3.576 \(\int x^2 (a+b x^n+c x^{2 n})^{3/2} \, dx\)

Optimal. Leaf size=149 \[ \frac{a x^3 \sqrt{a+b x^n+c x^{2 n}} F_1\left (\frac{3}{n};-\frac{3}{2},-\frac{3}{2};\frac{n+3}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{3 \sqrt{\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^n}{\sqrt{b^2-4 a c}+b}+1}} \]

[Out]

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

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Rubi [A]  time = 0.150804, antiderivative size = 149, 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} \[ \frac{a x^3 \sqrt{a+b x^n+c x^{2 n}} F_1\left (\frac{3}{n};-\frac{3}{2},-\frac{3}{2};\frac{n+3}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{3 \sqrt{\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^n}{\sqrt{b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [B]  time = 1.52096, size = 524, normalized size = 3.52 \[ \frac{x^3 \left (2 (n+3) \left (4 a^2 c \left (8 n^2+18 n+9\right )+a \left (3 b^2 n^2+2 b c \left (23 n^2+63 n+36\right ) x^n+4 c^2 \left (10 n^2+27 n+18\right ) x^{2 n}\right )+x^n \left (b+c x^n\right ) \left (3 b^2 n^2+2 b c \left (7 n^2+27 n+18\right ) x^n+4 c^2 \left (2 n^2+9 n+9\right ) x^{2 n}\right )\right )-3 b n^2 x^n \left (b^2 (n+6)-12 a c (n+2)\right ) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^n}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^n}{\sqrt{b^2-4 a c}+b}} F_1\left (\frac{n+3}{n};\frac{1}{2},\frac{1}{2};2+\frac{3}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}},\frac{2 c x^n}{\sqrt{b^2-4 a c}-b}\right )+2 a n^2 (n+3) \left (4 a c (2 n+3)-3 b^2\right ) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^n}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^n}{\sqrt{b^2-4 a c}+b}} F_1\left (\frac{3}{n};\frac{1}{2},\frac{1}{2};\frac{n+3}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}},\frac{2 c x^n}{\sqrt{b^2-4 a c}-b}\right )\right )}{24 c (n+1) (n+3)^2 (2 n+3) \sqrt{a+x^n \left (b+c x^n\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^3*(2*(3 + n)*(4*a^2*c*(9 + 18*n + 8*n^2) + x^n*(b + c*x^n)*(3*b^2*n^2 + 2*b*c*(18 + 27*n + 7*n^2)*x^n + 4*c
^2*(9 + 9*n + 2*n^2)*x^(2*n)) + a*(3*b^2*n^2 + 2*b*c*(36 + 63*n + 23*n^2)*x^n + 4*c^2*(18 + 27*n + 10*n^2)*x^(
2*n))) + 2*a*n^2*(3 + n)*(-3*b^2 + 4*a*c*(3 + 2*n))*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b - Sqrt[b^2 - 4*a
*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[3/n, 1/2, 1/2, (3 + n)/n, (-2*c
*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] - 3*b*n^2*(-12*a*c*(2 + n) + b^2*(6 + n))*x
^n*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b +
Sqrt[b^2 - 4*a*c])]*AppellF1[(3 + n)/n, 1/2, 1/2, 2 + 3/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b
+ Sqrt[b^2 - 4*a*c])]))/(24*c*(1 + n)*(3 + n)^2*(3 + 2*n)*Sqrt[a + x^n*(b + c*x^n)])

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Maple [F]  time = 0.048, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*x^n+c*x^(2*n))^(3/2),x)

[Out]

int(x^2*(a+b*x^n+c*x^(2*n))^(3/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^(2*n) + b*x^n + a)^(3/2)*x^2, x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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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(x**2*(a+b*x**n+c*x**(2*n))**(3/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^(2*n) + b*x^n + a)^(3/2)*x^2, x)