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

Optimal. Leaf size=31 \[ \frac {x (a+b x)^{n+1}}{b c (n+1) \sqrt {c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 32} \[ \frac {x (a+b x)^{n+1}}{b c (n+1) \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 30, normalized size = 0.97 \[ \frac {x^3 (a+b x)^{n+1}}{b (n+1) \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 37, normalized size = 1.19 \[ \frac {\sqrt {c x^{2}} {\left (b x + a\right )} {\left (b x + a\right )}^{n}}{{\left (b c^{2} n + b c^{2}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{n} x^{3}}{\left (c x^{2}\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 29, normalized size = 0.94 \[ \frac {x^{3} \left (b x +a \right )^{n +1}}{\left (n +1\right ) \left (c \,x^{2}\right )^{\frac {3}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.44, size = 31, normalized size = 1.00 \[ \frac {{\left (b \sqrt {c} x + a \sqrt {c}\right )} {\left (b x + a\right )}^{n}}{b c^{2} {\left (n + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.23, size = 42, normalized size = 1.35 \[ \frac {\left (\frac {x^2}{c\,\left (n+1\right )}+\frac {a\,x}{b\,c\,\left (n+1\right )}\right )\,{\left (a+b\,x\right )}^n}{\sqrt {c\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \frac {x^{4}}{a c^{\frac {3}{2}} \left (x^{2}\right )^{\frac {3}{2}}} & \text {for}\: b = 0 \wedge n = -1 \\\frac {a^{n} x^{4}}{c^{\frac {3}{2}} \left (x^{2}\right )^{\frac {3}{2}}} & \text {for}\: b = 0 \\\int \frac {x^{3}}{\left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )}\, dx & \text {for}\: n = -1 \\\frac {a x^{3} \left (a + b x\right )^{n}}{b c^{\frac {3}{2}} n \left (x^{2}\right )^{\frac {3}{2}} + b c^{\frac {3}{2}} \left (x^{2}\right )^{\frac {3}{2}}} + \frac {b x^{4} \left (a + b x\right )^{n}}{b c^{\frac {3}{2}} n \left (x^{2}\right )^{\frac {3}{2}} + b c^{\frac {3}{2}} \left (x^{2}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**4/(a*c**(3/2)*(x**2)**(3/2)), Eq(b, 0) & Eq(n, -1)), (a**n*x**4/(c**(3/2)*(x**2)**(3/2)), Eq(b,
0)), (Integral(x**3/((c*x**2)**(3/2)*(a + b*x)), x), Eq(n, -1)), (a*x**3*(a + b*x)**n/(b*c**(3/2)*n*(x**2)**(3
/2) + b*c**(3/2)*(x**2)**(3/2)) + b*x**4*(a + b*x)**n/(b*c**(3/2)*n*(x**2)**(3/2) + b*c**(3/2)*(x**2)**(3/2)),
True))

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