\(\int \frac {\sqrt {c x^2} (a+b x)^n}{x} \, dx\) [926]

Optimal result

Integrand size = 20, antiderivative size = 30 \[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x} \, dx=\frac {\sqrt {c x^2} (a+b x)^{1+n}}{b (1+n) x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, 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} \[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x} \, dx=\frac {\sqrt {c x^2} (a+b x)^{n+1}}{b (n+1) x} \]

[In]

Int[(Sqrt[c*x^2]*(a + b*x)^n)/x,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x} \, dx=\frac {c x (a+b x)^{1+n}}{b (1+n) \sqrt {c x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97

method	result	size
gosper	\(\frac {\left (b x +a \right )^{1+n} \sqrt {c \,x^{2}}}{b \left (1+n \right ) x}\)	\(29\)
risch	\(\frac {\left (b x +a \right ) \left (b x +a \right )^{n} \sqrt {c \,x^{2}}}{b \left (1+n \right ) x}\)	\(32\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x} \, dx=\frac {\sqrt {c x^{2}} {\left (b x + a\right )} {\left (b x + a\right )}^{n}}{{\left (b n + b\right )} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x} \, dx=\frac {{\left (b \sqrt {c} x + a \sqrt {c}\right )} {\left (b x + a\right )}^{n}}{b {\left (n + 1\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x} \, dx=-\sqrt {c} {\left (\frac {a^{n + 1} \mathrm {sgn}\left (x\right )}{b n + b} - \frac {{\left (b x + a\right )}^{n + 1} \mathrm {sgn}\left (x\right )}{b {\left (n + 1\right )}}\right )} \]

[In]

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

[Out]

-sqrt(c)*(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 = 31, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {c x^2} (a+b x)^n}{x} \, dx=\frac {\sqrt {c\,x^2}\,{\left (a+b\,x\right )}^n\,\left (a+b\,x\right )}{b\,x\,\left (n+1\right )} \]

[In]

int(((c*x^2)^(1/2)*(a + b*x)^n)/x,x)

[Out]

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