3.255 \(\int \frac{(d x)^m}{(a+b x^3+c x^6)^{3/2}} \, dx\)

Optimal. Leaf size=160 \[ \frac{(d x)^{m+1} \sqrt{\frac{2 c x^3}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^3}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{m+1}{3};\frac{3}{2},\frac{3}{2};\frac{m+4}{3};-\frac{2 c x^3}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^3}{b+\sqrt{b^2-4 a c}}\right )}{a d (m+1) \sqrt{a+b x^3+c x^6}} \]

[Out]

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

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Rubi [A]  time = 0.14082, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {1385, 510} \[ \frac{(d x)^{m+1} \sqrt{\frac{2 c x^3}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^3}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{m+1}{3};\frac{3}{2},\frac{3}{2};\frac{m+4}{3};-\frac{2 c x^3}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^3}{b+\sqrt{b^2-4 a c}}\right )}{a d (m+1) \sqrt{a+b x^3+c x^6}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1385

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

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{(d x)^m}{\left (a+b x^3+c x^6\right )^{3/2}} \, dx &=\frac{\left (\sqrt{1+\frac{2 c x^3}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^3}{b+\sqrt{b^2-4 a c}}}\right ) \int \frac{(d x)^m}{\left (1+\frac{2 c x^3}{b-\sqrt{b^2-4 a c}}\right )^{3/2} \left (1+\frac{2 c x^3}{b+\sqrt{b^2-4 a c}}\right )^{3/2}} \, dx}{a \sqrt{a+b x^3+c x^6}}\\ &=\frac{(d x)^{1+m} \sqrt{1+\frac{2 c x^3}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^3}{b+\sqrt{b^2-4 a c}}} F_1\left (\frac{1+m}{3};\frac{3}{2},\frac{3}{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 d (1+m) \sqrt{a+b x^3+c x^6}}\\ \end{align*}

Mathematica [A]  time = 0.176848, size = 221, normalized size = 1.38 \[ \frac{x (d x)^m \left (\sqrt{b^2-4 a c}-b-2 c x^3\right ) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^3}{b-\sqrt{b^2-4 a c}}} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x^3}{\sqrt{b^2-4 a c}+b}\right )^{3/2} F_1\left (\frac{m+1}{3};\frac{3}{2},\frac{3}{2};\frac{m+4}{3};-\frac{2 c x^3}{b+\sqrt{b^2-4 a c}},\frac{2 c x^3}{\sqrt{b^2-4 a c}-b}\right )}{(m+1) \left (\sqrt{b^2-4 a c}-b\right ) \left (a+b x^3+c x^6\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(d*x)^m*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^3)*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*
((b + Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]))^(3/2)*AppellF1[(1 + m)/3, 3/2, 3/2, (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])])/((-b + Sqrt[b^2 - 4*a*c])*(1 + m)*(a + b
*x^3 + c*x^6)^(3/2))

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Maple [F]  time = 0.008, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dx \right ) ^{m} \left ( c{x}^{6}+b{x}^{3}+a \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{{\left (c x^{6} + b x^{3} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{6} + b x^{3} + a} \left (d x\right )^{m}}{c^{2} x^{12} + 2 \, b c x^{9} +{\left (b^{2} + 2 \, a c\right )} x^{6} + 2 \, a b x^{3} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^6 + b*x^3 + a)*(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{\left (a + b x^{3} + c x^{6}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{{\left (c x^{6} + b x^{3} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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