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

Optimal. Leaf size=48 \[ -\frac {x (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )}{a c (n+1) \sqrt {c x^2}} \]

[Out]

-x*(b*x+a)^(1+n)*hypergeom([1, 1+n],[2+n],1+b*x/a)/a/c/(1+n)/(c*x^2)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 65} \[ -\frac {x (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )}{a c (n+1) \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((x*(a + b*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a])/(a*c*(1 + n)*Sqrt[c*x^2]))

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 47, normalized size = 0.98 \[ -\frac {x^3 (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b x}{a}+1\right )}{a (n+1) \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((x^3*(a + b*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (b*x)/a])/(a*(1 + n)*(c*x^2)^(3/2)))

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{c^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2)*(b*x + a)^n/(c^2*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (b x +a \right )^{n}}{\left (c \,x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^2\,{\left (a+b\,x\right )}^n}{{\left (c\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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