\(\int \frac {(c x^2)^{3/2} (a+b x)^n}{x^3} \, dx\) [934]

Optimal result

Integrand size = 20, antiderivative size = 31 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^3} \, dx=\frac {c \sqrt {c x^2} (a+b x)^{1+n}}{b (1+n) x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 32} \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^3} \, dx=\frac {c \sqrt {c x^2} (a+b x)^{n+1}}{b (n+1) x} \]

[In]

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

[Out]

(c*Sqrt[c*x^2]*(a + b*x)^(1 + n))/(b*(1 + 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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^3} \, dx=\frac {\left (c x^2\right )^{3/2} (a+b x)^{1+n}}{b (1+n) x^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94

method	result	size
gosper	\(\frac {\left (c \,x^{2}\right )^{\frac {3}{2}} \left (b x +a \right )^{1+n}}{b \,x^{3} \left (1+n \right )}\)	\(29\)
risch	\(\frac {c \sqrt {c \,x^{2}}\, \left (b x +a \right ) \left (b x +a \right )^{n}}{x b \left (1+n \right )}\)	\(33\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^3} \, dx=\frac {{\left (b c x + a c\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{{\left (b n + b\right )} x} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^3} \, dx=\begin {cases} \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{a x^{2}} & \text {for}\: b = 0 \wedge n = -1 \\\frac {a^{n} \left (c x^{2}\right )^{\frac {3}{2}}}{x^{2}} & \text {for}\: b = 0 \\\int \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{x^{3} \left (a + b x\right )}\, dx & \text {for}\: n = -1 \\\frac {a \left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{n}}{b n x^{3} + b x^{3}} + \frac {b x \left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{n}}{b n x^{3} + b x^{3}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^3} \, dx=\frac {{\left (b c^{\frac {3}{2}} x + a c^{\frac {3}{2}}\right )} {\left (b x + a\right )}^{n}}{b {\left (n + 1\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^3} \, dx=-c^{\frac {3}{2}} {\left (\frac {a^{n + 1} \mathrm {sgn}\left (x\right )}{b n + b} - \frac {{\left (b x + a\right )}^{n + 1} \mathrm {sgn}\left (x\right )}{b {\left (n + 1\right )}}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.45 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^3} \, dx=\frac {\left (\frac {c\,x\,\sqrt {c\,x^2}}{n+1}+\frac {a\,c\,\sqrt {c\,x^2}}{b\,\left (n+1\right )}\right )\,{\left (a+b\,x\right )}^n}{x} \]

[In]

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

[Out]

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