3.5.5 \(\int \frac {(d+e x^2)^q}{x (a+b x^2+c x^4)} \, dx\) [405]

Optimal. Leaf size=262 \[ \frac {c \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^{1+q} \, _2F_1\left (1,1+q;2+q;\frac {2 c \left (d+e x^2\right )}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{2 a \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1+q)}+\frac {c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^{1+q} \, _2F_1\left (1,1+q;2+q;\frac {2 c \left (d+e x^2\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 a \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+q)}-\frac {\left (d+e x^2\right )^{1+q} \, _2F_1\left (1,1+q;2+q;1+\frac {e x^2}{d}\right )}{2 a d (1+q)} \]

[Out]

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

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Rubi [A]
time = 0.37, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1265, 974, 67, 844, 70} \begin {gather*} \frac {c \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 a (q+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}+\frac {c \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 a (q+1) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {\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 d (q+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 67

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

Rule 70

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

Rule 844

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 974

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 1265

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

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Mathematica [A]
time = 0.60, size = 218, normalized size = 0.83 \begin {gather*} \frac {\left (d+e x^2\right )^{1+q} \left (\frac {c \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \, _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 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}+\frac {c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \, _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 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}-\frac {\, _2F_1\left (1,1+q;2+q;1+\frac {e x^2}{d}\right )}{d}\right )}{2 a (1+q)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (e \,x^{2}+d \right )^{q}}{x \left (c \,x^{4}+b \,x^{2}+a \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x^{2}\right )^{q}}{x \left (a + b x^{2} + c x^{4}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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