3.10.33 \(\int \frac {(c x^2)^{3/2} (a+b x)^n}{x^2} \, dx\) [933]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 45} \begin {gather*} \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} \end {gather*}

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 45

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*}

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Mathematica [A]
time = 0.01, size = 46, normalized size = 0.71 \begin {gather*} \frac {c^2 x (a+b x)^{1+n} (-a+b (1+n) x)}{b^2 (1+n) (2+n) \sqrt {c x^2}} \end {gather*}

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])

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Maple [A]
time = 0.11, size = 46, normalized size = 0.71

method	result	size
gosper	\(-\frac {\left (b x +a \right )^{1+n} \left (c \,x^{2}\right )^{\frac {3}{2}} \left (-b n x -b x +a \right )}{x^{3} b^{2} \left (n^{2}+3 n +2\right )}\)	\(46\)
risch	\(-\frac {c \sqrt {c \,x^{2}}\, \left (-b^{2} n \,x^{2}-a b n x -x^{2} b^{2}+a^{2}\right ) \left (b x +a \right )^{n}}{x \,b^{2} \left (2+n \right ) \left (1+n \right )}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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Maxima [A]
time = 0.32, size = 51, normalized size = 0.78 \begin {gather*} \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 {gather*}

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)

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Fricas [A]
time = 0.61, size = 68, normalized size = 1.05 \begin {gather*} \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 {gather*}

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {a^{n} \left (c x^{2}\right )^{\frac {3}{2}}}{2 x} & \text {for}\: b = 0 \\\int \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{x^{2} \left (a + b x\right )^{2}}\, dx & \text {for}\: n = -2 \\\int \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{x^{2} \left (a + b x\right )}\, dx & \text {for}\: n = -1 \\- \frac {a^{2} \left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{n}}{b^{2} n^{2} x^{3} + 3 b^{2} n x^{3} + 2 b^{2} x^{3}} + \frac {a b n x \left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{n}}{b^{2} n^{2} x^{3} + 3 b^{2} n x^{3} + 2 b^{2} x^{3}} + \frac {b^{2} n x^{2} \left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{n}}{b^{2} n^{2} x^{3} + 3 b^{2} n x^{3} + 2 b^{2} x^{3}} + \frac {b^{2} x^{2} \left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{n}}{b^{2} n^{2} x^{3} + 3 b^{2} n x^{3} + 2 b^{2} x^{3}} & \text {otherwise} \end {cases} \end {gather*}

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]

Piecewise((a**n*(c*x**2)**(3/2)/(2*x), Eq(b, 0)), (Integral((c*x**2)**(3/2)/(x**2*(a + b*x)**2), x), Eq(n, -2)
), (Integral((c*x**2)**(3/2)/(x**2*(a + b*x)), x), Eq(n, -1)), (-a**2*(c*x**2)**(3/2)*(a + b*x)**n/(b**2*n**2*
x**3 + 3*b**2*n*x**3 + 2*b**2*x**3) + a*b*n*x*(c*x**2)**(3/2)*(a + b*x)**n/(b**2*n**2*x**3 + 3*b**2*n*x**3 + 2
*b**2*x**3) + b**2*n*x**2*(c*x**2)**(3/2)*(a + b*x)**n/(b**2*n**2*x**3 + 3*b**2*n*x**3 + 2*b**2*x**3) + b**2*x
**2*(c*x**2)**(3/2)*(a + b*x)**n/(b**2*n**2*x**3 + 3*b**2*n*x**3 + 2*b**2*x**3), True))

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Giac [A]
time = 0.53, size = 119, normalized size = 1.83 \begin {gather*} {\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 {gather*}

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)

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Mupad [B]
time = 0.23, size = 88, normalized size = 1.35 \begin {gather*} \frac {{\left (a+b\,x\right )}^n\,\left (\frac {c\,x^2\,\sqrt {c\,x^2}\,\left (n+1\right )}{n^2+3\,n+2}-\frac {a^2\,c\,\sqrt {c\,x^2}}{b^2\,\left (n^2+3\,n+2\right )}+\frac {a\,c\,n\,x\,\sqrt {c\,x^2}}{b\,\left (n^2+3\,n+2\right )}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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