3.3.51 \(\int \frac {(d x)^m}{(a+b x^3+c x^6)^2} \, dx\) [251]

Optimal. Leaf size=315 \[ \frac {(d x)^{1+m} \left (b^2-2 a c+b c x^3\right )}{3 a \left (b^2-4 a c\right ) d \left (a+b x^3+c x^6\right )}+\frac {c \left (b^2 (2-m)+b \sqrt {b^2-4 a c} (2-m)-4 a c (5-m)\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{3};\frac {4+m}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}\right )}{3 a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) d (1+m)}-\frac {c \left (b^2 (2-m)-b \sqrt {b^2-4 a c} (2-m)-4 a c (5-m)\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{3};\frac {4+m}{3};-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{3 a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) d (1+m)} \]

[Out]

1/3*(d*x)^(1+m)*(b*c*x^3-2*a*c+b^2)/a/(-4*a*c+b^2)/d/(c*x^6+b*x^3+a)-1/3*c*(d*x)^(1+m)*hypergeom([1, 1/3+1/3*m
],[4/3+1/3*m],-2*c*x^3/(b+(-4*a*c+b^2)^(1/2)))*(b^2*(2-m)-4*a*c*(5-m)-b*(2-m)*(-4*a*c+b^2)^(1/2))/a/(-4*a*c+b^
2)^(3/2)/d/(1+m)/(b+(-4*a*c+b^2)^(1/2))+1/3*c*(d*x)^(1+m)*hypergeom([1, 1/3+1/3*m],[4/3+1/3*m],-2*c*x^3/(b-(-4
*a*c+b^2)^(1/2)))*(b^2*(2-m)-4*a*c*(5-m)+b*(2-m)*(-4*a*c+b^2)^(1/2))/a/(-4*a*c+b^2)^(3/2)/d/(1+m)/(b-(-4*a*c+b
^2)^(1/2))

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Rubi [A]
time = 0.48, antiderivative size = 315, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1380, 1524, 371} \begin {gather*} \frac {c (d x)^{m+1} \left (b (2-m) \sqrt {b^2-4 a c}-4 a c (5-m)+b^2 (2-m)\right ) \, _2F_1\left (1,\frac {m+1}{3};\frac {m+4}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}\right )}{3 a d (m+1) \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {c (d x)^{m+1} \left (-b (2-m) \sqrt {b^2-4 a c}-4 a c (5-m)+b^2 (2-m)\right ) \, _2F_1\left (1,\frac {m+1}{3};\frac {m+4}{3};-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{3 a d (m+1) \left (b^2-4 a c\right )^{3/2} \left (\sqrt {b^2-4 a c}+b\right )}+\frac {(d x)^{m+1} \left (-2 a c+b^2+b c x^3\right )}{3 a d \left (b^2-4 a c\right ) \left (a+b x^3+c x^6\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^m/(a + b*x^3 + c*x^6)^2,x]

[Out]

((d*x)^(1 + m)*(b^2 - 2*a*c + b*c*x^3))/(3*a*(b^2 - 4*a*c)*d*(a + b*x^3 + c*x^6)) + (c*(b^2*(2 - m) + b*Sqrt[b
^2 - 4*a*c]*(2 - m) - 4*a*c*(5 - m))*(d*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/3, (4 + m)/3, (-2*c*x^3)/(b -
Sqrt[b^2 - 4*a*c])])/(3*a*(b^2 - 4*a*c)^(3/2)*(b - Sqrt[b^2 - 4*a*c])*d*(1 + m)) - (c*(b^2*(2 - m) - b*Sqrt[b^
2 - 4*a*c]*(2 - m) - 4*a*c*(5 - m))*(d*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/3, (4 + m)/3, (-2*c*x^3)/(b + S
qrt[b^2 - 4*a*c])])/(3*a*(b^2 - 4*a*c)^(3/2)*(b + Sqrt[b^2 - 4*a*c])*d*(1 + m))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1380

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(d*x)^(m + 1))*(
b^2 - 2*a*c + b*c*x^n)*((a + b*x^n + c*x^(2*n))^(p + 1)/(a*d*n*(p + 1)*(b^2 - 4*a*c))), x] + Dist[1/(a*n*(p +
1)*(b^2 - 4*a*c)), Int[(d*x)^m*(a + b*x^n + c*x^(2*n))^(p + 1)*Simp[b^2*(m + n*(p + 1) + 1) - 2*a*c*(m + 2*n*(
p + 1) + 1) + b*c*(m + n*(2*p + 3) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[n2, 2*n] && NeQ[b^
2 - 4*a*c, 0] && IGtQ[n, 0] && ILtQ[p, -1]

Rule 1524

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_)))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Wi
th[{q = Rt[b^2 - 4*a*c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[(f*x)^m/(b/2 - q/2 + c*x^n), x], x] + Dist[e/
2 - (2*c*d - b*e)/(2*q), Int[(f*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[n2
, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.15, size = 78, normalized size = 0.25 \begin {gather*} \frac {x (d x)^m F_1\left (\frac {1+m}{3};2,2;\frac {4+m}{3};-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )}{a^2 (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m/(a + b*x^3 + c*x^6)^2,x]

[Out]

(x*(d*x)^m*AppellF1[(1 + m)/3, 2, 2, (4 + m)/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 -
4*a*c])])/(a^2*(1 + m))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (d x \right )^{m}}{\left (c \,x^{6}+b \,x^{3}+a \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m/(c*x^6+b*x^3+a)^2,x)

[Out]

int((d*x)^m/(c*x^6+b*x^3+a)^2,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((d*x)^m/(c*x^6+b*x^3+a)^2,x, algorithm="maxima")

[Out]

integrate((d*x)^m/(c*x^6 + b*x^3 + a)^2, 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((d*x)^m/(c*x^6+b*x^3+a)^2,x, algorithm="fricas")

[Out]

integral((d*x)^m/(c^2*x^12 + 2*b*c*x^9 + (b^2 + 2*a*c)*x^6 + 2*a*b*x^3 + a^2), x)

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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((d*x)**m/(c*x**6+b*x**3+a)**2,x)

[Out]

Timed out

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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((d*x)^m/(c*x^6+b*x^3+a)^2,x, algorithm="giac")

[Out]

integrate((d*x)^m/(c*x^6 + b*x^3 + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d\,x\right )}^m}{{\left (c\,x^6+b\,x^3+a\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m/(a + b*x^3 + c*x^6)^2,x)

[Out]

int((d*x)^m/(a + b*x^3 + c*x^6)^2, x)

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