3.948 \(\int \frac{x^2 (a+b x)^n}{\sqrt{c x^2}} \, dx\)

Optimal. Leaf size=59 \[ \frac{x (a+b x)^{n+2}}{b^2 (n+2) \sqrt{c x^2}}-\frac{a x (a+b x)^{n+1}}{b^2 (n+1) \sqrt{c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0157513, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {15, 43} \[ \frac{x (a+b x)^{n+2}}{b^2 (n+2) \sqrt{c x^2}}-\frac{a x (a+b x)^{n+1}}{b^2 (n+1) \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x)^n)/Sqrt[c*x^2],x]

[Out]

-((a*x*(a + b*x)^(1 + n))/(b^2*(1 + n)*Sqrt[c*x^2])) + (x*(a + b*x)^(2 + n))/(b^2*(2 + 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0206006, size = 43, normalized size = 0.73 \[ \frac{x (a+b x)^{n+1} (b (n+1) x-a)}{b^2 (n+1) (n+2) \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.002, size = 44, normalized size = 0.8 \begin{align*} -{\frac{ \left ( bx+a \right ) ^{1+n}x \left ( -bxn-bx+a \right ) }{{b}^{2} \left ({n}^{2}+3\,n+2 \right ) }{\frac{1}{\sqrt{c{x}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x+a)^n/(c*x^2)^(1/2),x)

[Out]

-(b*x+a)^(1+n)*x*(-b*n*x-b*x+a)/(c*x^2)^(1/2)/b^2/(n^2+3*n+2)

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Maxima [A]  time = 1.02931, size = 61, normalized size = 1.03 \begin{align*} \frac{{\left (b^{2}{\left (n + 1\right )} x^{2} + a b n x - a^{2}\right )}{\left (b x + a\right )}^{n}}{{\left (n^{2} + 3 \, n + 2\right )} b^{2} \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^n/(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

(b^2*(n + 1)*x^2 + a*b*n*x - a^2)*(b*x + a)^n/((n^2 + 3*n + 2)*b^2*sqrt(c))

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Fricas [A]  time = 1.70888, size = 134, normalized size = 2.27 \begin{align*} \frac{{\left (a b n x +{\left (b^{2} n + b^{2}\right )} x^{2} - a^{2}\right )} \sqrt{c x^{2}}{\left (b x + a\right )}^{n}}{{\left (b^{2} c n^{2} + 3 \, b^{2} c n + 2 \, b^{2} c\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^n/(c*x^2)^(1/2),x, algorithm="fricas")

[Out]

(a*b*n*x + (b^2*n + b^2)*x^2 - a^2)*sqrt(c*x^2)*(b*x + a)^n/((b^2*c*n^2 + 3*b^2*c*n + 2*b^2*c)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x+a)**n/(c*x**2)**(1/2),x)

[Out]

Exception raised: TypeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)^n/(c*x^2)^(1/2),x, algorithm="giac")

[Out]

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