2\right )^{q+1} \left (-\frac{c \left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d+\left (\sqrt{b^2-4 a c}-b\right ) e}\right )}{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}-\frac{c \left (\frac{2 a c-b^2}{\sqrt{b^2-4 a c}}+b\right ) \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}+\frac{a e \, _2F_1\left (2,q+1;q+2;\frac{e x^2}{d}+1\right )}{d^2}+\frac{b \, _2F_1\left (1,q+1;q+2;\frac{e x^2}{d}+1\right )}{d}\right )}{2 a^2 (q+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x^2)^(1 + q)*(-((c*(b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + q, 2 + q, (2*c*(d +
e*x^2))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)])/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)) - (c*(b + (-b^2 + 2*a*c)/
Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + q, 2 + q, (2*c*(d + e*x^2))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/
(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e) + (b*Hypergeometric2F1[1, 1 + q, 2 + q, 1 + (e*x^2)/d])/d + (a*e*Hypergeom
etric2F1[2, 1 + q, 2 + q, 1 + (e*x^2)/d])/d^2))/(2*a^2*(1 + q))

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Maple [F]  time = 0.063, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( e{x}^{2}+d \right ) ^{q}}{{x}^{3} \left ( c{x}^{4}+b{x}^{2}+a \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{q}}{{\left (c x^{4} + b x^{2} + a\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (e x^{2} + d\right )}^{q}}{c x^{7} + b x^{5} + a x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x^2 + d)^q/(c*x^7 + b*x^5 + a*x^3), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{q}}{{\left (c x^{4} + b x^{2} + a\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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