\(\int \frac {(c x^2)^{5/2} (a+b x)^n}{x^5} \, dx\) [943]

Optimal result

Integrand size = 20, antiderivative size = 33 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^5} \, dx=\frac {c^2 \sqrt {c x^2} (a+b x)^{1+n}}{b (1+n) x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, 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} \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^5} \, dx=\frac {c^2 \sqrt {c x^2} (a+b x)^{n+1}}{b (n+1) x} \]

[In]

Int[((c*x^2)^(5/2)*(a + b*x)^n)/x^5,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^5} \, dx=\frac {\left (c x^2\right )^{5/2} (a+b x)^{1+n}}{b (1+n) x^5} \]

[In]

Integrate[((c*x^2)^(5/2)*(a + b*x)^n)/x^5,x]

[Out]

((c*x^2)^(5/2)*(a + b*x)^(1 + n))/(b*(1 + n)*x^5)

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88

method	result	size
gosper	\(\frac {\left (c \,x^{2}\right )^{\frac {5}{2}} \left (b x +a \right )^{1+n}}{b \,x^{5} \left (1+n \right )}\)	\(29\)
risch	\(\frac {c^{2} \sqrt {c \,x^{2}}\, \left (b x +a \right ) \left (b x +a \right )^{n}}{x b \left (1+n \right )}\)	\(35\)

[In]

int((c*x^2)^(5/2)*(b*x+a)^n/x^5,x,method=_RETURNVERBOSE)

[Out]

1/b/x^5/(1+n)*(c*x^2)^(5/2)*(b*x+a)^(1+n)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^5} \, dx=\frac {{\left (b c^{2} x + a c^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{{\left (b n + b\right )} x} \]

[In]

integrate((c*x^2)^(5/2)*(b*x+a)^n/x^5,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^5} \, dx=\begin {cases} \frac {\left (c x^{2}\right )^{\frac {5}{2}}}{a x^{4}} & \text {for}\: b = 0 \wedge n = -1 \\\frac {a^{n} \left (c x^{2}\right )^{\frac {5}{2}}}{x^{4}} & \text {for}\: b = 0 \\\int \frac {\left (c x^{2}\right )^{\frac {5}{2}}}{x^{5} \left (a + b x\right )}\, dx & \text {for}\: n = -1 \\\frac {a \left (c x^{2}\right )^{\frac {5}{2}} \left (a + b x\right )^{n}}{b n x^{5} + b x^{5}} + \frac {b x \left (c x^{2}\right )^{\frac {5}{2}} \left (a + b x\right )^{n}}{b n x^{5} + b x^{5}} & \text {otherwise} \end {cases} \]

[In]

integrate((c*x**2)**(5/2)*(b*x+a)**n/x**5,x)

[Out]

Piecewise(((c*x**2)**(5/2)/(a*x**4), Eq(b, 0) & Eq(n, -1)), (a**n*(c*x**2)**(5/2)/x**4, Eq(b, 0)), (Integral((
c*x**2)**(5/2)/(x**5*(a + b*x)), x), Eq(n, -1)), (a*(c*x**2)**(5/2)*(a + b*x)**n/(b*n*x**5 + b*x**5) + b*x*(c*
x**2)**(5/2)*(a + b*x)**n/(b*n*x**5 + b*x**5), True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^5} \, dx=\frac {{\left (b c^{\frac {5}{2}} x + a c^{\frac {5}{2}}\right )} {\left (b x + a\right )}^{n}}{b {\left (n + 1\right )}} \]

[In]

integrate((c*x^2)^(5/2)*(b*x+a)^n/x^5,x, algorithm="maxima")

[Out]

(b*c^(5/2)*x + a*c^(5/2))*(b*x + a)^n/(b*(n + 1))

Giac [F]

\[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^5} \, dx=\int { \frac {\left (c x^{2}\right )^{\frac {5}{2}} {\left (b x + a\right )}^{n}}{x^{5}} \,d x } \]

[In]

integrate((c*x^2)^(5/2)*(b*x+a)^n/x^5,x, algorithm="giac")

[Out]

integrate((c*x^2)^(5/2)*(b*x + a)^n/x^5, x)

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^5} \, dx=\frac {\left (\frac {c^2\,x\,\sqrt {c\,x^2}}{n+1}+\frac {a\,c^2\,\sqrt {c\,x^2}}{b\,\left (n+1\right )}\right )\,{\left (a+b\,x\right )}^n}{x} \]

[In]

int(((c*x^2)^(5/2)*(a + b*x)^n)/x^5,x)

[Out]

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