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

Optimal. Leaf size=65 \[ \frac{c \sqrt{c x^2} (a+b x)^{n+2}}{b^2 (n+2) x}-\frac{a c \sqrt{c x^2} (a+b x)^{n+1}}{b^2 (n+1) x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0179092, antiderivative size = 65, 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} \[ \frac{c \sqrt{c x^2} (a+b x)^{n+2}}{b^2 (n+2) x}-\frac{a c \sqrt{c x^2} (a+b x)^{n+1}}{b^2 (n+1) x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.007752, size = 46, normalized size = 0.71 \[ \frac{c^2 x (a+b x)^{n+1} (b (n+1) x-a)}{b^2 (n+1) (n+2) \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.001, size = 46, normalized size = 0.7 \begin{align*} -{\frac{ \left ( bx+a \right ) ^{1+n} \left ( -bxn-bx+a \right ) }{{x}^{3}{b}^{2} \left ({n}^{2}+3\,n+2 \right ) } \left ( c{x}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.02159, size = 69, normalized size = 1.06 \begin{align*} \frac{{\left (b^{2} c^{\frac{3}{2}}{\left (n + 1\right )} x^{2} + a b c^{\frac{3}{2}} n x - a^{2} c^{\frac{3}{2}}\right )}{\left (b x + a\right )}^{n}}{{\left (n^{2} + 3 \, n + 2\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.60575, size = 136, normalized size = 2.09 \begin{align*} \frac{{\left (a b c n x - a^{2} c +{\left (b^{2} c n + b^{2} c\right )} x^{2}\right )} \sqrt{c x^{2}}{\left (b x + a\right )}^{n}}{{\left (b^{2} n^{2} + 3 \, b^{2} n + 2 \, b^{2}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: TypeError

________________________________________________________________________________________

Giac [A]  time = 1.08445, size = 161, normalized size = 2.48 \begin{align*}{\left (\frac{a^{2} a^{n} \mathrm{sgn}\left (x\right )}{b^{2} n^{2} + 3 \, b^{2} n + 2 \, b^{2}} + \frac{{\left (b x + a\right )}^{n} b^{2} n x^{2} \mathrm{sgn}\left (x\right ) +{\left (b x + a\right )}^{n} a b n x \mathrm{sgn}\left (x\right ) +{\left (b x + a\right )}^{n} b^{2} x^{2} \mathrm{sgn}\left (x\right ) -{\left (b x + a\right )}^{n} a^{2} \mathrm{sgn}\left (x\right )}{b^{2} n^{2} + 3 \, b^{2} n + 2 \, b^{2}}\right )} c^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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