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

Optimal. Leaf size=152 \[ -\frac{a \sqrt{a+b x^n+c x^{2 n}} F_1\left (-\frac{2}{n};-\frac{3}{2},-\frac{3}{2};-\frac{2-n}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{2 x^2 \sqrt{\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^n}{\sqrt{b^2-4 a c}+b}+1}} \]

[Out]

-(a*Sqrt[a + b*x^n + c*x^(2*n)]*AppellF1[-2/n, -3/2, -3/2, -((2 - n)/n), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (
-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(2*x^2*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 = 0.152111, antiderivative size = 152, 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} \[ -\frac{a \sqrt{a+b x^n+c x^{2 n}} F_1\left (-\frac{2}{n};-\frac{3}{2},-\frac{3}{2};-\frac{2-n}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{2 x^2 \sqrt{\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^n}{\sqrt{b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Sqrt[a + b*x^n + c*x^(2*n)]*AppellF1[-2/n, -3/2, -3/2, -((2 - n)/n), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (
-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(2*x^2*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 \frac{\left (a+b x^n+c x^{2 n}\right )^{3/2}}{x^3} \, dx &=\frac{\left (a \sqrt{a+b x^n+c x^{2 n}}\right ) \int \frac{\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}}{x^3} \, 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 \sqrt{a+b x^n+c x^{2 n}} F_1\left (-\frac{2}{n};-\frac{3}{2},-\frac{3}{2};-\frac{2-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 )}{2 x^2 \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 = 1.39259, size = 520, normalized size = 3.42 \[ \frac{2 (n-2) \left (16 a^2 c \left (2 n^2-3 n+1\right )+a \left (3 b^2 n^2+2 b c \left (23 n^2-42 n+16\right ) x^n+8 c^2 \left (5 n^2-9 n+4\right ) x^{2 n}\right )+x^n \left (b+c x^n\right ) \left (3 b^2 n^2+2 b c \left (7 n^2-18 n+8\right ) x^n+8 c^2 \left (n^2-3 n+2\right ) x^{2 n}\right )\right )-3 b n^2 x^n \left (4 a c (4-3 n)+b^2 (n-4)\right ) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^n}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^n}{\sqrt{b^2-4 a c}+b}} F_1\left (\frac{n-2}{n};\frac{1}{2},\frac{1}{2};2-\frac{2}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}},\frac{2 c x^n}{\sqrt{b^2-4 a c}-b}\right )-6 a (n-2) n^2 \left (4 a c (n-1)+b^2\right ) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^n}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^n}{\sqrt{b^2-4 a c}+b}} F_1\left (-\frac{2}{n};\frac{1}{2},\frac{1}{2};\frac{n-2}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}},\frac{2 c x^n}{\sqrt{b^2-4 a c}-b}\right )}{16 c (n-2)^2 (n-1) (3 n-2) x^2 \sqrt{a+x^n \left (b+c x^n\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(-2 + n)*(16*a^2*c*(1 - 3*n + 2*n^2) + x^n*(b + c*x^n)*(3*b^2*n^2 + 2*b*c*(8 - 18*n + 7*n^2)*x^n + 8*c^2*(2
 - 3*n + n^2)*x^(2*n)) + a*(3*b^2*n^2 + 2*b*c*(16 - 42*n + 23*n^2)*x^n + 8*c^2*(4 - 9*n + 5*n^2)*x^(2*n))) - 6
*a*(b^2 + 4*a*c*(-1 + n))*(-2 + n)*n^2*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[-2/n, 1/2, 1/2, (-2 + 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*(4*a*c*(4 - 3*n) + b^2*(-4 + n))*n^2*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[(-2 + n)/n, 1/2, 1/2, 2 - 2/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^
2 - 4*a*c])])/(16*c*(-2 + n)^2*(-1 + n)*(-2 + 3*n)*x^2*Sqrt[a + x^n*(b + c*x^n)])

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}} \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((a+b*x^n+c*x^(2*n))^(3/2)/x^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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