3.5.1 \(\int \frac {x^7 (d+e x^2)^q}{a+b x^2+c x^4} \, dx\) [401]

Optimal. Leaf size=313 \[ -\frac {(c d+b e) \left (d+e x^2\right )^{1+q}}{2 c^2 e^2 (1+q)}+\frac {\left (d+e x^2\right )^{2+q}}{2 c e^2 (2+q)}+\frac {\left (a-\frac {b^2}{c}+\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^{1+q} \, _2F_1\left (1,1+q;2+q;\frac {2 c \left (d+e x^2\right )}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{2 c \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1+q)}+\frac {\left (a-\frac {b^2}{c}-\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^{1+q} \, _2F_1\left (1,1+q;2+q;\frac {2 c \left (d+e x^2\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+q)} \]

[Out]

-1/2*(b*e+c*d)*(e*x^2+d)^(1+q)/c^2/e^2/(1+q)+1/2*(e*x^2+d)^(2+q)/c/e^2/(2+q)+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))))*(a-b^2/c+b*(-3*a*c+b^2)/c/(-4*a*c+b^2)^(1/2))/c/(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))))*(a-b^2/c-b*(-3*a*c+b^2)/c/(-4*a*c+b^2)^(1/2))/c/(1+q)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))

________________________________________________________________________________________

Rubi [A]
time = 0.65, antiderivative size = 313, 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 = {1265, 1642, 70} \begin {gather*} \frac {\left (\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}+a-\frac {b^2}{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 c (q+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}+\frac {\left (-\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}+a-\frac {b^2}{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 c (q+1) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {(b e+c d) \left (d+e x^2\right )^{q+1}}{2 c^2 e^2 (q+1)}+\frac {\left (d+e x^2\right )^{q+2}}{2 c e^2 (q+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^7*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x]

[Out]

-1/2*((c*d + b*e)*(d + e*x^2)^(1 + q))/(c^2*e^2*(1 + q)) + (d + e*x^2)^(2 + q)/(2*c*e^2*(2 + q)) + ((a - b^2/c
+ (b*(b^2 - 3*a*c))/(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*c*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(1 + q)) + ((a - b^2/c -
(b*(b^2 - 3*a*c))/(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*c*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(1 + q))

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 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]

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {align*} \int \frac {x^7 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x^3 (d+e x)^q}{a+b x+c x^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (\frac {(-c d-b e) (d+e x)^q}{c^2 e}+\frac {\left (\frac {b^2}{c^2}-\frac {a}{c}-\frac {b \left (b^2-3 a c\right )}{c^2 \sqrt {b^2-4 a c}}\right ) (d+e x)^q}{b-\sqrt {b^2-4 a c}+2 c x}+\frac {\left (\frac {b^2}{c^2}-\frac {a}{c}+\frac {b \left (b^2-3 a c\right )}{c^2 \sqrt {b^2-4 a c}}\right ) (d+e x)^q}{b+\sqrt {b^2-4 a c}+2 c x}+\frac {(d+e x)^{1+q}}{c e}\right ) \, dx,x,x^2\right )\\ &=-\frac {(c d+b e) \left (d+e x^2\right )^{1+q}}{2 c^2 e^2 (1+q)}+\frac {\left (d+e x^2\right )^{2+q}}{2 c e^2 (2+q)}-\frac {\left (a-\frac {b^2}{c}-\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\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 c}-\frac {\left (a-\frac {b^2}{c}+\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\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 c}\\ &=-\frac {(c d+b e) \left (d+e x^2\right )^{1+q}}{2 c^2 e^2 (1+q)}+\frac {\left (d+e x^2\right )^{2+q}}{2 c e^2 (2+q)}+\frac {\left (a-\frac {b^2}{c}+\frac {b \left (b^2-3 a c\right )}{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 c \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1+q)}+\frac {\left (a-\frac {b^2}{c}-\frac {b \left (b^2-3 a c\right )}{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 c \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+q)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.01, size = 272, normalized size = 0.87 \begin {gather*} \frac {\left (d+e x^2\right )^{1+q} \left (-\frac {c d+b e}{e^2 (1+q)}+\frac {c \left (d+e x^2\right )}{e^2 (2+q)}+\frac {c \left (a-\frac {b^2}{c}+\frac {b \left (b^2-3 a c\right )}{c \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 )}{\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) (1+q)}+\frac {c \left (a-\frac {b^2}{c}-\frac {b \left (b^2-3 a c\right )}{c \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 )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+q)}\right )}{2 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x]

[Out]

((d + e*x^2)^(1 + q)*(-((c*d + b*e)/(e^2*(1 + q))) + (c*(d + e*x^2))/(e^2*(2 + q)) + (c*(a - b^2/c + (b*(b^2 -
3*a*c))/(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)*(1 + q)) + (c*(a - b^2/c - (b*(b^2 - 3*a*c))/(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)*(1 + q))))/(2*c^2)

________________________________________________________________________________________

Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {x^{7} \left (e \,x^{2}+d \right )^{q}}{c \,x^{4}+b \,x^{2}+a}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(e*x^2+d)^q/(c*x^4+b*x^2+a),x)

[Out]

int(x^7*(e*x^2+d)^q/(c*x^4+b*x^2+a),x)

________________________________________________________________________________________

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(x^7*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate((x^2*e + d)^q*x^7/(c*x^4 + b*x^2 + a), x)

________________________________________________________________________________________

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(x^7*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

integral((x^2*e + d)^q*x^7/(c*x^4 + b*x^2 + a), x)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(e*x**2+d)**q/(c*x**4+b*x**2+a),x)

[Out]

Timed out

________________________________________________________________________________________

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(x^7*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

integrate((x^2*e + d)^q*x^7/(c*x^4 + b*x^2 + a), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^7\,{\left (e\,x^2+d\right )}^q}{c\,x^4+b\,x^2+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^7*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x)

[Out]

int((x^7*(d + e*x^2)^q)/(a + b*x^2 + c*x^4), x)

________________________________________________________________________________________